third year engineering math3602 mathematics (integral

University College Dublin

Third Year Engineering

math3602 Mathematics (Integral Calculus)http://mathsa.ucd.ie/courses/math3602

ftp://mathsa.ucd.ie/pub/courses/math3602

February 13, 2003

Dr. J. Brendan Quigley

prepared using LATEXrunning under Redhat Linuxdrawings prepared using gnuplot

comments to:-

Dr J.Brendan.QuigleyDepartment of MathematicsUniversity College Dublin, Belfieldph. 716–8276, 716–8265; fax 716–1196email [email protected].

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jbquig-UCD February 13, 2003

Part I

Lectures and Problem Sets

solutions in Part II

iii

Contents

I Lectures and Problem Sets

solutions in Part II iii

0 Organization of math3602 2000-2001 1

1 Integration from First Principles 31.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Closed Intervals, Hyper-Rectangloids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.2 measure µ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.3 span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.1 definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3.2 mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 introducing the Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4.1 Rieman Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4.2 the Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.5 Justifying the concept of Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 example of Integration from First Principles in R3 . . . . . . . . . . . . . . . . . . . . . . . 7

1.6.1 introduction to the tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6.2 the volume of T as an integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.6.3 partitions, mesh and span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.6.4 computing the Rieman Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6.5 taking the limit of the Rieman Sum . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.7 Integration of a discontinuous function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.8 Appendices:- Integrations performed by Computer Program . . . . . . . . . . . . . . . 121.9 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Fubini’s Theorem 152.1 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 statement and exposition of Fubini’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2.1 Decomposition of Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.2 first examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 example, integration over the ball in 3-space . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3.1 limits for the solid ball B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3.2 computing m for the solid ball B . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.3.3 computing M for the solid ball B . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.3.4 computing I for the solid ball B . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

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2.4 example, integration over the solid cone in R3 . . . . . . . . . . . . . . . . . . . . . . . . . 232.4.1 find limits for C, by two methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.4.2 computing m for the solid cone C . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.4.3 computing M for the solid cone C . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.4.4 computing I for the solid cone C . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.4.5 center of mass, radius of gyration . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.5 problem set:- Fubini’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.6 APPENDIX A—–PASCAL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.7 APPENDIX B—-C PROGRAMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3 Scalar and Vector Fields with Differential Operators 553.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2 Scalar and Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.2.1 Definition of Scalar Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2.2 Definition of Vector Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3.3 Differential Operators:- grad � curl � div and ∇2 . . . . . . . . . . . . . . . . . . . . . 563.3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.3.2 Composition of Differential Operators . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.4 Inverse Square Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 583.4.1 Differential Operators and the Inverse Square Law . . . . . . . . . . . . . . . . . . 583.4.2 Summary–Inverse Square Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.5 Physical meaning of grad � curl � div and ∇2 . . . . . . . . . . . . . . . . . . . . . . . . . 633.6 Dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.7 Computing the Dipole fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.7.1 Collecting these results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653.7.2 Differential Operators and the Dipole . . . . . . . . . . . . . . . . . . . . . . . . . 653.7.3 Summary–Dipole field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.8 Fields in R2, Vortex and Source . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.8.1 The Vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.8.2 The Two Dimensional Source/Sink . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.9 Advanced Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.9.1 Does an irrotational field have a scalar potential. . . . . . . . . . . . . . . . . . . . 683.9.2 Does an divergence free field have a vector potential. . . . . . . . . . . . . . . . . . 69

3.10 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4 Differential Equations with the Laplace Transform 734.1 Classical functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.2 Heavyside Step and Dirac Delta Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

4.2.1 Heavyside step function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 754.2.2 Dirac Delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

4.3 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 764.3.1 commutativity, associativity of � . . . . . . . . . . . . . . . . . . . . . . . . . . . . 774.3.2 Integration and Shift operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794.3.3 convolution with ua and δa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.4 The Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824.5 Laplace transform of classical functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.5.1 L�eat � . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.5.2 L�cosat � and L

�sinat � . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

4.5.3 L�tn � . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.6 Laplace transform with step and delta functions . . . . . . . . . . . . . . . . . . . . . . . . 854.6.1 L

�ua

� . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.6.2 L

�δa

� . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.7 Formulae for the Laplace Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

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4.7.1 L�f � g � � L � 1 and f � g � h . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4.7.2 L and differentiation L�f� � � L

�f�n � � and L

��f � and also L

�tn f � . . . . . . . . . . 88

4.7.3 L and shifting L�Sha f � and Shb

�L f � . . . . . . . . . . . . . . . . . . . . . . . . . 90

4.7.4 L and periodicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.8 Two difficult proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.8.1 full proof that L�f � g � �

s �� L�f � �

s �� L�g � �

s � . . . . . . . . . . . . . . . . . . . . 914.8.2 full proof that

�f � g � � h �� f � �

g � h � . . . . . . . . . . . . . . . . . . . . . . . . 934.9 Homogenous Differential equations and the Laplace Transform . . . . . . . . . . . . . . . . 95

4.9.1 characteristic polynomial of a differential equation . . . . . . . . . . . . . . . . . . 954.9.2 example HDE of degree 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954.9.3 example HDE of degree 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 954.9.4 example HDE of degree 2 with natural sympathy . . . . . . . . . . . . . . . . . . . 964.9.5 example HDE of degree 2, complex roots . . . . . . . . . . . . . . . . . . . . . . . 974.9.6 solving a general CCHDEwIC of order 4 . . . . . . . . . . . . . . . . . . . . . . . 974.9.7 solving the most general CCHDEwIC of order n . . . . . . . . . . . . . . . . . . . 984.9.8 natural response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

4.10 Inhomogeneous Differential equations and the Laplace Transform . . . . . . . . . . . . . . 994.10.1 example unsympathetically driven IHDE of degree 1 . . . . . . . . . . . . . . . . . 994.10.2 sympathetically driven IHDE of degree 1 . . . . . . . . . . . . . . . . . . . . . . . 1004.10.3 example unsympathetically driven IHDE of degree 2 . . . . . . . . . . . . . . . . . 1004.10.4 example sympathetically driven IHDE of degree 2 . . . . . . . . . . . . . . . . . . 1014.10.5 solving the most general CCIHDEwIC of order n . . . . . . . . . . . . . . . . . . . 103

4.11 Sympathy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1034.12 Table of Laplace Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1044.13 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

4.13.1 Convolution, Delta and Heavyside functions, Laplace Transform . . . . . . . . . . . 1064.13.2 Differential Equations and the Laplace Transform . . . . . . . . . . . . . . . . . . . 108

5 Change of Variable theorem and the Jacobean 1115.1 matrices and volume dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

5.1.1 determinant of a matrix whose columns are mutually � . . . . . . . . . . . . . . . . 1135.2 change of variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1145.3 differential matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1155.4 Jacobean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1155.5 change of variable theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.5.1 first examples with the c.o.v.th. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1185.5.2 proof of the change of variable theorem . . . . . . . . . . . . . . . . . . . . . . . . 120

5.6 spherical polar transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1215.7 toroidal polar transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

5.7.1 derivation of the toroidal polar transform . . . . . . . . . . . . . . . . . . . . . . . 1215.7.2 differential and Jacobian of the t.p.tf. . . . . . . . . . . . . . . . . . . . . . . . . . 1225.7.3 integration on the solid torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.8 ellipsoidal polar transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1255.8.1 derivation of the elliptical polar transform . . . . . . . . . . . . . . . . . . . . . . . 1255.8.2 differential and Jacobian of the e.p.tf. . . . . . . . . . . . . . . . . . . . . . . . . . 1265.8.3 integration on the solid ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

5.9 Further examples of integration using the c.o.v.th. . . . . . . . . . . . . . . . . . . . . . . . 1275.9.1 area under the bell shaped curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1275.9.2 hypervolume of hyperball and hyperellipsoid . . . . . . . . . . . . . . . . . . . . . 129

5.10 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

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6 Surface integrals, Divergence theorem of Gauss 1376.1 parametrization of a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

6.1.1 differential matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1376.1.2 S2 �

a � , the spherical surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1386.1.3 T 2 �

a � b � , the toroidal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1386.1.4 E2 �

a � b � c � , ellipsoidal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1396.2 area dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

6.2.1 area dilation of a linear mapping . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1406.2.2 area dilation under surface parametrization . . . . . . . . . . . . . . . . . . . . . . 141

6.3 integration of a scalar field over a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . 1426.4 the normal field of a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1436.5 vector flux integral across a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1446.6 examples of the surface vector flux integral . . . . . . . . . . . . . . . . . . . . . . . . . . 146

6.6.1 flux of field x �� x �� across spherical surface . . . . . . . . . . . . . . . . . . . . . . 1466.6.2 flux integral across ellipsoidal surface . . . . . . . . . . . . . . . . . . . . . . . . . 1476.6.3 flux integral across toroidal surface . . . . . . . . . . . . . . . . . . . . . . . . . . 147

6.7 divergence theorem of Gauss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1496.8 examples of the divergence theorem of Gauss . . . . . . . . . . . . . . . . . . . . . . . . . 149

6.8.1 flux integral across the spherical surface . . . . . . . . . . . . . . . . . . . . . . . . 1496.8.2 flux integral across the ellipsoid surface . . . . . . . . . . . . . . . . . . . . . . . . 1496.8.3 flux integral across the toroidal surface . . . . . . . . . . . . . . . . . . . . . . . . 150

6.9 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

7 Integration over Curves, Stokes theorem 1557.1 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1557.2 the curve, technicalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1557.3 Integration of a Scalar Field over a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 1567.4 Flow of a Vector Field along a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1577.5 Stokes Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1597.6 Potential, Work and Curve Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1597.7 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1607.8 more–what is this?? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

II Answers to problem sets of part I 201

1 answers:- integration by first principles 203

1.1 answer �� xy frpm first principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

1.1.1 drawing the partion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2031.1.2 the mesh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2031.1.3 integral as limit of Rieman sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

1.2 integration from first principles over the tetrahedron . . . . . . . . . . . . . . . . . . . . . . 2051.2.1 sketch of tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2051.2.2 the integral as limit of Rieman sum . . . . . . . . . . . . . . . . . . . . . . . . . . 205

2 solutions: problems with Fubini’s theorem 2092.1 solution: integration on the tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

2.1.1 limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2092.1.2 mass,moment, moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2112.1.3 center of mass, radius of gyration . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

2.2 solution: integration on the cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2132.2.1 different limits, hyperbolic conic section . . . . . . . . . . . . . . . . . . . . . . . . 2132.2.2 integration using the new different limits . . . . . . . . . . . . . . . . . . . . . . . 214

2.3 solution: integration on the solid ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

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2.4 solution: integration on the solid ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . 2162.4.1 sketch of the solid ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2162.4.2 finding the limits of integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2162.4.3 evaluation of the mass (volume) integral . . . . . . . . . . . . . . . . . . . . . . . . 217

2.5 volume of hyper-hedron T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

3 solutions: scalar and vector fields with differential operators 2193.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2193.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2193.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

3.3.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2193.3.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2203.3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2203.3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2203.3.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2203.3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

3.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2213.6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2223.6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2223.6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2223.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2233.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2233.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2233.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

4 solutions:- convolution, laplace transform, differential equations 2254.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2254.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

5 solutions:-change of variable th and the jacobean 2275.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

5.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2275.1.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2275.1.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2285.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

5.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2295.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2295.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

5.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2305.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

5.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2335.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2345.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2355.4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

5.5 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2355.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235


jbquig-UCD

x CONTENTS

5.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2355.5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2355.5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

5.6 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2365.6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2365.6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2365.6.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2375.6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

5.7 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2385.7.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2385.7.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2385.7.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

5.8 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2395.8.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2395.8.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2405.8.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2405.8.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

6 Answers surface integration, divergence theorem of Gauss 2436.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2436.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

6.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2436.2.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2446.2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247

6.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2486.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

6.4.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2506.4.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2516.4.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

6.5 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2536.5.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2536.5.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2536.5.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

6.6 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2576.6.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2576.6.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

6.7 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

c�


List of Figures

2.1 (i) the ball B , (ii) the (cutaway) hemiball H . . . . . . . . . . . . . . . . . . . . . . . 182.2 Ball (i) cut by a plane of constant height (ii) crossection in xz–plane . . . . . . . . . . . . . 192.3 line segment L, where y is constant, on disc D in xy–plane . . . . . . . . . . . . . . . . . . 192.4 (i) cone (ii) disc D of constant height . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.5 (i) section of cone in xz–plane (ii) the disc D in xy–plane . . . . . . . . . . . . . . . . . . . 24

4.1 (i) oscillation cos (ii) damped oscillation exp � cos . . . . . . . . . . . . . . . . . . . . . 744.2 (i) decay, eat � a � 0 (ii) growth, eat � a � 0 . . . . . . . . . . . . . . . . . . . . . . . . . . 744.3 (i) Heavyside step function ua (ii) and elbow functions ea . . . . . . . . . . . . . . . . . . 754.4 (i) finite pulse

�ua � ua � h

� � h (ii) delta function as limh � 0�ua � ua � h

� � h . . . . . . . . 764.5 (i) t2 (ii)

�t � 1 � 2,inappropriate (iii) Sh � t2 � � u1

�t � �

t � 1 � 2 . . . . . . . . . . . . . . . . 804.6 (i) t (ii) staircase function � t � . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 854.7 (i) sawtooth (ii) squarewave functions; t � � t � and

� � 1 �� t . . . . . . . . . . . . . . . . 864.8 the quadrant Q and the triangle A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.9 triangles (ii) T ; a b t (ii) D; 0 q t and 0 p t � q . . . . . . . . . . . . . . 93

5.1 (i)1� 2π

exp

� � x2

2 � (ii)1

2πexp

� � x2 y2

2 � . . . . . . . . . . . . . . . . . . . . 127

5.2 table of hypervolumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.1 parametrization of a surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1376.2 parametrization of spherical surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1386.3 parametrization of toroidal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1396.4 parametrization of ellipsoidal surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1406.5 (i) standard square S (ii) parallelogram A

�S � . . . . . . . . . . . . . . . . . . . . . . . . 141

6.6 scalar field over a parametrized surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1426.7 solid between spheroidal and ellipsoidal surfaces . . . . . . . . . . . . . . . . . . . . . . . 150

1.1 Partition for Function involving Square Roots . . . . . . . . . . . . . . . . . . . . . . . . . 203

2.1 (i) tetrahedron T, with triangle∆ of constant height (ii) ∆ in the xy–plane . . . . . . . . . . . 2092.2 (i) face of T in the xz-plane (ii) y fixed in ∆ in the xy–plane . . . . . . . . . . . . . . . . . . 2102.3 hyperbolic section of a (full) cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2132.4 the (cutaway) solid ellipsoid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

xi

xii LIST OF FIGURES

c�


Chapter 0

Organization of math3602 2000-2001

under preparation 19/sept/2001

1

2 CHAPTER 0. ORGANIZATION OF MATH3602 2000-2001

c�


Chapter 1

Integration from First Principles

1.1 Summary

We will define the concept closed interval in n � space, Rn. For n � 1 this is a line segment, for n � 2 arectangle, for n � 3 a box like object which you might call a rectangloid and for n � 4 you might call it ahyper-rectangloid. We will compute many types of integral in this course the easiest of which will be, theintegral of a scalar field over such a closed interval in Rn.

We define the span of a closed interval i.e.the length of the longest line segment that lies inside it. Wedefine the measure,µ, of a closed interval; for dimensions n � 1 � 2 or 3 this measure is length, area, orvolume; for n � 3 you might use the word hyper-volume.

We break an interval up into a collection of smaller intervals, like a box holding cube sugar. This is calledtaking a partition P of the interval. The largest span taken over all subintervals in the partition is called themesh of the partition.

Then we are ready to define and study �I

f

the integral of the scalar field f over the interval I � Rn. First we introduce the Riemann Sum

R�f � P �

of the function f over a partition P of I. �I

f will be defined to be the limit of the Riemann Sum as the

partition is made more and more refined and its mesh tends to zero. Details and examples occupy the rest ofthe chapter.

1.2 Closed Intervals, Hyper-Rectangloids

1.2.1 definition

Let

a �

��

a1

a2...a j...an

��

and b �

��

b1

b2...b j...bn

��

be vectors in Rn with a j b j � 1 j n. The set

I � � a � b �3

4 CHAPTER 1. INTEGRATION FROM FIRST PRINCIPLES

�

��

a1

a2...a j...an

��

�

��

b1

b2...b j...bn

��

�� a1 � b1 �� a2 � b2 �� a j � b j �� an � bn ��

n

∏j 1

� a j � b j �� x �� a j x j b j � 1 j n �

is known as the closed interval or hyper-rectangloid delineated in Rn by a and b.

(i) example in RIf p � q R then � p � q � is a closed interval or line segment in R.

(ii) example in R2

In f

p ��

ab � and q �

�cd �

then

I � � p � q � �� ab � �

�cd �� a � c �� b � d � �� x �

�xy � �� a x c

andb y d

� ��(iii) example in R3

If

p �� a

bc

�� and q �� d

ef

��then

I � � p � q � �� abc

��

�� def

�� a � d �� b � e �� c � f � � �� x �� x

yz

��

a x dand

b y eand

c z f

� ��1.2.2 measure µ

We define the measure µ�I � of the interval I � � a � b �� Rn as follows

µ�I �� ∏

1 � j � n

�b j � a j �

remark Of course in low dimensions n � 0 � 1 � 2 or 3 µ�I � denotes the length, area or volume respectively

of I. In higher dimensions one might refer to µ�I � as hyper-volume.

example in R3

µ

�� 134

��

�� 257

�� 2 � 1 ��

5 � 3 �� 7 � 4 � � 1 � 2 � 3 � 6

c�


1.3. PARTITIONS 5

remark The measure or hypervolume of an n-dimensional hyper-rectangloid is equal to the product ofthe lengths of all n sides. Part of the theory of integration is to extrapolate from this definition to obtain themeasure of any (reasonable) set in E � Rn � n � 1, as the integral of the constant function f � 1 over E. see(for example) problem 4.

1.2.3 span

We define the span of the subset S � Rn

span�S � � sup � � � w � v � � � v � w S

�remarkquad Thus span

�S � is the upper bound of distances taken between pairs of points in S. In the case which

particularly interests us, when the set S is an interval I � � a � b � � Rn, the span is of course the distancebetween diagonall opposed corners.

spanI � span � a � b � � � � b � a � � ��

n

∑j 1

�b j � a j � 2

example in R3

span

�� 134

��

�� 257

�� 2 � 1 � 2 �

5 � 3 � 2 �7 � 4 � 2 � � 1 4 9 � � 14

1.3 Partitions

1.3.1 definition

Let I � � a � b �� Rn be a closed interval. A collection P � � Iα � 1 Iα N�

of subintervals of I is knownas a partition of I iff

(i) I � �1 � α � N

Iα

(ii) Iα � Iβ � Bdry�Iα

� � Bdry�Iβ

� � for α �� β.

Thus I must be covered by the collection of subintervals which must overlap each other only trivially i.e.only meet along their boundaries. For the record we must say what is meant by the boundary of an intervalI � � a � b � � Rn .

Bdry�I � � x �� x j � a j or x j � b j � for at least one j � 1 j n �

1.3.2 mesh

definitionquad The mesh mesh

�P � of the partition P � � Iα � 1 Iα N

�is given by

mesh�P � � max � span

�Iα

� � 1 α N�

remark A partition of an interval in R2 looks like a net laid over a rectangle. The mesh, in this analogy,would be the size of the largest hole in the net, i.e. the size of the largest fish which could possibly escape byfinding and swimming through the hole of largest span.


jbquig-UCD


1.4 introducing the Riemann Integral

1.4.1 Rieman Sum

definition Letf : I � � a � b � � Rn � x �� f

�x � R

and let P � � Iα � 1 Iα N�

be a partition of I. For each α � 1 α N let xα Iα. The expression

R�f � P � �

N

∑α 1

f�xα

� µ�Iα

�

is known as a Riemann Sum of f over the partition P .remark The Riemann Sum is obtained by breaking the hypervolume up into very small pieces, computing—function value multiplied by measure— over each piece, and adding to a total.

1.4.2 the Riemann Integral

definition Letf : I � � a � b � � Rn � x �� f

�x � R

and take a sequence of partitons Pk � 1 k of I with the property that

limk � ∞

mesh�Pk

�� 0

then we define�

I f , the Rieman integral of f over the interval I � Rn, as

�I

f � limk � ∞

R�f � P �

remark The integral is the limit of Riemann sums taken over partitions of smaller and smaller mesh. Inother words granted that the concept of integral is valid one can obtain an arbitrary close approximation tothis integral by computing a Riemann sum over a partition of sufficiently fine (small) mesh.

1.5 Justifying the concept of Riemann Integral

remark Their are difficulties; is ”integral” well defined? Indeed given fixed f � I and P their are manyRiemann sums possible, indeed we can choose xα Iα in many possible ways for each subinterval Iα1 α N. However the supremum of these R

�f � P � is the Upper Sum

U�f � P � � ∑

1 � α � N

Mα � µ�Iα

�

and the infimum of these R�f � P � is the Lower Sum

L�f � P � � ∑

1 � α � N

mα � µ�Iα

�

Here Mα and mα are the supremum and infimum of values of the function f over the subinterval Iα � 1 α N.

Mα� sup � f

�x � � x Iα

�� mα

� in f � f�x � � x Iα

�since by definition

mα f�xα

� Mα � xα Iα � 1 α N

we deduceL

�f � P � R

�f � P � U

�f � P �

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1.6. EXAMPLE OF INTEGRATION FROM FIRST PRINCIPLES IN R3 7

The variation in R�f � P � can be measured, it is

U�f � P � � L

�f � P � � ∑

1 � α � N

�Mα � mα

� µ�Iα

�

for a fixed partition P . If this variation can be demonstrated to be small, in fact arbitrarily small, then therewill be no ambiguity in the definition of the integral. Now we toss in the indispensible requirement

f must be continuous over its domain I a closed bounded interval

In this case Mα � mα is the variation (wobble) of the value of f over the subinterval Iα. But if we takemesh

�P � to be small, say

mesh�P � δ

then span�Iα

� � δ for each α � 1 α N then appealing to CONTINUITY which is UNIFORM under thepresent hypotheses, we can assume that the variation of the value of f over Iα is small, say

Mα � mα ε

Thus

U�f � P � � L

�f � P � � ∑

1 � α � N

�Mα � mα

� µ�Iα

�

� ∑1 � α � N

εµ�Iα

�

� ε ∑1 � α � N

µ�Iα

�

� εµ��

1 � α � N Iα�

� εµ�I �

This latter εµ�I � can be made arbitrarily small by taking a Partion P of mesh less than δ a sufficiently small

number. To sum it all up

theorem 1 Let f : I � x � � f�x � R be continuous over its domain a closed bounded interval I � Rn then

(i) �I

f exists.

Let Pk � k � 1 be a sequence of partitions of I such that mesh�Pk

� � 0 as k � ∞, and for each k � 0 letR

�f � Pk

� be a Riemann sum of f over the partition Pk. then

(ii) �I

f � limk � ∞

R�f � Pk

�

remark Do not despair that this theory will go on much longer. We will soon reach an example. In fact,here is the example we will do first. We will compute the volume of a solid tetrahedron in R3. In detail:-

1.6 example of Integration from First Principles in R3

1.6.1 introduction to the tetrahedron

definition Let a � b � c � 0

T � T3 �a � b � c �� x �

�� xyz

�� x � y � z � 0 andxa

y

b zc

1

� ��February 13, 2003 c

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remark To understand this important solid T see figure 2.1. Also, recall that in R2 the equation of the lineL which intercepts the x � axis at x � a and the y � axis at y � b is

xa

yb� 1

It is not hard to prove that the equation of the plane P � R3 intercepting the x � y � z � axes respectively atx � a � y � b � z � c is

xa

yb

zc� 1

Indeed one checks that the three points �� a00

��

�� 0b0

��

�� 00c

�� satisfy this equation. Next note

that T is given by four inequalities

x � 0 � y � 0 � z � 0 � xa

yb

zc

1

i.e. lies on one side of each of four bounding planes, three of them being axial planes.remark Note that

T � I � � 0 � p � � �� 000

��

�� abc

�� R3

1.6.2 the volume of T as an integral

Let f be the characteristic function of T i.e. the scalar field on I which is identically 1 on T and identically 0off T.

f : I � � 0 � p � � � R � f�x � �

�1 � if x T;0 � otherwise

We will compute the volume or measure µ�T � as our first example of integration from first principles.

µ�T � � µ � T3 �

abc �� I

f

1.6.3 partitions, mesh and span

Let N � 0 be an integer, we begin by partitioning � 0 � a � � R into N subintervals of equal length, and similarlyfor � 0 � b � and � 0 � c � . From this we produce a partition of I � � 0 � p � � R3 into N3 sub-rectangloids of equalvolume. � 0 � a � � � N

1 Ii � where Ii� � a �

i � 1 �

N� aiN � � 1 i N

� 0 � b � � � N1 I j � where I j

� � a �j � 1 �N

� a jN � � 1 j N

� 0 � c � � � N1 Ik � where Ik

� � a �k � 1 �N

� akN � � 1 k N

and putting all three together define our partition PN of I � R3

I � � 0 � p � � � 0 � a �� 0 � b �� 0 � c � � �

1 � i � j � k � N Ii � j � kwhere

Ii � j � k � Ii � I j � Ik� �

��

a�i � 1 �N

a�j � 1 �N

a�k � 1 �N

� ��

�� aiNa jNakN

�� 1 i � j � k N

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1.6. EXAMPLE OF INTEGRATION FROM FIRST PRINCIPLES IN R3 9

Next we compute spans and measures of subintervals and the mesh of PN

span�Ii � j � k � �

�� ai

Na jNakN

��

a�i � 1 �N

a�j � 1 �N

a�k � 1 �N

� ��

�aiN � a

�i � 1 �N � 2 �

b jN � b

�j � 1 �N � 2 �

ckN � c

�k � 1 �N � 2 � 1 � 2

�� aN � 2 �

bN � 2 �� c

N � 2 � 1 � 2� � a2 b2 c2

N

All N3 subintervals have the same span which thus equals the mesh of the partition.

mesh�PN

�� a2 b2 c2

N

Again all N3 subintervals have the same measure µ, indeed given i � j and k � 1 i � j � k N

µ�Ii � j � k � �

�aiN � a

�i � 1 �N � � �

b jN � b

�j � 1 �N � � �

ckN � c

�k � 1 �N � �� a

N � � �bN � � � c

N � � abcN3

1.6.4 computing the Rieman Sum

Next let us pick a convenient point in each subinterval of PN , perhaps the best point to pick would be at thecenter but that is messy, the extreme corner of a subinterval is easy to work with, thus put,

xi � j � k �� ai

Nb jNckN

�� Ii � j � k � 1 i � j � k N

By now we have assembled all the details necessary to compute the Riemann sum.

R�f � PN

� � ∑ ∑1 � i � j � k � N

∑ f�xi � j � k � µ

�Ii � j � k �

� ∑ ∑1 � i � j � k � N

∑ f�xi � j � k � abc

N3

� abcN3 ∑ ∑

1 � i � j � k � N∑ f

�xi � j � k �

Now we take into account that f takes only the values 0 and 1. We remove from the sum all cases wheref � 0. Thus we sum only when 1 i j k N and in these cases f � 1.

� abcN3 ∑ ∑

1 i � j � k N1 i � j � k N ∑ 1

We simplify the indexing a little by removing a superfluous “ N”.

� abcN3 ∑ ∑

1 i � j � k1 i � j � k N ∑ 1

Consider now the statement “ Tom has T � 1 punt, Dick has D � 1 punt, Harry has H � 1 punt and togetherthey have at most 100 punt, i.e. T D H 100 ”. This statement is equivalent to, “ Tom has T punt where1 T 100 and Dick has D punt where 1 D 100 � T and Harry has H punt where 1 H 100 � T � D ”


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� abcN3 ∑

1 � i � N

�∑

1 � j � N � i

�∑

1 � k � N � i � j

1 � �

� abcN3 ∑

1 � i � N

�∑

1 � j � N � i

N � i � j �

� abcN3 ∑

1 � i � N

�∑

0 � J � N � i � 1J �

� abcN3 ∑

1 � i � N

�N � i � 1 � �

N � i �2

� abc2N3 ∑

1 � i � N

�N � i � 1 � �

N � i �

� abc2N3 ∑

1 � i � N

�I � 1 � �

I �

� abc2N3 ∑

1 � i � N

�I2 � I �

� abc2N3� ∑

1 � i � N

I2 � ∑1 � i � N

I�

� abc2N3

� N �N 1 � �

2N 1 �6 � N

�N 1 �

2 �� abc

12N3 � N�N 1 � �

2N 1 � � 3N�N 1 � �

� abc12N3 � N

�N 1 � �

2N 1 � � 3N�N 1 � �

� abc12

� 1�1 1 � N � �

2 1 � N � � 3�1 1 � N � �

1 � N � �To sum up, we have proven that given the partition PN of I

mesh�PN

� � abcN3 � � 0 � asN � � ∞

and that

R f PN� abc

12� 1

�1 1 � N � �

2 1 � N � � 3�1 1 � N � �

1 � N � �1.6.5 taking the limit of the Rieman Sum

Thus the integral may be obtained by passing to the limit as N � ∞

µ�T �

� limN � ∞

R�f � PN

�

� limN � ∞

abc12

� 1�1 1 � N � �

2 1 � N � � 3�1 1 � N � �

1 � N � �� abc

12� 1

�1 0 � �

2 0 � � 3�1 0 � �

0 � �� abc

12� 1 � 1 � � 3 � 1 � 0 �

� abc6

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1.7. INTEGRATION OF A DISCONTINUOUS FUNCTION 11

1.7 Integration of a discontinuous function

more theory So far in the theory of integration we have asserted the existence of the integral only forcontinuous functions and only where the domain of integration is a closed bounded interval. We must relaxboth of these restrictions for our theory to be useful. Both weaknesses of the theory appear already in our

first example above. We computed the volume of the tetrahedron T as �I

f when we really want to write

�T

1, it would be convenient to have the concept of integral over non rectangular domains. The artificial

way we got round this was to extend the function 1 on T to the characterictic function f on I � T � f beingzero outside T . But this brings us up against the second difficulty, the constant continuous function 1 on Thas been replaced by f on I which unhappily is not continuous. Indeed as x passes from the interior to theexterior of T then f

�x � drops instantly from 1 to 0, thus f (the characteristic function of T ) is discontinuous

at each point on the boundary of T .

In general if f : S � Rn � R is a function on a bounded set S we define �S

f to be�

I g where I � S is an

interval and g is simply f extended to be 0 outside S. Now to prove that �S

f � �I

g exists we study the set

D of discontinuities of g. This set D can be broken into two parts D � A�

B, here A is the set of intrinsicdiscontuities of f and B lies in the boundary set of S and contains discontinuities of g caused by instantaneoszeroing. After this discussion we see there is really only one problem, can the integral exist even if f isdiscontinuous? The answer is, f : I � R exists if f is bounded and its set of discontinuities has measurezero, the details follow.

definition A subset S is said to be bounded iff S � I for some closed interval I.

definitionquad The bounded set B � I is said to be of measure zero iff given any number ε � 0 no matter how small wecan find a partition P � “I �

� N1 Iα” of the interval I such that the sum of the measures of those subintervals

which intersect B is less than ε.example A circle curve has measure zerol in R2 as does a line segment, indeed so does any reasonablecurve.example The spherical surface, the toroidal surface, indeed most bounded surfaces in R3 have measurezero.remark Now we are ready for the formal statement of our main existence theorem for integrals.

theorem 2 Let f : I � Rn � � R be bounded and continuous except possibly on a set of measure zero.Then

(i) �I

f exists.

In that case if � Pk � k � 1�, is a sequence of partitions s.t. mesh

�Pk

� � 0 as k � ∞ then

(ii)�

I f � limk � ∞ R�Pk � f �

i.e. the integral can be approximated arbitrarily closely by a Riemann Sum of sufficiently small mesh.

corollary If f : S � Rn � R is a function whose domain S is bounded, then �S

f exists if (i) f is

bounded, (ii) f is continuous except possibly on a set of measure zero and (iii) the boundary of S is a set ofmeasure zero.

quasi proof Consider the crucial term mentioned above in this context.

N

∑α 1

�Mα � mα

� µ�Iα

�


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as before for the continuous case we must prove that with apt choice of the partition this sum can be madearbitraily small. Let D � I denote the set of discontinuities of f and put� α � 1 α N

� � A�

CA

whereA �� α � 1 α N and Iα

�D �� φ

�Then

∑α � A

�Mα � mα

� µ�Iα

�� ∑α � C A

�Mα � mα

� µ�Iα

� ∑α � A

�Mα � mα

� µ�Iα

�

We will ignore the second sum on the RHS it concerns subintervals where f is continuous and can be madearbitrarily small by apt choice of the partition P just as in the continuous case, see above.

Next recall that f is assumed bounded on I, by which we mean that we can find M and m R such that

m f�x � M � for all x I

Thus the other summand

∑α � A

�Mα � mα

� µ�Iα

� ∑α � A

�M � m � µ

�Iα

�

� �M � m � ∑

α � A

µ�Iα

�

Again by apt choice of the partition this term can be made arbitrarily small, indeed the factor ∑α � A

µ�Iα

� can

be made arbitrarily small, since D is a set of measure zero.Q.E.D.

The question sheet on this material follows the appendices

1.8 Appendices:- Integrations performed by Computer Program

remark The method of computing integrals from first principles, i.e. using Riemann Sums can be carriedout by a computer. I include some programs in Pascal, see appendix A. By request, but only for those whoare interested, I have produced an integration program written in C, see Appendix B.

1.9 problem set

Integration from First Principles

1. Let I � � 0 � 1 �� 0 � 1 � � R2 and f�x � y � � � xy for

�x � y � I. Let

Pn�

� � � i � 1n � 2

��

in � 2 � � � � j � 1

n � 2

��

jn � 2 � �� 1 i � j n �

(i) Sketch the partition Pn of I in the case n � 4 and indicate the mesh on your sketch.

(ii) Compute mesh�Pn

� and prove that mesh�Pn

� tends to zero as n tends to infinity.

(iii) Compute ��I

f�x � y � d

�x � y � by taking the limit as n tends to infinity of a Riemann sum of f

over the partition Pn.

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1.9. PROBLEM SET 13

2. Let I � � 0 � 1 �� 0 � 1 �� 0 � 1 � � R3 , ∆ � � x I � x y z 1�

and f ; I � R3 � R be such thatf

�x � � 1 if x I and f

�x � � 0 otherwise.

(i) Sketch ∆ � I � R3.

(ii) Find the volume of ∆ by computing a Riemann sum of f , over the partition Pn of I, and takingthe limit as n tends to infinity.

Pn�

� � i � 1n

� in�� j � 1

n� jn�� k � 1

n� kn� �� 1 i � j � k n �


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Chapter 2

Fubini’s Theorem

2.1 summary

To compute an integral over a domain in Rn, four methods spring to mind

by first principlesSee chapter1. This method is not really practical being too difficult.

by first principles using a computerSee � 1.8. This makes it easy to work out the Riemann Sum; but only yields an approximate result.Also decimals can appear meaningless. For example who would spot that 4 � 1888 � 4π � 3.

Fubini’s TheoremThis method reduces a single (difficult) integral over n � space to n (easier) integrals each of one vari-able, our task in this chapter will be to study this method in detail.

Change of Variable TheoremThis is the most useful method; it relies on Fubini’s theorem, see chapter 5.

2.2 statement and exposition of Fubini’s theorem

We begin our study of Fubini’s Theorem.

2.2.1 Decomposition of Intervals

Let I1� � a � b � � Rk and I2

� � c � d � � Rl and let I be the product set, also a closed interval as follows

I � I1 � I2� � a � b �� c � d � � � � a

c � ��

bd � � � Rk � Rl � Rn � where n � k l

Consider next a scalar field f over I, taking into account the method by which I was constructed we use thefollowing notation for f .

f : I � I1 � I2 � Rk � Rl � Rn � � R�xy � � � � f

� � xy � � � f

�x � y �

15

16 CHAPTER 2. FUBINI’S THEOREM

��

x1

x2

...xk

y1

y2

...yl

� ��

� � � f�x1 � x2 � � � � � xk � y1 � y2 � � � � � yl �

Next, given an arbitrary but fixed x I1 � Rk we define a function whose name is “ f�x � � � ” acting on the

variable y I2 � Rl , the details are,

f�x � � � : I2 � Rl � � R

y � � � f�x � � � �

y � � f�x � y �

If the original function f is continuous it is easy to prove that f�x � � � is, for each x I1, a continuous function

of y I2 and so that f�x � � � can be integrated over I2 w.r.t. dy, yielding a new function F of x over I1.

F : I1 � Rk � � R

x � � � F�x � � �

I2 � Rlf

�x � � � � �

I2 � Rlf

�x � y � dy

The original function f being continuous one can readily prove that the new function F is also continuous.But then the integral �

I1 � RkF � �

I1 � RkF

�x � dx

is well defined. We are ready for the informal statement of Fubini’s Theorem.

�I � Rn

f � �I1 � Rk

F

Let us replace F by its full definition.

�I � Rn

f � �I1 � Rk

� �I2 � Rl

f�x � � � �

With notation given in detail this becomes

�I � Rn

f�x � y � d

�x � y � � �

I1 � Rk

� �I2 � Rl

f�x � y � dy � dx

Now we are ready for the formal statement of Fubini’s Theorem for continuous functions.

theorem 3 Let I1 � Rk and I2 � Rl be closed bounded intervals and I � I1 � I2 � Rn where n � k l. Let

f : I � �x � y � � � � f

�x � y � R

be a continuous function. Then all is well defined and equality holds in

�I � Rn

f�x � y � d

�x � y ��

I1 � Rk

� �I2 � Rl

f�x � y � dy � dx

Fubini’s theorem and discontinuous functions

RemarkThe requirement that f is continuous everywhere may be relaxed. Fubini’s theorem holds if f is everywherecontinuous except on a set of measure zero, this is the version for which we will have the most use.

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2.2. STATEMENT AND EXPOSITION OF FUBINI’S THEOREM 17

2.2.2 first examples

example in R3 Let

g : I �� a

bc

��

�� def

�� R

then applying Fubini’s Theorem

� ��I

g�x � y � z � d

�x � y � z � � � �� a

b � � � de ��

� � f

cg

�x � y � z � dz � d

�x � y �

and applying Fubini’s theorem again

� � d

a

� � e

b

� � f

cf

�x � y � z � dz � dy � dx

Of course, according to the order in which we disassemble I

dxdydz � dxdzdy � dydxdz � dydzdx � dzdxdy or dzdydx

we can obtain six variants, here is one other in full detail

� � f

c

� � e

b

� � d

af

�x � y � z � dx � dy � dz

In difficult examples an apt choice of order minimizes computation.

example in R3

We now do a very easy straight forward example. Note the tricks, for use in harder examples, which follow.

� �� 1 � 3 �� 3 � 4 �� 2 � 7 xy3z2 d�x � y � z � � � 7

2

� � 4

3

� � 3

1xy3z2 dx � dy � dz

The decomposition can be carried out in six different orders, in general one would take care to choose thatorder which minimizes subsequent computation, in this case it does not matter. Note that the inner integralmust be carried out first. Since y3z2 is constant w.r.t. dx we move this expression across the inner integralsign.

� � 7

2

� � 4

3y3z2

� � 3

1xdx � dy � dz

� � 7

2 � 4

3y3z2 x2

2�� 31 dy dz

� � 7

2

� � 4

3y3z2

�82 � dy � dz

� � 7

2

� � 4

34y3z2 dy � dz

The expression 4z2 is constant w.r.t. dy and can be moved across the (new) inner integral sign

� � 7

24z2

� � 4

3y3 dy � dz

� � 7

24z2 y4

4�� 43 dz


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x

y

z

x

y

z

Figure 2.1: (i) the ball B , (ii) the (cutaway) hemiball H

� � 7

24

256 � 814

z2 dz

� � 7

2175z2 dz

Next we move the constant 175 across the remaining integral sign.

� 175 � 3

1z2 dz

� 175z3

3�� 31

� 17527 � 1

3

� 17579

2.3 example, integration over the ball in 3-space

Consider the region in R3

B � B3 �a �� x �� x2 y2 z2 a2 �

being the solid ball of radius a � 0 and center 0 in R3. We will compute three integrals over B. We assumethat B is made of a homogeneous material of density ρ � 1. The first integral m is the mass=volume of B,the second M is the moment of B about the xy—plane and the third I is the moment of inertia of B about thez–axis. Actually the second computation is not interesting, by symmetry M � 0; we change the domain ofintegration to the hemiball H � H3 �

a �� x �� x2 y2 z2 a2 and z � 0 � . The integrals are

m � m�B3 �

a � � � � � �B

1d�x � y � z �

M � M�H3 �

a � � � � � �H

zd�x � y � z �

I � I�B3 �

a � � � � � �B

x2 y2 d�x � y � z �

We mention the interesting quantities,

z � Mm

and r ��

Im � 2

the z coordinate of the center of mass of H and the radius of gyration of B respectively. We next compute allthree integrals using Fubini’s theorem.

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2.3. EXAMPLE, INTEGRATION OVER THE BALL IN 3-SPACE 19

2.3.1 limits for the solid ball B

Setting up limits is usually the hardest part of an application of Fubini’s theorem. The current example isnot particularly difficult, nonetheless we study it in great detail. One could use this example as a template inmore demanding situations, say where dimension might be � 3 and drawing impossible.

limits for B, obtained from drawings

The six limits for B can be obtained from drawings, This method is slow and instructive but often preferred(especially by beginners) to the fast (too) slick method of inequalities used next. In figure 2.2(i) we see

x

y

z

D

a

x

z

A

z a

Figure 2.2: Ball (i) cut by a plane of constant height (ii) crossection in xz–plane

B � B3 �a � , we see the extremes that z can obtain on B which give us the limits for z.� a z a

We also see that a plane on which z is constant cuts the solid ball B in a disc D. Figure 2.2(ii) shows thecross-section of B in the xz–plane: from the right angled triangle, using Pythagoras’ theorem, we see that

disc D has radius A � � a2 � z2. This disc D is the range of

�xy � with z fixed.

x

y

y A

L

Figure 2.3: line segment L, where y is constant, on disc D in xy–plane

Figure 2.3 shows the disc D in xy–space. On D the extreme values attained by y are � the radius A of thedisc, i.e.

� �a2 � z2 y �

a2 � z2

We also see that y is constant on the disc D along a line segment L; Apply Pythagoras theorem to the rightangled triangle in figure 2.3 to see that on L


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� �a2 � z2 � y2 � � �

A2 � y2 x �A2 � y2 � �

a2 � z2 � y2

We have found all six limits for B

limits for B, obtained from inequalities

Step 1Consider the inequality which defines the B3 �

a � , (a solid region in R3 is described by one or more inequali-ties, i.e. not by equations) x2 y2 z2 a2 . Since x2 y2 � 0 the range of z is maximum when x � 0 � y,i.e. on the z-axis and this maximum range is

z2 a2 � � a z a

Let us fix z in this range. Then

�xy � moves governed by the inequality x2 y2 a2 � z2 obtained by

moving z now fixed and constant to the R.H.S. of the inequality which defines B. Thus

�xy � wanders

through the disc

D � Dz� B2 � �

a2 � z2 �

center 0, radius A � � a2 � z2 , in R2. If f is any function on the ball B we can now apply Fubini’s theoremobtaining.

��B

f�x � y � z � d

�x � y � z � � � a

z � a

� ��D

f�x � y � z � d

�x � y � � dz

Step 2 Now z has been fixed and

�xy � has obeys the inequality x2 y2 a2 � z2. Thus y attains its

maximum range when x � 0 and this range is

y2 a2 � z2 � � �a2 � z2 y �

a2 � z2

Fixing y in this interval, and moving y now constant to the R.H.S. of the inequality which defines D we obtainthe range of x

x2 a2 � z2 � y2 � � �a2 � z2 � y2 x �

a2 � z2 � y2

Applying Fubini’s theorem a second time we obtain.

� ��B

f�x � y � z � d

�x � y � z � � � a

z � a � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y2f

�x � y � z � d

�x � dy dz

2.3.2 computing m for the solid ball B

We have twice found all six limits for the solid ball B. We are ready to apply Fubini’s theorem, using theselimits, to compute the integral m

m � � ��B

1d�x � y � z �

� � a

z � a � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y21dx dy dz

� � a

z � a � � a2 � z2

y � � a2 � z2x�� a2 � z2 � y2

� � a2 � z2 � y2dy dz

� � a

z � a � � a2 � z2

y � � a2 � z22

�a2 � z2 � y2 dy dz

c�


2.3. EXAMPLE, INTEGRATION OVER THE BALL IN 3-SPACE 21

We started by carrying out the inner integral w.r.t. dx. Next we must carry out the new inner integral, i.e.w.r.t. dy. Since z is constant in the dy integral we write A � �

a2 � y2 constant, for the present only.

� a

z � a

� � A

y � A2

�A2 � y2 dy � dz � � r

z � r

� � π � 2θ � π � 2 2 � Acosθ � Acosθdθ � dz

Here we used the substitution y � Asinθ so that

dy � Acosθdθ ��

A2 � y2 � Acosθ � y � � A � θ � � π � 2

� a

z � aA2

� � π � 2θ � π � 2 2cos2 θdθ � dz � � r

z � rA2

� � π � 2θ � π � 2 1 cos2θdθ � dz

Here we moved the constant A2 (w.r.t. dy) across the inner integral sign and applied the double angle formulacos2θ � 2cos2 θ � 1. Note that cos2θ integrated across a full period π yields zero.

� � a

z � aA2

� � π � 2θ � π � 2 1dθ � dz

� � a

z � aA2 � θ � π � 2

� π � 2 � dz

� � a

z � aπA2 dz

� π � a

z � aa2 � z2 dz

Here we moved the constant π across the integral sign and put back the value of A.

� π�a2z � z3 � 3 � �� az � a

� π � �a3 � a3 � 3 � � � � a3z a3 � 3 � �

� 43

πa3

m � �� B

1d�x � y � z � � 4

3πa3

2.3.3 computing M for the solid ball B

Next we compute the second of the three required integrals.

M � � � �H

zd�x � y � z �

� � a

z 0 � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y2zdx dy dz

The integrand z is constant w.r.t dx and so we can move it outside the core integral, but z is also constantw.r.t. dy and so we can move it outside the dy integral. Our two inner integrals will be simplified, in factthey become the same as in the previous computation and so we can go right to the outer dz integral. Thisshows that the order we choose in our decomposition dxdydz was apt. Unhappily this will not be the case inour third computation to follow. One more thing, do not forget that we are working over the hemiball H: theouter limits are no longer z � � a to z � a but z � 0 to z � a.

� a

z 0z � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y21d

�x � dy dz


jbquig-UCD


� π � a

0z

�a2 � z2 � dz see above

� π � a

0a2zdz � π � a

0z3 dz

� πa2 z2

2�� a0 � π

z4

4�� a0

� πa4

2 � πa4

4

� πa4

4

� 12

�43

πa3 � �38

a �Thus

M � � ��H

zd�x � y � z � � 3a

8m2

The center of gravity of the solid hemiball is z � 38

a above the xy-plane.

2.3.4 computing I for the solid ball B

Next the last of our three integrals.

I � �� B

�x2 y2 � d

�x � y � z �

If we attempt to us Fubini’s theorem with the limits above we will have to compute

I � � a

z 0z � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y2x2 y2 d

�x � dy dz

Horrors! the integrals at the core are messy, and the outer will be messier still. We use a clever trick whichpostpone all work to the last (outer) integral By symmetry of the solid ball B

� ��B

x2 d�x � y � z � � � � �

By2 d

�x � y � z � � � ��

Bz2 d

�x � y � z �

and so

I � � � �B

�x2 y2 � d

�x � y � z �

� 2 � � �B

z2 d�x � y � z �

� 2 � a

z � a � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y2z2 d

�x � dy dz

� 2 � a

z � az2 � � a2 � z2

y � � a2 � z2 � � a2 � z2 � y2

� � a2 � z2 � y21d

�x � dy dz

� 2 � a

z � az2 � π

�a2 � z2 � � dz

We have mimicked the argument for M, how fast and easy it has been.

� 2π � a

� aa2z2 � z4 dz

c�


2.4. EXAMPLE, INTEGRATION OVER THE SOLID CONE IN R3 23

� 2 π�

a2z3

3 � z5

5 � �� a� a

� 2π�

2a5

3 � 2a5

5 �� 4πa5

�13 � 1

5 �� 4πa5

�5 � 3

15 �� 8

15πa5

��

25

a2 � �43

a3 �Thus the moment of inertia I of the solid ball B3 �

a � is 25 ma2 from which we deduce that the radius of

gyration about a diameter is ρ ��

25

a .

I � � � �B3 �

a ��x2 y2 � d

�x � y � z � � m �

25

a 2

2.4 example, integration over the solid cone in R3

The solid right circular cone, whose base of radius r � 0 lies in the xy-plane, with axis of symmetry the z-axisand height h � 0 is the set

C ��

x�� 0 z h 1 � �

x2 y2

r �We will soon see how to obtain these inequalities which define C. We will compute the mass m, the momentM about the xy-plane, and the moment of inertia I of C about the z-axis, the axis of symmetry, from thesewe will compute the height, z, of the center of gravity above the base and the radius, r, of gyration aboutthe z–axis. Of course we will assume that C is made of a homogeneous material of density ρ � 1. We mustcompute

m � �� C

1d�x � y � z � � M � ��

Czd

�x � y � z � � I � � � �

C

�x2 y2 � d

�x � y � z �

z � Mm

� r ��

Im

Here is how we arrive at the inequality which defines the solid cone C. Consider the line L (see figure 2.5)in the xz–plane with axial intercepts x � r � 0 and z � h � 0. The equation of L is

xr

zh� 1

Rotating this line about the z axis we obtain a cone surface Σ � R3, see figure 2.4(ii) and 2.4(i). The equationof Σ is �

x2 y2

r z

h� 1 � z � h 1 � �

x2 y2

r being obtained from that of L by substitution of

�x2 y2 for x (note the trick, used to obtain equation of

a surface of revolution). We are interested in the solid cone C, see figure 2.4, being the region above thexy � plane and under the surface Σ. Thus

C ��

x�� 0 z h 1 � �

x2 y2

r �February 13, 2003 c

�jbquig-UCD


x

y

z

xr

y

z

h

D

Figure 2.4: (i) cone (ii) disc D of constant height

As in any application of Fubini’s theorem the tricky part is finding the limits of integration.

2.4.1 find limits for C, by two methods

limits from drawingsyr z

h� 1 is the equation of the line L in the yz-plane with intercepts at z � h and y � r on the y and z axes

respectively. Revolving the line L about the z-axis one obtains the solid cone C, see Fig.2.4(i). We see thaton C

0 z h

x

z

z

h-z

r

R

L

x

y

y R

Figure 2.5: (i) section of cone in xz–plane (ii) the disc D in xy–plane

With z fixed in this range

�xy � lies in a disc D, see figure2.5. Using Euclid and similar right angled

triangles in 2.5(i) we obtain the radius R of the disc D.

Rr� h � z

h� R � r

�1 � z

h�

The inequality which describes the disc D is.�

x2 y2 r � 1 � zh �

Thus

� ��C

f�x � y � z � d

�x � y � z � � � h

0��

Df

�x � y � z � d

�x � y � dz where D �� x �� x2 y2 r2 � 1 � z

h ��c

�jbquig-UCD February 13, 2003


On D, see Fig.2.5(ii), y lies between � R where R � r�1 � z

h� is the radius of D.

� r�1 � z

h� y r

�1 � z

h�

Using Pythagoras theorem, see figure2.5(ii) if y is fixed in this range on D the range of x is between � R

� �r2 � 1 � z

h � 2 � y2 x �r2 � 1 � z

h � 2 � y2

We have found all six limits

� � �C

f�x � y � z � d

�x � y � z � � � h

0� r

�1 � z

h�

� r�1 � z

h� � � r2 � 1 � z � h � 2 � y2

� � r2 � 1 � z � h � 2 � y2f

�x � y � z � dxdydz

limits from inequalities

The solid cone C is determined by the inequality

0 z h 1 � �x2 y2

r The range of z on C is given by this inequality This range is maximum when

�x2 y2 � 0, i.e. when

x � 0 � y, i.e. on the z-axis. This maximum range is

0 z h

With z fixed in this range, rearrange the inequality for C with z on the R.H.S..�

x2 y2 r � 1 � zh �

If x � 0 in the latter inequality the range of y is maximizum and is given by

�y2 � � x � r � 1 � z

h � � � r � 1 � zh � y r � 1 � z

h �If y is fixed in this range we obtain the inequality governing x by rearranging the original equality for C sothat both y and z are on the R.H.S.

x2 r2 � 1 � zh � 2 � y2 � r

� � 1 � zh � 2 � y2 x r

� � 1 � zh � 2 � y2

2.4.2 computing m for the solid cone C

Now that the question of the limits has been dealt with we are ready to compute the first of 3 required integrals

m � �� C

1d�x � y � z �

� � h

0 � r�1 � z � h �

� r�1 � z � h � � � r2

�1 � z � h � 2 � y2

� � r2�1 � z � h � 2 � y2

1dx dy dz

� � h

0

� � r�1 � z � h �

� r�1 � z � h �

�2 � r2

�1 � z � h � 2 � y2 � dy � dz

by evaluation of the easy core integral. Next write A � r�1 � z � h � which is constant w.r.t. dy

� � h

0

� � A

� A2

�A2 � y2 dy � dz


jbquig-UCD


Next write y � Asinθ as in an earlier computation above

dy � Acosθdθ � A � � A � θ � � π � 2 ��

A2 � y2 � Acosθ

� � h

0

� � π � 2� π � 2 2Acosθ � Acosθdθ � dz

� � h

0

� � π � 2� π � 2 2A2 cos2 θdθ � dz

� � h

0

� � π � 2� π � 2 A2 �

1 cos2θ � dθ � dz

� � h

0A2

� � π � 2� π � 2 �

1 cos2θ � dθ � dz

� � h

0A2

� � π � 2� π � 2 1dθ � dz

� � h

0πA2 dz

� � h

0π

�r2 �

1 � z � h � 2 � dz

� πr2 � h

0

�1 � z � h � 2 dz

� πr2 � h

0

�1 � 2

zh z

h

2 � dz

� πr2�

z � 2z2

2h z3

3h2 � �� h0� πr2

�h � 2

h2 h

3 �� πr2 � h

3�

The mass or volume of the homogeneous cone is

m � � ��C

1d�x � y � z � � 1

3πr2h

2.4.3 computing M for the solid cone C

We will next compute the second of the 3 required integrals

M � ��C

zd�x � y � z �

� � h

0 � r�1 � z � h �

� r�1 � z � h � � � r2

�1 � z � h � 2 � y2

� � r2�1 � z � h � 2 � y2

zdx dy dz

� � h

0z � r

�1 � z � h �

� r�1 � z � h � � � r2

�1 � z � h � 2 � y2

� � r2 � 1 � z � h � 2 � y21dx dy dz

� � h

0zπr2 �

1 � z � h � 2 dz

where we moved z to the outer integral and the two integrals at the core were seen to be the same as incomputation of m above.

� πr2 � h

0

�z � 2

z2

h z3

h2 � � dz

c�



� πr2

�z2

2 � 2z3

3h z4

4h2 � �� h0� πr2

�h2

2 � 2h2

3 h2

4 �� πr2h2

�12 � 2

3 1

4 �� πr2h2

�6 � 8 3

12 � 23 1

4 �� πr2h2

�1

12 ��

�h4 � �

13

πr2h �Thus the moment M of the cone about its base and the height z of its cog above the base are

M � h4

m �� π12 � r2h2 � z � h

4

2.4.4 computing I for the solid cone C

We will compute the third of 3 required integrals.

I � � ��C

�x2 y2 � d

�x � y � z �

� 2 � � �C

y2 d�x � y � z �

since the cone C is symmetrical w.r.t. x and y

This move will greatly reduce future computation, read on.

� 2 � h

0 � r�1 � z � h �

� r�1 � z � h � � � r2

�1 � z � h � 2 � y2

� � r2�1 � z � h � 2 � y2

y2 dx dy dz

by two uses of Fubini’s th.; limits as above

� 2 � h

0 � r�1 � z � h �

� r�1 � z � h � y2 � � r2

�1 � z � h � 2 � y2

� � r2�1 � z � h � 2 � y2

1dx dy dz

since y2 is independent of dx

� 2 � h

0

� � r�1 � z � h �

� r�1 � z � h � y2

�2 � r2

�1 � z � h � 2 � y2 � dy � dz

� 4 � h

0

� � A

� Ay2 � �

A2 � y2 � dy � dz

by writing A for r�1 � z � h �

� 4 � h

0

� � π � 2� π � 2 A2 sin2 θ � Acosθ � Acosθdθ � dz

write y � Asinθ; y � � A � θ � � π � 2; dy � Acosθ;�

A2 � y2 � Acosθ

� � h

0

� � π � 2� π � 2 A4 �

2sinθcosθ � 2 dθ � dz

anticapate use of sin2θ � 2sinθcosθ

� � h

0A4

� � π � 2� π � 2 sin2 �

2θ � dθ � dz

anticipate use of cos�2x � � 1 � 2sin2 �

x �


jbquig-UCD


� 12� h

0A4

� � π � 2� π � 2 1 � cos

�4θ � dθ � dz

�cos

�4θ � � 0, being over 2 full periods

� 12� h

0A4

� � π � 2� π � 2 1dθ � dz

� π2� h

0A4 dz

recall definition of A above

� π2� h

0r4 �

1 � z � h � 4 dz

� π2

r4 � h

0

�1 � z � h � 4 dz

� π2

r4

� � h5 � �

1 � z � h � 5 �� h0� π

2r4

� � h5 � � � 1 �

��

13

πhr2 � �3

10r2 �

We have found the moment of inertia I and the radius of gyration r of the solid cone C.

I � � ��C

�x2 y2 � d

�x � y � z � � mr2 �

�13

πhr2 � �3

10r2 � � 1

10πhr4 � r �

�3

10r

2.4.5 center of mass, radius of gyration

Writing z and r for height of the center of mass above the base the radius of gyration of the solid cone C, wehave

z � Mm

� h4

and r ��

Im

��

310

r

c�


2.5. PROBLEM SET:- FUBINI’S THEOREM 29

2.5 problem set:-Fubini’s theorem

1. Let a � b � c � 0 and consider the solid tetrahedron

T � �� xyz

�� x � y � z � 0 andxa y

b z

c 1

� ��(i) Sketch T � R3.

(ii) Find limits p � q � r�z � � s

�z � � t

�y � z � and u

�y � z � for T so that

�� T

f�x � y � z � d

�x � y � z � � � q

p� s�z �

r�z �

� v�y � z �

u�y � z � f


for any scalar function f defined over T.

(iii) Compute,

m � �� T

1d�x � y � z � � M � � ��

Tzd

�x � y � z � � I � ��

T

�x2 y2 � d

�x � y � z �

Here m � M and I are respectively the mass, the moment about the base and moment of inertiaabout the z–axis of T ( assuming T is of uniform density ρ � 1).

(iv) Find z � M � m the height of the center of mass of T above the base. Find r ��

I � m, the radiusof gyration of T about the z � axis.

2. Let r� h � 0 and consider

C � �� xyz

��

x2 y2

r z

h 1

� �� R3

the solid right circular cone of base radius r � 0, height h � 0 and axis of symmetry the z–axis.

(i) Sketch C � R3.


�z � � t


�y � z � for C so that

� ��C

f�x � y � z � d

�x � y � z � � � q

p� s�z �

r�z �

� v�y � z �

u�y � z � f


for any scalar function f defined over C.

(iii) Compute,

m � �� T

1d�x � y � z � � M � � ��

Tzd

�x � y � z � � I � ��

T

�x2 y2 � d

�x � y � z �

Here m � M and I are respectively the mass, the moment about the base and moment of inertiaabout the z–axis of C (assuming C is of uniform density ρ � 1).

(iv) Find z � M � m the height of the center of mass of C above the base. Find r ��

I � m, the radiusof gyration of C about the z � axis.

(v) (*hard; changing order of integration on C)Find p � q � r

�x � � s

�x � � t

�x � z � and u

�x � z � so that

� ��C

f�x � y � z � d

�x � y � z � � � q

p� s�x �

r�x �

� v�x � z �

u�x � z � f

�x � y � z � dydzdx

(vi) (** very hard) Compute all the above quantities again using this different order of integration inFubini’s theorem.


jbquig-UCD


3. Consider the solid ball of radius a � 0

B � �� xyz

�� x2 y2 z2 a2

� ��(i) Sketch B � R3. Sketch the hemiball H � � x � x B and z � 0

�(ii) Find limits p � q � r

�z � � s

�z � � t


�y � z � for B so that

��B

f�x � y � z � d

�x � y � z � � � q

p� s�z �

r�z �

� v�y � z �

u�y � z � f


for any scalar function f defined over B.

(iii) Compute,

m � � ��B

1d�x � y � z � � M � � � �

Hzd

�x � y � z � � I � ��

B

�x2 y2 � d

�x � y � z �

Here m and I are respectively the mass and moment of inertia of B about the z � axis (assumingB is of uniform density ρ � 1). M is the moment of the hemiball H about the xy–plane (if westudied B we would have M � 0 by symmetry; that is boring).

(iv) Find z � M � �m � 2 � the height of the center of mass of H above the base. Find r �

�I � m, the

radius of gyration of B about the z � axis.

4. Consider the solid ellipsoidal ball with semi-axes of lengths a � b and c � 0.

E � �� xyz

�� x2

a2 y2

b2 z2

c2 1

� ��(i) Sketch E � R3.


�z � � t


�y � z � for E so that

��E

f�x � y � z � d

�x � y � z � � � q

p� s�z �

r�z �

� v�y � z �

u�y � z � f


for any scalar function f defined over E.

(iii) Compute,

m � �� B

1d�x � y � z �

Here m are is the mass ( � volume) E, assuming constant density ρ � 1. Other standard integrals,on E, are too difficult to compute.

5. Let a � b � c � d � e � 0 and consider the solid hyper-hedron

T ��

��

vwxyz

�� v � w � x � y � z � 0 and

va w

b x

c y

d z

e 1

� ��(i) Sketch T � R5.

(ii) Find limits p � q�z � � r

�y � z � � s

�x � y � z � � t

�w � x � y � z � for T so that

�� T

f�v � w � x � y � z � d

�v � w � x � y � z � � � p

0� q

0� r

0� s

0� t

0f

�v � w � x � y � z � dvdwdxdydz

for any scalar function f defined over T.

(iii) Compute the hyper volume in 5-space of T

m � �� T

1d�x � y � z �

c�


2.6. APPENDIX A—–PASCAL 31

2.6 APPENDIX A—–PASCAL

Next we will compute some more integrals from first principles but this time we use a computer program todo the calculations. The method is essentially this, to reduce every integral to have domain an interval, topartition the interval very finely, to compute the Riemann Sum over such a partition and to trust that we havein this Riemann sum a reasonably good approximation to the integral. Actually we have above the means tocompute how accurate our estimate is, its better than

N

∑α 1

�Mα � mα

� µ�Iα

�

but we will not persue matters that far, Computing is an appendage to this course, recall we are doingmathematics here. We write in Pascal which reads like Fortran.


jbquig-UCD


}program rs1;(* Computes the integral from a to b of the function x*x. *)

consta = 1.0 ; (* integrate from here *)b = 2.0 ; (* to here *)N = 1000 ; (* num subints in partition*)

varx:real; (* chosen pt in each subint of Riemann sum *)dx:real; (* the span and the mesh and the measure *)RS:real; (* the Riemann sum *)i:integer; (* loop control *)

function f(r:real):real;beginf:=r*r;end;

begindx:=(b-a)/N; (* get mu *)RS:=0; (* start Riemann sum at zero *)for i:= 1 to N do (* traverse the partition *)

beginx:=a+i*dx; (* the choose one end pt of subint *)RS:=RS+ f(x)*dx; (* build up the Riemann Sum *)end;

WriteLn(’’);WriteLn(’Output of program RS1!’);Writeln(’a :- ’, a:5:10);Writeln(’b :- ’, b:5:10);Writeln(’Num Subints:- ’, N:5);Writeln(’Function :- y=x*x’);Writeln(’Riemann Sum:- ’,RS:5:10);

end.{

c�



}Output of program RS1!a :- 1.0000000000b :- 2.0000000000Num Subints:- 1000Function :- y=x*xRiemann Sum:- 2.3348335000

{


jbquig-UCD


}program RS1_1;(*

Computes the integral from a to b of the arbitrary function f. Theactual function f(x)=x*x is used.

*)

(*This program is a variant on program RS1. It computes x byincrements in the main loop. This is faster but not by much.Howeverin future programs in high dimensional spacesignificant gains in speed will result by using this method.The price paid is in complication of the logic.

*)

consta = 1.0; (* integrate from here *)b = 2.0; (* to here *)N = 1000; (* num subints in partition*)

varx:real; (* chosen pt in each subint of Riemann sum *)dx:real; (* the span and the mesh and the measure *)RS:real; (* the Riemann sum *)i:integer; (* loop control *)

function f(r:real):real;beginf:=r*r;end;

begindx:=(b-a)/N; (* get mu *)RS:=0; (* start Riemann sum at zero *)x:=a; (* start x at suitable position *)for i:= 1 to N do (* traverse the partition *)

beginx:=x+dx; (* increment previous x *)RS:=RS+ f(x)*dx; (* build up the R-Sum, note function call *)end;

WriteLn(’’);WriteLn(’Output of program RS1_1!’);Writeln(’a :- ’, a:5:10);Writeln(’b :- ’, b:5:10);Writeln(’Num Subints:- ’, N:5);Writeln(’Function :- y=x*x’);Writeln(’Riemann Sum:- ’,RS:5:10);

end.{

c�



}Output of program RS1_1!a :- 1.0000000000b :- 2.0000000000Num Subints:- 1000Function :- y=x*xRiemann Sum:- 2.3348335003

{


jbquig-UCD


}program rs4;(*Computes the integral of the arbitrary f over the integral J in four space.J=[a,e]x[b,f]x[c,g]x[d,h]. We give the characteristic function of thesolid ball in four space.*)

consta = -1.0; b = -1.0; c = -1.0; d = -1.0; (* one corner of J *)p = 1.0; q = 1.0; r = 1.0; s = 1.0; (* the other *)N1= 20 ; N2= 20 ; N3= 20 ; N4= 20 ; (* partition numbers*)

varw,x,y,z : real;dw,dx,dy,dz : real; (* edges of subinterval *)span,mesh : real;mu : real;i,j,k,l : integer; (* loop control *)RS : real; (* Riemann sum *)

function f(w,x,y,z:real):real;beginif (w*w+x*x+y*y+z*z>1) then f:=0 else f:=1;end;

begindw:=(p-a)/N1;dx:=(q-b)/N2;dy:=(r-c)/N3;dz:=(s-d)/N4;span:=sqrt(dw*dw+dx*dx+dy*dy+dz*dz);mesh:=span;Writeln(’’);Writeln(’’);WriteLn(’Output of program RS4!’);Write(’I am a snail’);mu:=dw*dx*dy*dw;RS:=0;for i:= 1 to N1 do

begin write(i:3); (* Has it crashed or is it running? *)for j:= 1 to N2 do

for k:= 1 to N3 dofor l:= 1 to N4 do

beginw:=a+i*dw;x:=b+j*dx;y:=c+k*dy;z:=d+l*dz; (* slow ugh! *)RS:=RS+f(w,x,y,z)*mu;end;

end;WriteLn(’’);Writeln(’’);Writeln(’’);Writeln(’a,b,c,d :- ’,’ ’, a:3:2,’ ’,b:3:2,’ ’,c:3:2,’ ’,d:3:2);Writeln(’p,q,r,s :- ’,’ ’, p:3:2,’ ’,q:3:2,’ ’,r:3:2,’ ’,s:3:2);Writeln(’N1,N2,N3,N4 :- ’, N1:10,N2:10,N3:10,N4:10);Writeln(’Function :- Characteristic of Ball in R4’);Writeln(’Riemann Sum :- ’,RS:5:10);Writeln(’Riemann Sum/PI:- ’,RS/pi:5:3);

c�



end.{


jbquig-UCD


}

Output of program RS4!I am a snail 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

a,b,c,d :- -1.00 -1.00 -1.00 -1.00p,q,r,s :- 1.00 1.00 1.00 1.00N1,N2,N3,N4 :- 20 20 20 20Function :- Characteristic of Ball in R4Riemann Sum :- 4.9345999930Riemann Sum/PI:- 1.571

{

c�



}program rs4_1;(*Computes the integral of the arbitrary f over the integral J in four space.J=[a,e]x[b,f]x[c,g]x[d,h]. We give the characteristic function of thesolid ball in four space. This program is the same as rs4.pasbut runs faster. Any computation deep in imbedded loops is moveda shallower level if possible, see how w,x,y,z are incremented ratherthan computed directly, see where "*mu" appears.*)

consta = -1.0; b = -1.0; c = -1.0; d = -1.0; (* one corner of J *)p = 1.0; q = 1.0; r = 1.0; s = 1.0; (* the other *)N1= 20 ; N2= 20 ; N3= 20 ; N4= 20 ; (* partition numbers*)

varw,x,y,z : real;dw,dx,dy,dz : real; (* edges of subinterval *)span,mesh : real;mu : real;i,j,k,l : integer; (* loop control *)RS : real; (* Riemann sum *)

function f(w,x,y,z:real):real;beginif (w*w+x*x+y*y+z*z>1) then f:=0 else f:=1;end;

begindw:=(p-a)/N1;dx:=(q-b)/N2;dy:=(r-c)/N3;dz:=(s-d)/N4;span:=sqrt(dw*dw+dx*dx+dy*dy+dz*dz);mesh:=span;Writeln(’’);Writeln(’’);WriteLn(’Output of program RS4_1!’);Write(’I am a fast snail’);mu:=dw*dx*dy*dw;RS:=0.0;w:=a;for i:= 1 to N1 do

beginwrite(i:3); (* Has it crashed or is it running? *)w:=w+dw;x:=b;for j:= 1 to N2 do

beginx:=x+dx;y:=c;for k:= 1 to N3 do

beginy:=y+dy;z:=d;for l:= 1 to N4 do

beginz:=z+dz;


jbquig-UCD


RS:=RS+f(w,x,y,z);end;

end;end;

end;RS:=RS*mu;WriteLn(’’);Writeln(’’);Writeln(’’);Writeln(’a,b,c,d :- ’,’ ’, a:3:2,’ ’,b:3:2,’ ’,c:3:2,’ ’,d:3:2);Writeln(’p,q,r,s :- ’,’ ’, p:3:2,’ ’,q:3:2,’ ’,r:3:2,’ ’,s:3:2);Writeln(’N1,N2,N3,N4 :- ’, N1:10,N2:10,N3:10,N4:10);Writeln(’Function :- Characteristic of Ball in R4’);Writeln(’Riemann Sum :- ’,RS:5:10);Writeln(’Riemann Sum/PI:- ’,RS/pi:5:3);

end.{

c�



}

Output of program RS4_1!I am a fast snail 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

a,b,c,d :- -1.00 -1.00 -1.00 -1.00p,q,r,s :- 1.00 1.00 1.00 1.00N1,N2,N3,N4 :- 20 20 20 20Function :- Characteristic of Ball in R4Riemann Sum :- 4.9288000000Riemann Sum/PI:- 1.569

{


jbquig-UCD


2.7 APPENDIX B—-C PROGRAMS

But it seems that engineers and scientists, to find employement nowadays, must know C, or even C++.Bowing to popular request I include a C program “RS.C” which integrates any function in any dimension.Note the core code consist of 13 lines only, the remaining 120 lines are there merely to exercise this core code.The important code is the definition of the C struct nLatticeInNSpace (5 lines) and the function RiemannSum(8 lines). The program is a puzzle in terse C. It uses pointers, dynamic memory allocation, and two powerfultechniques (i) recursion, (ii) functions as parameters of functions. Also the program has been modularized.The important code (13 lines excluding comments) lies in files RS.H and RS.C. Then there is a libraryFUNCTS.C of functions, useful if you wish to exercise the RS module, i.e. ie you may integrate thesefunctions. Then there are 3 main programs, each of which exercises the module RS and the FUNCT library.The first vball4.c computes the volume of the ball in 4 space, the partition is fine 50x50x50x50 subintervals,it takes some time to run. The second VBALL ALL.C computes the volumes of balls in dimensions 1,23 and 4. The third INT 3 SP.C carries out certain integrals in 3-space, associated with the hemiball andthe tetrahedro. The output of these program has been routed to files VBALL4.OPD,VBALL ALL.OPD andINT 3 SP.OPD. The task of input in C has been avoided, input is introduced by the quickfix of a loop at thestart of the main code.

As I am tiring of typing I will not explain how RS integrates, nor is it easy understand. For now a puzzle,later an explaination!

c�


2.7. APPENDIX B—-C PROGRAMS 43

//

/*FILE RS.H by JBQ-UCD 1993

*/

#ifndef RS_H#define RS_H

typedef struct nLatticeInNSpace{int n,N;int * num_dvs;double * v,*dv;

};

double RiemannSum( nLatticeInNSpace L, double f(int,double *) );

#endif//


jbquig-UCD


//

/*RS.C by JBQ-UCD 1993

*/#include"rs.h"

/* ***************************** The RS function, ** it is recursive, ** it is terse, ** it is a puzzle for you. ***************************** */

double RiemannSum( nLatticeInNSpace L, double f(int,double *) ){/*

Approximates the integral of f, a function with dim real variablesas a Riemann Sum over a homogenous lattice, the computationof the Riemann Sum is carried out by recursion.

*/int d=(--L.n);double tot=0.0,dv = L.dv[d],hold = L.v[d];int i;for(i=0;i<L.num_dvs[d];i++,L.v[d]+=dv)tot+=(d?RiemannSum(L,f):f(L.N,L.v));L.v[d]=hold;return tot*dv; // its all over, what happened?

};

//

c�



//

/*FUNCTS.H by JBQ-UCD 1993

*/

#ifndef FUNCTS_H#define FUNCTS_H

double OnTheTetrahedron(int dim,double * x);double OnTheBall(int dim, double * x);double xLastOnTheHemiBall(int dim, double * x);double xLastOnTheTetrahedron(int dim, double * x);

#endif

//


jbquig-UCD


//

/*FUNCTS.C by JBQ-UCD 1993

*/#include"functs.h"

/* ************************* Some Functions ** write more ** yourself ************************* */

double OnTheBall(int dim, double * x){// characteristic function of ball of radius 1,any dim

double s , r = 0.0;int i;if (x[dim-1]<0.0) return 0.0;for(i=0;i<dim;i++){

s = x[i] ;if( ( r+= s*s ) >1.0 ) return 0.0;}

return 1.0;};

double OnTheTetrahedron(int dim,double * x){// characteristic function of the standard tetrahedron, any dim// if dimension 3 given by x,y,z>=0,x+y+z<=1

double s , r = 0.0;int i;for(i=0;i<dim;i++){

s = x[i] ;if( (( r+= s ) >1.0)||s<0.0 ) return 0.0;}

return 1.0;}

double xLastOnTheHemiBall(int dim, double * x){// in 3 dimensions f(x,y,z)=z, used to get cog

double s , r = 0.0;int i;for(i=0;i<dim;i++){

s = x[i] ;if(s<0.0) return 0.0;if( ( r+= s*s ) >1.0 ) return 0.0;}

return s;};

double xLastOnTheTetrahedron(int dim, double * x){// in 3 dimensions f(x,y,z)=z, used to get cog

double s , r = 0.0;int i;

c�



for(i=0;i<dim;i++){s = x[i] ;if( (( r+= s ) >1.0)||s<0.0 ) return 0.0;}

return s;}//


jbquig-UCD


//

/*volb4.c compute volume of 4 ball JBQ-UCD 30/11/93

*/#include <stdio.h>#include <alloc.h>#include"rs.h"#include"functs.h"#define pi 3.1416

int main() // usefully exercises the function RiemannSum{

int j; // loop controlint dim=4; // dimension, user inputdouble v1,v2; // rectangle, to be input by userint ChopNum; // to be input by usernLatticeInNSpace L; // the key data structuredouble RS; // Riemann Sum and auxillary

printf("\n OutPut volb_4.C, C Integration 30/11/93 by JBQ-UCD\n\n");L.v = (double *)malloc(dim*sizeof(double));//allocate space forL.dv = (double *)malloc(dim*sizeof(double));//lattice vectorsL.num_dvs = ( int *)malloc(dim*sizeof(int));if((!L.v)||(!L.dv)||(!L.num_dvs)){ // crash if alloc failsprintf("memory allocation failure");return -1;

}L.N = dim; // fill fields of LL.n = dim; // controls recursionfor(j=0;j<dim;j++){

v1 = -1.0, v2 = 1.0;ChopNum = 50; // user inputL.dv[j] = (v2-v1)/ChopNum; // fill L outL.v[j] = v1+L.dv[j]/2;L.num_dvs[j] = ChopNum;

} // Lattice is now ready

RS=RiemannSum(L,OnTheBall); // INTEGRATEprintf("\n volume %i-ball/pi:= %f",dim,RS/pi);free(L.v);free(L.dv);free(L.num_dvs); // clean up, ie deallocatereturn 0;

};//

c�



OutPut volb_4.C, C Integration 30/11/93 by JBQ-UCD

volume 4-ball/pi:= 1.569596


jbquig-UCD


//

/*VOLB4.C compute volume balls dims 1 to 4, JBQ-UCD 30/11/93

*/#include <stdio.h>#include <alloc.h>#include"rs.h"#include"functs.h"#define pi 3.1416

int main(){ // usefully exercises the function RiemannSum

int j; // loop controlint dim; // dimension, to be input by userdouble v1,v2; // rectangle, to be input by userint ChopNum; // to be input by usernLatticeInNSpace L; // the key data structuredouble RS; // Riemann Sum and auxillary

printf("\n OutPut volb_all.C C Integration 30/11/93 by JBQ-UCD\n\n");

// we will perform varios integrations in dimension 1 to 4for(dim=1;dim<5;dim ++){

L.v = (double *)malloc(dim*sizeof(double));//allocate space forL.dv = (double *)malloc(dim*sizeof(double));//lattice vectorsL.num_dvs = ( int *)malloc(dim*sizeof(int));if((!L.v)||(!L.dv)||(!L.num_dvs)){ // crash if alloc fails

printf("memory allocation failure");return -1;


v1 = -1.0, v2 = 1.0;ChopNum = 96/dim; // user inputL.dv[j] = (v2-v1)/ChopNum; // fill L outL.v[j] = v1+L.dv[j]/2;L.num_dvs[j] = ChopNum;


RS=RiemannSum(L,OnTheBall); // INTEGRATEif (dim==1) printf("\n volume 1-ball := %f",RS);else printf("\n volume %i-ball/pi:= %f",dim,RS/pi);free(L.v);free(L.dv);free(L.num_dvs); // clean up, ie deallocate

}return 0;

};//

c�



OutPut volb_all.C C Integration 30/11/93 by JBQ-UCD

volume 1-ball := 2.000000volume 2-ball/pi:= 0.996927volume 3-ball/pi:= 1.341002volume 4-ball/pi:= 1.568458


jbquig-UCD


//

/*int_3_sp.c integrations in 3 space JBQ-UCD 30/11/93

*/

#include <stdio.h>#include <alloc.h>#include"rs.h"#include"functs.h"#define pi 3.1416

int main() // usefully exercises the function RiemannSum{

int j; // loop controlint dim=3; // dimension, to user inputdouble v1,v2; // rectangle, to be input by userint ChopNum; // to be input by usernLatticeInNSpace L; // the key data structuredouble RS,RS1; // Riemann Sum and auxillary

printf("\n OutPut int_3_sp.C, C Integration 30/11/93 by JBQ-UCD\n\n");

// we will perform various integrations in 3 space

L.v = (double *)malloc(dim*sizeof(double));//allocate space forL.dv = (double *)malloc(dim*sizeof(double));//lattice vectorsL.num_dvs = ( int *)malloc(dim*sizeof(int));if((!L.v)||(!L.dv)||(!L.num_dvs)){ // crash if alloc fails

printf("memory allocation failure");return -1;


v1 = -1.0, v2 = 1.0;ChopNum = 50; // user inputL.dv[j] = (v2-v1)/ChopNum; // fill L outL.v[j] = v1+L.dv[j]/2;L.num_dvs[j] = ChopNum;


RS=RiemannSum(L,OnTheBall); // INTEGRATEprintf("\n volume 3-ball/pi :- %f ",RS/pi);RS1=RiemannSum(L,xLastOnTheHemiBall);printf("\n hght of cog of HemiBall ?3/8? :- %f",2*RS1/RS);RS=RiemannSum(L,OnTheTetrahedron);printf("\n vol standard tetrahedron ?1/6? :- %f",RS);RS1=RiemannSum(L,xLastOnTheTetrahedron);printf("\n hght c.o.g. std Tetrahedron ?1/4? :- %f",RS1/RS);

free(L.v);free(L.dv);free(L.num_dvs); // clean up, ie deallocatereturn 0;

};//

c�



OutPut int_3_sp.C, C Integration 30/11/93 by JBQ-UCD

volume 3-ball/pi :- 1.339486hght of cog of HemiBall ?3/8? :- 0.3.75605vol standard tetrahedron ?1/6? :- 0.166400hght c.o.g. std Tetrahedron ?1/4? :- 0.250000


jbquig-UCD


c�


Chapter 3

Scalar and Vector Fields withDifferential Operators

3.1 Summary

To be added

3.2 Scalar and Vector Fields

3.2.1 Definition of Scalar Field

A scalar valued function

f : A � R3 � � R

x � xi yj zk �� x

yz

�� f�x � � f

�x � y � z � � f

�xi yj zk �

is known as a scalar field over its domain (usually an open set) A � R3.

3.2.2 Definition of Vector Field

A vector valued function

v : A � R3 � � � R3

x � xi yj zk �� x

yz

�� v�x �� M

�x � y � z � i N

�x � y � z � j P

�x � y � z � k �

�� MNP

��is known as a vector field on A.

3.2.3 Notation

If all partial derivatives of f (resp. M � N and P) exist and are continuous up to and including the n � th orderthen we say that f (resp. v � Mi Nj Pk) is a C n scalar (resp. vector) field.

55

56 CHAPTER 3. SCALAR AND VECTOR FIELDS WITH DIFFERENTIAL OPERATORS

3.3 Differential Operators:- grad � curl � div and ∇2

3.3.1 Definitions

Remark F�A � denotes the set of all scalar fields and V

�A � denotes the set of all vector fields, over the

domain (open set) A � R3. We study three first order differential operators (FODO s) grad � curl and div andthe second order differential operator (SODO ) ∆ � ∇2 known as the Laplacian.

grad the gradient, is a FODO which acts on a C n scalar field to yield a C n � 1 vector field.

grad : F�A � � � V

�A �

f � � � grad f �

��

∂ f∂x∂ f∂y∂ f∂z

� �� ∂ f

∂xi ∂ f

∂yj ∂ f

∂zk

curl is a FODO which acts on a C n vector field to yield a C n � 1 vector field.

curl : V�A � � � V

�A �

v � Mi Nj Pk �� M

NP

�� curlv ��

i j k∂∂x

∂∂y

∂∂z

M N P

��where��

i j k∂∂x

∂∂y

∂∂z

M N P

��

∂P∂y � ∂N

∂z � i �∂M∂z � ∂P

∂x � j �∂N∂x � ∂M

∂y � k �

��

∂P∂y � ∂N

∂z∂M∂z � ∂P

∂x∂N∂x � ∂M

∂y

� ��

div the divergence, is a FODO which acts on a C n vector field to yield a C n � 1 scalar field.

div : F�A � � � F

�A �

v � Mi Nj Pk �� M

NP

�� div f � ∂M∂x

∂N∂y

∂P∂z

∆ � ∇2 the Laplacian, is a SODO which acts on a C n scalar field to yield a C n � 2 scalar field.

∆ : F�A � � � F

�A �

f � � � ∆ f � ∇2 f � ∂2 f∂x2

∂2 f∂y2

∂2 f∂z2

3.3.2 Composition of Differential Operators

RemarkThe two diagrams illustrate the next theorem and also serve to give an overview of the theory thus far.

F�A �

grad//

0))

V Acurl

//

0**

V�A �

div// F

�A � � F

�A �

grad//

∆ ∇2

**V

�A �

div// F

�A �

c�


3.3. DIFFERENTIAL OPERATORS:- GRAD CURL DIV AND ∇2 57

theorem 4Let A � R3 � f F

�A � and v V

�A � be C 2 fields, then

(i) curl � grad f � 0 , (ii) div � curlv � 0 , (iii) div � grad f � ∇2 f

proof(i)

curl � grad f � curl�grad

�f � �

� curl � ∂ f∂x

i ∂ f∂y

j ∂ f∂z

k ��

��i j k∂∂x

∂∂y

∂∂z

∂ f∂x

∂ f∂y

∂ f∂z

��

��

∂∂y

∂ f∂z � ∂

∂z∂ f∂y

∂∂z

∂ f∂x � ∂

∂x∂ f∂z

∂∂x

∂ f∂y � ∂

∂y∂ f∂x

� ��

�� 0

00

��because, f being C 2,

∂∂x

∂ f∂y

� ∂∂x

∂ f∂y

(and two similar results).

proof(ii)

div � curlv � div�curlv �

� div � � ∂P∂y � ∂N

∂z � i �∂M∂z � ∂P

∂x � j �∂N∂x � ∂M

∂y � k �� ∂

∂x

� �∂P∂y � ∂N

∂z � � ∂∂y

� �∂M∂z � ∂P

∂x � � ∂∂z

� �∂N∂x � ∂M

∂y � �� ∂2P

∂x∂y � ∂2N∂x∂z

∂2M∂y∂z � ∂2P

∂y∂x ∂2N

∂z∂x � ∂2M∂z∂y

��

∂2M∂y∂z � ∂2M

∂z∂y � �∂2N∂z∂x � ∂2N

∂x∂z � �∂2P∂x∂y � ∂2P

∂y∂x �� 0 0 0

as in (i).proof(iii)

div � grad f � div � ∂ f∂x

i ∂ f∂y

j ∂ f∂z

k �� ∂

∂x

�∂ f∂x � ∂

∂y

�∂ f∂y � ∂

∂z

�∂ f∂z �

� ∂2 f∂x2

∂2 f∂y2

∂2 f∂z2

� ∇2 f


jbquig-UCD


3.4 Inverse Square Law

The three fields defined next are important and must be memorized.

f A classical scalar field.

f : R3 � � 0 � � � R3

x �� x

yz

�� f�x �� f

�x � y � z � � 1

�x2 y2 z2 � 1 � 2

v A classical vector field

v : R3 � � 0 � � � R3

x �� x

yz

�� xi yj zk

�x2 y2 z2 � 3 � 2 �

��

x�x2 y2 z2 � 3 � 2

y�x2 y2 z2 � 3 � 2

z�x2 y2 z2 � 3 � 2

� ��

w Another classical vector field

w : R3 � Lz � � R3

x �� x

yz

�� z

� � yi xj ��x2 y2 � �

x2 y2 z2 � 1 � 2 �

��

� zy�x2 y2 � �

x2 y2 z2 � 1 � 2zx

�x2 y2 � �

x2 y2 z2 � 1 � 20

�x2 y2 � �

x2 y2 z2 � 1 � 2

� ��

Here Lz� � x � x � 0 and y � 0

� � x �� x2 y2 � 0 � denotes the z � Axis. Because of the factorx2 y2 in the denominator the field is, of course, not defined on Lz.

3.4.1 Differential Operators and the Inverse Square Law

Soon we will see the importance of these field, but first let us perform a few computations involving themand the operators grad � curl � div and ∇2. We will calculate.

grad f and ∇2 f

curlw and divw

curlv and divv

Computing: grad f

grad f � ∂ f∂x

i ∂ f∂y

j ∂ f∂z

k

� ∂∂x

1�x2 y2 z2 � 1 � 2 i ∂

∂y1

�x2 y2 z2 � 1 � 2 j ∂

∂z1

�x2 y2 z2 � 1 � 2 j

� ∂∂x

� x2 y2 z2 � � 1 � 2i ∂

∂y� x2 y2 z2 � � 1 � 2

j ∂∂z

� x2 y2 z2 � � 1 � 2k

c�


3.4. INVERSE SQUARE LAW 59

We need only compute the first of these partial derivatives, the other two are similar. Thus

∂∂x

� x2 y2 z2 � � 1 � 2 � � 12

�x2 y2 z2 � � 3 � 2 ∂

∂x

�x2 y2 z2 �

� � 12

�x2 y2 z2 � � 3 � 2 �

2x �

� � x � x2 y2 z2 � � 3 � 2� � x

�x2 y2 z2 � 3 � 2

Similarly∂∂y

� x2 y2 z2 � � 1 � 2 � � y�x2 y2 z2 � 3 � 2

Similarly∂∂z

� x2 y2 z2 � � 1 � 2 � � z�x2 y2 z2 � 3 � 2

Putting it all together

grad f � � x � x2 y2 z2 � � 3 � 2i � y � x2 y2 z2 � � 3 � 2

j � z � x2 y2 z2 � � 3 � 2k

� � xi yj zk�x2 y2 z2 � 3 � 2

� grad f � v

Computing: curlw

curlw ��

i j k∂∂x

∂∂y

∂∂z

M N P

��

∂P∂y � ∂N

∂z � i �∂M∂z � ∂P

∂x � j �∂N∂x � ∂M

∂y � k

where

M � � zy�x2 y2 � �

x2 y2 z2 � 1 � 2 � N � zx�x2 y2 � �

x2 y2 z2 � 1 � 2 � P � 0�x2 y2 � �

x2 y2 z2 � 1 � 2Since P � 0,

curlw � � � ∂N∂z

� i � ∂M∂z

� j � ∂N∂x � ∂M

∂y� k

Only four partial differentiations are needed. Since∂N∂z

is similar to∂M∂z

and∂M∂y

to∂N∂x

we need only

compute two partial derivatives,∂M∂z

and∂M∂y

!

∂M∂z

� ∂∂z� � zy

�x2 y2 � �

x2 y2 z2 � 1 � 2 �� y

x2 y2

∂∂z

�z

�x2 y2 z2 � � 1 � 2 �

� � yx2 y2

� ∂z∂z� �

x2 y2 z2 � � 1 � 2 z �∂∂z

�x2 y2 z2 � � 1 � 2 �

� � yx2 y2

� 1 � �x2 y2 z2 � � 1 � 2 z �

� � 12 � �

x2 y2 z2 � � 3 � 2 ∂∂z

�x2 y2 z2 � �


jbquig-UCD


� � yx2 y2

� �x2 y2 z2 � � 1 � 2 z �

� � 12 � �

x2 y2 z2 � � 3 � 2 �2z � �

� � yx2 y2

� �x2 y2 z2 � � 1 � 2 � z2 �

x2 y2 z2 � � 3 � 2 �� y�

x2 y2 � �x2 y2 z2 � 3 � 2 � �

x2 y2 z2 � � z2 �� y�

x2 y2 � �x2 y2 z2 � 3 � 2 � x2 y2 �

� � y�x2 y2 z2 � 3 � 2

Summary, to date

∂M∂z

� � y�x2 y2 z2 � 1 � 2 � ∂N

∂z� x

�x2 y2 z2 � 1 � 2 (similarly)

We next compute∂M∂y

(and∂N∂x

).

∂M∂y

� ∂∂y� � zy

�x2 y2 � �

x2 y2 z2 � 1 � 2 �� z � ∂

∂y

�y

�x2 y2 � � 1 �

x2 y2 z2 � � 1 � 2 �� z �� 1 �

�x2 y2 � � 1 �

x2 y2 z2 � � 1 � 2 � � z � y � � �x2 y2 � � 2 �

2y � � � �x2 y2 z2 � � 1 � 2 � � z � y

�x2 y2 � � 1 �

� � 12

�x2 y2 z2 � � 3 � 2 �

2y � �� z

�x2 y2 � � 1 �

x2 y2 z2 � � 1 � 2 2zy2 �x2 y2 � � 2 �

x2 y2 z2 � � 1 � 2 zy2 �x2 y2 � � 1 �

x2 y2 z2 � � 3 � 2Obtain

∂N∂x

similarly and summarize.

∂M∂y

� � z�x2 y2 � � 1 �

x2 y2 z2 � � 12 2zy2 �

x2 y2 � � 2 �x2 y2 z2 � � 1

2 zy2 �x2 y2 � � 1 �

x2 y2 z2 � � 12

∂N∂x

� z�x2 y2 � � 1 �

x2 y2 z2 � � 12

� 2zx2 �x2 y2 � � 2 �

x2 y2 z2 � � 12� zx2 �

x2 y2 � � 1 �x2 y2 z2 � � 1

2

For the third component of curlw we need

∂N∂x � ∂M

∂y� 2z

�x2 y2 � � 1 �

x2 y2 z2 � � 1 � 2� 2z

�x2 y2 � �

x2 y2 � � 2 �x2 y2 z2 � � 1 � 2� z

�x2 y2 � �

x2 y2 � � 1 �x2 y2 z2 � � 1 � 2

c�


3.4. INVERSE SQUARE LAW 61

The first two terms cancel, leaving

∂N∂x � ∂M

∂y� � z

�x2 y2 z2 � 3 � 2


curlw � � � ∂N∂z

� i � ∂M∂z

� j � ∂N∂x � ∂M

∂y� k

� � xi yj zk�x2 y2 z2 � 3 � 2� curlw � v

Computing: curlv

curlv � curl� � grad f � see 3.4.1

� � curl � grad�f �

� 0 see 4(ii)

curlv � 0

Computing: divv

divv � div� � grad f � see 3.4.1

� � div � grad�f �

� � ∇2 f see 4� 0 see 3.4.1

divv � 0

Computing: ∇2 f

∇2 f � div � grad f since ∇2 � div � grad� � divv since � grad f � v� 0 see 3.4.1

∇2 f � 0

Computing: divw

divw � ∂∂x � zy

�x2 y2 � �

x2 y2 z2 � 1 � 2 ∂∂y zx

�x2 y2 � �

x2 y2 z2 � 1 � 2 ∂∂z

�0 �

� ∂∂x� � zy � x2 y2 � � 1 � x2 y2 z2 � � 1 � 2 � ∂

∂y� zx � x2 y2 � � 1 � x2 y2 z2 � � 1 � 2 � ∂

∂z

�0 �

� � zy∂∂x

� � x2 y2 � � 1 � x2 y2 z2 � � 1 � 2 � zx∂∂y

� � x2 y2 � � 1 � x2 y2 z2 � � 1 � 2 � 0

� � zy� � � 2x � � x2 y2 � � 2 � x2 y2 z2 � � 1 � 2 � x2 y2 � � 1 � � x � � x2 y2 z2 � � 3 � 2 � zx

� � � 2y � � x2 y2 � � 2 �x2 y2 z2 � � 1 � 2 � x2 y2 � � 1 � � y � �

x2 y2 z2 � � 3 � 2 �


jbquig-UCD


� 2xyz�x2 y2 � 2 �

x2 y2 z2 � 1 � 2 xyz�x2 y2 � �

x2 y2 z2 � 3 � 2

� 2xyz�x2 y2 � 2 �

x2 y2 z2 � 1 � 2 � xyz�x2 y2 � �

x2 y2 z2 � � 3 � 2� 0

divw � 0

3.4.2 Summary–Inverse Square Law

In the above computations we have found that with

f � 1�x2 y2 z2 � 1 � 2 � v � xi yj zk

�x2 y2 z2 � 3 � 2 � w � z

� � yi xj ��x2 y2 � �

x2 y2 z2 � 1 � 2� grad f � v � ∇2 f � 0� curlw � v � divw � 0

curlv � 0 � divv � 0� f

grad ��>>>

>>>>

>∆2

// 0

vdiv

@@��

curl

��>>>

>>>>

w

curl??��

div

��>>>

>>>>

> 0

0

Each fact tabulated is shown in the diagram.The centerpiece is the vector field v, which is known as the inverse square law field.

v�x � � x�� x �� 3

At the point x the field v points from the origin 0 to x, i.e. v radiates out fom the origin in R3. Also

�� v �x � �� x �� x �� 3 � 1�� x �� 2

so the magnitude of the field v falls off as the reciprocal of the square of distance from the origin.

In gravitation � v is the law of attraction to a point mass. In electrostatics v is Coulomb’s law. Inmagnetism v expresses the force field of the monopole. In fluid flow � v is the velocity field of a pointsource/sink.

In the physical sciences one meets the following terminology.

� grad f � v; f is conservative with scalar potential v� curlw � v; w is the vector potential of v

curlv � 0; v is irrotational

divv � 0; v is incompressible

∇2 f � 0; f is harmonic

divw � 0; w is incompressible

c�


3.5. PHYSICAL MEANING OF GRAD CURL DIV AND ∇2 63

3.5 Physical meaning of grad � curl � div and ∇2

Note that div and curl measure dilation and rotation in a field, it is not on this course, but the student mightlike to convince himher self on this point, see for example, volume I of Feynmann(? reference).

3.6 Dipole

We take the point of view that v � f and w are associated with the field, the scalar potential and the vectorpotential, respectively, of a monopole of strength +1 located at 0 R3: the monopole might be electrical,magnetic or fluid source according to physical context. Let ε � 0 be small and consider two monopoles one

of strength � 1ε

at 0 and the other of strength 1ε

but located at � εk, (at distance ε from 0 along the negative

z � axis), this pair is known as a dipole of moment1ε� ε � 1. or rather it is called a dipole in the limit as

ε � 0, i.e. pole strength tends to infinity, seperation tends to 0, but throughout, the moment stays at fixed 1.Thus

F�x � y � z � � lim

ε � 0

f�x � y � z ε � � f

�x � y � z �

ε� ∂

∂zf

�x � y � z �

V�x � y � z � � lim

ε � 0

v�x � y � z ε � � v

�x � y � z �

ε� ∂

∂zv

�x � y � z �

W�x � y � z � � lim

ε � 0

w�x � y � z ε � � w

�x � y � z �

ε� ∂

∂zw

�x � y � z �

are the scalar potential,actual field and vector potential of the unit dipole.

3.7 Computing the Dipole fields

computing F

F � ∂ f∂z

� ∂∂z

1�x2 y2 z2 � 1 � 2

� ∂∂z

�x2 y2 z2 � � 1 � 2

�� 1

2 � �x2 y2 z2 � � 3 � 2 ∂

∂z

�x2 y2 z2 �

�� 1

2 � �x2 y2 z2 � � 3 � 2 �

0 0 2z �

� � z�x2 y2 z2 � � 3 � 2

F�x � y � z � � � z�

x2 y2 z2 � � 3 � 2computing V

v � Mi Nj Pk � xi yj zk�x2 y2 z2 � 3 � 2

By V � ∂v∂z

we mean

V � ∂M∂z

i ∂N∂z

j ∂P∂z

k �


jbquig-UCD


But

∂M∂z

� ∂∂z

x�x2 y2 z2 � 3 � 2

� ∂∂z

�x

�x2 y2 z2 � � 3 � 2 �

� x∂∂z

�x2 y2 z2 � � 3 � 2

� x

� � 32 � �

x2 y2 z2 � � 5 � 2 �0 0 2z �

� � 3xz�x2 y2 z2 � � 5 � 2

Similarly we may obtain∂N∂z

.

∂M∂z

� � 3xz�x2 y2 z2 � 5 � 2 � ∂N

∂z� � 3yz�

x2 y2 z2 � 5 � 2Next

∂P∂z

� ∂∂z

z�x2 y2 z2 � 3 � 2

� ∂∂z

�z

�x2 y2 z2 � � 3 � 2 �

� 1 ��x2 y2 z2 � � 3 � 2 z �

� � 32 � �

x2 y2 z2 � � 5 � 2 �0 0 2z �

� �x2 y2 z2 � � 3 � 2 � 3z2 �

x2 y2 z2 � � 5 � 2� x2 y2 z2 � 3z2

�x2 y2 z2 � 5 � 2

� x2 y2 � 2z2�x2 y2 z2 � 5 � 2

∂P∂z

� x2 y2 � 2z2�x2 y2 z2 � 5 � 2

Putting all the pieces together

V � � 3xzi � 3yzj �x2 y2 � 2z2 � k�

x2 y2 z2 � 5 � 2computing W

w � Mi Nj Pk � � zyi zxj�x2 y2 � �

x2 y2 z2 � 1 � 2W � ∂w

∂z� ∂M

∂zi ∂N

∂zj ∂P

∂zk � ∂M

∂zi ∂N

∂zj

since P � 0.In this

∂M∂z

� ∂∂z

� yz�x2 y2 � �

x2 y2 z2 � 1 � 2� � y�

x2 y2 �∂∂z

z�x2 y2 z2 � 1 � 2

c�


3.7. COMPUTING THE DIPOLE FIELDS 65

� � y�x2 y2 �

∂∂z

�z

�x2 y2 z2 � � 1 � 2 �

� � y�x2 y2 � � 1 � �

x2 y2 z2 � � 1 � 2 z �� 1

2 � �x2 y2 z2 � � 3 � 2 �

0 0 2z � �� y�

x2 y2 ��

x2 y2 z2 � � 1 � 2 � z2 �x2 y2 z2 � � 3 � 2 �

� � y�x2 y2 �

�x2 y2 z2 � � z2

�x2 y2 z2 � 3 � 2

� � y�x2 y2 �

x2 y2�x2 y2 z2 � 3 � 2

� � y�x2 y2 z2 � 3 � 2

Similarly we may obtain∂N∂z

∂M∂z

� � y�x2 y2 z2 � 3 � 2 � ∂N

∂z� x�

x2 y2 z2 � 3 � 2Putting it all together

W � � yi xj�x2 y2 z2 � 3 � 2

3.7.1 Collecting these results

F�x � y � z � � ∂

∂z� 1

�x2 y2 z2 � 1 � 2 � � � z�

x2 y2 z2 � � 3 � 2V

�x � y � z � � ∂

∂z� xi yj zk

�x2 y2 � �

x2 y2 z2 � 1 � 2 � � � 3xzi � 3yzj �x2 y2 � 2z2 � k�

x2 y2 z2 � 5 � 2W

�x � y � z � � ∂

∂z� z

� � yi xj ��x2 y2 z2 � 3 � 2 � � � yi xj�

x2 y2 z2 � 3 � 2Examination of denominators shows that R3 � � 0 � is domain of definition of all three fields.

3.7.2 Differential Operators and the Dipole

As in section ? the following six results hold� gradF � V � ∇2 F � 0� curlW � V � divW � 0curlV � 0 � divV � 0

The reader may, as in section ?, prove these by using partial differentiation. All six can be obtained withouteffort by varying the next argument.

� gradF � � grad

�∂ f∂z �

� � �grad �

∂∂z � �

f �

� � �∂∂z

� grad � �f �


jbquig-UCD


� � ∂∂z

�grad f �

� � ∂∂z

� � v �

� ∂v∂z

� V

This proof uses the fact that the differential operators∂∂z

and grad commute when applied to a C 2 function

and the fact that � ∂ f∂z

� v.

3.7.3 Summary–Dipole field

Define F � V and W by

F�x � y � z � � ∂ f

∂z� V

�x � y � z � � ∂v

∂z� W

�x � y � z � � ∂w

∂z

where v � f and w are the monopole field with its scalar potential and vector potential respectively, see 3.4.Over the domain R3 � � 0 �

F�x � y � z � � � z�

x2 y2 z2 � � 3 � 2V

�x � y � z � � � 3xzi � 3yzj �

x2 y2 � 2z2 � k�x2 y2 z2 � 5 � 2

W�x � y � z � � � yi xj�

x2 y2 z2 � 3 � 2Applying grad � curl � div and ∇2

� gradF � V � ∇2 F � 0� curlW � V � divW � 0curlV � 0 � divV � 0

� F

grad @@@

@@@@

@∆2

// 0

Vdiv

??��

curl

��>>>

>>>>

W

curl>>~~~~~~~~

div

@@@

@@@@

@ 0

0

The diagram illustrates each fact tabulated.The centerpiece is the vector field V, which is known as the dipole field. more??

In the physical sciences one meets the following terminology.� gradF � V; V is conservative with scalar potential of F� curlW � V; W is the vector potential of V

curlV � 0; V is irrotational

divV � 0; V is incompressible

∇2 F � 0; F is harmonic

divW � 0; W is incompressible

c�


3.8. FIELDS IN R2, VORTEX AND SOURCE 67

3.8 Fields in R2, Vortex and Source

3.8.1 The Vortex

θ�x � y � z � � arctan

yx

� x R3 � H

p�x � y � z � � � yi xj

x2 y2 � x R3 � Lz

q�x � y � z � � xi j

x2 y2 � x R3 � Lz

The reader may prove (see exercise 3.10) that

gradθ � p ∇2 θ � 0

curlq � p divq � 0

curlp � 0 divp � 0

In all this the centerpiece is the vector field p, which is known as the vortex field.At the point x the field v is perpendicular to the line joining the z-axis to x, i.e. v rotates about the z-axis.Also

� � v �x � � � �

�x2 y2

x2 y2� 1�

x2 y2

so the magnitude of the field v increases from 0 to ∞ as x passes from infinity to the z-axis, increasing as thereciprocal of distance.

In magnetism p is essentially the magnetic field caused by steady current flowing in a long straight infinitewire lying along the z-axis. In fluid flow � p represents rotation about the z-axis with a central vortex.

In the physical sciences one says.� gradθ � p; θ is the scalar potential of p� curlq � p; q is the vector potential of p

curlp � 0; p is irrotational

divp � 0; p is incompressible

∇2 θ � 0; θ is harmonic

divq � 0; q is incompressible

3.8.2 The Two Dimensional Source/Sink

ρ�x � y � z � � 1

2 log�x2 y2 � � x R3 � Lz

q�x � y � z � � xi yj

x2 y2 � x R3 � Lz

p�x � y � z � � � yi xj

x2 y2 � x R3 � L

Here the the set Lz is the z-axial line and the set H is a semi infinite plane.

Lz� x R3 �� x � 0 and y � 0 � � H � x R3 �� y � 0 and x 0 �


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The reader may check (see ??) that

gradρ � q ∇2 ρ � 0

curlq � 0 divq � 0

curlp � q divp � 0

As before the centerpiece is the vector field q, which is known as the Line Source in R3 or the (2-dimensional)point source field.At the point x direction of the field q is perpendicularly from the z-axis to x, i.e. q radiates from the z-axis.Also

� � q �x � � � � �

x2 y2

x2 y2� 1�

x2 y2

so the magnitude of the field q rises from 0 to ∞ as x passes from infinity to the z-axis, increasing as thereciprocal of distance from the axis.

In electrostatics q is the field due to a line of charge of constant line density distributed along the z-axis.In fluid flow q is a line source along the z-axis (or a constant point source at 0 in 2 dimensions).

In the physical sciences one says.� gradρ � q; ρ is the scalar potential of q� curlp � q; p is the vector potential of q

curlq � p; q is rotational

divq � 0; q is incompressible

∇2 ρ � 0; ρ is harmonic

divp � 0; p is incompressible

3.9 Advanced Observations

3.9.1 Does an irrotational field have a scalar potential.

Given a vector field v of the form v � grad f , for some scalar field f , it is certainly true that curlv � 0. Letus ask conversely; if curlv � 0 does there exist a scalar field f such that grad f � v? In other words; grantedthat a gradient field is irrotational, is it true that an irrotational field has a scalar potential? The answer isyes, with reservations. To go further, we define the concept of a hole of dimension 1 and we rephrase thequestion.

If there is a closed loop in the domain A which cannot be shrunk to a point then we say that there is ahole of dimension 1 in A.

For example there is such a 1-hole in R3 � Lz, (R3 with the z-axis removed); this hole is manifest whenwe try to shrink the loop Γ � x �� z � 0 and x2 y2 � 1 � to a point. There is no 1-hole in R3 � � 0 � ; indeedany loop in this domain can be shrunk away, the absence of 0 presenting no difficulty to this process; you cancatch a long pole but not a fish with a lassoo.

Now we rephrase the question and give a correct answer.

Question: Let v : A � R3 � R3 be a vector field defined over the domain A � R3, with curlv � 0 onA. Does v � grad f for some scalar field f : A � R3 defined over all of A?

Answer: If the line integral�

Γ v � ds � 0 for all closed curves in A then the answer is yes. Moreover allsuch line integrals are zero if there no holes of dimension 1 in A. See Chap?Sect? for details.

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3.9. ADVANCED OBSERVATIONS 69

Example

Consider the vector field

p � � yi xjx2 y2 note curlp � 0

If we view p as defined over A � R3 � � x � x � 0 � y � 0�

(there are no 1-holes in A) then gradθ � p where

θ � arctanyx

is defined over all of A.

If however we consider p as defined over the domain B � R3 � Lz , (there is a 1-hole in B), then one canprove that there is no scalar field defined over B whose gradient is p, see ??Note θ is not well defined over all of B.

3.9.2 Does an divergence free field have a vector potential.

Given a vector field v of the form v � gradw, for some vector field w, it is certainly true that divv � 0. Letus ask conversely; if divv � 0 does there exist a vector field w such that curlw � v? In other words; grantedthat a curl is incompressible (divergence free), is it true that a incompressible vector field has always a vectorpotential? Let us rephrase the question.

Question

Given a vector field v : A � R3 � R3 with divv � 0 on A does there exist a vector field w : A � R3 � R3 (ie.defined over all of A) with curlw � v on A.

Before we answer this question let us look at an examples

Example

Consider the Inverse Square Law field v of section 3.4. If the domain of v is considered to be R3 � Lz then was in that section has the property that curlw � v and indeed w is defined over all of the domain R3 � Lz.

However if we consider the domain of v to be the larger set R3 � � 0 � then w fails to be defined over allof this domain. Indeed see section3.4 where it is asserted that there is no field defined over all of R3 � � 0 �whose curl is v.

Answer

If there is a sphere or (indeed any closed surface) in the domain A � R3 which cannot be filled in then wesay that there is a 2-hole in A. In the above example there is a 2 � hole in R3 � � 0 � but not in R3 � Lz. Ifthere are 2-holes in a domain then one can find divergence free fields which have no vector potential.


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3.10 problem set

Scalar and Vector Fields with Differential Operators

1. (Memorize f � v and w and the six formulae concerning these three important fields.)Let

f�x � y � z � � 1�

x2 y2 z2x �� 0

v�x � y � z � � xi yj zk�

x2 y2 z2 � 3 � 2 x �� 0

w�x � y � z � � z

� � yi xj ��x2 y2 � �

x2 y2 z2 � 1 � 2 x2 y2 �� 0

Prove all six formulae � grad f � v � ∇2 f � 0curlw � v � divw � 0curlv � 0 � divv � 0

See ?? for the physical significance of these fields and formulae. (Memorize F � V and W and the sixformulae concerning these three important fields.)Let

F�x � y � z � � � z�

x2 � y2 � z2 � 3 � 2 x �� 0

V�x � y � z � � � 3xzi � 3yzj � � x2 � y2 � 2z2 � k�

x2 � y2 � z2 � 5 � 2 x �� 0

W�x � y � z � �

�� yi � xj ��

x2 � y2 � z2 � 3 � 2 x �� 0

Prove all six formulae � gradF � V � ∇2F � 0curlW � V � divW � 0curlV � 0 � divV � 0

See ?? for the physical significance of these fields and formulae.

2. With f � v � w � F � V and W as defined in the preceeding two questions,

(i) Prove∂ f∂z

� F � ∂v∂z

� V � ∂w∂z

� W

(ii) Explain the physical significance of these formulae.

3. Let


yx

� x R3 � � x � x 0 � y � 0�

u�x � y � z � � � yi xj

x2 y2 � x R3 � x �� x2 y2 � 0 �φ

�x � y � z � � log

�x2 y2 � x R3 � x �� x2 y2 � 0 �

v�x � y � z � � xi yj

x2 y2 � x R3 � x �� x2 y2 � 0 �(i) Prove grad θ � u.

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3.10. PROBLEM SET 71

(ii) Compute divu � curlu � ∇2θ.

(iii) Find a vector potential p of the vector field u; ie. find a vector field p with curl p � u.

(iv) Prove grad φ � v.

(v) Compute divv � curlv � ∇2φ.

(vi) Find a vector potential q of the vector field v; ie. find a vector field q with curlq � v.

(vii) Explain the physical significance of all this.

4. (i) Find a domain B � R3 and a vector field u defined over B such that curlu � 0 and yet there isno scalar field f defined over B with grad f � u .

(ii) Prove that no such f exists.

(iii) Find a domain A � B and scalar field g defined over A such that gradg � u on A only.

(iv) Find a domain C � R3 and a vector field w defined over C such that divw � 0 and yet there is novector field v defined over C with curlv � w. Prove that no such v exists. Find a domain D � Cand vector field r defined over C such that curlr � w on C only.

5. Letrn �

x � y � z � � �x2 y2 z2 � n � 2 for each integer n

(i) Compute gradrn .

(ii) Compute ∇2rn .

(iii) Prove that, ∇2 rn � 0 , for two values of n only: find these values.

6. Let f and g be scalar fields and v and w be vector fields. Prove

(i)

grad�f g �� f gradg ggrad f

(ii)

div�f v � � f divv � grad f � v �

(iii)

div�v � w � � � curlv � w � � � v � curlw �

(iv)

curl�f v � � �

grad f � � v f curlv

7. Let

f � x2 y2 z2

2�x2 y2 � and F � � yi xj

be a scalar and a vector field respectively. Prove that

curl�f F � � � xzi � yzj �

x2 y2 � kx2 y2

[Hint: Use previous question .]

8. (i) Sketch a surface where f�x � y � z � � � �

x2 y2 � b � 2 z2 is constant.

(ii) Sketch a surface where g�x � y � z � � y

xis constant.


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(iii) Find a vector field w such that

w � grad f � w � gradg and curlw � 0

9. (i) Provecurl

�f v �� f curlv �

grad f � � v

(ii) Compute

� grad�x2 y2 z2 � � � � � yi xj �

(iii) Find, with proof, a vector field u such that

curlu � � xzi � yzj �x2 y2 � k

x2 y2

10. Let

r�x � y � z � � �

x2 y2 z2 � x R3

φ�x � y � z � � arctan � �

x2 y2�

z � � 0 φ π � x R3 � � 0 �θ

�x � y � z � � arctan

�y � x � � � π � θ � π � x R3 � H

where H �� x � x 0 and y � 0� � R3, being half of the xz–plane .

(i) Compute u � gradr � v � gradφ � w � gradθ.

(ii) Compute divu � divv and divw.

(iii) Compute curlu � curlv and curlw.

(iv) Compute ∇2r� ∇2φ and ∇2θ.

11. Let b � 0. Let

r�x � y � z � � � � �

x2 y2 � b � 2 z2 � for all x R3

φ�x � y � z � � arctan � z � �

x2 y2 � b � � � π � φ � π for all x R3 � D


�y � x � � � π � θ � π for all x R3 � H

Above, D � x �� z � 0 and x2 y2 b2 � (a disc) and H � � x � x 0 and y � 0�

(half of the xz–plane).Compute

(i)

u � gradr � v � gradφ and w � gradθ

(ii)

divu � divv and divw

(iii)

curlu � curlv and curlw

(iv)

∇2r � ∇2φ and ∇2θ

c�


Chapter 4

Differential Equations with the LaplaceTransform

remark In this chapter we will meet only real valued functions defined for positive time, i.e of the form,

f : � 0 � ∞ � � R

t �� f�t �

C n � 0 � ∞ � will denote the (vector space) of such functions which, in addition, are differentiable with continu-ous derivatives, up to and including the n–th order. Note that

f C n � 0 � ∞ � � f� C n � 1 � 0 � ∞ � � � f C n � 1 � 0 � ∞ �

remark We will study two simple but very important classes of differential equations, both of the secondorder with constant coefficients, see � 4.9 and � 4.10.

SOCCLHODE A typical second order constant coefficient linear homogeneous differential equationis of the form.

ax�t � bx

�t � cx

�t �� 0 � t � 0

where the constants a � b � c R are called coefficients.

SOCCLIHODE A typical second order constant coefficient linear inhomogeneous differential equa-tion is of the form.

ax�t � bx

�t � cx

�t �� h

�t � � t � 0

where the constants a � b � c R are called coefficients and h : � 0 � ∞ � � t �� f�t � R is called the driving

function. These differential equations will be solved using L the Laplace transform, see � 4.4. Thegeneral solution will involve convolution, see � 4.3. The differential equations describing mechanicaldamped oscillation and the all important electronic LRC circuit are SOCCLIHODE’s

4.1 Classical functions

Solutions of the differential equations above will generally involve the following classical functions.

oscillations The functions cos�at � and sin

�at � are viewed as oscillations, see figure 4.1.

decay and growth The functions eat is viewed as decay or growth according a a � 0 or a � 0, seefigure 4.2.

sympathy The functions t � t2 and t3 appear when sympathy is present, see � 4.11.

73

74 CHAPTER 4. DIFFERENTIAL EQUATIONS WITH THE LAPLACE TRANSFORM

-1

-0.5

0

0.5

1

-3p -2p -p 0 p 2p 3p

x

y

x

y

y=1

y=-1

Figure 4.1: (i) oscillation cos (ii) damped oscillation exp � cos

x

y

y=1

x

y

y=1

Figure 4.2: (i) decay, eat � a � 0 (ii) growth, eat � a � 0

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4.2. HEAVYSIDE STEP AND DIRAC DELTA FUNCTIONS 75

t

y

1

0

a

t

y

t-a

a

0

Figure 4.3: (i) Heavyside step function ua (ii) and elbow functions ea

4.2 Heavyside Step and Dirac Delta Functions

4.2.1 Heavyside step function

The Heavyside step function, for a � 0, is denoted ua and represents turnon of unit voltage at time t � a, itdescribes one stroke of a Morse key in (19’th century) telegraphy. The graph can be seen in figure 4.3 andthe concise definition is

ua : � 0 � ∞ � � R

t �� ua�t ��

�0 : 0 t � a;1 : a t;

When integrated ua yields the elbow function, see the graph in figure 4.3,

ea : � 0 � ∞ � � R

t �� ea�t �� t

0ua

�α � dα �

�0 : 0 t � a;

t � a : a t;

ex. ea is continuous everywhere but is not differentiable everywhere, indeed�ea

� � does not exist at t � a(it does exist at every other point).ex. ua is not continuous, indeed ua is not continuous at t � a (it is continuous at every other point).remark

ea� � t

0ua

� ua� �

ea� �

Consequentlyea C 0 � 0 � ∞ � � ua C � 1 � 0 � ∞ �

4.2.2 Dirac Delta function

The Dirac Delta function, for a � 0, is denoted δa and represents a unit pulse at time t � a. The concisedefinition follows and consist of a limiting process involving Heavyside step functions. Let a � 0 and h � 0be small; consider the function,

ua � ua � h

h: � 0 � ∞ � � R

t ��ua

�t � � ua � h

h� �� 0 : 0 t � a;

1 � h : a t � a h;0 : a h t


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t

y

h 1/h t

y

Figure 4.4: (i) finite pulse�ua � ua � h

� � h (ii) delta function as limh � 0�ua � ua � h

� � h

The graph can be viewed in figure 4.4 which shows of a pulse of height 1 � h and time of duration h at timet � a: the area under the curve is

� a

00 � a � h

a

�1 � h � � ∞

a � h0 � � ∞

0

ua�α � � ua � h

h

�α � dα � 1

Now we are ready to define the Dirac delta (quasi)function

δa� lim

h � 0

ua � ua � h

h

This limit process is sketched in figure 4.4. An alternative definition of this (quasi) function is given by

δa�t �� 0 : 0 t � a;

∞ : t � a;0 : a � t

and also � ∞

0δa

�α � dα � 1

Next we consider the Antiderivative function or integral

G�t �� t

0δa

�α � dα

G�t � is the area under the graph of δa between 0 and t. Thus G

�t � � 0 if 0 t � a and G

�t � � 1 if t � a (all

area is concentrated under the pulse at t � a). We have that

� t

0δa

� ua� �

δa� � � ua

Consequentlyua C � 1 � 0 � ∞ � � δa C � 2 � 0 � ∞ �

4.3 Convolution

definition Let f and g be two real valued functions defined over positve time, t � 0.

f : � 0 � ∞ � � R � g : � 0 � ∞ � � Rt �� g

�t � � t �� f

�t �

We define a new real valued function over positive time, denoted

f � g : � 0 � ∞ � � Rt ��

�f � g � �

t � � � t0 f

�α � g

�t � α � dα

c�


4.3. CONVOLUTION 77

Note that

� t

0f

�α � g

�t � α � dα � � 0

tf

�t � β � g

�β � � � 1 � dβ

�� put β � t � α � α � t � βthen dβ � � dα � dα � � dβalso α � 0 � β � t � α � t � β � 0

�� t

0f

�t � β � g

�β � dβ

� � t

0f

�t � α � g

�α � dα switch dummy variables α and β

summary

f � g�t � � � t

0f

�t � α � g

�α � dα � � t

0f

�α � g

�t � α � dα � � t � 0

and this implies thatf � g � g � f

4.3.1 commutativity, associativity of �

theorem 5 Let f � g and h : � 0 � ∞ � � R be functions.

�i � f � g � g � f �

�ii � f � �

g � h � � �f � g � � h

proofs

proofs The proof of (i) is above. We will assume and use (ii) from now on. The proof is quite strenousbut is postponed until theorem 6. Assuming (ii) we can write f � g � h without ambiguity.remark Trivial results about convolution will be clearly stated if needed and left as easy exercises; suchasex. f � �

g h � � f � g f � h.

ex. Compute, eat � ebt ,i.e. if a � b R and f�t � � eat and g

�t � � ebt for all t � 0 find f � g

�t � for all t � 0.

f � g�t � � � t

0f

�α � g

�t � α � dα

� � t

0eaαeb

�t � α � dα

� � t

0ebt e

�a � b � α dα use next, ebt independent w.r.t. dα

� ebt � t

0e�a � b � α dα beware!, special case a � b, see below

� ebt 1a � b

e�a � b � α �� tα 0

since�

ekx dx � ekx � k

� ebt 1a � b

� e � a � b � t � 1 �� eat � ebt

a � b

We now take up the case a � b: continuing from the warning line above

� ebt � t

0e�a � b � α dα now we look at the case a � b

� ebt � t

01dα

� tebt

conclusion

eat � ebt � eat � ebt

a � b� if a �� b � eat � eat � teat


jbquig-UCD


In the latter case we are convoluting two sympathetic decays. As promised in � 4.1, where there is sympathythere is t.eg. Now would be a good time to attempt the following example which appears inproblem sheet, � 4.13.Comput the convolution of unsympathetic and of sympathetic oscillation

cosat � cosbt � a �� b and cosat � cosat

remark on the concept of convolutionThe value

f � g�t � � � t

0f

�α � g

�t � α � dα � � t

0f

�t � α � g

�α � dα

of the convolute f � g at the instant t involves the entire history of both f and g with time running forwardfor one and backward for the other.

eg. Let n � m � 0 be integers, compute tn � tm.

tn � tm � � t

0αn �

t � α � m dα prepare for an int. by parts

� � t

0

�t � α � m 1

n 1d � αn � 1 � next integrate by parts

� 1n 1

�t � α � m αn � 1 �� tα 0

� � t

α 0

1n 1

αn � 1 d�t � α � m after integration by parts

� � 0 � 0 � � mn 1

� t

α 0αn � 1 �

t � α � m � 1 � � 1 � dα

� mn 1

� t

α 0αn � 1 �

t � α � m � 1 dα� m

n 1tn � 1 � tm � 1

summary to date tn � tm � mn 1

tn � 1 � tm � 1 if m � 1.

Applying this process repeatedly, until m � 1 is reduced to 0

tn � t � m � mn 1

tn � 1 � tm � 1

� m � �m � 1 �

�n 1 ��

n 2 � tn � 2 � tm � 2

� m � �m � 1 ��

m � 2 ��n 1 ��

n 2 �� n 3 � tn � 3 � tm � 3

...

...

� m � �m � 1 ��

m � 2 � � � � 2 � 1�n 1 ��

n 2 �� n 3 � � � � �

n m � 1 �� n m � tn � m � t0

� � 1 � 2 � 3 � 3 �n � 1 � n � � � � m � �

m � 1 � � �m � 2 � � � � 2 � 1 �� 1 � 2 � 3 � 3 �

n � 1 � n � � � � �n 1 ��

n 2 �� n 3 � � � � �

n m � 1 �� n m � � tn � m � 1

� n!m!�n m � !

tn � m � 1

� n!m!n m!

� t

0αn � m 1

�t � α � dα

� n!m!�n m � !

� t

0αn � m 1dα

� n!m!�n m � !

1n m 1

tn � m � 1

� n!m!�n m 1 � !

tn � m � 1

c�


4.3. CONVOLUTION 79

Conclusion, for integers n � m � 0

tn � tm � n!m!�n m 1 � !

tn � m � 1

eg.

1 � t � t0 � t1 � t2 � 2 � 1 � t2 � t0 � t2 � t3 � 3 � 1 � tn � t0 � tn � tn � 1 � �n 1 �

t2 � t3 � t6 � 60 � t4 � t5 � 4!5! t6 � 10! �

remark The first three seem to indicate that the process of integration can be carried out by convolutionwith 1 � u0, see � 4.4.

ex. Now would be a good time to compute (see problems � 4.13)

eat � cos�bt � � t � cos

�at � � t � eat

4.3.2 Integration and Shift operators

Recall that C � 2 � 0 � ∞ � or C � 2 � for short denotes the vector space of functions defined over positive time � 0 � ∞ �and which become continuous if integrated twice. This space contains at least the continuous functions,the step and delta functions. We study two operations on this space, the integration operator and the shiftoperator.

integration operator Let a � 0 be arbitrary

� : C � 2 � C � 1

f �� F

Here

F�t � � � t

0f

�α � dα � � t � 0

shift operator

Sha : C � 2 � C � 2

f �� Sha�f �

Here

Sha�f � �

t � � ua�t �� f

�t � a � �

�0 : 0 t � a;

f�t � a � : a t;

� � t � 0

The shift operator simply shifts the origin from 0 to a. Naively one expects the definition to beSha

�f � �

t � � f�t � a � , but this gives wrong results in examples. The difficulty is caused by the fact

that f is not defined for t � a so that f�t � a � is undefined for 0 � t � a. Using the correct defini-

tion Sha�f � �

t � � ua�t � f

�t � a � clearly states that Sha

�f � �

t � � 0 in the range 0 � t � a, which is correct.example

Sh1t2 � u1�t � � �

t � 1 � 2 ��

0 : 0 t � 1;�t � 1 � 2 : 1 t;

Contrast this with the naive wrong result Sh1t2 � �t � 1 � 2. See figure 4.5

4.3.3 convolution with ua and δa

Convolution of standard classical functions is an exercise in elementary integration, albeit at times strenous.Soon we will have a new easy method to perform convolutions by using the Laplace transform L , see section


jbquig-UCD


t

y

t

y

t

y

Figure 4.5: (i) t2 (ii)�t � 1 � 2,inappropriate (iii) Sh � t2 � � u1

�t � �

t � 1 � 2

� 4.4. Now we study convolution with the Heavyside Step and the Dirac Delta functions. Let f � g : � 0 � ∞ � � Rbe well behaved C 0 functions; we need to fill out the table.

� ub δb gua (vi) (iv) (ii)δa (iv) (v) (iii)f (ii) (iii) (i)

Since � is commutative there are really only six not nine computations to be carried out. Also entry (i) ismerely the definition of convolution

(i) f � g � � t

0f

�α � g

�t � α � dα .

(ii) ua� f

ua� f

�t � � � t

0 ua�α � f

�t � α � dα

� �� t

00dα : if t a;

� a

00dα � t

a1 � f

�t � α � dα : if a � t

by defn. of ua

� �� 0 : if t a;

0 � t

af

�t � α � dα : if a � t

next put β � t � α

� �� 0 : if t a;

� 0

t � af

�β � � � 1 � dβ : if a � t

dα � � dβ

� �� 0 : if t a;

� t � a

0f

�β � dβ : if a � t

��

0 : if t a;F

�t � a � : if a � t

where F�t � � � t

0f

�α � dα

� ua�t � F

�t � a �

conclusion Convoluting with ua can be performed by (α) integrating, followed by (β) shifting; inother words

ua� f � ua

�t � F

�t � a � � Sha

� � t

0f

�α � dα � (4.1)

In particular, if a � 0

u0� f � u0

�t � F

�t � 0 � � F

�t ��

� � t

0f

�α � dα � (4.2)

c�


4.3. CONVOLUTION 81

(ii) δa� f Given a � 0 and t � 0 � ∞ �

δa� f

�t � �

�limh � 0

ua � ua � h

h� f � �

t � by defn. of δa

� limh � 0

�ua

� f � �t � � �

ua � h� f � �

t �h

� f

In the latter fraction, there are three cases to consider

t a In this case t a and t � a h, by two applications of 4.1, ua� f

�t � � 0 � ua � h

� f�t � . We

conclude that δa� f

�t � � 0 for all t a.

a � t a h In the limit as h � 0 eventually a h will be less than t. We may ignore this case, itbecomes the next case.

t � a h In this case

δa� f

�t � � lim

h � 0

�ua

� f � �t � � �

ua � h� f � �

t �h

recall F�t � � � t

0f

� limh � 0

F�t � a � � F

�t � a � h �

hsince ua

� f�t � � F

�t � a � � t � a

� limk � 0

F�t � a � � F

�t � a k �� k

since k � � h � 0 as h � 0

� limk � 0

F�t � a k � � F

�t � a �

kthis is differentiation from first principles

� F� �

t � a � next use the fundamental theorem of calculus� f

�t � a �

Gathering all this we have proven

�δa

� f � �t �� Sha

�f � �

t � ��

0 : if t a;f

�t � a � : if a � t

In other words, convolution with δa is the shift (by a) operator.

(iv) δa� ub

δa� ub

� Sha�ub

�� ua � b

(v) δa� δb

δa� δb

� Shaδb� δa � b

(vi) ua� ub

ua� ub

� �ua

� � �ub

��

δa� u0

� � �δb

� u0�

� �δa

� δb� � �

u0� u0

��

δa � b� � �

u0� 1 �

� �δa � b

� �� t

01 �

� �δa � b

� � t� Sha � b t� ua � b

�t � � �

t � a � b �

��

0 : if t a b;t � �

a b � : if a b � t


jbquig-UCD


4.4 The Laplace Transform

definition Given a real valued function f : � 0 � ∞ � � t �� f�t � R of positive time, we determine a number

a R called the abscissa of convergence of f and define a new function called the Laplace transform of fand denoted

L�f � :

�a � ∞ � � R

s : �� L�f � �

s � � � ∞

0e � st f

�t � dt for s � a only

revision Recall that � ∞

0e � st f

�t � dt is called an improper integral (infinite range of integration), and is

defined by a limit process

� ∞

0e � st f

�t � dt � lim

α � ∞ � α

0e � st f

�t � dt

Generally there is a number a such that if s � a this limit process converges (and if s � a it diverges); such ais called the abscissa of convergence of f ; enough for now, see the the next few examples.

4.5 Laplace transform of classical functions

4.5.1 L�eat �

L�eat � �

s � � � ∞

0e � st eat dt

� limα � ∞ � α

0e � steat dt

� limα � ∞ � α

0e�a � s � t dt

� limα � ∞

e�a � s � t

a � s

�� α0� lim

α � ∞ e�a � s � α � 1a � s

� limα � ∞

1s � a

� 1 � e�a � s � α �

� 1s � a � 1

s � a

�limα � ∞ e

�a � s � α �

The latter limit does not exist if a � s � 0 and exists and equals 0 if a � s � 0. Finally

L�eat � �

s �� 1s � a

� if s � a only

The abscissa of convergence of f�t � � eat is a.

c�


4.5. LAPLACE TRANSFORM OF CLASSICAL FUNCTIONS 83

A less concise but faster and acceptable way to write this proof is

L�eat � �

s � � � ∞

0e � steat dt

� � ∞

0e�a � s � t dt

� e�a � s � t

s � a

�� ∞0� 1

a � s� e � a � s � t � �� ∞0

� 1a � s

� e � a � s � ∞ � e0 �� 1

a � s� e � ∞ � e0 � but only if s � a

� 1a � s

�0 � 1 � but only if s � a

� 1s � a

but only if s � a

4.5.2 L�cosat � and L

�sinat �

The direct method is correct but leads to onerous computation

L�cos

�at � � �

s � � � ∞

0e � st cos

�at � dt

This integration can be carried out by two integration by parts, followed by a clever trick (you will find thisin a first course on integration). There is quicker slicker way to proceed.

L�cos

�at � � �

s � iL�sin

�at � � �

s � � L�cos

�at � isin

�at � � �

s �� L

�eiat � �

s �� ∞

0e � st eiat � dt

� � ∞

0e�ia � s � � t � dt

� e�ia � s � � t

ia � s

�� ∞0� 1

s � ia� e0 � e

�ia � s � � ∞ �

� 1s � ia

�1 � 0 � but only for s � 0, see note below

� 1s � ia

but only for s � 0

� s ias2 a2 but only for s � 0

Extracting real and imaginary parts

L�cos

�at � � �

s �� ss2 a2 � L

�sin

�at � � �

s �� as2 a2 � but only for s � 0

For both cos and sin the abscissa of convergence is 0.note Here we used �� e � ia � s � ∞ �� lim

α � ∞�� e � ia � s � α ��

� limα � ∞ �� eiaαe � sα ��

� limα � ∞ �� eiaα �� e � sα ��


jbquig-UCD


� limα � ∞ 1 � �� e � sα ��

� limα � ∞ �� e � sα ��

� 0 but only for s � 0

4.5.3 L�tn �

Let n � 0 be an integer

L�tn � �

s � � � ∞

0e � st tn dt

� � 1s

� ∞

0tn d

�e � st � prepare for integration by parts

� � 1s

tne � st �� ∞0 � � 1s

� ∞

0e � st d

�tn � used integration by parts

� � 1s

� ∞ne � ∞ � 0ne0 � � � 1s

� ∞

0e � std

�tn � but only for s � 0

� � 1s

�0 � 0 � � � n

s� ∞

0e � st tn � 1 dt ∞ne � ∞ � 0, see note below

� � ns

L�tn � 1 � �

s �

Summing up to date

L�tn � �

s � � � ns

L�tn � 1 � but only for s � 0

Repeating this step

L�tn � �

s � � � � 1 �� ns

L�tn � 1 �

� � 1 � � n�n � 1 �s2 L

�tn � 2 �

� � � 1 � � n�n � 1 � �

n � 2 �

s3 L�tn � 3 �

� ...

� ...

� � � 1 � n � n!sn L

�t0 �

� � � 1 � n � n!sn L

�1 �

� � � 1 � n � n!sn L

�u0

�

� � � 1 � n � n!sn

1s

� � � 1 � n n!sn � 1

Finally

L�tn � �

s � � � � 1 � n n!sn � 1 but only if s � 0

The abscissa of convergence of tn is 0 for any integer n � 0.example L

�coshat � and L

�sinhat �

left to reader, see section � 4.13

c�


4.6. LAPLACE TRANSFORM WITH STEP AND DELTA FUNCTIONS 85

4.6 Laplace transform with step and delta functions

4.6.1 L�ua�

L�ua

� �s � � � ∞

0ua

�t �� e � st dt

� � a

00 � e � st dt � ∞

a1 � e � st dt by defn. of ua

� � ∞

ae � st dt

� e � st� s�� ∞a

� 1s

� e � as � e � s∞ �� 1

s� e � as � 0 � but only if s � 0

To sum up

L�ua

� �s � � e � as

sbut only for s � 0

The abscissa of convergence for ua is 0.

exercise L�t � � L

� � t � � � L�t � � t � � � L

� � � 1 � � t � Compute L�t � � L

� � t � � � L�t � � t � � � L

� � � 1 � � t � , the last threeare the staircase, sawtooth and squarewave functions , read the explanation on the next few lines, and viewthe graphs.

The standard function t ,see figure 4.6(i),can be broken into integer and fractional parts � t � and t � � t � ; theseare called the staircase and sawtooth functions, see figure 4.6(ii) and 4.7(i). The staircase function � t � can beviewed as a digital version of t. The function

� � 1 �� t is known as the square wave, see figure 4.7(ii) To be

-5

-3

-1

1

3

5

-4 -2 0 2 4

t

y

-5

-3

-1

1

3

5

-4 -2 0 2 4

t

y

Figure 4.6: (i) t (ii) staircase function � t �concise � t � denotes the largest integer less than or equal to t, ( � 3 � 5 � � 3 � � 3 � � 3 � � � 3 � � � 3 � � � 3 � 5 � � � 4),thus � t � � u1

u2 u3

� � � un � � � �

∞

∑n 1

un� n; n t � n 1 � n Z

t � � t � � t � ∞

∑n 1

un� t � n ; n t � n 1 � n Z

� � 1 � � t � u0 � 2u1 2u2 � 2u3

� � � � � u0 2

∞

∑n 0

� � 1 � nun�

�1; 2n t � 2n 1 � n Z� 1; 2n 1 t � 2n 2 � n Z


jbquig-UCD


t

y1

t

y1

-1

Figure 4.7: (i) sawtooth (ii) squarewave functions; t � � t � and� � 1 �� t

4.6.2 L�δa�

Lδa�s � � L

�limh � 0

ua � ua � h

h� �

s �

� limh � 0

L�ua

� �s � � L

�ua � h

� �s �

h

� limh � 0

e � as � s � e ��a � h � s � s

h

� � � 1 � 1s

limh � 0

e ��a � h � s � e � as

hthe limit of a Newton Quotient

� � � 1 � 1s

ddx

e � xs �� x a i.e. a derivative

� � � 1 � 1s

� � s � exs � x � a i.e. a derivative� e � as

FinallyL

�δa

� �s � � e � as for s � 0 only

The abscissa of convergence of δa is 0.

4.7 Formulae for the Laplace Transform

4.7.1 L�f � g �� L � 1 and f � g � h

We begin with a two part theorem; one part has already been mentioned, see � � 4.3.1

theorem 6 Let f � g � h : � 0 � ∞ � be real valued functions of positive time.

(i) L�f � g � � L

�f �� L

�g �

(ii) f � �g � h ��

f � g � � h

for proofs see � 4.8, see also equation 4.3. We will use these results right away. The very important part (i)says that the Laplace transform of a convolution is the (ordinary) product of Laplace transforms.

corollary Denote the inverse Laplace transform by L � 1. One may rewrite theorem 6(i) as

f � g � L � 1 �L

�f �� L

�g � �

c�


4.7. FORMULAE FOR THE LAPLACE TRANSFORM 87

remark This gives a fast easy way to convolute (up to now a strenous business). There is an integrationformula for the inverse Laplace transform, but we will not use it. Instead to compute L � 1 of any givenfunction we rely on a table, see � 4.12, of known Laplace transforms and formulae which we wield in reverse.

It is known that L is a special case of a more general transform F called the Fourier transform. The inverseFourier transform is easily described in terms of integration. We do not pursue the question of L � 1 beyondthese few remarks.

remark Here is a proof of

f � �g � h � � �

f � g � � h (4.3)

being part(ii) of theorem 6. This proof uses part(i), but see also � 4.8.2.

f � �g � h � � L � 1 �

L�f � �

g � h � � �� L � 1 �

L�f �� L

�g � h � � � by part(i)

� L � 1 �L

�f ��

L�g �� L

�h � � � by part(i) again

� L � 1 � �L

�f �� L

�g � �� L

�h � � by associativity of ordinary multiplication

� L � 1 � �L

�f � g �� L

�h � � by part(i)

� L � 1 �L

� �f � g � � h � � � by part(i) again

� �f � g � � h �

example

eat � ebt � L � 1L�eat � ebt �

� L � 1 �L

�eat �� L

�ebt � �

� L � 1 � 1s � a

� 1s � b

�

� L � 1�

1�s � a � �

s � b � �� L � 1

�A

s � a B

s � b �� AL � 1

�1

s � a � BL � 1

�1

s � b �� Aeat Bebt

Above the theory of partial fractions (as used in integration of fractional polynomials in FY calculus) hasbeen used.

1�s � a � �

s � b � � As � a

Bs � b

1 � A�s � b � B

�s � a �

Putting s � b we get B � 1 � b � a Putting s � a we get A � 1 � a � b Putting it all together

�eat � ebt � �

s � � eat � ebt

a � b

But all this works for a �� b only. If there is sympathy a � b we instead obtain.

�eat � ebt � �

s � � L � 1

�1�

s � a � 2 �See ahead � � 4.7.3 to deal with this very important atomic (it can’t be broken up or simplified) partial fraction.


jbquig-UCD


4.7.2 L and differentiation L�f� �� L

�f�n � � and L

��f � and also L

�tn f �

To wield L and solve SOCCLHDE’s one must compute the Laplace transform of the integral and of deriva-tives of all orders � f � L

�f� � � L

�f� � � � L

�f� � � � � � � L

�f�n � � � �

of functions f : � 0 � ∞ � � R.L

�f� � �

s � � sL�f � �

s � � f�0 � (4.4)

proof

L�f� � �

s � � � ∞

0e � st f

� �t � dt

� � ∞

t 0e � st d f

� e � st f�t � �� ∞0 � � ∞

t 0f

�t � d

�e � st � after integration by parts

� � e � s∞ f�∞ � � e0 f

�0 � � � � ∞

t 0f

�t � d

�e � st � see below for meaning of e � s∞ f

�∞ �

� � 0 � f�0 � � � � ∞

t 0f

�t � d

�e � st � but only for s � a, the abs. of cgce. of f

� � f�0 � � � ∞

t 0f

�t � e � st � � � s � dt

� s � ∞

t 0f

�t � e � st � dt � f

�0 �

� sL�f � �

s � � f�0 �

remark e � s∞ f�∞ � denotes the limit

limt � ∞

e � st f�∞ � if this limit actually exists

definition f�t � is said to be exponentially bounded by eAt iff for large t � 0 � f �

t � � � eAt .

In that case it is easy to see (exercise) that the abscissa of convergence of f is no larger than A. Moreover,since �� e � st f

�t � �� e

�A � s � t for large t � 0

the latter limit exists (and equals 0) for s � A.

example Let f�t �� eat � t � 0 . Then f

� � a f . apply L to both sides of this equation. L�f� � � aL

�f � .

Thus sL�f � � f

�0 � � aL

�f � . But f

�0 � � 1. Thus

�s � a � L

�f � � f

�0 � � 1

L�eat � �

s � � L�f � �

s � � 1s � a

L�f� � � �

s � � s2L�f � �

s � � s f�0 � � f

� �0 � (4.5)

proofL

�f� � � �

s � � L� �

f� � � � �

s �� sL

�f� � � �

s � � f� �

0 � by 4.4� s � sL

�f � � �

s � � f�0 � � � f

� �0 � again by 4.4

� s2L�f � � �

s � � s f�0 � � f

� �0 �

L�f�n � � �

s � � snL�f � �

s � � sn � 1 f�0 � � sn � 2 f

� �0 � � sn � 2 f

� � �0 � � � � � � � s2 f

�n � 3 � �

0 � � s f�n � 2 � �

0 � � f�n � 1 � �

0 �(4.6)

� snL�f � �

s � � n � 1

∑0

sn � j � 1 f�j � �

0 � (4.7)

textbfexample. Using the L�f� � � formula one can calculate. L

�cos

�at � � � L

�sin

�at � � � L

�cosh

�at � � and L

�sinh

�at � � .

c�


4.7. FORMULAE FOR THE LAPLACE TRANSFORM 89

Three of these are left as exercises, here we calculate only L�cosh

�at � � . Note that cosh

� �at � � asinh

�at �

and cosh� � �

at � � a2 cosh�at � . Applying the Laplace transform to both sides of the latter equation we obtain

L�cosh

� � �at � �� a2L

�cosh

�at � � . Thus

s2L�cosh

�at � � �

s � � scosh�0 � � cosh

� �0 �� L

�cosh

�at � �

s � � (4.8)�s2 � a2 � L

�cosh

�at � � �

s � � scosh�0 � cosh

� �0 �� scosh

�0 � asinh

�0 �� s 0 (4.9)

ThusL

�cosh

�at � � �

s �� ss2 � a2 (4.10)

proof This is a more elaborate version of formula 4.5. The proof is left as an exercise, do it for n=3 and 4.Then use induction to obtain a proof for a general integer n � 0

Writing F�t �� t

0f

L�F � �

s � � 1s

L�f � �

s � (4.11)

proof Since F� � f by formula 4.4

L�f � �

s � � sL�F � �

s � F�0 �� sL

�F � �

s � 0 � sL�F � �

s �

The reader may prefer the alternative proof

L�F � �

s � � L� � t

0f � �

s � � L�u0

� f � �s � � L

�u0

� �s �� L

�f � �

s � � 1s� L

�f � �

s �

Now we study L and differentiation in the imaginary frequency s we study

L�t f

�t � � �

s � � � ∞

0e � stt f

�t � dt

� � ∞

0

dds

� e � st f�t � � dt

� � ∞

0� d

ds� e � st f

�t � � dt

� � dds

� � ∞

0e � st f

�t � dt �

� � dds

L�f � �

s �

Applying this result once, twice or even n times we have

L�t f � �

s � � � dds

L�f � �

s �

L�t2 f � �

s � � d2

ds2 L�f � �

s �

L�tn f � �

s � �� d

ds � n

L�f � �

s �

example

L�tn � �

s � � L�tn � 1 � �

s �

�� d

ds � n

L�1 � �

s �

�� d

ds � n 1s

� n!sn 1


jbquig-UCD


4.7.3 L and shifting L�Sha f � and Shb

�L f �

There are two shift formulae involving the Laplace transform. One involves shift in real time t, discussed indetail above. The second involves shifting in the imaginary frequency s: fortunately there are no difficulties(as experienced when shifting in t). In fact, let G be a function of s and b R.

Shb�G � �

s � � G�s � b � de f inition

�L

�Sha f � � �

s � � �L

�δa

� f � � �s � � e � asL

�f � �

s � (4.12)

proof �L

�δa

� f � � �s � � L

�δa

� �s �� L

�f � �

s � � e � asL�f � �

s ��L

�eat f

�t � � � �

s � � Shb�L

�f � � �

s � � L�f � � �

s � a � (4.13)

proof

�L

�eat � f

�t � � � �

s � � � t

0e � st eat f

�t � dt � � t

0e�s � a � t f

�t � dt � �

L f � �s � b � � �

Shb�L f � � �

s �

4.7.4 L and periodicity

Let f : � 0 � ∞ � � t �� f�t � R be periodic with period p � 0, i.e. f

�t p � � f

�t � for all t � 0, then

L�f � �

s � �� p

0 e � st f�t � dt

1 � e � sp (4.14)

proof

L�f � �

s � � � ∞

0e � st f

�t � dt

�� p

0

� 2p

p

� � � � �k � 1 � p

kp

� � � � e � st f�t � dt

�∞

∑k 0

� �k � 1 � p

kpe � st f

�t � dt put t � τ kp

�∞

∑k 0

� p

0e � s

�τ � kp � f

�τ kp � dτ by periodicity f

�τ kp � � f

�τ �

�∞

∑k 0

� p

0e � s

�τ � kp � f

�τ � dτ use e � s

�τ � kp � � e � staue � kp

�∞

∑k 0

� p

0e � sτe � skp � f

�τ � dτ use e � kp � �

e � p � k in indepent of dτ

�∞

∑k 0

� e � p � k � p

0e � sτ f

�τ � dτ

� ∞

∑k 0

� e � p � k � p

0e � sτ f

�τ � dτ

��

11 � e � sp � � p

0e � sτ f

�τ � dτ recall for � x � � 1,

∞

∑0

xn � 1 � �1 � x �

�� p

0 e � sτ f�τ � dτ

1 � e � sp being the sum of a geometric progression

example, Laplace transform of square wave Recall the square wave function� � 1 � � t , see � � 4.6.1 and

figure 4.7. f is periodic with period 2

L � � � 1 � � t � �� 2

0e � st � � 1 � � t dt

1 � e � 2t

c�


4.8. TWO DIFFICULT PROOFS 91

� 11 � e � 2t

� � 1

0

� 2

1 � e� � st � � � 1 � � t

� 11 � e � 2t

� � 1

0

�1 � e � st dt � 2

1

� � 1 � e � st dt �� 1

1 � e � 2t

� � 1

0e � st dt � 2

1

� � 1 � e � st dt �� 1

1 � e � 2t e � st� s�� 10 � e � st� s

�� 21 � 1

s�1 � e � 2t � � �

1 � e � t � � �e � t � e � 2t ��

� 1 � 2e � t e � 2t

s�1 � e � 2t �

��1 � e � t � 2

s�1 � e � t � �

1 e � t �

� 1 � e � t

s�1 e � t �

exercise The reader might now apply the same formula to compute L�t � � t � � , see figure4.7 and � 4.6.1. The

sawtooth function t � � t � has period 1.

4.8 Two difficult proofs

4.8.1 full proof that L�f � g �

�s �� L

�f ��s �� L

�g ��s �

Let f and g be real valued functions of positive time t, i.e. functions from � 0 � ∞ � to R. We will here provethe, (up to now missing ) proof of theorem 6 part (i), i.e. that

L�f � g � �

s �� L�f � �

s �� L�g � �

s � for suitable s

proof Let s R. We expand the L.H.S.

a

b

p

t

Figure 4.8: the quadrant Q and the triangle A


jbquig-UCD


L�f � g � �

s � � � ∞

0e � st �

f � g � �t � dt by defn. of L

� � ∞

0e � st � t

0f

�p � g

�t � p � dpdt by defn. of convolution �

� � ∞

0� t

0e � st f

�p � g

�t � p � dpdt since e � st is independent of p

� ��A

e � st f�p � g

�t � p � d

�p � t � here A � R2 is the infinite triangle

A ��

pt � �� 0 p t � ∞ �

see figure 4.8

Next we expand the R.H.S.

L�f � �

s �� L�g � �

s � � � ∞

0e � sa f

�a � da � � ∞

0e � sbg

�b � db by defn. of L , used twice

� � ∞

0

� � ∞

0e � sa f

�a � da � e � sbg

�b � db since the da integral, being

constant is independent of b

� � ∞

0e � sbg

�b � � ∞

0e � sa f

�a � dadb after rearrangement

� � ∞

0� ∞

0e � sa f

�a �� e � sbg

�b � dadb e � sbg

�b � is indep. of a

� � �Q

e � s�a � b � f

�a � g

�b � d

�a � b � where Q � R2 is the quadrant

Q ��

ab � �� a � b � 0 �

see figure 4.8

For convenience writeG

�p � t �� e � st f

�p � g

�t � p � (4.15)

Define the bijective mapping or transformation h from R2 to R2 by

h : Q � A�ab � ��

�p

�a �

t�a � b � � �

�1 01 1 � �

ab � �

�a

a b �This h is linear (i.e. a matrix mapping) with det

�h � � 1 �� 0 and so h is bijective as claimed.

h carries Q onto A, i.e.h

�Q �� A

Indeeda � 0 and b � 0 � 0 a a b � ∞ � 0 p t � ∞

Since h is linear, Dh � h , and so the Jacobian of h is

det�Dh �� det

�h �� 1

Finally

L�f � g � �

s �

� ��A

e � st f�p � g

�t � p � d

�p � t �

� ��A h

�Q �

G�p � t � d

�p � t �

c�


4.8. TWO DIFFICULT PROOFS 93

� � �Q

G � h�a � b �� det

�Dh � �

a � b � � d �a � b � by the COV theorem

� � �Q

G�p

�a � b � � t

�a � b � � � 1d

�a � b �

� � �Q

e � st�a � b � f �

p�a � b � � g

�t

�a � b � � p

�a � b � � d

�a � b �

� � �Q

e � s�a � b � f

�a � g

�a b � a � d

�a � b �

� � �Q

e � s�a � b � f

�a � g

�b � d

�a � b �

� L�f � �

s �� L�g � �

s �

4.8.2 full proof that�f � g � � h � � f �

�g � h �

Let f � g and g be real valued functions of positive time t, i.e. functions from � 0 � ∞ � to R. We will here givethe (skipped over) proof that

� �f � g � � h � �

t � � �f � �

g � h � � �t � for t � 0

see theorem 6, � � 4.3.1 and � 4.3.

a

b b=t

p

qt

t

Figure 4.9: triangles (ii) T ; a b t (ii) D; 0 q t and 0 p t � q

proof Let t R. We expand the L.H.S.

� �f � g � � h � �

t � � � t

0

�f � g � �

b � h�t � b � db by defn of �

� � t

0

� � b

0f

�a � g

�b � a � da � h

�t � b � db by defn of � again

� � t

0h

�t � b �

� � a

0f

�a � g

�b � a � da � db after a slight rearrangement

� � t

0� a

0f

�a � g

�b � a � h

�t � b � dadb since h

�t � b � is independent of da

� �� t

Tf

�a � g

�b � a � h

�t � b � d

�a � b � by Fubini’s theorem: here the triangle

T ��

ab � �� 0 a b t � � R2

see figure 4.9


jbquig-UCD


Next we expand the R.H.S.

�f � �

g � h � � �t � � � t

0f

�q � �

g � h � �t � q � dq by defn of �

� � t

0f

�q �

� � t � q

0g

�p � h

� �t � q � � p � dp � dq by defn of � again

� � t

0� t � p

0f

�q � g

�p � h

�t � p � q � dpdq since f

�q � is indep. of dp

� � �D

f�q � g

�p � h

�t � p � q � d

�p � q � by Fubini’s theorem: here the triangle

D ��

pq � �� 0 q t and 0 p t � q �

see figure 4.9

Using these expansions of both the R.H.S and L.H.S., we must prove

� �T

f�a � g

�b � a � h

�t � b � d

�a � b � � � �

Df

�q � g

�p � h

�t � p � q � d

�p � q � (4.16)

For convenience writeH

�p � q �� f

�q � g

�p � h

�t � p � q � (4.17)

Define the bijective mapping or transformation h from T to R2 D

h : T � D�ab � ��

�p

�a � b �

q�a � b � � �

� � 1 11 0 � �

b � aa � �

�b � a

a �This h is linear (i.e. a matrix mapping) with det

�h � � � 1 �� 0 and so h is bijective as claimed.

h carries T onto D, i.e.h

�T �� D

Indeed

0 a b t � 0 a t and 0 b � a t � a � 0 q t and 0 p t � q

Since h is linear, Dh � h , and so the Jacobian of h is

� det�Dh � � � � det

�h � � � 1

Finally

f � �g � h � �

t �

� � �D

f�q � g

�p � h

�t � p � q � d

�p � q �

� � �D

H�p � q � d

�p � q �

� � �T

H�p

�a � b � � q

�a � b � � � detDh � d

�p � q � by the COV theorem

� � �T

H�p

�a � b � � q

�a � b � � d

�p � q � since the Jacobean is � 1

� � �T

f�q

�a � b � � g

�p

�a � b � � h

�t � p

�a � b � � q

�a � b � � d

�p � q � by definition of H

� � �T

f�a � g

�b � a � h

�t � b � d

�p � q �

� � � t

Tf

�a � g

�b � a � h

�t � b � d

�a � b �

�f � g � � h

�t �

c�


4.9. HOMOGENOUS DIFFERENTIAL EQUATIONS AND THE LAPLACE TRANSFORM 95

4.9 Homogenous Differential equations and the Laplace Transform

4.9.1 characteristic polynomial of a differential equation

The most general CCHDE of any order n takes the form

(HDE) a0 x�n � �

t � a1 x�n � 1 � �

t � � � � ak x�n � k � �

t � � � � an � 1 x�t � anx

�t � � 0 (4.18)

here ak R � 0 k n are called coefficients Suitable general initial conditions are

(IC) x�k � �

0 � given for 0 k n � 1 (4.19)

definition The characteristicpolynomial associated with this HDE is

p�λ ��

n

∑k 0

akλn � k � a0λn a1λn � 1 � � � akλn � k � � � an � 1λ an (4.20)

This, polynomial (of degree n) will appear frequently below.

4.9.2 example HDE of degree 1

Solve

(HDE) x�t � 3x

�t �� 0 with (IC) x

�0 �� 7 (4.21)

Apply the L , ignore IC for now

L�x � �

s � 3L�x � �

s � � L�0 �

� � sL�x � �

s � � x�0 � � 3L

�x � �

s � � 0 see � 4.12 formula 4.69� �

s 3 � L�x � �

s � � 7 after rearranging and using the IC� L

�x � �

s � � 7s 3

divide across by p�s � � s 3

� x�t � � L � 1

�7

s 3 � inverse L applied

� x�t � � 7L � 1

�1

s � � � 3 � �� x

�t � � 7e � 3t see � 4.12 formula 4.53

4.9.3 example HDE of degree 2

Solve

(HDE) x�t � 5x

�t � 6x

�t �� 0 with (IC) x

�0 �� 2 � x

�0 � � 9 (4.22)


L�x � �

s � 5L�x � �

s � 6L�x � �

s � � L�0 � next use � 4.12 formulae 4.69 and 4.70

� � s2L�x � �

s � � sx�0 � � x

�0 � � 5 � sL

�x � �

s � � x�0 � � 6 � L �

x � �s � � � 0 next use the IC

� � s2L�x � �

s � � � � 2 � s � 9 � 5 � sL�x � �

s � � � � 2 � � 6 � L �x � �

s � � � 0 rearrange� � s2 5s 6 � L �

x � �s � � � 2s � 1 next divide by p

�s � � s2 5s 6

� L�x � �

s � � � 2s � 1s2 5s 6

next apply the inverse Laplace transform


jbquig-UCD


Thus

x�t � � L � 1

� � 2s � 1s2 5s 6 �

� L � 1� � 2s � 1�

s 3 � �s 2 � � next we split up the partial fraction

� L � 1

�A

1s 2

B1

s 3 � where A and B soon will be calculated

� AL � 1

�1

s � � � 2 � � BL � 1

�1

s � � � 3 � � next use � 4.12 formula 4.53

� Ae � 2t Be � 3t

We calculate the constants A and B using� 2s � 1�s 2 � �

s 3 �� A

s 2 B

s 3� � 2s � 1 � A

�s 3 � B

�s 2 �

Put s � � 3 and obtain B � � 5. Put s � � 2 and obtain A � 3. Finally

x�t � � 3e � 3t � 5e � 2t

4.9.4 example HDE of degree 2 with natural sympathy

Solve

(HDE) x�t � 10x

�t � 25x

�t � � 0 with (IC) x

�0 �� 2 � x

�0 � � 9 (4.23)


L�x � �

s � 10L�x � �

s � 15L�x � �

s � � L�0 � next use formulae � 4.12 formulae 4.69 4.70

� � s2L�x � �

s � � sx�0 � � x

�0 � � 10 � sL

�x � �

s � � x�0 � � 25 � L �


� � s2L�x � �

s � � � � 2 � s � 9 � 10 � sL�x � �

s � � � � 2 � � 25 � L �x � �

s � � � 0 rearrange� � s2 10s 25 � L �

x � �s � � � 2s � 11 next divide by p

�s � � s2 10s 25

� L�x � �

s � � � 2s � 11s2 10s 25


� x�t � � L � 1

� � 2s � 11s2 10s 25 �

� x�t � � L � 1

� � 2s � 11�s 5 � 2 � get ready for the second shift formula

� x�t � � L � 1

� � 2�s 5 � � 1�s 5 � 2 � use � 4.12 formula 4.52

� x�t � � e � 5t L � 1

� � 2s � 1s2 � break up the partial fraction

� x�t � � e � 5t L � 1

�� 1 � 1

s2 � 21s � see � 4.12 formulae 4.57, 4.58

� x�t � � e � 5t � � 1 � 2t �

Finallyx

�t �� e � 5t �

1 2t �

The important thing is to note the term te � 5t ; “where there is sympathy there is t”.

c�


4.9. HOMOGENOUS DIFFERENTIAL EQUATIONS AND THE LAPLACE TRANSFORM 97

4.9.5 example HDE of degree 2, complex roots

Solve

(HDE) x�t � 8x

�t � 25x

�t � � 0 with (IC) x

�0 �� 4 � x

�0 � � 7 (4.24)


L�x � �

s � 8L�x � �

s � 25L�x � �

s � � L�0 � next use � 4.12 formulae 4.69, 4.70

� � s2L�x � �

s � � sx�0 � � x

�0 � � 8 � sL

�x � �

s � � x�0 � � 25 � L �


� � s2L�x � �

s � � �4 � s � 7 � 8 � sL

�x � �

s � � �2 � � 25 � L �

x � �s � � � 0 rearrange

� � s2 8s 25 � L �x � �

s � � 4s 23 next divide by p�s � � s2 8s 25

� L�x � �

s � � 4s 23s2 8s 25


x�t � � L � 1

�4s 23

s2 8s 25 � irreducible quadratic denominator

� L � 1�

4�s 4 � 7�

s 4 � 2 32 � completed the square

� e � 4t L � 1

�4s 7s2 32 � used � 4.12 formula 4.52

� e � 4t

�43L � 1 s

s2 32 7

3L � 1 3

s2 32 � prepare to use trig formulae

� e � 4t

�43cos

�3t � 7

3sin

�3t � � used � 4.12 formulae 4.54, 4.55

Finally

x�t �� 43e � 4t cos

�3t � 7

3e � 4t sin

�3t �

The solution will involve damped vibration if the characteristic polynomial of a HDE has complex roots.

4.9.6 solving a general CCHDEwIC of order 4

After these examples it is easy to solve the general HDE of any order n. First consider the most generalCCHOLDEwIC of order four only.

�HDE � ax

�4 � �

t � bx�3 � �

t � cx�t � dx

�t � ex

�t � � 0; t � 0�

IC � x�0 �� α � x

�0 �� β � x

�0 � � γ � x

�3 � �

0 �� δ

here the coefficients a � b � c � d and e are assumed constant as are the initial conditions α � β � γ and δ. ApplyL to the HDE,

aL x�4 � �

s � bL x�3 � �

s � cL x�s � dL x

�s � eLx

�s � � 0 (4.25)

Use the differentiation formulae for L , � 4.12 formula 4.56

a � s4L x�s � s3x

�0 � s2x

�0 � sx

�0 � x

�3 � �

0 � � b � s3L x�s � s2x

�0 � sx

�0 � x

�0 � � c � s2L x

�s � sx

�0 � x

�0 � � d � sL x

�s � x

�0 � � e � Lx

�s � �

� 0


jbquig-UCD


Use the IC, initial conditions

a � s4L x�s � αs3 βs2 γs δ � b � s3L x

�s � αs2 βs γ � c � s2L x

�s � αs β � d � sL x

�s � α � e � Lx

�s � �

� 0

Rearrange � as4 bs3 cs2 e � L x�s � � As3 Bs2 Cs D (4.26)

where the constants A � b � C and D are easily calculated as expressions in terms of the initial conditions. TheR.H.S can be written

As3 Bs2 Cs D (4.27)� α

�as3 bs2 cs d � β

�bs2 cs d γ

�cs d � δd (4.28)

� �p

�s � � s �� x

�0 � �

p�s � � s2 �� x

�0 � �

p�s � � s3 �� x

�0 � �

p�s � � s4 �� x

�3 � �

0 � (4.29)

Dividing across

Lx�s � �

�p

�s � � s �� x

�0 � �

p�s � � s2 �� x

�0 � �

p�s � � s3 � � x

�0 � �

p�s � � s4 �� x

�3 � �

0 �p

�s � (4.30)

Apply the inverse Laplace transform

x�t �� L � 1

��

3

∑k 0

p�s �

sk � 1 x�k � �

0 �

p�s �

� �� (4.31)

4.9.7 solving the most general CCHDEwIC of order n

The general case of a HDE of order n is no harder than the for order 4: here is a terse statement of the HDE,t � 0, IC and solution

(HDE)n

∑j 0

a jx�n � j � �

t �� 0 with given (IC) xk �0 � for 0 k n � 1 (4.32)

(solution) x�t �� L � 1

��

n � 1

∑k 0

p�s �

sk � 1 x�k � �

0 �

p�s �

�� ; t � 0 (4.33)

4.9.8 natural response

A particularly important case is when all IC are zero except that x�n � 1 � �

0 � � 1. The solution in that case iscalled the natural response of the undriven (homogenous) system

(HDE)n

∑j 0

a jx�n � j � �

t � � 0 with (IC) xk �0 �� 0 for 0 k n � 2 but x

�n � 1 � �

0 � � 1 (4.34)

(solution) x�t � � L � 1

�1

p�s � � ; t � 0 (4.35)

c�


4.10. INHOMOGENEOUS DIFFERENTIAL EQUATIONS AND THE LAPLACE TRANSFORM 99

4.10 Inhomogeneous Differential equations and the Laplace Trans-form

4.10.1 example unsympathetically driven IHDE of degree 1

Solve

(IHDE) x�t � 3x

�t �� e � 2t cos

�t � with (IC) x

�0 �� 7 (4.36)


L�x � �

s � 3L�x � �

s � � L�e � 2t cos

�t � �

� � sL�x � �

s � � x�0 � � 3L

�x � �

s � � Sh � 2Lcos�t � used formula � 4.12 4.68

� �s 3 � L

�x � �

s � � 7 Sh � 2s

s2 1used � 4.12 4.54

� �s 3 � L

�x � �

s � � 7 s 2�s 2 � 2 1

used the � 4.12 4.52

� L�x � �

s � � 7s 3

s 2�

s 3 � �s2 4s 5 � divided across by p

�s � � s 3

� x�t � � L � 1

�7

s 3 � L � 1

�s�

s 3 � �s2 4s 5 � � L � 1 applied

� x�t � � L � 1

�7

s 3 � L � 1

�1

s 3� s 2s2 4s 5 �

� x�t � � L � 1

�7

s 3 � L � 1

�1

s 3 � � L � 1

�s 2

s2 4s 5 � used � 4.12 4.66� x

�t � � 7e � 3t e � 3t � e � 2t cos

�t � see � 4.12

The solution of the IHDEwIC is formed by summing the solution of the correcponding HDEwIC and thenatural response convoluted with the driving (RHS) function. The natural response (a decay) is not in sym-pathy with the driving function (a decaying oscillation). But this is all too abstract. To hack out the answergo back to line

x�t � � L � 1

�7

s 3 � L � 1

�1

s 3� s � 2s2 4s 5 � next decompose the PF

� L � 1�

7s 3 � L � 1

�A

s 3 Bs C

s2 4s 5 � see below for A, B, C

� L � 1

�7

s 3 � L � 1

�A

s 3 B

�s 2 � �

C � 2B ��s 2 � 2 1 � completed the square

� e � 3t L � 1

�AL � 1 1

s 3 B

s 2�s 2 � 2 1

�C � 2B � 1�

s 2 � 2 1 � use � 4.12 4.52

� e � 3t L � 1

�AL � 1 1

s 3 e � 2tB

ss2 1

�C � 2B � e � 2t 1

s2 1 � see table � 4.12

� e � 3t � Ae � 3t Be � 2t cos�t � �

C � 2B � e � 2t sin�t �� see table � 4.12


jbquig-UCD


The constants A � B and C are found from

s�s 3 � �

s2 4s 5 � � A1

s 3 Bs C

s2 4s 5� s � A

�s2 4s 5 � �

Bs C � �s 3 �

Putting s � � 3 obtain A � � 3 � 2. Equating the coeff of s2 on RHS and LHS obtain 0 � A B, B � � A � 3 � 2.Equating the constant term on RHS and LHS obtain 0 � 5A 3C, 0 � � 15 � 2 3C, C � 5 � 2.

4.10.2 sympathetically driven IHDE of degree 1

Solve

(IHDE) x�t � 3x

�t �� e � 3t with (IC) x

�0 � � 7 (4.37)


L�x � �

s � 3L�x � �

s � � L�e � 3t �

� � sL�x � �

s � � x�0 � � 3L

�x � �

s � � 1s 3

� �s 3 � L

�x � �

s � � 7 1s 3

rearranged and used the IC

� L�x � �

s � � 7s 3

1�s 3 � 2 divided across by p

�s � � s 3

� x�t � � L � 1

�7

s 3 � L � 1

�s�

s 3 � 2 � inverse L applied

� x�t � � L � 1

�7

s 3 � L � 1

�1

s 3� 1s 3 �

� x�t � � L � 1

�7

s 3 � L � 1

�1

s 3 � � L � 1

�1

s 3 � used � 4.12 formula 4.66� x

�t � � 7e � 3t e � 3t � e � 3t

The solution of the IHDEwIC is formed by summing the solution of the correcponding HDEwIC and thenatural response convoluted with the driving (RHS) function. The natural response (a decay) is fully insympathy with the driving function (the same decay). But this is all too abstract. To hack out the answer goback to line

� x�t � � L � 1

�7

s 3 � L � 1

�1�

s 3 � 2 � next use � 4.12 4.68

� x�t � � 7e � 3tL � 1

�1s � e � 3tL � 1

�1s2 �

� x�t � � 7e � 3tL � 1

�1s � e � 3t �

7 t �

4.10.3 example unsympathetically driven IHDE of degree 2

Solve

(IHDE) x�t � 8x

�t � 25x

�t � � e � 2t cos

�t � with (IC) x

�0 � � 4 � x

�0 � � 7 (4.38)

c�


4.10. INHOMOGENEOUS DIFFERENTIAL EQUATIONS AND THE LAPLACE TRANSFORM 101


L�x � �

s � 8L�x � �

s � 25L�x � �

s � � L�e � 2t cos

�t � � use � 4.12 4.68, 4.69, 4.70

� � s2L�x � �

s � � sx�0 � � x

�0 � � 8 � sL

�x � �

s � � x�0 � � 25 � L �

x � �s � � � sh � 2L

�cos

�t � � �

s � use IC and � 4.12 4.54� � s2L

�x � �

s � � �4 � s � 7 � 8 � sL

�x � �

s � � �2 � � 25 � L �

x � �s � � � sh � 2

1s2 1

rearrange

� � s2 8s 25 � L �x � �

s � � 4s 23 1�s 2 � 2 1

divide by p�s � � s2 8s 25

� L�x � �

s � � 4s 23s2 8s 25

1�

s2 8s 25 � �s2 4s 5 �

Apply the inverse Laplace transform to obtain

x�t �

� L � 1�

4s 23s2 8s 25 � L � 1

�1�

s2 8s 25 � �1�

s2 4s 5 � �� L � 1

�4s 23

s2 8s 25 � L � 1

�1�

s2 8s 25 � � � L � 1

�1�

s2 4s 5 � �� L � 1

�4

�s 4 � 7�

s 4 � 2 32 � L � 1

�1� �

s 4 � 2 32 � � � L � 1

�1� �

s 2 � 2 1 � �� e � 4tL � 1 4s 7

s2 32 �

e � 4t L � 1 1�s2 32 � � �

�e � 2tL � 1 1�

s2 1 � �� e � 4t

�4cos

�3t � � 7

3 � sin�3t � � �

e � 4t � 13 � sin

�3t � � � � e � 2t sin

�t ��

The driving force (a damped oscillation) is not sympathetic with the natural response (also a damped oscil-lation but a different one). The solution of the IHDEwIC is the sum of the solution of the correspondingHDEwIC and the driving function convoluted with the natural response. The expression of the solution interms of a convolution is too abstract. We now hack out the details of solution. Continuing from a few linesback

x�t �

� L � 1

�4s 23

s2 8s 25 � L � 1

�1� �

s 8s 25 � �s 4s 5 � � next decompose the PF

� L � 1

�4s 23

s2 4s 5 � L � 1

�As B

s2 8s 25 Cs D

s2 4s 5 � ex. work out A � B � C and D

� L � 1�

4�s 4 � 7�

s 2 � 2 1 � L � 1

�A

�s 4 � �

B � 4s ��s 4 � 2 32

C�s 2 � �

D � 2c ��s 2 � 2 1 � by completing the square

� e � 2tL � 1

�4s 7s2 1 � e � 4t L � 1

�As �

B � 4A �s2 32 �

e � 2tL � 1

�C

�s 2 � �

D � 2C ��s 2 � 2 1 � by � 4.124 � 68

� e � 2t �4cos

�t � 7sin

�t � � e � 4t �

Acos3t � �B � 4A � � 3 � sin

�3t � �

e � 2t �C cos

�t � �

D � 2C � sin�t � �

4.10.4 example sympathetically driven IHDE of degree 2

Solve

(IHDE) x�t � 8x

�t � 25x

�t � � e � 4t cos

�3t � with (IC) x

�0 � � 4 � x

�0 � � 7 (4.39)


jbquig-UCD


Apply the L , ignore IC for now.

L�x � �

s � 8L�x � �

s � 25L�x � �

s � � L�e � 4t cos

�3t � � use � 4.12 4.68, 4.71

� � s2L�x � �

s � � sx�0 � � x

�0 � � 8 � sL

�x � �

s � � x�0 � � 25 � L �

x � �s � � � sh � 4L

�cos

�3t � � �

s � use IC� � s2L

�x � �

s � � �4 � s � 7 � 8 � sL

�x � �

s � � �2 � � 25 � L �

x � �s � � � sh � 4

1s2 32 rearrange

� � s2 8s 25 � L �x � �

s � � 4s 23 1�s 4 � 2 32 divide by p

�s � � s2 8s 25

� x�t � � L � 1

�4s 23

s2 8s 25 � L � 1

�1�

s2 8s 25 � 2 �� x

�t � � L � 1

�4s 23

s2 8s 25 � L � 1

�1

s2 8s 25 � �

L � 1

�1

s2 8s 25 �The driving force (a damped oscillation) is fully sympathetic with the natural response (a damped oscillationwith the same frequency and rate of decay). The solution of the IHDEwIC is the sum of the solution ofthe corresponding HDEwIC and the driving function convoluted with (itself) the natural response. Theexpression of the solution in terms of a convolution is too abstract. We now hack out the details of solution.Continuing from a few lines back

x�t � � L � 1

�4

�s 4 � 7�

s 4 � 2 25 � L � 1

�1� �

s 4 � 2 32 � 2 � by completing the square

� e � 4tL � 1

�4s 7s2 25 � e � 4tL � 1

�1�

s2 32 � 2 � by � 4.12 4.68

We leave the reader to find the former expression which has been, in fact, calculated above. The latterexpression is very important, and we will use a trick (NB, and see � 4.12, formulae 4.77 and 4.78) to computeit. Note that, as expected (when sympathy is present), t appears in the solution.

L � 1�

1�s2 32 � 2 � � L � 1

�1�

s2 a2 � 2 � �� a 3

� L � 1

� � � 12a � d

da1�

s2 a2 � � �� a 3

�� 1

2a � dda

L � 1�

1�s2 a2 � � �� a 3

�� 1

2a � dda

�1a

sin�at � � �� a 3

�� 1

2a � � � 1a2 sin

�at � t

acos

�at � � �� a 3

��

12a3 sin

�at � � t

2a2 cos�at � � �� a 3

� 154

sin�3t � � t

18cos

�3t �

c�


4.11. SYMPATHY 103

4.10.5 solving the most general CCIHDEwIC of order n

The general solution of a IHDEwIC of order n is easily derived from that of the corresponding HDEwIC hereis a terse statement of the HDE, t � 0, IC and solution.

(IHDE)n

∑j 0

a jx�n � j � �

t �� h�t � with given (IC) xk �

0 � for 0 k n � 1 (4.40)

(solution) L�x � �

t � �

n � 1

∑k 0

p�s �

sk � 1 x�k � �

0 �

p�s � L

�h � �

s �p

�s � ; t � 0 (4.41)

(solution)�x � �

t � � L � 1

��

n � 1

∑k 0

p�s �

sk � 1 x�k � �

0 �

p�s �

�� L � 1

�L

�h � �

s �� 1p

�s � � ; t � 0 (4.42)

(solution)�x � �

t � � L � 1

��

n � 1

∑k 0

p�s �

sk � 1 x�k � �

0 �

p�s �

� �� h

�t � � L � 1

�1

p�s � � ; t � 0 (4.43)

The solution of the general solution of the IHDEwIC is formed by taking the sum of the solution of thecorresponding HDEwIC and the natural response (of the HDE) convoluted with the driving function (RHS).

4.11 Sympathy

Sympathy is present in a differential equation in x�t � , when one finds a factor ( of either of the forms�

s � a � n or� �

s � a � 2 b2 � n, with n � 1 ), in the denominator of L�x � �

s � . This is called natural sympathyif the term is present as a factor of the characteristic polynomial p

�s � . However the most important form

of sympathy occurs when p�s � and the denominator of L

�h � �

s � have a common factor. When sympathy ispresent, powers of t appear in the solution (eg. terms such as tne � at � tne � at cos

�bt � ; n � 1 appear).


jbquig-UCD


4.12 Table of Laplace Transforms

�f � g � �

t � � � t

0f

�α � g

�t � α � dα (4.44)

f � g � g � f (4.45)

f � �g � h � � �

f � g � � h (4.46)

ua�t � �

�0; 0 t � a1; a t

(4.47)

δa�t � � limh � 0

ua � ua � h

h

�t �� 0; 0 t � a

∞; t � a0; a � t

and � ∞

0δa

� 1 (4.48)

ua� � δa

� ua� � t

0δa (4.49)

L�f � �

s � � � ∞

0e � st f

�t � dt (4.50)

Sha f�t � � �

δ � f � �t � � ua

�t � f

�t � a � (4.51)

ShbG�s � � G

�s � b � (4.52)

L�eat � �

s � � 1s � a

(4.53)

L�cosat � �

s � � ss2 a2 (4.54)

L�sinat � �

s � � as2 a2 (4.55)

lt�tn � �

s � � n!sn � 1 (4.56)

lt�1 � �

s � � 1s

(4.57)

lt�t � �

s � � 1s2 (4.58)

lt�t2 � �

s � � 2s3 (4.59)

L�coshat � �

s � � ss2 � a2 (4.60)

L�sinhat � �

s � � as2 � a2 (4.61)

L�δa

� �s � � e � as (4.62)

L�δ0

� �s � � 1 (4.63)

L�ua

� �s � � e � as

s(4.64)

L�u0

� �s � � 1

s(4.65)

L�f � g � �

s � � L�f � �

s �� L�g � �

s � (4.66)

L�Sha f � �

s � � e � asL�f � �

s � (4.67)

L�ebt f

�t � � �

s � � L�f � �

s � b � (4.68)

L�f� � �

s � � sL�f � �

s � � f�0 � � (4.69)

L�f� � � �

s � � s2L�f � �

s � � s f�0 � � � f

� �0 � � (4.70)

L�f�n � � �

s � � snL�f � �

s � � n � 1

∑k 0

sn � k � 1 f�k � �

0 � � (4.71)

L� � t

0f � �

s � � 1s

L�f � �

s � (4.72)

c�


4.12. TABLE OF LAPLACE TRANSFORMS 105

L�t f � �

s � � � dds

L�f � �

s � (4.73)

L�t2 f � �

s � � d2

ds2 L�f � �

s � (4.74)

L�tn f � �

s � �� d

ds � n

L�f � �

s � (4.75)

L�f � �

s � �� p

0f

�t � dt

1 � e � sp if f is periodic with period p (4.76)

L � 1 �AS Bs2 a2 � 2 � � 1

2adda

�L � 1 As B

s2 a2 � (4.77)

L � 1

� �AS Bs2 a2 � n � � � � 1 � n � 1 1�

n � 1 � !

�12a

dda � n � 1 �

L � 1 As Bs2 a2 � (4.78)


jbquig-UCD


4.13 problem set

Differential Equations and the Laplace transformDo not attempt all these; there are too many; I will give a selection. jbq

4.13.1 Convolution, Delta and Heavyside functions, Laplace Transform

1. On Heavyside and Dirac Functions.Let f be the square wave function f

�t � � � � 1 � � t � t � 0 , here � t � denotes the greatest whole integer t.

(i) Draw a graph of f .

(ii) f �∞

∑1

bnuan , find an and bn � n � 1 i.e. express f in terms of Heavyside step functions.

(iii) Compute f�. Express the result in terms of Dirac Delta functions.

(iv) Compute the antiderivate F�t � � � t

0f

�x � dx � u0

� f .

(v) Draw graphs of f�and F , can you put names on these functions?

(vi) f Cα, f� Cβ and F Cγ Find α � β and γ. Here Cn denotes the class of functions on � 0 � ∞ �

which are every where continuous with all derivarives up to order n (an integer possibly negative,would you believe it ?).

2. On ConvolutionCompute f � g

�t � � t � 0 where

(i) f�t � � exp

�at � � g

�t � � exp

�bt � � t � 0 � for both cases a � b and a �� b.

(ii) f�t � � sin

�at � � g

�t � � sin


(iii) f�t � � sin

�at � � g

�t � � cos


(iv) f�t � � cos

�at � � g

�t � � cos


(v) f�t � � tn � g

�t � � tm � t � 0.

3. On Convolution

(i) Compute 1 � 1 and 1 � 1 � 1 and finally 1 � 1 � 1 � � � � � 1 � n times.

(ii) Hence compute tn � tm � n � m � 0.

4. On ConvolutionCompute

�6 3t sin � t

2 � u5 δ2

� � �11 t2 cost δ3

�

5. Laplace Transform.Compute the Laplace transform, L

�f � �

s � , for each case f�t � � t � 0 � below. In each case find the

abscissa of convergence.�i � cosat

�ii � sinat

�iii � expat�

iv � 1�v � t

�vi � t2

�vii � tn �

viii � ua�t � �

ix � sinhat�x � coshat

�xi � t sin2t

�xii � sin2 t�

xiii � t2 cos3t

6. Use the L�f� � formula to obtain

c�



(i) L�cosat � from L

�sinat � and vica versa.

(ii) L�expat � , recall exp

�at � aexpat.

(iii) L�cosat � , recall cos

� �at � � a2 cosat.

(iv) L�sinat � , recall sin

� �at � � a2 sinat.

(v) L�t3 � from L

�t2 � from L

�t � from L

�1 � .

(vi) L�coshat � from L

�sinhat � and vica versa.

(vii) L�coshat � recall cosh

� �at � � a2 coshat.

(viii) L�sinhat � recall sinh

� �at � � a2 sinhat.

7. Study of first Shift Formula L � ua�t � f

�t � a � � .

(i) Compute L � ua�t � sin

�t � a � � � t � 0.

(ii) Compute L � ua�t � sin

�t � � � t � 0.

(iii) Compute L � sin�t � a � � � t � 0.

(iv) Compute L�f � where f

�t � � t � 0 t � 2 � f

�t � � 2 � 2 t � 4 �

f�t � � t � 4 t � 6 � f

�t � � t � 6 t . Also draw the graph of f .

(v) Compute L�f � where f

�t � � sin t � 0 t � 2π � f

�t � � 0 � 2π t .

8. Use the formula L � tn f�t � � to compute

�i � L

�t � �

ii � L�t2 � �

iii � L�t3 �

�iv � L

�t sin2t � �

v � L�t2 cos3t � �

vi � L�tn � � n � 0

9. Use the second Shift formula L�exp

�at � f

�t � � to compute

�i � L

�exp

�at � � given L

�1 � �

ii � L�exp

�2t � sin t � �

iii � L�exp

� � 3t � cos2t ��iv � L

�t2 expt cos t � �

v � L�exp

� � 3t � cos�2t 4 � �

10. Use the formula L�f � g � � L

�f � � L

�g � or f � g � L

�� 1 � � L �

f �� L�g � � to compute

�i � tm � tn �

ii � exp�at � � exp

�bt � � a � b and a �� b�

iii � sin�at � � sin

�bt � � a � b and a �� b

�iv � cos

�at � � cos

�bt � � a � b and a �� b�

v � cos�at � � sin

�bt � � a � b and a �� b

�vi � 1 � 1�

vii � 1 � 1 � 1�viii � 1 � 1 � 1 � � � � 1 � 1 � n-times�

ix � �6 3t sin

� t2

� u5 δ2

� � �11 t2 cost δ3

� �x � ua

� ub�xi � ua

� δb�xii � δa

� δb

11. Formula for the Laplace Transform of a periodic function.

(i) Use to compute L�cosat � and L

�sinat � .

(ii) Also to compute L�f � where f

�t � � � � 1 � � t � t � 0 , here � t � denotes the greatest whole integer t.

(iii) And L�f� � and L

�F � where F

�t � � � t

0f

�x � dx.

(iv) F above is a Sawtooth wave, so is g defined below. Compare the graphs of F and g. ComputeL

�g � where g

�t � � n 1 � t � n � t � n 1.

12. Let h�t � � n 1 � n � t � n 1. This h is a staircase function, graph it. Compute L

�h � both from

the point of view h � ∑∞0 un and from the point of view that g � t � h is periodic.


jbquig-UCD


4.13.2 Differential Equations and the Laplace Transform

1. Solve x � 3x 2x � t expt �u3 � u1

� cos t δ2 with initial conditions x�0 � � 1 � x

�0 � .

2. Solve x � 2x x � t2 δ3 with initial conditions x�0 � � 2 � x

�0 � � 0 .

3. Solve x x x � expt cos t �t � 3 � u3

�t � with initial conditions x

�0 � � 2 � x

�0 � � 3 .

4. Solve x x x � expt cos t �t � 3 � u3

�t � with initial conditions x

�0 � � 2 � x

�0 � � 3 .

5. Solve mx λx � 2mδ1 17mδ10 Asinωt with initial conditions x

�0 � � a � x

�0 � � 0 .

This is the case of the undamped mechanical oscillator with periodic driving force (and also irregularblows). Also comment on the analogous electrical system.

Hint:- first work the example with only the delta functions on the right hand side, a routine piece ofwork. Next work it with Asinωt on the right hand side, this is more complicated. The full solution isobtained by adding these two together.

The case os the sinusoidal driving force breaks down into two cases�i � ω � �

λ � m and�ii � ω ��

λ � maccording as the frequency of the driving force equals or does not equal the frequency of the oscillator.In the second case the solution is the sum of two oscillations (frequencies 2πω and 2π

�λ � m) each with

finite amplitude. In the first case the solution will be of the form A�t � sin

�ωt φ � where the amplitude

A�t � � ∞ as t � ∞. Thus the motion gets more and more violent and the spring breaks (maybe).

6. From our studies it is clear that L � 1 �p

�s � � q

�s � � (the inverse Laplace transform of the general ratio-

nal polymnomial) needs to be worked out in solution of differential equations. By the method of partial

fractions, this task can be reduced to computation of L � 1 1�s � a � n � n � 1 and of L � 1 as b�

cs2 ds e � n � n �1, here cs2 ds e is the general irreducible quadratic in s.

Compute both expressions for n � 1 then for n � 2 finally for general n.[ Hint:-

1�s � a � n

� 1n � 1!

dn � 1

dan

1s � a

For the second expression, use a similar differentiation.]

7. Given the system of differential equations

d2xdt2

d2ydt2 � dx

dt 2x � y � δ1

d2xdt2 � 4

dxdt

2dydt

4x 2y � δ2

and with initial conditionsx

�0 � � x

�0 �� y

�0 �� y

�0 �� 0

(i) Prove that

L�x � �

s � ��s � 1 � exp

� � 2s � � 2exp� � s �

s�s � 2 � �

s � 5 �

(ii) Compute x�t � � t � 0 �

8. Find the Laplace transform of t1 � 2.

9. Find the inverse Laplace transform of

3s 1s2 4s

c�



10. Using Laplace transforms, or otherwise, solve the coupled differential equations

x � 2y � 1 x y � x � 0

for x�t � � y

�t � , when x

�0 � � y

�0 � � 0.

11. Assuming that all transforms exist, show that

L � y � � s2L � y � � sy�0 � � y

�0 �

and

L � ty � � � dds

L � y �12. Using Laplace transforms, solve the differential equation

ty �1 � t � y 3y � 0

where y�0 � � 1.

13. Given the differential equation

d2xdt2

2dxdt

2x � g�t � � t � 0

with initial conditionsx

�0 �� x

�0 �� 0

where g�t �� 1 � 0 t 1 and g

�t �� 0 � t � 1 �

(i) Prove that

L�x � �

s � � 1 � exp� � s �

s�s2 2s 2 �

(ii) Compute x�t � � t � 0 �


jbquig-UCD


c�


Chapter 5

Change of Variable theorem and theJacobean

introduction We have seen three methods for evaluation of an integral I � �A

f�x � d

�x � of a scalar

field (i.e. scalar function of several variables) f over a region A � Rn, of Euclidean n-space. These are

� From first principles as the limit of a Rieman Sum, see chapter 1.

� From first principles but using a computer to calculate a Riemann sum which closely approximates theintegral, see � 1.8.

� Use Fubini’s theorem, chapter 2.

Now we turn to the fourth and most sophisticated method, by change of variable (c.o.v.) and the Jacobean. Aregion of integration such as A above might be quite much more complicated than a simple rectangloid Take,for example, the region to be a solid ball B of radius a � 0. Each point on B is given by three Cartesian co-ordinates x � y and z which unhappily run between very complicated limits, as indeed we have seen in chapter2. But each point on B is also determined by 0 � r � a, the radial distance from 0, by 0 � θ � 2π longitudeand 0 � φ � π latidude. In different words, there is a bijective mapping, or transformation, or change ofvariable

h :�0 � a � � � 0 � 2π �� 0 � π � � B

The change of variable theorem tells us, in this example, that I can be expressed as an integral over the simplerectangloid

�0 � a � � � 0 � 2π �� 0 � π � . But there is a complication. Each integral is built up from function value

multiplied by volume and volume is dilated (or distorted) by a change of variable mapping. To compensatefor this the famous Jacobean term (read about it below � 5.4 and � 5.5) appears in the transformed version ofthe integral. We begin with the easiest of transformations, the matrix or linear mapping.

5.1 matrices and volume dilation

remark A 3 � 3 matrix or linear mapping

A : R3 � R3

x �� x

yz

�� Ax � A

�� xyz

��changes or dilates volume. The next theorem shows that the volume dilation factor under the action of A is� det A � . (In passing we mention that, according as det A is positive or negative, A preserves Flemings’s RightHand Rule or changes it into the left hand Rule.). Here we have the very meaning (in geometric terms) ofdeterminant. A similar result holds in each dimension n. In dimensions n � 1 � 2 or 3 � det A � is the length,

111

112 CHAPTER 5. CHANGE OF VARIABLE THEOREM AND THE JACOBEAN

area, volume dilation factor respectively, under the action of the n � n matrix A: in these cases if det�A � � 0

then A reverses direction, changes clockwise to anticlockwise or F.R.H.R. to F.L.H.R. respectively.

theorem 7 volume dilation of matrixLet A : R3 � R3 be a linear mapping or 3 � 3 matrix

A �� a d g

b e hc f i

��Then the volume dilation factor under the action of A is

v � d � f � �A �� detA � � � det

�� a d gb e hc f i

�� proofThe basic vectors i � j � k R3 are carried by the matrix A to the three column vectors

A i �� a b c

d e fg h i

�� 100

�� a

bc

��

A1

A j �� a b c

d e fg h i

�� 010

�� d

ef

��

A2

Ak �� a b c

d e fg h i

�� 001

�� g

hi

��

A3

respectively.Consider the basic solid cube C, delineated by the three basic vectors i � j and k. The eight vertices of C are

0 � i � i j � j � k � i k � i j k and j k

and the volume of C is 1.The cube C is carried by the matrix mapping A to P � A

�C � the parallelopiped delineated by the column

vectors�

A1 ��

A2 and�

A3. The eight vertices of P are�

0 ��

A1 ��

A1 �

A2 ��

A2 ��

A3 ��

A1 �

A3 ��

A1 �

A2 �

A3 ��

A2 �

A3

and the volume (a positive quantity) of P is (recall first year vector geometry or see the next paragraph) thepositivised vector triple product

� �

A1 � �

A2� �

A3 � � �� a

bc

�� def

�� g

hi

�� detA �The volume dilation factor under the action of the matrix A is

volume Pvolume C

� � �

A1 � �

A2� �

A3 �1

� � detA �but there remains:-

volume of the parallelopiped revised The base of the parallelopiped P is the parallelogram delineatedby the two vectors

�

A1 and�

A2, i.e. with vertices�

0 ��

A1 ��

A1 �

A2 and�

A2. Writing θ for the angle between thesetwo vectors the area of this base is��

A1��

A2�� sin

�θ � � ��

A1 � �

A2�� by very definition of the crossproduct

c�


5.1. MATRICES AND VOLUME DILATION 113

The height of P is the positivised inner product

� n � �

A3 � � � ��

A3�� cos

�φ � �

where n is the unit normal vector to the base plane and φ is the angle between�

A3 and n. Since both�

A1 and�

A2

lie in the base, n is the vector product, normalised for unit length

n ��

A1 � �

A2��

A1 � �

A2��

Finally the volume of P is base area by height��

A1 � �

A2�� n � �

A3 � � � ��

A1 � �

A2��

A1 � �

A2��

A1 � �

A2��

A3 � � � � �

A1 � �

A2 � � �

A3 �5.1.1 determinant of a matrix whose columns are mutually

�

If a n � n matrix A has mutually perpendicular columns�

Ai ��

A j � 1 i � j n � i �� j

there is an easy way to compute � detA � which is very useful especially if the entries of A are difficulttrigonometric expressions.

theorem 8 If the columns of the n � n matrix A are mutually perpendicular then

�detA � 2 � ��

A1�� 2 ��

A2�� 2 ��

A3�� 2 ��

A j�� 2 � � � ��

An � 1�� 2 ��

An�� 2 �

n

∏i 1

��

Ai�� 2

i.e. The square of the determinant is the product of the length squared of the matrix columns.

proof�detA � 2 � �

detA � �detA �

� �detAt � �

detA ��

detAtA �

� det

��

�

At1�

At2� � �

�

Atj� � �

�

Atn

� ��

� �

A1 ��

A2 � � � � ��

A j � � � � ��

An�

� det

��

�

At1

�

A1

�

At1

�

A2� � � �

At1

�

A j� � � �

At1

�

An�

At2

�

A1

�

At2

�

A2� � � �

At2

�

A j� � � �

At2

�

An...

.... . .

.... . .

...�

Ati

�

A1

�

Ati

�

A2� � � �

Ati

�

A j� � � �

Ati

�

An...

.... . .

.... . .

...�

Atn

�

A1

�

Atn

�

A2� � � �

Atn

�

A j� � � �

Atn

�

An

� ��

� det

��

� �

A1 ��

A1 � � �

A1 ��

A2 � � � � � �

A1 ��

A j � � � � � �

A1 ��

An ��

A2 ��

A1 � � �

A2 ��

A2 � � � � � �

A2 ��

A j � � � � � �

A2 ��

An �...

.... . .

.... . .

...� �

Ai ��

A1 � � �

Ai ��

A2 � � � � � �

Ai ��

A j � � � � � �

Ai ��

An �...

.... . .

.... . .

...� �

An ��

A1 � � �

An ��

A2 � � � � � �

An ��

A j � � � � � �

An ��

An �

� ��


jbquig-UCD


� det

��

��

A1�� 2 0 � � � 0 � � � 0

0 ��

A2�� 2 � � � 0 � � � 0

......

. . ....

. . ....

0 0 � � � ��

Ai�� 2 � � � 0

......

. . ....

. . ....

0 0 � � � 0 � � � ��

An�� 2

��

�n

∏i 1

��

Ai�� 2

5.2 change of variable

Let

h : A � Rn � B � Rn

x �

��

x1

x2

...x j

...xn

� ��

�� h�x1 � x2 � � � � � x j � � � � � xn � � x

�x1 � x2 � � � � � x j � � � � � xn �

�

��

y1

y2

...y j

...yn

� ��

�

��

y1 �x1 � x2 � � � � � x j � � � � � xn �

y2 �x1 � x2 � � � � � x j � � � � � xn �

...y j �

x1 � x2 � � � � � x j � � � � � xn �...yn �

x1 � x2 � � � � � x j � � � � � xn �

� ��

be a bijective C 1 mapping from the region A � R3 to the region B � R3. Such a mapping ic called atransformation or a change of variable mapping on the region B, from co-ordinates x � y � z to new co-ordinatesp � q � r.example Spherical Polar TransformLet B � B3 �

a � � x �� x2 y � z2 a2 � be the solid three dimensional ball of radius a � 0 center 0. Wegive (for the present without explanation, but see ahead � 5.6) the important Spherical Polar Transform whichchanges from Cartesian to polar co-ordinates on B.

h :�0 � a � � � 0 � 2π �� 0 � π � � B � x �� x2 y2 z2 a2 �� r

θφ

�� h�r� θ � φ � � x

�r � θ � φ � �

�� xyz

�� r sin

�φ � cos

�θ �

r sin�φ � sin

�θ �

r cos�φ �

��

c�


5.3. DIFFERENTIAL MATRIX 115

5.3 differential matrix

Let h : A � Rn � B � Rn be C 1 as above in � 5.2 Recall from Second Year Mathematics the Differentialmatrix

Dh�x �� Dh

�x1 � x2 � � � � � x j � � � � � xn � �

��

∂y1

∂x1

∂y1

∂x2� � � ∂y1

∂x j� � � ∂y1

∂xn

∂y2

∂x1

∂y2

∂x2� � � ∂y2

∂x j� � � ∂y2

∂xn...

.... . .

.... . .

...∂yi

∂x1

∂yi

∂x2� � � ∂yi

∂x j� � � ∂yi

∂xn...

.... . .

.... . .

...∂yn

∂x1

∂yn

∂x2� � � ∂yn

∂x j� � � ∂yn

∂xn

� ��

This matrix, by definition, is the linear mapping which best approximates h (after change of origin) near x.More concisely, if p � x A then, near p,

h�x � � h

�p � � Dh

�p � �

x � p � � h�x � � h

�p � Dh

�p � �

x � p �

example For the spherical polar transform above the differential matrix is,

Dh�r� θ � φ � �

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

� ��

�� sin�φ � cos

�θ � � r sin

�φ � sin

�θ � r cos

�φ � cos

�θ �

sin�φ � sin

�θ � r sin

�φ � cos

�θ � r cos

�φ � sin

�θ �

cos�φ � 0 � r sin

�φ �

��

5.4 Jacobean

Given a change of variable mapping h : A � Rn � B � Rn as in � 5.2. Near any point p A the differentialmatrix Dh

�p � , see � 5.3 closely approximates h near p. The volume dilation factor of h at p is therefore equal

to the volume dilation factor of the linear mapping Dh�p � and is, by theorem 7,

� detDh�p � � � � det

��

∂y1

∂x1

∂y1

∂x2� � � ∂y1

∂x j� � � ∂y1

∂xn

∂y2

∂x1

∂y2

∂x2� � � ∂y2

∂x j� � � ∂y2

∂xn...

.... . .

.... . .

...∂yi

∂x1

∂yi

∂x2� � � ∂yi

∂x j� � � ∂yi

∂xn...

.... . .

.... . .

...∂yn

∂x1

∂yn

∂x2� � � ∂yn

∂x j� � � ∂yn

∂xn

� ��

�

this factor is called the Jacobean of the transformation h.

example We will compute the Jacobean of the spherical polar transform. From the example in S5.3 weneed

� detDh�r� θ � φ � � � � det

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

� ��


jbquig-UCD


� � det


�θ � � r sin

�φ � sin

�θ � r cos

�φ � cos

�θ �

sin�φ � sin

�θ � r sin

�φ � cos

�θ � r cos

�φ � sin

�θ �


�φ �

�� It is possible to slug this out but it is much easier to apply theorem 8. Indeed

Dh1 � Dh2 � Dh1 � Dh3 and Dh2 � Dh3

Two of these involve Dh2 and are trivial. Let us check the remaining one, which is a little more complicated.

Dh1 � Dh3� � Dh1 � Dh3 � � 0

� � �� sin�φ � cos

�θ �

sin�φ � sin

�θ �

cos�φ �

��

�� r cos�φ � cos

�θ �

r cos�φ � sin

�θ �� r sin

�φ �

�� 0

� �sin

�φ � cos

�θ � �

r cos�φ � cos

�θ � � �

sin�φ � sin

�θ � � �

r cos�φ � sin

�θ � � �

cos�φ � � � r sinφ � � 0

� r sin�φ � cos

�phi � � cos2 �

θ � sin2 �θ � � 1 �


�phi � � 1 � 1 �


�phi � � 1 � 1 � � 0

� 0 � 0

which is true. Thus, by theorem 8

� detDh�r� θ � φ � � 2

� � � Dh1 � � 2 � � � Dh2 � � 2 � � � Dh3 � � 2� ��


�θ �

sin�φ � sin

�θ �

cos�φ �

�� 2 � �� r sin

�φ � sin

�θ �

r sin�φ � cos

�θ �

0

�� 2 � �� r cos

�φ � cos

�θ �

r cos�φ � sin

�θ �� r sin

�φ �

�� 2� � sin2 �

φ � cos2 �θ � sin2 �

φ � sin2 �θ � cos2 �

φ � �� r2 sin2 �

φ � sin2 �θ � r2 sin2 �

φ � cos2 �θ � 02 �

� � r2 cos2 �φ � cos2 �

θ � r2 cos2 �φ � sin2 �

θ � r2 sin2 �φ � �

� � sin2 �φ � �

cos2 �θ � sin2 �

θ � cos2 �φ � �

� � r2 sin2 �φ � �

sin2 �θ � cos2 �

θ � � �� r2 cos2 �

φ � �cos2 �

θ � sin2 �θ � � r2 sin2 �

φ � �� sin2 �

φ � �1 � cos2 �

φ � � � � r2 sin2 �φ � �

1 � � � � r2 cos2 �φ � �

1 � r2 sin2 �φ � �

� � 1 � � � r2 sin2 �φ � � � � r2 �

cos2 �φ � sin2 �

φ � � �� 1 � � � r2 sin2 �

φ � � � � r2 �1 � �

� r4 sin2 �φ �

We conclude that the Jacobean of the Spherical transform is

� det Dh�r� θ � φ � � � r2 sin

�φ �

5.5 change of variable theorem

The change of variable theorem (for evaluation of an integral of a scalar field over a domain in Rn) will bestated and proven in the present section and is the central topic in the present chapter.

c�


5.5. CHANGE OF VARIABLE THEOREM 117

theorem 9 change of variable theorem, c.o.v.th.Let h : AsubsetRn � B � Rn be a C 1 bijection or change of variable mapping between two domains in Euclideann � space Rn.

h : A � Rn � B � Rn

x �

��

x1

x2

� � �x j

� � �xn

�� y

�x � �

��

y1 �x1 � x2 � � � � � x j � � � � � xn �

y2 �x1 � x2 � � � � � x j � � � � � xn �

...yi �

x1 � x2 � � � � � x j � � � � � xn �...yn �

x1 � x2 � � � � � x j � � � � � xn �

��

Let f : B � Rn � R be a C 0 scalar field defined over the region B.

f : B � Rn � R

y �

��

y1

y2

� � �yi

� � �yn

�� f

�y �� f

� �y1 � y2 � � � � � yi � � � � � yn � �

Then � � � � � �B

f�y � d

�y � � � � � � � �

Af � h

�x � � � detDh

�x � � d

�x �

Written in full detail this becomes

� � � � � �B

f�y1 � y2 � � � � � yi � � � � � yn � d

�y1 � y2 � � � � � yi � � � � � yn �

� � � � � � �A

f�

y1 �x1 � x2 � � � � � x j � � � � � xn � �

y2 �x1 � x2 � � � � � x j � � � � � xn � �

� � � �yi �

x1 � x2 � � � � � x j � � � � � xn � �� yn �

x1 � x2 � � � � � x j � � � � � xn �

�

� det

��

∂y1

∂x1

∂y1

∂x2� � � ∂y1

∂x j� � � ∂y1

∂xn

∂y2

∂x1

∂y2

∂x2� � � ∂y2

∂x j� � � ∂y2

∂xn...

.... . .

.... . .

...∂yi

∂x1

∂yi

∂x2� � � ∂yi

∂x j� � � ∂yi

∂xn...

.... . .

.... . .

...∂yn

∂x1

∂yn

∂x2� � � ∂yn

∂x j� � � ∂yn

∂xn

��

� d �x1 � x2 � � � � � x j � � � � � xn �

proof For the proof see � � 5.5.2 below. But first we give a diagram and sketch which illustrate thehypotheses, then subsection � � 5.5.1 with first examples of the use of the theorem and finally, as promised,subsection � � 5.5.2 containing the proof.DIAGRAM TO BE INSERTED jbqSKETCH TO BE INSERTED jbq


jbquig-UCD


5.5.1 first examples with the c.o.v.th.

We will repeat five calculations already carried out in chapter 2 using Fubini’s theorem; note how mucheasier these are when carried out, as here, using the c.o.v.th.. Let

B � B3 �a �� x �� x2 y2 z2 a2 �

be the solid ball of radius a � 0. we will compute

�i � m � ��

B1d

�x � y � z � �

�ii � M � ��

Hzd

�x � y � z � �

�iii � I � ��

Bx2 y2 d

�x � y � z �

(recall in (ii) that H � � x B � z � 0�

the upper solid hemiball) and

�iv � z � M

m � 2and

�v � r �

�Im

Recall the s.p.t.f. from � 5.2

h : R3 � �0 � a � � � 0 � 2π �� 0 � π � �

�� rθφ

��

�� xyz

�� r sin

�φ � cos

�θ �

r sin�φ � sin

�θ �

r cos�φ �

�� B � R3

and from � 5.4 the Jacobean of the s.p.t.f.

� detDh�r� θ � φ � � � r2 sin

�φ �

Applying the c.o.v.th, theorem 9

(i)

m � � � � �0 � a � � � 0 � 2π � � 0 � π 1 �

�r2 sinφ � d

�r� θ � φ �

(ii)

M � �� 0 � a � � � 0 � 2π � � 0 � π � 2 �

r cos�φ � � � �

r2 sin�φ � � d

�r� θ � φ �

note that, (whereas for the ball B latitude φ runs from 0 at the North to π at the South pole), for theHemiball H latitude runs from 0 at the North pole only to π � 2 at the equator.

(iii)

I � � �� 0 � a � � � 0 � 2π � � 0 � π r2 sin2 �

φ � �cos2 �

θ � sin2 �θ � � � �

r2 sin�φ � � d

�r� θ � φ �

It remains to carry out the three integrals

c�


5.5. CHANGE OF VARIABLE THEOREM 119

(i)

m� � a

0� π

0� π

� πr2 sinφ � dθdφdr by Fubini’s theorem

� � a

0� π

0r2 sin

�φ � � π

� π1 dθ dφdr since r2 sin

�φ � is independent of dθ

� � a

0� π

02πr2 sin

�φ � dφdr since � 2π

01dθ � 2π

� � a

02πr2 � π

0sin

�φ � dφdr since 2πr2 is independent of dφ

� � a

02πr2

� � cos�φ � � �� πφ 0

dr

� � a

02πr2 � � cos

�π � cos

�0 � � dr

� � a

04πr2 dr

� 43

πr3 �� a0� 4

3πa3

(ii)

M� � a

0� π � 2

0� π

� πr3 sin

�φ � cos

�φ � dθdφdr by Fubini’s theorem

� � a

0� π � 2

0r3 sin

�φ � cos

�π � � π

� π1 dθdφdr since r3 sin

�φ � cos

�φ � is indep of dθ

� � a

0� π � 2

02πr3 sin

�φ � cos

�φ � dφdr since

� 2π0 1dθ � 2π

� � a

0πr3 � π � 2

0sin

�2φ � dφdr since πr3 indep of dφ, and D.A.F.

� � a

0πr3 � � 1

2cos

�2φ � � �� π � 2φ 0

dr

� � a

0πr3 1

2� � cos

�π � cos

�0 � � dr

� π � a

0r3 dr

� 14

πa4

� 12

�43

πa3 � � 38

a �

� � m2 � � 3

8a �


jbquig-UCD


(iii)

I� � a

0� π

0� π

� πr4 sin3 �

φ � dθdφdr by Fubini’s theorem

� � a

0� π

0r4 sin3 �

φ � � π

� π1 dθ dφdr since r4 sin3 �

φ) indep. of dθ

� 2π � a

0� π

0r4 sin3 �

φ � dφdr since � 2π

0

� 2π is constant

� 2π � a

0r4 � π

0sin3 �

φ � dφdr since r4 indep. of dφ

� 2π � a

0r4 � π

0sin2 �

φ � sin�φ � dφdr prepare for c.o.v u � cos

�φ �

� 2π � a

0r4 � π

φ 0

�1 � cos2 �

φ � � � � 1 � d�cos

�φ � � dr since d

�cos

�φ � � � � � 1 � sin

�φ � d

�φ �

� 2π � a

0r4 � � 1

u 1

�1 � u2 � � � 1 � dudr put u � cos

�φ �

� 2π � a

0r4 � 1

u � 1

�1 � u2 � dudr

� 2π � a

0r4 �

u � u3 � 3 � �� 1u � 1 dr

� 2π � a

0

43

r4 dr

� 83

π � a

0r4 dr

� 83

π�

a5

5 �� 8

15πa5

��

43

πa3 � �25

a2 �� m �

25

a 2

(iv) We deduce that

z � Mm � 2

� 38

a

(v) We deduce that

r ��

Im

��

25

a

5.5.2 proof of the change of variable theorem

Refer to section � 5.5. The proof will be heuristic rather than rigorous and will be valid in Rn for anydimension n; drawings will be for the case n � 2. Partition the region B.

A � � Nj 1 I j (5.1)

into small subdomains. For each subdomain I j pick a point

x j I j � 1 j N (5.2)

Writey j

� h�x j

� J j� h

�I j

� � B (5.3)

Thus we have a partition of BB � � N

j 1 J j (5.4)

c�


5.6. SPHERICAL POLAR TRANSFORM 121

Now take the partition of A, and thus also that of B, to be finer and finer.

�� B

f�y � d

�x �

�N

∑j 1

f�y j

� µI j by definition of Rieman integral

�N

∑j 1

f�h

�x j

� ��

µI j

µJ j � µJ j since h�x j

� � y j

�N

∑j 1

f � h�x j

��

µI j

µJ j � µJ j note µI j � µJ j is a ratio of hypervolumes

�N

∑j 1

f � h�x j

� det�Dh

�x j

� � µJ j since Jacobean is volume dilation factor

� �� A

f � h�x � d

�x � by definition of Riemann integral

Now taking the partition of A to be finer and finer, in the limit, approximation becomes equality. We haveproven the c.o.v. theorem � � � � � �

Bf

�y � d

�x ��

Af � h

�x � d

�x � (5.5)

5.6 spherical polar transform

Already we have seen the s.p.tf. (see � 5.2) , its differential matrix (see � 5.3), its Jacobean (see � 5.4) andexamples applying the s.p.tf. with the c.o.v.th.. (see � � 5.5.1).BUT MISSING, TO BE INSERTED:-DESCRIPTION OF DERIVATION OF THE SPTF ILLUSTRATEDBY SEVERAL SKETCHES.

5.7 toroidal polar transform

5.7.1 derivation of the toroidal polar transform

Here is a parametrization of C � R2, the circle with center 0 and radius a � 0 lying in x–z space.

� 0 � 2π � � φ ��

�x

�φ �

z�φ � � �

�asin

�φ �

acos�φ � � C � R2 (5.6)

Let b � a. Alterating these formula we obtain a parametrization of the circle C, center

�xz � �

�0b � of

radius a. � 0 � 2π � � φ ��

�x

�φ �

z�φ � � �

�b asin

�φ �

acos�φ � � C � R2 (5.7)

If the circle C is rotated about the z-axis it sweeps out a toroidal surface T � T2 � T2 �a � b � � R3 in x � y � z

space. Here is a parametrization of the suface T. If the vector

�xz � is spun by the longitude angle θ, z

remains fixed but x is broke into two components x and y.

� 0 � 2π �� 0 � 2π � ��

θφ � ��

�� x�θ � φ �

y�θ � φ �

z�θ � φ �

��

b asin�φ � cos

�θ � �

�b asin

�φ � � sin

�θ �

acos�φ �

�� T2 �a � b � � R3 (5.8)

Note that the implicit equation of C is x2 z2 � a2.The implicit equation of C is

�x � b � 2 z2 � a2.

The implicit equation of T2 �a � b � is � �

x2 y2 � b � 2 z2 � a2.


jbquig-UCD


As T is the surface of revolution of C about the z axis the latter equation can be obtained from the former bysubstituting

�x2 y2 for x, ie. the distance in R3, from the z-axis to a vector x, for the similar distance in

R2.

The reader should check that the parmetric expressions satisfy the implicit equation, in all three cases.

Finally if the radius a is replaced by a varying radius 0 r a the toroidal surface T2 �a � b � sweeps out a

toroidal solid

T3 �a � b ��

�x��

x2 y2 � b � 2 z2 a2 � (5.9)

with parametrization, the toroidal polar transform

h : R3 ��0 � a � � � 0 � 2π �� 0 � 2π � � T3 �

a � b � � R3 (5.10)�� rθφ

�� h�r� θ � φ � � x

�r� θ � φ � �

�� x�r� θ � φ �

y�r� θ � φ �

z�r� θ � φ �

�� (5.11)

��

b r sin�φ � � cos

�θ �

�b r sin

�φ � � sin

�θ �

r cos�φ �

�� (5.12)

5.7.2 differential and Jacobian of the t.p.tf.

From � 5.7.1 the toroidal polar transform is

h :�0 � a � � � 0 � 2π �� 0 � 2π � � T �� x ��

x2 y2 � b � 2 z2 a2 �� rθφ

�� h�r� θ � φ � � x

�r� θ � φ � �

�� xyz

��


�θ �

�b r sin

�φ � � sin

�θ �

r cos�φ �

��The differential matrix is Dh

�r� θ � φ �

�

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

� ��


�θ � � �

b r sin�φ � � sin

�θ � r cos

�φ � cos

�θ �

sin�φ � sin

�θ � �


�θ � r cos

�φ � sin

�θ �


�φ

��

The Jacobean of the t.p.tf.. is

� detDh�r� θ � φ � � � � det

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

� �� det


�θ � � �


�θ � r cos

�φ � cos

�θ �

sin�φ � sin

�θ � �


�θ � r cos

�φ � sin

�θ �


�φ

��

We note that the columns of Dh are mutually perpendicular as required by theorem 8, which we will use.

Dh1 � Dh2 � Dh1 � Dh3 and Dh2 � Dh3

Two of these involve Dh2 and are trivial. Let us check the remaining one, which is a little more complicated.

Dh1 � Dh3� � Dh1 � Dh3 � � 0

c�


5.7. TOROIDAL POLAR TRANSFORM 123

� � �� sin�φ � cos

�θ �

sin�φ � sin

�θ �

cos�φ �

��

�� r cos�φ � cos

�θ �

r cos�φ � sin

�θ �� r sin

�φ

�� 0

� �sin

�φ � cos

�θ � � �

r cos�φ � cos

�θ � � �

sin�φ � sin

�θ � � �

r cos�φ � sin

�θ � � �

cos�φ � � � � r sin

�φ � � 0


�φ � � cos2 �

θ � sin2 �θ � � � r sin

�φ � cos

�φ � � 0


�φ � � 1 � � r sin

�φ � cos

�φ � � 0

� 0 � 0

which is true. Thus by theorem 8

� detDh�r� θ � φ � � 2

� � � Dh1 � � 2 � � � Dh2 � � 2 � � � Dh3 � � 2� ��


�θ �

sin�φ � sin

�θ �

cos�φ �

�� 2 � ��


�θ �

�b r sin

�φ � � cos

�θ �

0

�� 2 � �� r cos

�φ � cos

�θ �

r cos�φ � sin

�θ �� r sin

�φ

�� 2� � sin2 �

φ � cos2 �θ � sin2 �

φ � sin2 �θ � cos2 �

φ � ��

b r sin�φ � � 2 sin2 �

θ � �b r sin

�φ � � 2 cos2 �

θ � 02 �� r2 cos2 �

φ � cos2 �θ � r2 cos2 �

φ � sin2 �θ � r2 sin2 �

φ � �� sin2 �

φ � �cos2 �

θ � sin2 �θ � cos2 �

φ � ��

b r sin�φ � � 2 �

sin2 �θ � cos2 �

θ � � �� r2 cos2 �

φ � �cos2 �

θ � sin2 �θ � � r2 sin2 �

φ � �� sin2 �

φ � �1 � cos2 �

φ � � � � �b r sin

�φ � � 2 �

1 � � � � r2 cos2 �φ � �

1 � r2 sin2 �φ � �

� � 1 � � � �b r sin

�φ � � 2 � � � r2 �

cos2 �φ � sin2 �

φ � � �� 1 � � � �

b r sin�φ � � 2 � � � r2 �

1 � �� r2 �

b rsin�φ � � 2

We conclude that the Jacobean of the toroidal polar transform is

� detDh�r� θ � φ � � � r

�b r sin

�φ � �

5.7.3 integration on the solid torus

It is very difficult to use Fubin’s theorem for evaluation of integrals over the solid torus, the limits are toocomplicated. We next computed two such integrals 2 using using the c.o.v.th. and the t.p.tf.. Let

T � T3 �a � b �� x ��

x2 y2 � b � 2 z2 a2 �be the solid torus. We will compute

�i � m � � � �

T1d

�x � y � z � �

�ii � I � ��

Tx2 y2 d

�x � y � z � and

�v � r �

�Im

Recall the t.p.t.f. see � 5.7.1

h : R3 � �0 � a � � � 0 � 2π �� 0 � 2π � �

�� rθφ

��

�� xyz

��


�θ �

�b r sin

�φ � � sin

�θ �

r cos�φ �

�� T � R3


jbquig-UCD


its Jacobean, see � 5.7.2

� detDh�r� θ � φ � � � r

�b r sin

�φ � �

Applying the c.o.v.th, theorem 9

(i)

m � � �� 0 � a � � � 0 � 2π � � 0 � 2π 1 � r

�b r sinφ � d

�r� θ � φ �

(ii)

I � � �� 0 � a � � � 0 � 2π � � 0 � 2π �

b r sin�φ � � 2 �


θ � �� r�b r sin

�φ � � d

�r� θ � φ �

We next evaluate both integrals.

(i)

m

� � a

0� 2π

0� 2π

0r

�b r sinφ � � dθdφdr by Fubini’s theorem

� � a

0� 2π

0r

�b r sin

�φ � � � 2π

01 dθdφdr since r

�b r sin

�φ � � is indep. of dθ

� 2π � a

0� 2π

0br r2 sin

�φ � dφdr since � 2π

01 � 2π is constant

� 2π� � a

0� 2π

0br dφdr � a

0� 2π

0r2 sin

�φ � dφdr �

� 2π� � a

0br � 2π

01dφdr � a

0r2 � 2π

0sin

�φ � dφdr � since br and r2 are indep of dφ

� 2π� � a

0br

�2π � dr � a

00dr � � since

�sin over a full period is 0

� 4π2b � a

0r dr 0 2π is constant

� 4π2b � r2

2� �� a0� 2π2ba2

� �πa2 � �

2πb �

c�


5.8. ELLIPSOIDAL POLAR TRANSFORM 125

(ii)

I� � a

0� π

0� 2π

0r

�b r sin

�φ � � 3 dθdφdr by Fubini’s theorem

� � a

0� π

0r

�b r sin

�φ � � 3 � 2π

01 dθdφdr since r

�b r sin φ 3 is indep. of dθ

� 2π � a

0� 2π

0r

�b r sin

�φ � � 3 dφdr since

� 2π0 1 � 2π is constant

� 2π � a

0� 2π

0b3r 3b2r2 sinφ 3br3 sin2 φ r4 sin3 φdφdr by the binomial theorem

� 2π � a

0

�b3r � 2π

01dφ 3b2r2 � 2π

0sinφdφ 3br3 � 2π

0sin2 φdφ r4 � 2π

0sin3 φdφ � dr

� 2π � a

0

�b3r

�2π � 3b2r2 �

0 � 3br3 � 2π

0sin2 φdφ r4 �

0 � � dr � by footnotes (a) and (b), two trig integrals are 0

� 2π � a

0

�b3r

�2π � 3br3 � 2π

0sin2 φdφ � dr

� 4π2b3 � a

0r dr 6πb � a

0r3 � 2π

0sin2 φdφdr

� 4π2b � r2

2 � �� a0 6πb � a

0r3 �

π � dr � by footnote(c),� 2π

0 sin2 � π

� 2π2b3a2 6π2b � r4

4 � �� a0� 2π2b3a2 3

2π2ba4

� �πa2 � �

2πb � � b2 32

a2 �� m �

b2 32

a2 2

llll

footnote(a) � 2π

0sinφdφ � 0 the range of integration being one full periods.

footnote(b) � 2π

0sin3 φdφ � � π

� πsin3 φdφ � 0 the integrand being an odd function and the range

of integration being of the form � � α � α � .footnote(c) � 2π

0sin2 φdφ � 1

2� 2π

01 � cos

�2φ � dφ � 1

2� 2π

01dφ � π use

� 2π0 cos

�2φ � � 0 the

range of integration being two full periods.

(iii) We deduce that

r ��

Im

��

b2 32

a2

5.8 ellipsoidal polar transform

5.8.1 derivation of the elliptical polar transform

Let a � b � c � 0, the solid ellipsoid

E � E3 � E3 �a � b � c ��

�x�� x2

a2 y2

b2 z2

c2 1 � � R3 (5.13)

might be viewed as a sort of rugby ball, only worse. A rugby ball has two but the ellipsoid has three differentradii in the directions x � y and z. The ellipsoidal polar transform is a variant on the spherical polar transform.The idea is to let r run from 0 not to a but to a number m symmetrical in a � b � c;

�a b c � � 3or

�abc � 1 � 3


jbquig-UCD


might seem a sensible choice for m, but why not 1. Then scale up the x � y � z formulae by a � b � c respectivelyand you have the parametrization.

h :�0 � 1 � � � 0 � 2π �� 0 � π � � E3 �

a � b � c � � R3 (5.14)�� rθφ

�� h�r� θ � φ � � x

�r � θ � φ � �

�� x�r� θ � φ �

y�r� θ � φ �

z�r � θ � φ �

�� ar sin

�φ � cos

�θ �

br sin�φ � sin

�θ �

cr cos�φ �

��(5.15)

As a check, show that x � y � z with r � 1 satisfy the equation of the ellipsoidal surface

x2

a2 y2

b2 z2

c2� 1

5.8.2 differential and Jacobian of the e.p.tf.

From � 5.8.1 the elliptical polar transform is

h :�0 � 1 � � � 0 � π �� 0 � 2π � � E � E3 �

a � b � c ��

x�� x2

a2 y2

b2 z2

c2 1 �� rθφ

�� h�r� θ � φ � � x

�r � θ � φ � �

�� xyz

�� ar sin

�φ � cos

�θ �

br sin�φ � sin

�θ �

cr cos�φ �

��The differential matrix is

Dh�r� θ � φ � �

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

��

�� asin�φ � cos

�θ � � ar sin

�φ � sin

�θ � ar cos

�φ � cos

�θ �

bsin�φ � sin

�θ � br sin

�φ � cos

�θ � br cos

�φ � sin

�θ �

ccos�φ � 0 � cr sin

�φ

��

The Jacobean of the e.p.tf. is � detDh�r� θ � φ � �

� � det

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

� �� det



�φ � sin

�θ � ar cos

�φ � cos

�θ �

bsin�φ � sin

�θ � br sin

�φ � cos

�θ � br cos

�φ � sin

�θ �


�φ

��

The columns of Dh are not mutually perpendicular , it seems we cannot use theorem 8. But

� detDh�r� θ � φ � �

� � det



�φ � sin

�θ � ar cos

�φ � cos

�θ �

bsin�φ � sin

�θ � br sin

�φ � cos

�θ � br cos

�φ � sin

�θ �


�φ

��

a � �b � �

c � det


�θ � � r sin

�φ � sin

�θ � r cos

�φ � cos

�θ �

sin�φ � sin

�θ � r sin

�φ � cos

�θ � r cos

�φ � sin

�θ �


�φ �

�� used row operations to extract a,b,c from rows 1,2,3 resp.

� abcr2 sin�φ � � Jacobean of s.p.tf.,see example � 5.4

We conclude that the Jacobean of the elliptical polar transform is

� detDh�r� θ � φ � � � abcr2 sin

�φ �

c�


5.9. FURTHER EXAMPLES OF INTEGRATION USING THE C.O.V.TH. 127

5.8.3 integration on the solid ellipsoid

We compute the mass( =volume if ρ � 1) of the solid ellipsoid by using the c.o.v.th. and the e.p.tf.. Let

E � E3 �a � b � c ��

�x�� x2

a2 y2

b2 z2

c2 1 �be the solid ellipsoid. Then

m� ��

E1d

�x � y � z �

� �� 0 � 1 � � � 0 � π � � 0 � 2π 1 � abcr2 sin

�φ � d

�r� θ � φ � � used c.o.v.th and e.p.tf. with Jacob abcr2 sinφ

� abc � � � �0 � 1 � � � 0 � π �� 0 � 2π r2 sin

�φ � d

�r� θ � φ �

� 43

πabc

The latter integral is known, see example � � 5.5.1 being the volume of the solid ball of radius 1.

5.9 Further examples of integration using the c.o.v.th.

5.9.1 area under the bell shaped curve

In statistics and probability the standard normal density function

f�t � � 1� 2π

exp� � x2 � 2 �

(whose graph is the oft quoted bell-shaped-curve ) plays a central role and it is important to be able tocompute the area total under this curve, see figure 5.1. We will find this area

I � 1� 2π� ∞

� ∞exp

� � x2 � 2 � dx

Although this looks like an easy first year integral the only well known way to calculate it is by a clever trickinvolving the c.o.v.th.. We remark that

x

y

x

y

z

Figure 5.1: (i)1� 2π

exp

� � x2

2 � (ii)1

2πexp

� � x2 y2

2 �February 13, 2003 c

�jbquig-UCD


2π I2

� 2π�

1� 2π� ∞

� ∞exp

� � x2 � 2 � dx ��

1� 2π� ∞

� ∞exp

� � y2 � 2 � dy ��

� � ∞

� ∞exp

� � x2 � 2 � dx � � � ∞

� ∞exp

� � y2 � 2 � dy �� ∞

� ∞exp

� � y2 � 2 �� ∞

� ∞exp

� � x2 � 2 � dx � dy moved constant I � � ∞

� ∞dx inside the dy integral

� � ∞

� ∞

� � ∞

� ∞exp

� � x2 � 2 � exp� � y2 � 2 � dx � dy moved

�exp

� � y2 � 2 � , constant w.r.t. dx

� � ∞

� ∞

� � ∞

� ∞exp

� � x2 y2

2 � dx � dy

� � �R2

exp

� � x2 y2

2 � d�x � y � by Fubini’s rheorem

Now we introduce the polar transform on R2.

h :�0 � ∞ � � � 0 � 2π � � R2�

rθ � ��

�xy � �

�x

�r� θ �

y�r� θ � � �

�r cos

�θ �

r sin�θ � �

with differential matrix

Dh�r� θ ��

�� ∂x

∂r∂x∂θ

∂y∂r

∂y∂θ

� ��

�cos

�θ � � r sin

�θ �

sin�θ � r cos

�θ � �

and with Jacobean

� detDh�r� θ � � � �� r cos2 �

θ � r sin2 �θ � �� r �


θ � � �� r �1 � � � r

Now apply the c.o.v.th to the integral above, use the polar transform on R2 with Jacobean r; note that

x2 y2 � r2 �cos2 �

θ � sin2 �θ � � � r2

2π I2

� ��R2

exp

� � x2 y2

2 � d�x � y �

� �� 0 � ∞ � � � 0 � 2π exp

� � r2

2 � � �r � d

�r� θ � used, c.o.v.th, polar tf. and x2 y2 � r2

� � ∞

0� 2π

0r exp

� � r2

2 � dθdr by Fubini’s theorem

� � ∞

0r exp

� � r2

2 � � 2π

01dθdr since r exp

� � r2

2 � is constant w.r.t. dθ

� 2π � ∞

0exp

� � r2

2 � rdr since � 2π

0

� 2π is constant

� 2π � ∞

r 0exp

� � r2

2 � d

�r2

2 � since d�r2 � 2 � � rdr

� 2π � ∞

u 0exp

� � u � du substitute u � r2 � 2

� 2π � � exp� � u � � �� ∞0� 2π � � exp

� � ∞ � exp�0 � �

� 2π � � 0 1 �� 2π

c�



We conclude that 2π I2 � 2π or that the total area under the bell shaped curve is

I � 1� 2π� ∞

� ∞exp

� � x2

2 � dx � 1

5.9.2 hypervolume of hyperball and hyperellipsoid

n2n

µBn � 2 �1 � � µBn �

1 � µBn �a � µEn

1 2 2a 2a2 π πa2 πab

343

2π43

πa2 43

piabc

42π4

π � 12

π2 12

π2a4 12

πabcda4

52π5

43

π � 815

π2 815

πa4 815

πabcde

6789

Figure 5.2: table of hypervolumes

definition of hyperball Let n � 0 be an integer,

B � Bn �a �� x Rn ��

x1 � �x2 � 2 � � � �

x j � 2 � � � �xn � 2 a2 �

is the hyperball of dimension n � with center at 0 and radius a � 0. When a � 1 � B � Bn �1 � is called the

standard hyperball of dimension n.

µ�Bn �

a � � � �� Bn�a �

1d�x ��

Bn�a �

1d�x1 � x2 � � � � � x j � � � � � xn �

is the hypervolume or measure, µ, of the hyperball. For low dimensions, n � 1 � 2 � 3

B1 �a � � � � a � a � � R , a line segment

B2 �a � �

� �xy � �� x2 y2 a2 � � R2 , a disc of radius a

B3 �a � � �� x

yz

�� x2 y2 z2 a2

� �� R3 , a solid ball of radius a

and “µ” denotes length, area and volume respectively.

definition of hyperellipsoid Let n � 0 be an integer, and a1 � a2 � � � � � a j � � � � � an be positive

E � En � En �a1 � a2 � � � � � a j � � � � � an � �

�x Rn

�� x1

a1 � 2 �x2

a2 � 2 � � � �x j

a j � 2 � � � �xn

an � 2 1 �is the hyperellipsoid of dimension n � with center at 0 and semiradii a j � 1 j n. When a j � a � 1 j nthen En �

a1 � a2 � � � � � a j � � � � � an � � Bn �a � , i.e. the ball is a special case of the ellipsoid.

µ�En ��

En1d

�x ��

En1d

�x1 � x2 � � � � � x j � � � � � xn �


jbquig-UCD


is the hypervolume or measure of the hyperellipsoid. For low dimensions, n � 1 � 2 � 3

E1 �a � � � � a � a � � R , an interval or line segment

E2 �a � b � �

� �xy � �� x2

a2 y2

b2 1 � � R2 , an elliptical disc

E3 �a � b � c � � �� x

yz

�� x2

a2 y2

b2 z2

c2 1

� �� R3 , a solid ellipsoid

again “µ” denotes length, area and volume respectively.

Our goal is, in each dimension n � 0, to calculate hypervolumes

µEn �a1 � a2 � � � � � a j � � � � � an � � µBn �

a � and µBn �1 �

As it turns out, the nub ofthe matter is to find the hypervolume of the standard hyperball µBn �1 � , indeed

µEn �a1 � a2 � � � � � a j � � � � � an � � � a1 � a2 � � � a j � � � an � Bn �

1 � � n

∏j 1

a j Bn �1 � (5.16)

µBn �a � � anBn �

1 � (5.17)

proofs (ii) is a special case of (i); we need only prove (i).Here is a linear (matrix) transform h which maps the standard ball Bn �

1 � bijectively onto the hyperellipsoidEn �

a1 � a2 � � � � � a j � � � � � an � . The transform performs its task by stretching the n-basic vectors each of length 1by factors a1 � a2 � � � � a j � � � � � an respectively.

h : Bn �1 � � En � �

a1 � a2 � � � � � a j � � � � � an �

x �

��

x1

x2

...x j

...xn

� ��

�� h�x � �

��

y1

y2

...yi

...yn

� ��

�

��

a1x1

a2x2

...a jx j

...anxn

� ��

In fact h is the linear matrix mapping

h �

��

a1 0 � � � 0 � � � 00 a2 � � � 0 � � � 0

0 0. . .

.... . . 0

0 0 � � � a j � � � 0

0 0. . .

.... . . 0

0 0 � � � 0 � � � an

a 0 � � � 0 � � � 0

��

� Diag�a1 � a2 � � � � � a j � � � � � an �

Since h is linear the differential Dh � h and the Jacobean is

� detDh � � � deth � � a1 � a2 � � � a j � � � an �n

∏j 1

a j

c�



Finally applying the c.o.v.th using the transform h

µEn �a1 � a2 � � � � � a j � � � � � an �

� � � � � � �En�a1 � a2 � � � � � a j � � � � � an �

1d�y1 � y2 � � � � � y j � � � � � yn �

� � � � � � �Bn�1 �

1 � n

∏j 1

a j d�y1 � y2 � � � � � y j � � � � � yn �

� n

∏j 1

a j �� Bn�1 �

1 � d�y1 � y2 � � � � � y j � � � � � yn �

� n

∏j 1

a j µBn �1 �

The next inductive formula 5.18 together with formulae 5.17 and 5.16 yields the measure or hypervolume ofboth Bn �

a � and En �a1 � a2 � � � � � a j � � � � � an � for any dimension n.

µBn �1 �� π

nµBn � 2 �

1 � (5.18)

proof

µBn �1 �

� � � � � � �Bn�1 �

1d�x �

� � � � � � �Bn�1 �

1d�x1 � x2 � � � � � x j � � � � � xn �

� � �B2�1 ��

Bn � 2� � 1 �

�xn � 1 � 2 �

�xn � 2 �

1d�x1 � x2 � � � � � x j � � � � � xn � 2 � d

�xn � 1 � xn �

by Fubini’s theorem: indeed x moves in Bn �a � governed by the inequality

�x1 � 2 �

x2 � 2 �x3 � 2 � � � �

x j � 2 � � � �xn � 2 � 2 �

xn � 1 � 2 �xn � 2 1

the range of

�xn � 1

xn � is maximum when

x1 � x2 � x3 � � � � � x j � � � � � xn � 2 � 0

and is�xn � 1 � 2 �

xn � 2 1 �

�xn � 1

xn � B2 �1 �

With

�xn � 1

xn � fixed in B2 �1 � the vector

��

x1

x2

...x j

...xn � 2

� ��

moves governed by the inequality

�x1 � 2 �

x2 � 2 �x3 � 2 � � � �

x j � 2 � � � �xn � 2 � 2 1 � �

xn � 1 � 2 � �xn � 2

i.e. ��

x1

x2

...x j

...xn � 2

� �� Bn � 2

� � 1 � �xn � 1 � 2 � �

xn � 2 �


jbquig-UCD


Thus

µBn �1 �

� � �B2�1 ��

Bn � 2� � 1 �

�xn � 1 � 2 �

�xn � 2 �

� 1d�x1 � x2 � � � � � x j � � � � � xn � 2 � d

�xn � 1 � xn �

� � �B2�1 �

µBn � 2

� � 1 � �xn � 1 � 2 � �

xn � 2 � d�xn � 1 � xn �

� � �B2�1 �

� � 1 � �xn � 1 � 2 � �

xn � 2 � n � 2

µBn � 2 �1 � d

�xn � 1 � xn � by formula 5.17

� µBn � 2 �1 � � �

B2�1 ��

1 � x2 � y2 � n � 2d

�x � y �

�write simply x � y for xn � 1 � xn

move constant µBn � 2 outside integral �� µBn � 2 �

1 � � �B2 � 1 �

� 1 � x2 � y2 � n � 22 d

�x � y �

Now we apply the c.o.v.theorem using the polar transform

h :�0 � a � � � 0 � 2π � �

�rθ � ��

�xy � �

�r cos

�θ �

r sin�θ � �

on R2 with Jacobean r (see � � 5.9.1).

µBn �1 �

� µBn � 2 �1 � � �

B2�1 � � 1 � x2 � y2 � n � 2

2 d�x � y �

� µBn � 2 �1 � � � �

0 � 1 � � � 0 � 2π � 1 � r2 cos2 �θ � � r2 sin

�θ � � n � 2

2 r d�r� θ � by c.o.v.th. 9 and polar tf.

� µBn � 2 �1 � � 1

0

� 2π0 r � 1 � r2 � n � 2

2 dθdr by Fubini’s theorem� µBn � 2 �

1 � � 10 r � 1 � r2 � n � 2

2� 2π

0 1dθdr since r�1 � r2 � n � 2

2 is indep of dθ� 2πµBn � 2 �

1 � � 10 r � 1 � r2 � n � 2

2 dr since� 2π

0 1 � 2π is constant� 2πµBn � 2 �

1 � � 1r 0 � 1 � r2 � n � 2

2� � 1 � 1

2 d�1 � r2 � since d

�1 � r2 � � � 2rdr

� πµBn � 2 �1 � � 0

1

�u � n � 2

2� � 1 � d

�u � write u for 1 � r2

� πµBn � 2 �1 � � 1

0

�u � n � 2

2 d�u � removed

� � 1 � and switched limits� πµBn � 2 �

1 � � 2n

�u � n

2 � �� 10� πµBn � 2 �1 � � 2

n�

� 2πn µBn � 2 �

1 �

µBn �1 �� 2π

nBn � 2 �

1 �

Using the inductive formula 5.18 and given that µB�1 � � 2 and µB2 �

1 � � π we deduce that

µB3 �1 �� 2π

3µB1 �

1 � � 43

π, (this agrees with the computation in � � 5.5.1)

µB4 �1 �� 2π

4µB2

�1 � � 1

2π,

µB5 �1 �� 2π

5µB3

�1 � � 8

15π

We have filled out the first 5 rows of column 1 of table 5.2. The other columns of this table can be filled outby applying formulae 5.17 and 5.16 to the first column. Several rows have been left blank as an exercise forthe reader.

c�



5.10 problem set

change of variable and the jacobeanThis problem set has been gleaned from an earlier document without much editing. There is repetition and nodiagrams but still much useful material. Will not be further edited this academic year, jbquig 03–02–2000.

1. Let a � 0 � B � x : x2 y2 z3 a2 � the solid ball of radius a. Let Q � � x � x 0 and y � 0� �

R3 .

(i) Write down the spherical polar transform h;�r � φ � θ � �

0 � ∞ � � �0 � π � � � � π � π � �� R3 � Q and

its inverse.

(ii) Compute the Jacobian of h. Does h preserve F.R.H.R.?

(iii) Compute the volume of B, the moment of inertia (ρ � 1) about the z-axis and the radius ofgyration.

(iv) Let H � B be the hemiball where z � 0. Compute the moment of H about the xy-plane and locatethe center of gravity of H.

2. Let E � Eabc the solid elliptical ball with semi-axes of lengths a � b � c � o. Let Q � � x � x 0 and y � 0� �

R3 .

(i) Write down the elliptical polar transform k;�r� φ � θ � �

0 � ∞ � � �0 � π � � � � π � π � �� R3 � Q and

its inverse.

(ii) Compute the Jacobian of k. Does k preserve F.R.H.R.?

(iii) Compute the volume of E, the moment of inertia (ρ � 1) about the z-axis and the radius ofgyration.

3. Let C � � x : 0 z h�1 � �

x2 y2 � a � � � R3 the solid right circular cone with base radius a andheight h.

(i) Use cylindrical polar coordinates to compute the volume of C , the moment of inertia (ρ � 1)about the z-axis, the radius of gyration and the height of the center of gravity above the xy-plane.

4. For 0 � a � b � let T � � x :� �

x2 y2 � b � 2 z2 a2 � the solid torus obtained by rotating the discD � � x :

�x � b � 2 z2 a2 � y � 0

�, which lies in the xz-plane, about the z-axis.

(i) Write down the toroidal polar transformation

h :�0 � a � � �

0 � 2π � � �0 � 2π � �

�� rφθ

��

�� xyz

�� T � R3

and its inverse transformation. Does h preserve Fleming’s Right Hand Rule?

(ii) Compute the Jacobian of h.

(iii) Compute the volume of T. Compute the moment of inertia (ρ � 1) about the z-axis and the radiusof gyration.

(iv) Sketch S � T where � � �x2 y2 � b � z � �

x2 y2 � b � .

(v) Compute the volume of S. Compute the moment of inertia (ρ � 1) about the z-axis and the radiusof gyration.


jbquig-UCD


5. (i) Sketch the region B � R3 given by the inequalities x2 y2 z2 1 � � x y x andx2 y2 z2.

(ii) Compute the volume of B.

(iii) Sketch the region C � R3 given by the inequalities X2

4 Y 2

9 Z2

25 1 � � 3X 2Y 3X andX2

4 Y 2

9 Z2

25 .

(iv) Compute the volume of C.

6. (i) Sketch the region K � R3 given by the inequalities x2 y2 1 � and

arctanyx

z 2π.

(ii) Compute the volume of K.

(iii) Sketch the region L � R3 given by the inequalities� X � Z

2� 2 � Y � Z

3� 2 1 and

arctan

�2

�Y � Z �

3�X � Z � � Z 2π .

(iv) Compute the volume of L.

7. Let V � � x : x2 y2 a2 and 0 z arctan�y � x � 2π

�, being the volume inside a cylinder under a

screw surface and let W � � x : x2 y2 z2 and 0 z arctan�y � x � 2π

�, being the volume inside

a cone under the same surface.

(i) Sketch the domains V and W � � R3

(ii) Use cylindrical polar coordinates to compute the volumes of V and W .

8. Let

Bn �a �� x Rn :

n

∑j 1

�x j � 2 a2 � � Rn

the solid ball of radius a � 0 in Euclidean n-space and let

En �a1 � a2 � � � � � a j � � � � � an

� � � x Rn :n

∑j 1

� x j

a j

� 2 1� � Rn

the solid elliptic ball with semi-axial lengths a j � 0 � 1 j n , in Euclidean n-space.

(i) Use a change of variable to prove, (where µ stands for measure or hypervolume ),µEn �

a1 � a2 � � � � � a j � � � � � an� � � ∏n

j 1 a j � µBn �1 � and µBn �

a � � anµBn �1 � , here Bn �

1 � is called thestandard unit ball in Rn.

(ii) Use Fubini’s theorem to prove that µBn �1 � � 2π

nµBn � 2 �

1 � .

(iii) Give formulae for µBn �1 � � µBn �

a � and µEn �a1 � a2 � � � � � a j � � � � � an

� , both for the cases n even and nodd.

(iv) The core of the above is the computation of µBn �1 � . Here is another method to compute the

hypervolume of the standard n-ball. Consider the n-dimensional spherical polar transform

h : Rn � I �

��

rθ1

θ2��

θn � 1

� ��

��

��

x1

x2

��

x j ��

xn

� ��

�

��

r cosθ1

r sinθ1 cosθ2

r sinθ1 sinθ2 cosθ3

r sinθ1 sinθ2 sinθ3 cosθ4��

r sinθ1 sinθ2� � � sinθn � 2 cosθn � 1

r sinθ1 sinθ2� � � sinθn � 2 sinθn � 1

� �� Bn �

1 � � Rn

c�



whereI � �

0 � 1 � � �0 � π � � �

0 � π � � � � � � �0 � π � � �

0 � 2π � � Rn

Prove that the Jacobian of h is rn � 1 sinn � 2 θ1 sinn � 3 θ2 sinn � 4 θ3� � � sin2 θn � 3 sinθn � 2. Compute

µBn �1 � as an multiple integral over the domain of h.


jbquig-UCD


c�


Chapter 6

Surface integrals, Divergence theorem ofGauss

introduction A solid D � R3 might be described by an inequality, see chapter5. A surface Σ � R3 isoften described by a single (implicit) equation; a solid is a three , whereas a surface is a two, dimensionalobject.

6.1 parametrization of a surface

See lecture notes See lecture notes

Figure 6.1: parametrization of a surface

Parametrization of a surface, see figure 6.1 takes the form

h : A � R2 � Σ � R3�rs � �� h

�r� s � � x

�r� s ��

�� x�r� s �

y�r� s �

z�r� s �

��The domain of parametrization A lies in two-space R2 and there are two parameters r and s.

6.1.1 differential matrix

The differential matrix of a surface parametrization mapping h is the 3 � 2 matrix

Dh�r� s ��

�� ∂x � ∂r ∂x � ∂s∂y � ∂r ∂y � ∂s∂z � ∂r ∂z � ∂s

�� (6.1)

137

138 CHAPTER 6. SURFACE INTEGRALS, DIVERGENCE THEOREM OF GAUSS

Dh�r � s � is a linear mapping from R2 to R3. Amongst linear mappings Dh

�r� s � is the best approximation to

the original parameterization mapping h near

�rs � A.

6.1.2 S2 � a � , the spherical surface

The ball B3 �a � is described by the inequality x2 y2 z2 a2 The boundary surface of this ball is the sphere

S2 �a �� x �� x2 y2 z2 � a2 � (6.2)

described here by an implicit equation. S2 �a � can be parameterized, see figure 6.2, by fixing r � a in the

spherical polar transform ( � 5.6).

rectanglemissingsee lecturenotes

spheremissingsee lecturenotes

Figure 6.2: parametrization of spherical surface

h : � 0 � 2π �� 0 � π � � S�a � � R3�

θφ � �� x

�θ � φ �� h

�θ � φ �

�θφ � ��

�� x�θ � φ �

y�θ � φ �

z�θ � φ �

�� asin

�φ � cos

�θ �

asin�φ � sin

�θ �

acos�φ �


Dh�θ � φ ��

�� ∂x � ∂θ ∂x � ∂φ∂y � ∂θ ∂y � ∂φ∂z � ∂θ ∂z � ∂φ

�� asin

�φ � sin

�θ � acos

�φ � cos

�θ �

asin�φ � cos

�θ � acos

�φ � sin

�θ �

0 � asin�φ �

�� (6.3)

6.1.3 T 2 � a � b � , the toroidal surface

The boundary surface T 2 �a � b � of the solid torus T 3 �

a � b � is called the toroidal surface,

T 2 �a � b ��

�x��

x2 y2 � b � 2 z2 � a2 � (6.4)

described here by an implicit equation, (for derivation of this equation see lecture notes). T 2 �a � b � can be

parameterized, see figure 6.3, by fixing r � a in the toroidal polar transform 5.7.

c�


6.1. PARAMETRIZATION OF A SURFACE 139


torus surfacemissingsee lecturenotes

Figure 6.3: parametrization of toroidal surface

h : � 0 � 2π �� 0 � 2π � � T 2 �a � b � � R3�

θφ � �� x

�θ � φ � � h

�θ � φ �

�θφ � ��

�� x�θ � φ �

y�θ � φ �

z�θ � φ �

��

b asin�φ � � cos

�θ �

�b asin

�φ � � sin

�θ �

acos�φ �


Dh�θ � φ � �

�� ∂x � ∂θ ∂x � ∂φ∂y � ∂θ ∂y � ∂φ∂z � ∂θ ∂z � ∂φ

��

b asin�φ � � sin

�θ � acos

�φ � cos

�θ �

�b asin

�φ � � cos

�θ � acos

�φ � sin

�θ �

0 � asin�φ �

�� (6.5)

6.1.4 E2 � a � b � c � , ellipsoidal surface

The outer surface E2 �a � b � c � of the solid ellipsoid E3 �

a � b � c � is called the ellipsoidal surface,

E2 �a � b � c ��

�x�� x2

a2 y2

b2 z2

c2� 1 � (6.6)

described here by an implicit equation. E2 �a � b � c � can be parameterized, see figure 6.4, as a variant of the

parameterization of the spherical surface.

h : � 0 � 2π �� 0 � π � � E2 �a � b � c � � R3�

θφ � �� x

�θ � φ � � h

�θ � φ �

�θφ � ��

�� x�θ � φ �

y�θ � φ �

z�θ � φ �

�� asin

�φ � cos

�θ �

bsin�φ � sin

�θ �

ccos�φ �


Dh�θ � φ � �

�� ∂x � ∂θ ∂x � ∂φ∂y � ∂θ ∂y � ∂φ∂z � ∂θ ∂z � ∂φ

�� asin

�φ � sin

�θ � acos

�φ � cos

�θ �

bsin�φ � cos

�θ � bcos

�φ � sin

�θ �

0 � csin�φ �

�� (6.7)


jbquig-UCD



ellipsoidal surfacemissingsee lecturenotes

Figure 6.4: parametrization of ellipsoidal surface

6.2 area dilation

Parametrization, h : A � Rn � B � Rn, of a solid is treated in section � 5.4. The volume dilation factoror Jacobean played an important role. For parametrization h : A � R2 � Σ � R3 of a surface, the areadilation factor or area Jacobean plays a similar role. We begin with the easiest case.

6.2.1 area dilation of a linear mapping

theorem 10 The linear mapping or 3 � 2 matrix

A �� a d

b ec f

�� : R2 � R3 (6.8)

dilates area by the factor

ADF�A � � ��

A1 � �

A2 �� i j ka b cd e f

�� b e

c f�� 2 �� c f

a d�� 2 �� a d

b e�� 2 (6.9)

proof In R2 the two standard unit basic vectors, i and j , delineate the standard square S whose fourvertices are

0 � i � i j and j

The area of S is 1, see figure 6.5(i). A carries i and j R2 to column vectors�

A1 and�

A2 R3 respectively.

Ai � A

�10 � �

�� abc

��

A1 � Aj � A

�01 � �

�� def

��

A2 (6.10)

Thus A carries the square S � R2 to the parallelogram A�S � � R3 delineated by the vectors

�

A1 and�

A2, i.e.with vertices

A�0 �� 0 � A

�i � � �

A1 � A�i j ��

A1 �

A2 and A�j ��

A2

see figure 6.5(ii). The area of the parallelogram A�S � is

µ�A

�S � � � ��

A1 ��

A2 �� sin�α � α being the angle between vectors A1 and A2

� �� A1 � A2 �� by definition of the � -vector

c�


6.2. AREA DILATION 141

standard square missingsee lecture notes

parallelogram missingsee lecture notes

Figure 6.5: (i) standard square S (ii) parallelogram A�S �

6.2.2 area dilation under surface parametrization

remark The area dilation factor under the surface parameterization mapping h : A ��

rs � �� x

�rs Σ �

varies from point to point in the domain A of paramertization. But at a particular point,

�rs � , the area

dilation factor of h is equal to the area dilation factor of the linear approximate mapping Dh�r� s � . We have

proven theorem 11.

theorem 11 The area dilation factor under

h : A � R2 � Σ � R3�rs � �� h

�r� s � � x

�r� s � �

�� x�r� s �

y�r� s �

z�r � s �

��a C 1 parametrization of the surface Σ � R3 is

ADF�h ��

Dh1 � �

Dh2 ��

��i j k∂x∂r

∂y∂r

∂z∂r

∂x∂s

∂y∂s

∂z∂s

��

∂y∂r

∂y∂s

∂z∂r

∂z∂s

��2 ��

∂z∂r

∂z∂s

∂x∂r

∂x∂s

��2 ��

∂x∂r

∂x∂s

∂y∂r

∂y∂s

��2

(6.11)

computational trick Direct computation of the ADF of h is onerous but, if the columns of the differentialmatrix are orthogonal, i.e.

�

Dh1 ��

Dh2, then

��

Dh1 � �

Dh2 ��

Dh1 ��

Dh2 �� sin�α � � ��

Dh1 ��

Dh2 �� (6.12)

since the angle between vectors�

Dh1 and�

DH2 is α � π � 2.exercise Find ��

Dh1 � �

Dh2 �� , where

� If h is the parameterization in � � 6.1.2 of the spherical surface S2 �a � then, ��

Dh1 � �

Dh2 �� a2 sin�φ � .

� If h is the parameterization in � � 6.1.3 of the toroidal surface T 2 �a � b � then, ��

Dh1 � �

Dh2 �� a�b

a � sin�φ �

� If h is the parameterization in � � 6.1.4 of the ellipsoidal surface E2 �a � b � c � , the columns of Dh are not

orthogonal and computation of ��

Dh1 � �

Dh2 �� is long and messy.


jbquig-UCD


6.3 integration of a scalar field over a surface

remark The actual definition of the integral,� �

Σ f�x � y � z � dσ, of a vector field f over a surface Σ � R3, was

covered in lectures (and will be typed up when time permits). There is a formula for evaluation of thisintegral, in terms of a parametrization of the surface, see theorem 12.

theorem 12 Letf : Σ � R3 � R

x �� x

yz

�� f�x � � f

�x � y � z �

be a C 0 scalar vector field over the surface Σ � R3. Let

h : A � R2 � Σ � R3�rs � �� h

�r � s � � x

�r� s � �

�� x�r� s �

y�r� s �

z�r� s �

��be a C 1 parameterization, over the domain A � R2, of the surface Σ, see figure 6.6, then

domain ofparametrization missing,see lecture notes

surface missingsee lecturenotes

targetR missing seelecture notes

Figure 6.6: scalar field over a parametrized surface

��Σ � R3

f�x � y � z � dσ

� ��A � R2

f � h�r� s � � ��

Dh1 � �

Dh2 �� d �r� s �

� ��A � R2

f � h�r� s � � ��

��i j k∂x∂r

∂y∂r

∂z∂r

∂x∂s

∂y∂s

∂z∂s

�� d�r� s �

� ��A � R2

f � h�r� s � �

��

∂y∂r

∂y∂s

∂z∂r

∂z∂s

��2 ��

∂z∂r

∂z∂s

∂x∂r

∂x∂s

��2 ��

∂x∂r

∂x∂s

∂y∂r

∂y∂s

��2

d�r� s �

proof Reread the proof of the Change of Variable theorem 9. Make appropriate alterations, in particularreplace the volume dilation factor (volume Jacobean) with the area dilation factor (area Jacobean).

example Find the surface area

c�


6.4. THE NORMAL FIELD OF A SURFACE 143

�

dsµ�S2 �

a � �� S2�a � 1 � dσ of the spherical surface S2 �

a � .

�

dsµ�T 2 �

a � b � �� T 2�a � b � 1 � dσ of the toroidal surface T 2 �

a � b � .

6.4 the normal field of a surface

remark There are three ways to think of a surface Σ � R3 and consequently three different forms ofthe equation of the surface. You may have met the first two methods in an earlier course: they are notso important in the theory of integration. The third and final method, which uses parametrization of thesurface, plays an important role in integration. At each point x Σ there is a unique (up to � 1) unit normalvector n. This leads to the concept of the normal vector field, n : σ � R3, to the surface. Each of the threeequations for the surface leads to a different expression for the vector field n. The expression for n in termsof parametrization of Σ plays an important role in the theory of integration.

� We have a scalar function f : R3 � R, the set on which f has constant value d R is known as thelevel surface, Σ, or contour on which f has constant value d. Thus

Σ � �� x �� x

yx

�� f �x � y � z � � d

� ��Here we see the implicit equation of the surface Σ. From earlier courses in advanced calculus, a normalfield to Σ in terms of the implicit equation is

�

n � grad�f � � ∂ f

∂xi ∂ f

∂yj ∂ f

∂zk

example Let f�x � y � z � � x2 y2 z2 and d � a2

S2 �a �� x �� x2 y2 z2 � a2 � �

�

n � 2xi 2yj 2zk

and here we have the implicit equation of the spherical surface, S2 �a � � R3, center 0, radius a � 0 and

the normal field derived from the implicit equation.

� Given a subset A � R2 and a scalar function g : A � R2 � R the graph of g is a surface

Σ � Gr�g � � �� x �

�� xyz

�� z � g�x � y �

� ��

n � ∂g∂x

i ∂g∂y

j � 1

Here we see the explicit equation of the surface Σ. and the corresponding formula for the normal vectorfield.example Let

z � g�x � y � � �

a2 � x2 � y2 for

�xy � B �

� �xy � �� x2 y2 a2 � � R2

S2 �a � � � x B �� z � g

�x � y � � �

a2 � x2 � y2 � ��

n � � x�a2 � x2 � y2

i � y�a2 � x2 � y2

j � k

Here we have the explicit equation of the spherical surface, S2 �a � (rather, only get the upper hemi-

sphere, this method has its limitations) and the normal vector field derived from the explicit equation.


jbquig-UCD


� Now we come to the third and most important way to describe a surface Σ � R3. We use a parametriza-tion h mapping from a domain of parametrization, A � R2

h : A � R2 � Σ � R3�rs � �� h

�r� s � x

�r� s � �

�� x�r� s �

y�r� s �

z�r� s �

��Here r and s are called parameters. The differential matrix Dh

�r� s � of the parameterization mapping h

is the 3 � 2 matrix

Dh�r� s � �

�� ∂x � ∂r ∂x � ∂s∂y � ∂r ∂y � ∂s∂z � ∂r ∂z � ∂s

��and is that linear mapping Dh

�r� s � : R2 � R3 which most closely approximates the parametrization

mapping, h : A � R2 � Σ � R3, close to

�rs � A. See lecture notes for the derivation of derived

the formula for the unit normal vector field n to Σ, in terms of the parametrization mapping

n ��

Dh1 � �

Dh2��

Dh1 � �

Dh2 �� (6.13)

example The parametrization mapping for the spherical surface S2 �a � is derived from the spherical

polar transform by fixing r � a

h : � � π � π �� 0 � π � � S2 �a � � R3�

θφ � �� h

�r � s � � x

�r� s � �

�� x�r� s �

y�r� s �

z�r� s �

�� asin

�φ � cos

�θ �

asin�φ � sin

�θ �

acos�φ �

��

Dh�θ � φ � �

�� ∂x � ∂θ ∂x � ∂φ∂y � ∂θ ∂y � ∂φ∂z � ∂θ ∂z � ∂φ

�� asin

�φ � sin

�θ � acos

�φ � cos

�θ �

asin�φ � cos

�θ � acos

�φ � sin

�θ �

0 � asin�φ �

��We leave the reader to derive the normal vector field.

6.5 vector flux integral across a surface

Let v be a vector field defined over the surface Σ � R3.

v : Σ � x �� x

yz

�� v�x � �

�� M�x � y � z �

N�x � y � z �

P�x � y � z �

��Consider also the unit normal vector field n : Σ � R3 over Σ. Taking the inner product of these two we obtaina scalar field v � n � � v � n � over the surface Σ. This scalar field can be integrated over Σ, we have

��Σ

v � ndσ

called the total flux of the vector field v across the surface Σ. Here dσ can be considered to be an infinitesimalpiece of surface area; v � n is the (scalar) resolution of the vector field v in the normal direction at each pointof the surface. The integral is sometimes written as

� �Σ

v � dσ

Here dσ � ndσ is a piece of directed area, i.e. with magnitude dσ and direction n.

c�


6.5. VECTOR FLUX INTEGRAL ACROSS A SURFACE 145

Given a parametrization h : A � R2 � Σ � R3 we can compute � �Σ

v � ndσ using theorem 13.

��Σ

v � ndσ � ��A � R2

�v � n � � h ��Dh1 � Dh2 �� dσ

Use the expression 6.13 for n in terms of h to obtain

��Σ

v � ndσ � ��A � R2

�v �

Dh1 � Dh2��Dh1 � Dh2 �� Dh1 � Dh2 �� dσ

Cancelling the area dilation factor, above and below, we obtain

� �Σ

v � ndσ � � �A � R2

�v � Dh1 � Dh2

� d�s � t �

On the right hand side we see a scalar triple product of vectors which is a determinant. We have arrived atthe evaluation theorem, using surface parametrization, of a vector flux integral, across a surface. The above

discourse is the proof. The integral � �Σ

v � ndσ can be expressed in two form notation. This notation is

suggested by writing the scalar vector triple as a determinant and expanding along the first row

� �A

�v � Dh1 � Dh2

� d�s � t �

� � �A

det

��

M�x

�s � t � � y

�s � t � � z

�s � t � � N

�x

�s � t � � y

�s � t � � z

�s � t � � P

�x

�s � t � � y

�s � t � � z

�s � t � �

∂x∂s

∂y∂s

∂z∂s

∂x∂t

∂y∂t

∂z∂t

� ��

� �� d

�s � t �

� � �A

det

�� M

�� ∂N

∂s∂P∂s

∂N∂t

∂P∂t

�� N

�� ∂P

∂s∂M∂s

∂P∂t

∂M∂t

�� P

�� ∂M

∂s∂N∂s

∂M∂t

∂N∂t

��

�� d

�s � t �

Now the two form notations suggests itself and is

��Σ

v � nσ � � �Σ

M dy � dz N dz � dx Pdx � dy

We have arrived at the evaluation theorem (by surface parametrization) for a vector flux integral across asurface.

theorem 13 Let

v : R3 � σ � x �� x

yz

�� x � �

�� M�x � y � z �

N�x � y � z �

P�x � y � z �

�� R3

be a C 0 vector field on the surface Σ � R3. Let

h : R2 � A ��

st � �� h

�s � t � � x

�s � t � �

�� x�s � t �

y�s � t �

z�s � t �

�� Σ � R3

be a C 1 parameterization of the surface. Then

� �Σ

v � ndσ

� � �Σ

v � dσ

� � �Σ

M dy � dz N dz � dx Pdx � dy


jbquig-UCD


� � �A

�v � Dh1 � Dh2

� d�s � t �

� � �A

det

��

��

M�x

�s � t � � y

�s � t � � z

�s � t � � N

�x

�s � t � � y

�s � t � � z

�s � t � � P

�x

�s � t � � y

�s � t � � z

�s � t � �

∂x∂s

∂y∂s

∂z∂s

∂x∂t

∂y∂t

∂z∂t

� �� d

�s � t �

6.6 examples of the surface vector flux integral

6.6.1 flux of field x��

x��

across spherical surface

We will compute

I � � �S2 � a �

�xi yj zk�

x2 y2 z2 � 3 � 2 � � ndσ � � �S2 � a �

xdy � dz ydz � dx zdx � dy�x2 y2 z2 � 3 � 2

This is the most basic flux integral of all, being the flux of the inverse square law field across a sphericalsurface of radius a, S � S2 �

a � � x �� x2 y2 z2 � a2 � .Use spherical polar parameterization of the surface S � S2 �

a � , see � 6.3, and the formula for evaluation ofsurface flux integrals by parameterization, see theorem13.

� �S

xdy � dz ydz � dx zdx � dy�x2 y2 z2 � � 3 � 2

� � � � � π � π �� 0 � π det

�� M N P∂x � ∂θ ∂y � ∂θ ∂z � ∂θ∂x � ∂φ ∂y � ∂φ ∂z � ∂φ

�� d�θ � φ �

� � � � � π � π �� 0 � π det

��

x�x2 y2 z � 3 � 2 y�

x2 y2 z � 3 � 2 z�x2 y2 z � 3 � 2� asinφsinθ asinφcosθ 0

acosφcosθ acosφsinθ � asinφ

� �� d

�θ � φ �

� � � � � π � π �� 0 � π det

��

x�a2 � 3 � 2 y�

a2 � 3 � 2 z�a2 � 3 � 2� asinφsinθ asinφcosθ 0


� �� d

�θ � φ �

� � � � � π � π �� 0 � π det

�� asinφcosθ � a3 asinφsinθ � a3 acosφ � a3� asinφsinθ asinφcosθ 0acosφcosθ acosφsinθ � asinφ

�� d�θ � φ �

� � � � � π � π �� 0 � π � aa3 � � �

a �� a � det

�� sinφcosθ sinφsinθ cosφ� sinφsinθ sinφcosθ 0cosφcosθ cosφsinθ � sinφ

�� d�θ � φ �

But we recognize this determinant as essentially the Jacobian determinant of the well known spherical polartransform with well known value sinφ.

I � a3

a3�� π � π � � 0 � π sinφd

�θ � φ �

� � π

0� π

� πsinφdθdφ

� 2π � π

0sinφdφ

� 2π� � cosφ � π0 �

� 2π� � � cosπ � � � � cos

�0 � � �

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6.6. EXAMPLES OF THE SURFACE VECTOR FLUX INTEGRAL 147

� 2π� � � � � 1 � � � � 1 � � �

� 2π�2 �

� 4π

6.6.2 flux integral across ellipsoidal surface

We will attempt to compute, the flux of the inverse square law field but this time across the ellipsoidal surface

E � E2 �a � b � c ��

�x R3 �� x2

a2 y2

b2 z2

c2� 1 �

Thus we seek

I � � �E

�xi yj zk�

x2 y2 z2 � 3 � 2 � � ndσ � � �E


Use elliptical polar parameterization of the surface E � E2 �a � b � c � , see 6.7, and the formula for evaluation of

surface flux integrals by parameterization, see theorem 13.

��E


� �� π � π � � 0 � π det


�� d�θ � φ �

� �� π � π � � 0 � π det

��

x�x2 y2 z � 3 � 2 y�

x2 y2 z � 3 � 2 z�x2 y2 z � 3 � 2� asinφsinθ bsinφcosθ 0

acosφcosθ bcosφsinθ � csinφ

� �� d

�θ � φ �

� �� π � π � � 0 � π det

��

x

� a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � 3 � 2 y

� a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � 3 2

z

a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ 3 2� asinφsinθ bsinφcosθ 0acosφcosθ bcosφsinθ csinφ

d θ φ

� �� π � π � � 0 � π 1

� a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � 3 � 2 det

�� asinφcosθ bsinφsinθ ccosφ� asinφsinθ bsinφcosθ 0acosφcosθ bcosφsinθ � csinφ

�� d�θ � φ

� �� π � π � � 0 � π abc

� a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � 3 � 2 det


�� d�θ � φ

But we recognize this determinant as essentially the Jacobian determinant of the well known spherical polartransform with well known value sinφ.

I � �� 0 � π �� π � π abcsinφ

� a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � 3 � 2 d�θ � φ �

It is too too dificult to finish this computation, we give up: but see � � 6.8.2, we will succeed with a differentapproach using the divergence theorem of Gauss.

6.6.3 flux integral across toroidal surface

Let 0 � a � b and Σ be the toroidal surface obtained by rotating the circle C � � �x � z � :

�x � b � 2 z2 � a2 �

which lies in the xz-plane, about the z-axis. Evaluate

� �Σ x � bx�

x2 y2 dy � dz y � by�x2 y2 dz � dx zdx � dy


jbquig-UCD


both directly by using a parametrization of Σ. This integral is again computed in � � 6.8.3 by a differentmethod, the famous divergence theorem of Gauss, see � 6.7.

In the problem we are asked to compute the flux

I � � �Σ

v � ndσ � � �Σ

M dy � dz N dz � dx Pdz � dx

across the surface Σ of the vector field

v � Mi Nj Pk � x � bx�x2 y2 i y � by�

x2 y2 j zk

Recall the parameterization of the toroidal surface, 6.5. and, using this, the formula for evaluation of a vectorflux integral across a surface, theorem 13. Thus

��Σ

v � ndσ � ��Σ

M dy � dz N dz � dy Pdx � dz � � � � 0 � 2π �� 0 � 2π det


��Now we must slug out all nine matrix entries.

M � x � 1 � b � �x2 y2 �

� �b asinφ � cosθ

�1 � b � �

b asinφ � ��

b asinφ � cosθ� �

asinφ � � �b acosφ � �

� asinφcosθ

By similar methods we obtain N � asinφsinθ and of course P � z � acosφ . Next hack out six partialderivatives

I � �� π � π � � � � π � π � det

�� asinφcosθ asinφsinθ acosφ� �b asinφ � sinθ

�b sinφ � cosθ 0


�� d�θ � dφ �

� �� π � π � � � � π � π � a �

�b asinφ �� adet

�� sinφcosθ sinφsinθ cosφ� sinθ cosθ 0cosφcosθ cosφsinθ � sinφ

�� d�θ � φ �

where we have divided constants out of all three rows. The remaining determinant has mutually perpendicularrows and so is � �� row �� row 2 �� row 3 �� Thus

I � �� π � π � � � � π � π � a �

�b asinφ �� a

�� 1 � 1 � 1d

�θ � φ �

� �� π � π � � � � π � π � a2 �

b asinφ � d�θ � φ �

� a2 � π

� π� π

� π

�b asinφ � dθdφ by Fubini’s Theorem

� a2 � π

� π� π

� πbdθdφ � since � π

� πsin � 0

� a2b � π

� π� π

� π1dθdφ

� 4π2a2b

We remind you that the above computation can also be performed as a scalar integral over the solid torus, byuse of the divergence theorem of Gauss, see � 6.8.3.

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6.7. DIVERGENCE THEOREM OF GAUSS 149

6.7 divergence theorem of Gauss

remark If Σ is a surface which bounds a solid region D and if v is a vector field defined over D thedivergence theorem of Gauss expresses the vector flux integral of v across Σ as the integral of the divergencescalar field divv over the solid region D. The statement of the theorem follows.

theorem 14 Let Σ be a surface bounding the solid region D � R3. Let v be a vector field defined over all ofthe region D. Then � �

Σv � ndσ � � � �

Ddivvd

�x � y � z �

Using two form notation and expanding the divergence

��Σ

M dy � dz N dz � dx Pdx � dy � �� D

∂M∂x

∂N∂y

∂P∂z

d�x � y � z �

6.8 examples of the divergence theorem of Gauss

6.8.1 flux integral across the spherical surface

We try to apply the divergence theorem of Gauss to compute the flux integra, I, of the inverse square lawfield v � x � �� x �� 3 across the spherical surface S � S2 �

a � . From chapter3 � � 3.4.1 we know that div�v � � 0. S

is the boundary surface of the solid ball B � B3 �a � � x R3 �� x2 y2 z2 a2 � . From � � 6.6.1 we have

prior knowledge that I � 4π. Thus

4π � I

� ��S

�xi yj zk�

x2 y2 z2 � � 3 � 2 � � ndσ

� ��S


� ��B

div�v � d

�x � y � z �

� ��B

0d�x � y � z �

� 0

We have proven that 4π � 0! What has gone wrong? This error has been included so that the reader may bewarned to take care when using the divergence theorem of Gauss, which requires that v be defined over ALLof the solid region B. However v

�0 � is undefined (∞). The mistake lies with us, not in the Gauss theorem.

The next section � � 6.8.2 shows a valid and important use of this theorem in a similar context.

6.8.2 flux integral across the ellipsoid surface

Let E � E2 �a � b � c � be the ellipsoidal surface

E � E2 �a � b � c ��

�x�� x2

a2 y2

b2 z2

c2� 1 �

Recall failure in � � 6.6.2 of an attempt to compute the flux integral of the inverse square law field across thesurface E. We try again using the divergence theorem of Gauss to compute

I � ��S

�x i y j zk�

x2 y2 z2 � 3 � 2 � � ndσ � ��S


Let 0r � � a � b � c. Let S � S2 �r � be tha small spherical surface enclosed by the ellipsoidal surface E, see

figure 6.7. Let D denote the solid region between the spherical surface S and the elliptical surface E, see


jbquig-UCD


z

x

y

Figure 6.7: solid between spheroidal and ellipsoidal surfaces

figure 6.7. The boundary surface, S�

E, of D consists of two separate pieces. Applying the divergencetheorem of Gauss

� �E� � �

S

�x i y j zk�

x2 y2 z2 � 3 � 2 � � ndσ

� � �E � S

�x i y j zk�

x2 y2 z2 � 3 � 2 � � ndσ

� � ��D

divxi yj zk�

x2 y2 z2 � 3 � 2 d�x � y � z �

� � ��D

0d�x � y � z �

� 0

In passing we mention that the inverse square law vector field is defined everywhere on D since 0 � D andthat the divergence of this field zero, see chapter3 � 3.4.1. The minus in the first line is because the normal outof D on S is the inward normal on S. Thus

��E

xdy � dz ydz � dx zdx � dy�x2 y2 z2 � 3 � 2 � � �

S


� 4π

remark See � � 6.6.2; we have computed a very difficult integral

� � � 0 � π �� π � π abcsinφ

� a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � 3 � 2 d�θ � φ � � 4π

6.8.3 flux integral across the toroidal surface

Let 0 � a � b and Σ be the toroidal surface obtained by rotating the circle, C ��

xz � ��

x � b � 2 z2 � a2 � ,

which lies in the xz-plane, about the z-axis. Denote by T the solid torus bounded by the surface of Σ. Thus

Σ � � x R3 �� x2 y2 � b � 2 z2 � a2 � � T � � x R3 ��

x2 y2 � b � 2 z2 a2 �Evaluate

� �Σ x � bx�


c�


6.8. EXAMPLES OF THE DIVERGENCE THEOREM OF GAUSS 151

using the divergence theorem of Gauss (this computation has already been carried out by direct evaluation ofthe surface integral I, see � � 6.8.3).

� �Σ

v � ndσ � � � �T

divvd�x � y � z �

or in more detail

� �Σ

M dy � dz N dz � dx Pdz � dx � � ��T

�∂M∂x

∂N∂y

∂P∂z � d

�x � y � z �

We must slug out

divv � ∂M∂x

∂N∂y

∂P∂z

� ∂∂x� x � bx � x2 y2 � � 1 � 2 � ∂

∂y� y � by � x2 y2 � � 1 � 2 � ∂

∂z

�z �

� � 1 � b � x2 y2 � � 1 � 2 bx2 � x2 y2 � � 3 � 2 � � 1 � b � x2 y2 � � 1 � 2 by2 � x2 y2 � � 3 � 2 � 1

� 3 � 2b � x2 y2 � � 1 � 2 b � x2 y2 � � x2 y2 � � 3 � 2� 3 � b � x2 y2 � � 1 � 2

We return to the main computation

I � �� T

� 3 � b � x2 y2 � � 1 � 2 � d�x � y � z �

To evaluate this scalar integral over the solid region T we of course appeal to the COV theorem 9 and use theToroidal Polar transform chapter 5 � 5.7

h :�0 � a � � � � π � π � � � � π � π � �

�θφ � ��

�� xyz

��

b r sinφ � cosθ�b r sinφ � sinθ

r cosφ

�� T

with its Jacobian det�Dh

�θ � φ � � � r

�b r sinφ � .

I � �� T

� 3 � b � x2 y2 � � 1 � 2 � d�x � y � z �

� �� 0 � a � � � � π � π � � � � π � π � � 3 � b � x2 y2 � � 1 � 2 � � detDhd

�r� θ � φ �

� �� 0 � a � � � � π � π � � � � π � π � � 3 � b � �

b r sinφ � 2 � � 1 � 2 � � r �b r sinφ � d

�r� θ � φ �

� �� 0 � a � � � � π � π � � � � π � π � � 3 � b

�b r sinφ � � 1 � � r �

b r sinφ � d�r� θ � φ �

� �� 0 � a � � � � π � π � � � � π � π �

�3r

�b r sinφ � � br � d

�r� θ � φ �

� � a

0� π

� π� π

� π

�2rb r sinφ � dθdφdr � by Fubini’s th

� 2π � a

0� π

� π

�2rb r sinφ � dφdr

� 2π � a

0� π

� π

�2rb � dφdr since � π

� πsin � 0

� 4π2 � a

02rbdr

� 4π2a2b

Do check back, that we have obtained the same result in section � � 6.8.3.


jbquig-UCD


6.9 problem set

Surface integration and the divergence theorem of GaussThis problem set has been gleaned from an earlier document without much editing. There is repetition and nodiagrams but still much useful material. Will not be further edited this academic year, jbquig 08–02–2000.

1. Let B, a solid body with surface Σ, be totally immersed in a liquid of density ρ, whose surface isthe plane z � 0. Let g denote the acceleration of gravity. The pressure at a point p Σ is ρgzn, thusthe magnitude of pressure is proportional to depth and the direction is that of n the unit normal to Σ atp. The bouyancy force per unit area at p Σ is ρgzn � k (the pressure resolved in the vertical direc-tion). By taking a surface integral of the bouyancy force over the surface Σ and using the divergencetheorem prove Archimedes Principle, ” The total buoyancy force on the body B equals the weight ofthe displaced liquid”.

2. Let 0 � a � b and let A be that region in the xz-plane where � �x � b � z �

x � b � and�x � b � 2

z2 a2. Let B � R3 be that volume obtained by revolution of A about the z axis. Let Σ be the surfacebounding B and

I � ��Σ

x� �

x2 y2 � b � dy � dz y� �

x2 y2 � b � dz � dx z�

x2 y2 dx � dy

(i) Sketch A � R2. Sketch Σ � B � R3.

(ii) Compute I, directly as a surface integral .

(iii) Use the Divergence theorem to compute I as an integral over B.

3. Let 0 � a � b and T be the solid torus obtained by rotating the discD � � �

x � z � :�x � b � 2 z2 a2 � , which lies in the xz-plane, about the z-axis. Denote by Σ the

surface of T . Evaluate

� �Σ x � bx�


both directly using a parametrization of Σ and by using the divergence theorem.

4. Let D � R3 be the region given by z � 0 � x2 y2 z2 and x2 y2 z2 9 . Let Σ be the surfacex2 y2 � z2 � 0 z 3 � � 2 being one of the two surfaces which together form the boundary of D.Let

I � � �Σ

xzdy � dz yzdz � dx � �x2 y2 � dx � dy

(i) Sketch D showing Σ.

(ii) Use the divergence theorem to prove, I � � ��D

2zd�x � y � z � . Compute I by evaluation of this

volume integral.

(iii) Compute I again by parametrization of the surface Σ.

5. Let 0 � a � b and T be the solid torus obtained by rotating the disc

D �� xz � :

�x � b � 2 z2 a2 � y � 0

�about the z-axis and let Σ be the toroidal surface which bounds T. Consider the vector field

v � � x � bx�x2 y2

�i � y � by�

x2 y2

�j zk

c�



(i) Parametrize Σ.

(ii) Compute

I � ��Σ

v�� v �� 2 � ndσ

Show that the integral I does not depend on a.

(iii) Prove that, (even though divv�� v �� 2 is infinite on the circle x2 y2 � b2 � z � 0 ), the integral

� ��T

divv�� v �� 2 d

�x � y � z �

exists and is zero. You might consider the region between T and a smaller torus where a isreplaced by ε which tends to zero.

6.

E � �� xyz

�� :x2

a2 y2

b2 z2

c2� 1

� �� R3 F ��

xy � :

x2

a2 y2

b2 1 � � R2

v � xi yj zkx2 y2 z2 I � ��

E� v � n � dσ

(i) Compute divv.

(ii) By manipulating I in two ways, first using a parametrization of the elliptic surface E and secondusing the divergence theorem prove

� 2π

0� π

0

abcsinφa2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ

dθdφ

� � �F

2�x2 y2

arctan c�

1 � x2 � a2 � y2 � b2�

x2 y2 d�x � y �

7. Let λ be a fixed positive real number. Define a vector function F on R3 � � 0 � by

F�x � y � z � � λ

�xi yj zk �

�x2 y2 z2 � 3 � 2 �

(i) Show that div F � 0.

(ii) Suppose that S is a piecewise smooth orientable surface in three dimensional Euclidean space,and that the origin is an interior point of S. Let n be the outward unit normal to S. Use thedivergence theorem to show that ��

S

F � n dσ � 4πλ

regardless of the shape of S.


jbquig-UCD


c�


Chapter 7

Integration over Curves, Stokes theorem

7.1 summary

We will study curves, one dimensional sets, lying in Euclidean three space R3. You may already have meta curve as the intersection of two surfaces. For example see figure ?? which shows a cylindrical surfaceC meeting a screw surface S in a curve Γ � C � S called the cylindrical helix. From this point of viewtwo equations (one for each surface) are needed to describe a curve. For example x2 y2 � 1 and z � θ �arctan

�y � x � describe the helix Γ. Each equation removes one degree of freedom from a point x R3 which

satisfies it. Thus a curve is a 3 � 2 � 1-dimensional set in R3.

The above approach is not suitable for our purposes; we wish to integrate scalar and especially vector fieldsalong a curve. We will therefore use an alternative description, parametrization.

�

γ : � a � b � � t ��

γ�t � � x

�t � �

�� x�t �

y�t �

z�t �

�� Γ � R3

From such a parametrization we will obtain useful technicalities such as the velocity and speed of the pointx

�t � and the unit tangent vector at the point x

�t � Γ.

The length dilation factor, LDF, of the parametrization mapping plays an important role in integration along acurve. This factor is conceptually simple. It is the ratio of distance on the curve to time in the parametrizationdomain and is of course speed. Using the LDF we obtain evaluation formulae for both scalar and vectorintegrals along a curve. Both can be expressed as ordinary integrals of a function (involving the LDF) of onevariable over the interval of parametrization.

7.2 the curve, technicalities

definition Given an injective mapping

�

γ : � a � b � � t � � ��

γ�t �� x

�t � R3

from a closed interval � a � b � � R into R3, the image set

Γ � Im� �

γ � � �

γ�t � � x

�t � �� t � a � b � � R3 �

is called a curve and�

γ is called a parametrization of Γ. The mapping�

γ is bijective, if regarded as a functionfrom � a � b � onto the curve Γ. The interval � a � b � is called the domain of parametrization and t � a � b � theparameter. The point p �

�

γ�a � Γ is called the start and q � γ

�b � the end of the curve. If p � q then we

say that the Γ is closed;�

γ is then bijective only on � a � b � . One curve Γ may admit many parametrizations: ifthere exists a C 1 parametrization

�

γ then the curve is said to be C 1 or smooth of the first order.

155

156 CHAPTER 7. INTEGRATION OVER CURVES, STOKES THEOREM

If one relaxes the requirement that�

γ : � a � b � � Γ be injective one can admit a curve which crosses itself inone or several places, see figure ??, but we will not pursue this matter further.

remark Given a C 1 parametrization

γ : � a � b � � t �� vecgamma�t � � x

�t ��

�� x�t �

y�t �

z�t �

�� Γ � R3

of the C 1 curve Γ, one may think a point x traveling along the curve Γ being at position x�t � Γ at time

t � a t b. Denote differentiation w.r.t. time t by means of a “˙”. Then

x�t �� x

�t � i y

�t � j z

�t � k �

�� x�t �

y�t �

z�t �

��is the velocity vector of the point

�

γ�t � � x

�t � at time t under the given parametrization. The velocity vector

x�t � is a vector tangential to the curve Γ at the point x

�t � . The magnitude of this velocity vector

�� x �t � �� x2

�t � y2

�t � z2

�t �

is the speed of the point x at time t. As mentioned above, the parametrization mapping�

γ carries time toposition and the factor by which it dilates length is called the LDF. The speed �� x �� and the length dilationfactor, LDF, are one and the same. This is no surprize since, distance divided by time equals speed. We willneed to know the unit tangent vector t at each point x of the curve Γ. But

t�t �� x

�t �

�� x �t � �� x

�t � i y

�t � j z

�t � k�

x2�t � y2

�t � z2

�t �

is the (unique up to � ) unit tangent vector to the curve Γ at the point x�t � . Later we will talk of the vector

field t of unit tangent vectors to the curve Γ.

7.3 Integration of a Scalar Field over a Curve

We take the general theorem of chapter 4.2, and recast it for the particular case of integration over a curvein R3. We use the DMF factor for parametrization of curves which has been worked out in chapter 4.5.. Wecan immediately write down the main theorem for evaluation of an integral of a scalar field over a curve.

Theorem Evaluation of Scalar Integral on a Curve, by Parametrization Let

f : Γ � x

�� xyz

�� f�x � y � z � R

be a C 0 scalar field over the C 1 curve Γ � R3. Let

γ : � � a � b � � t � � � γ�t �� x

�t ��

�� x�t �

y�t �

z�t �

�� Γ

be a C 1 parametrization of Γ over the interval � a � b � � R. Then

�Γ

f�x � ds � � b

af � γ

�t � ��Dγ

�t � �� dt

c�


7.4. FLOW OF A VECTOR FIELD ALONG A CURVE 157

in full detail,

�Γ

f�x � y � z � ds � � b

af

�x

�t � � y

�t � � z

�t � � � x2

�t � y2

�t � z2

�t � dt

Remark We are not primarily interested in integration of scalar fields over curves but rather in integrationof vector fields along curves which topic is covered below. However we give two examples; (i) computingthe curve length of the cylindrical helix (ii) ?? of the conical helix.

Example Curve Length of the Cylindrical Helix CurveRecall the parametrization of the once wound cylindrical helix to height h at radius r, the details are in

Chapter 4.5.

γ : � 0 � 2π � � t � � � γ�t � � x

�t � �

�� x�t �

y�t �

z�t �

�� r cos t �

r sin th

2π t

��The curve length mu

�Γ � is found by integrating the constant function f � 1 over the curve.

µ�Γ �

� �Γ

1ds

� � 2π

0f � γ

�t � ��Dγ

�t � �� ds

� � 2π

0f

�x

�t � � y

�t � � z

�t � � � x2

�t � y2

�t � z2

�t � dt

� � 2π

01 �� r sin t2 r cos t2 �

h2π � 2

dt

� � 2π

0

�r2

�sin2t cos2 t � �

h2

4π2 � dt

� � 2π

0

�r2 �

h2

4π2 � dt

��

r2 �h2

4π2 � t � 2π0

� 2π

�r2 �

h2

4π2 ��

2πr � 2 h2

7.4 Flow of a Vector Field along a Curve

Remark Let Γ � R3 be a curve and

v : Γ � x �� x

yz

�� v�x � �

�� MNP

��be a C 0 vector field defined over the curve. Denote the unit tangent field on Γ by t. Then the inner product of

vector fields � v � t � is a scalar field on Γ and so the integral �Γ� v � t � ds is well defined and is known as

the flow of the vector field v along the curve Γ. This integral is often denoted�

Γ v � ds, here “ � ” denotes inner


jbquig-UCD


product and ds denotes the vector�ds � t with magnitude ds an infinitesimal piece of length and tangential

direction t.Given a parametrization

γ : � � a � b � � t � � � γ�t �� x

�t ��

�� x�t �

y�t �

z�t �

�� Γ

of the curve Γ we apply the evaluation formula for scalar integrals to evaluate the flow.

�Γ

v � ds

� �Γ� v � t � ds

� � b

a� v

�x

�t � � � t

�x

�t � � � ��Dγ

�t � �� dt

� � b

a� v

�x

�t � � � t

�x

�t � � � ��Dγ

�t � �� dt

Now we use Dγ�t � � �� x �

t � and replace t at the point x�t � with its value as given above.

� b

a� v

�x

�t � � � x

�t �

�� x �t � �� x �

t � �� dt

� � b

a� v

�x

�t � � � x

�t � � dt

� � b

a� �� M

�x

�t � �

N�x

�t � �

P�x

�t � �

��

�� x�t � �

x�t � �

x�t � �

�� dt

� � b

aM

�x

�t � � x

�t � N

�x

�t � � y

�t � P

�x

�t � � z

�t � dt

� � b

aMx Ny Pzdt

The original notation�

Γ � v � t � ds reminds us of the geometry and how the field v is involved. The finalline of the latter computation suggests a new less geometric notation

�Γ

M dx N dy N dy Pdz

for the integral of flow. Finally we can sum it all up with the formal statement, of the evaluation byparametrization formula, for an integral of a vector field along a curve.

Theorem Evaluation of the Flow integral of a Vector Field along a CurveLet

v : Γ � x �� x

yz

�� v�x � �

�� MNP

�� R3

be a C 0 vector field over the C 1 curve Γ � R3. Let

γ : � � a � b � � t � � � γ�t �� x

�t ��

�� x�t �

y�t �

z�t �

�� Γ

be a C 1 parametrization of Γ over the interval � a � b � � R. Then

�Γ� v � t � ds � � b

a� v � x � dt

c�


7.5. STOKES THEOREM 159

Changing to a different notation on the LHS and expanding the RHS

�Γ

M dx N dy N dy Pdz � � b

aM

dxdt

Ndydt

Pdzdt

dt

7.5 Stokes Theorem

Remark We gave a preliminary mention of Stokes theorem at in section 6.?. We have now done all thepreliminaries for a fuller treatment. We will state the theorem and illustrate it with worked examples but donot give a proof. Consider the Hemisphere surface

Σ � x �� x2 y2 z2 � a2 and z � 0 �the boundary set of this surface is a curve Γ in fact the equatorial circle lying in the xy-plane.

Γ � x �� x2 y2 � a2 and z � 0 �If we choose the unit normal field on the Hemisphere Σ to be the outward unit normal field then using FRHRwe are forced to pick a direction of circulation on the closed boundary curve Γ. In Stokes theorem theunderlying geometry is a general situation of this type. Given a Surface with a continuous unit normal fieldwhose boundary boundary consists of one or more closed curves, the direction of circulation on these beingdetermined by the given normal field according to FRHR, then the circulation integral of a vector field varound the boundary curve(s) can be expressed a the flux across the surface of the curl v. If it is possible tofind such a continuous unit normal field on a surface then the surface is said to be orient-able. Nonorientablesurfaces exist, see sketch of the Mobius strip, a circular ribbon but with a half twist.

We now formally state Stokes theorem.

Theorem of Stokes

Let

v : Σ � x �� x

yz

�� v�x ��

�� MNP

�� R3

be an C 1 vector field over an orient-able surface Σ � R3 with chosen C 0 normal field n whose boundary set Γconsists of one or more closed curves, each with direction of circulation chosen to agree with n under FRHR.Then �

Γ� v � t � ds � � � Σ � curl v � dσ

Using differential form notation

�Γ

M dx N dy Pdz � ��Σ

� ∂P∂y � ∂N

∂z� dy � dz � ∂M

∂z � ∂P∂x

� dz � dx � ∂N∂x � ∂M

∂y� dx � dy

7.6 Potential, Work and Curve Integrals

The divergence theorem links integration in dimensions 2 and 3 and Stokes theorem links integration indimensions 1 and 2; there is a third theorem of this type linking integration in dimensions 0 and 1. By a setof dimension 0 we mean a subset K � R3 with finite cardinality i.e. has only a finite number of elements.The formal statement of the theorem

Theorem


jbquig-UCD


Let Γ � R3 be a C 1 curve with beginning and end points p and q respectively. Let Γ � U � R3 whereU is an open set and let

f : U � x �� x

yz

�� f�x � � f

�x � y � z � R

be a scalar field defined over U , then

f�q � � f

�p � � �

Γ� grad f

�x � � t � ds

or using differential form notation

f�q � � f

�p � � �

Γ

d fdx

dx d fdy

dy d fdz

dz

Remark The LHS above can be regarded as a sort of integral over the finite set � p � q�

. The secondversion suggests that the notation

�Γ d f for the RHS. The physical use...... Of the three similar theorems this

is the only one which can be quickly proven

proof

7.7 Overview

7.8 more–what is this??

Example Parametrization of a helix curve in 3-space.Consider the helical curve Γ which ascends to a height h � 0 while winding once ACW around the

cylinder of radius r � 0. The world is full of such helices, threads of screws, coiled springs, banisters ofcircular staircases. paths of charged particles in magnetic fields. To make such a helix take a Heinz bean canof height h and radius b. Remove the label which will be seen to be a rectangle. Draw one diagonal line onthe label. Glue the label back on the can. Hey Presto a helix. Alternatively tightening a piece of slipperythread held against a cylinder will yield a helix.

To be more technical the Helix Γ is the curve of intersection of two surfaces in R3. The first surface Γ1

is a Cylinder with radius r � 0, its equation is

x2 y2 � r2

. The second surface is the tricky Screw surface with equation

x tan� 2π

hz � � y � 0 � z � θ � h

2πarctan

�y � x �

. As we will be studying surfaces soon we omit details here. The parking ramp at Dublin airport is a Screwsurface. Thus one description of the helix is

Γ �� x : x2 y2 � r2 and x tanz � y and 0 z h�

However the preceding is just a build up. Now you know what a helix is. Our main goal is to parameterizeΓ, here is how its done.

γ : R � � 0 � 2π � � t � � � γ�t � �

�� x�t �

y�t �

z�t �

�� r cos t �

r sin th

2π t

��One can check that this is correct by plugging the vector γ

�t � into the equations of both surfaces. A better

way is to break the formula into two parts (i) x�t � i y

�t � j � r cos ti r sintj runs round in a circle of radius

r as t runs from 0 to 2π. (ii) z�t � � h

2π t simultaneously runs from height 0 to height h.

c�



c�


Part II

Answers to problem sets of part I

201

Chapter 1

answers:- integration by first principles

1.1 answer��

xy frpm first principles

First attempt this exercise using the standard partition. The presence of square roots in the Riemann Sumcreates difficulties. You will then see the reason for strange choice of partition.

1.1.1 drawing the partion

x

y

0

1/16

4/16

9/16

1

0 1/16 4/16 9/16 1

Figure 1.1: Partition for Function involving Square Roots

1.1.2 the mesh

The span of the partition element Ii j� � � i � 1

n� 2 �

� in

� 2 �� j � 1n

� 2 �� jn

� 2 � is the distance between the twocorners

d��

1 � nj � n � �

� �i � 1 � � n�j � 1 � � n � �

�

�� in � 2 � �

i � 1n � 2 2 �

jn � 2 � �

j � 1n � 2 2

� 1n2� �

2i � 1 � 2 �2 j � 1 � 2

203

204 CHAPTER 1. ANSWERS:- INTEGRATION BY FIRST PRINCIPLES

The mesh of the partition Pn is the largest of these spans 1 i � j n and occurs when i � j � n. Thus

mesh�Pn

� � 1n2� �

2n � 1 � 2 �2n � 1 � 2

� 1n2

�8n2 � 8n 2

��

8n2 � 8

n3 2

n4

which tends to zero as n tends to infinity.

1.1.3 integral as limit of Rieman sum

First compute the measure (area) of the subinterval Ii j

µ�Ii j

��

i2

n2 � �i � 1 � 2

n2 � ��

j2

n2 � �j � 1 � 2

n2 � ��2i � 1 � �

2 j � 1 �n4

We pick an (easy) point

xi j�

�i2 � n2

j2 � n2 � Ii j for each i � j; 1 i � j n �

and form the Riemann Sum

R�f � Pn

�� ∑

1 � i � j � n

f�xi j

� µ�Ii j

�

� ∑1 � i � j � n �

i2 j2

n4 ��

2i � 1 � �2 j � 1 �

n4 �� 1

n6 ∑1 � i � j � n

i�2i � 1 � j

�2 j � 1 �

� 1n6

n

∑i 1

n

∑j 1

i�2i � 1 � j

�2 j � 1 �

� 1n6 n

∑i 1

i�2i � 1 � n

∑j 1

j�2 j � 1 �

� 1n6 n

∑i 1

2i2 � i 2

� 1n6 2

n

∑i 1

i2 � n

∑i 1

i 2

� 1n6

�2

n�n 1 � �

2n 1 �6 � n

�n 1 �

2 � 2

� n2 �n 1 � 2

n6

� �2n 1 �

3 � 12 � 2

� n2 �n 1 � 2 �

4n � 1 � 2

62n6

��1 1 � n � 2 �

4 � 1 � n � 2

36

c�


1.2. INTEGRATION FROM FIRST PRINCIPLES OVER THE TETRAHEDRON 205

Thus

� �I� xyd

�x � y �

� limn � ∞

R�f � Pn

�

� limn � ∞

�1 1 � n � 2 �

4 � 1 � n � 2

36

��1 0 � 2 �

4 � 0 � 2

36

� 49

1.2 integration from first principles over the tetrahedron

1.2.1 sketch of tetrahedron

See � 1.6.1

1.2.2 the integral as limit of Rieman sum

Write

Ii jk� � i � 1

n� in�� j � 1

n� jn�� k � 1

n� kn� � 1 i � j � k n

Partition Pn may be writtenI � �

1 � i � j � k � n

Ii � j � kWe compute the measure of a subinterval;

µ�Ii jk

��

in � i � 1

n � �jn � j � 1

n � �kn � k � 1

n � ��

1n � �

1n � �

1n � � 1

n3

as expected all subintervals have the same measure.Next we compute the span of a subinterval;

span�Ii jk

� � d�� i � n

j � nk � n

��

�� i � 1 �

�j � 1 �

�k � 1 �

��

� �in � i � 1

n � 2 �jn � j � 1

n � 2 �kn � k � 1

n � 2 1 � 2� �

1n � 2 �

1n � 2 �

1n � 2 1 � 2

��

3n

as expected all subintervals have the same span which thus equals the mesh of the partition (recall mesh=maxspan ).

mesh�Pn

� ��

3n

As the theory requires, mesh�Pn

� tends to zero as n tends to infinity.


jbquig-UCD


As a final preliminary to forming a Riemann sum lets pick a point in each subinterval; put

xi jk�

�� i � nj � nk � n

�� Ii jk � 1 i � j � k n

We are ready to form and compute a Rieman Sum for the function f over the partition Pn of the rectangloidI.

R�f � Pn

� � ∑1 � i � j � k � n

f�xi jk

� µ�Ii jk

�

� ∑1 � i � j � k � n

f�xi jk

�� 1 � n3 �

��

1n3 � ∑

1 � i � j � k � n

f�xi jk

�

But f�xi jk

� � 0 unless

xi jk ∆ � in j

n k

n 1 � i j k n

in which case f�xi jk

� � 1 .

R�f � Pn

� ��

1n3 � ∑

1 � i � j � k ; i � j � k � n

1

��

1n3 � ∑

1 � k � n∑

1 � j � n � k∑

1 � i � n � j � k

1

Note the manipulation of the index range, see lecture notes. We compute the easy inner sum.

��

1n3 � ∑

1 � k � n∑

1 � j � n � k

�n � j � k �

To simplify the current inner sum put J � n � k � j; j � 1 � J � n � k � 1; j � n � k � j � 0.

��

1n3 � ∑

1 � k � n∑

0 � J � n � k � 1

J

��

1n3 � ∑

1 � k � n

�n � k � 1 � �

n � k �2

To simplify current inner sum substitute K � n � k; k � 1 � K � n � 1; k � n � K � 0.

��

1n3 � ∑

0 � K � n � 1

�K � 1 � K

2

��

12n3 � ∑

0 � K � n � 1

K2 � ∑0 � K � n � 1

K �

�1

2n3 � � �n � 1 � n

�2n � 1 �

6 � n�n � 1 �

2 ��

�1

2n3 � � �n � 1 � n

�2n � 1 � � 3n

�n � 1 �

6 ��

�1

2n3 � � �n � 1 � n

�2n � 4 �

6 ��

�n � 1 � n

�2n � 4 �

12n3

� 112

�1 � 1

n� 1

�2 � 4

n�

c�


1.2. INTEGRATION FROM FIRST PRINCIPLES OVER THE TETRAHEDRON 207

Thus

µ�∆ ��

∆1 � lim

n � ∞

112

�1 � 1

n� 1

�2 � 4

n� � 1

12

�1 � 0 � 1

�2 � 0 � � 1

6


jbquig-UCD


c�


Chapter 2

solutions: problems with Fubini’stheorem

2.1 solution: integration on the tetrahedron

sketch

z

0

a

b

c

x

y

z

0a

b

A

B

line L

x

y

Figure 2.1: (i) tetrahedron T, with triangle∆ of constant height (ii) ∆ in the xy–plane

2.1.1 limits

�� T

f�x � y � z � d

�x � y � z � � � c

0� b

�1 � z � c �

0� a

�1 � z � c � y � b �

0f


These limits can be established by two methods, from drawings or from inequalities.

limits from drawings

Figure 2.1 shows that the range of z on T is

0 z c

Moreover with z fixed in this range

�xy � wanders across a triangle ∆, which depends on z. Applying

Fubini’s theorem we have

� ��T

f�x � y � z � d

�x � y � z � � � c

0� �

∆f

�x � y � z � d

�x � y � dz

209

210 CHAPTER 2. SOLUTIONS: PROBLEMS WITH FUBINI’S THEOREM

The triangle ∆ lies in the xy–plane, and is determined by the line segment L which intercepts the x and y axes

0a

A

z

c-z

x

z

0 A

Bline L

x=0 x=A(1-y/B)

x

y

Figure 2.2: (i) face of T in the xz-plane (ii) y fixed in ∆ in the xy–plane

at A and B respectively, see figure 2.1(ii).

∆ ��

xy � �� x � y � 0 and

xA

yB

1 �We must compute A and B (if we find A we can write down B by similarity). Using Euclid and the similartriangles in figure 2.2(i), which shows the face of T which lies in the xz–plane we have

Ac � z

� ac

� A � a�1 � z � c �

SimilarlyB

c � z� b

c� B � b

�1 � z � c �

From figure 2.2(ii) we see that with z fixed the range of y is

0 y B � b�1 � z � c �

Solving the equation x � A y � B � 1 of L for x we get x � A�1 � y � B � � a

�1 � z � c � y � b � , we see that the

range of x on ∆ with y fixed is

0 x a�1 � z � c � y � b �

We have found all six limits for the tetrahedron T .

limits from inequalities

On T each of x � y � z is positive. The range of z is determined by the inequalities

0 z andxa y

b z

c 1

In this the range of z is maximum when x � y � 0, i.e. along the z–axis, being

0 zc

1 � 0 z c

we have our first two limits.

If z is kept constant in this range, the range of

�xy � is governed by (move z to RHS of inequality)

0 x � y andxa y

b 1 � z

c

Here the range of y is maximum when x � 0 is 0, being given by

c�


2.1. SOLUTION: INTEGRATION ON THE TETRAHEDRON 211

0 y andyb

1 � zc

� 0 y b � 1 � zc �

we have two more limits Finally if both y and z are kept constant in these ranges, the range of x is governedby (move y and z to R.H.S. of inequality)

0 x andxa

1 � zc � y

b

We have our final two limits

0 x a � 1 � zc � y

b �2.1.2 mass,moment, moment of inertia

mass

m � � c

0� b

�1 � z � c �

0� a

�1 � z � c � y � b �

01dxdydz

� � c

0� b

�1 � z � c �

0x � a � 1 � z � c � y � b �

x 0 dydz

� � c

0� b

�1 � z � c �

0a

�1 � z � c � y � b � dydz

� a � c

0y � yz � c � y2 � 2b �� b � 1 � z � c �

y 0 dz

� a � c

0y

�1 � z � c � � y2 � 2b �� b � 1 � z � c �

y 0 dz

� a � c

0b

�1 � z � c � 2 � b2 �

1 � z � c � 2 � 2bdz

� ab2� c

0

�1 � z � c � 2 dz

� ab2�� c

3� 1 � z

c � 3 �� c0 �� ab

2� � c

3

�0 � 1 �

� abc6

moment

M � � c

0� b

�1 � z � c �

0� a

�1 � z � c � y � b �

0zdxdydz

� � c

0z � b

�1 � z � c �

0� a

�1 � z � c � y � b �

01dxdydz

� proceed as in (ii) until

� ab2� c

0z

�1 � z � c � 2 dz

� ab2� c

0z � 2z2

c z3

c2 dz

� ab2

�z2

2 � 2z3

3c z4

4c3 � �� c0February 13, 2003 c

�jbquig-UCD


� ab2

�c2 �

�12 � 2

3 1

4 �� ab

2c2

12

� abc6

c4

moment of inertia

I � � ��T

�x2 y2 � d

�x � y � z � � � c

0� b

�1 � z � c �

0� a

�1 � z � c � y � b �

0

�x2 y2 � dxdydz

Evaluation of the latter is straightforward, but a long slog. Indeed x2 is involved from the first or coreintegration. A lazy way (lazyness is a virtue) to proceed is to note that if you know one only of the followingthree integrals, the other two can be written down by similarity

��T

x2 d�x � y � z � � ��

Ty2 d

�x � y � z � � � ��

Tz2 d

�x � y � z �

The easiest of these three (using our known limits above) is

� ��T

z2 d�x � y � z �

� � c

0� b

�1 � z � c �

0� a

�1 � z � c � y � b �

0z2 dxdydz

� � c

0z2 � b

�1 � z � c �

0� a

�1 � z � c � y � b �

01dxdydz

� � c

0z2

�ab2� 1 � z

c � 2 � dz

� ab2

� c

0

�z2 � 2z3

c z4

c2 � dz

� ab2 z3

3 � z4

2c z5

5c2�� h0

� abc3

2

�13 � 1

2 1

5 �� abc3

60

��

abc6 � �

c2

10 �By similarity

��T

x2 d�x � y � z � �

�abc6 � �

a2

10 � � ��T

y2 d�x � y � z � �

�abc6 � �

b2

10 � � �� T

z2 d�x � y � z � �

�abc6 � �

c2

10 �Putting the first two together

I � � � �T

�x2 y2 � d

�x � y � z � �

�abc6 � �

a2 b2

10 � � m

�a2 b2

10 �2.1.3 center of mass, radius of gyration

z � Mm

� c4

� r ��

Im

��

a2 b2

10

c�


2.2. SOLUTION: INTEGRATION ON THE CONE 213

2.2 solution: integration on the cone

This problem on integration over the cone has been covered (except for the last two very difficult parts) inlectures, see � 2.4. Solutions of these difficult parts follow

2.2.1 different limits, hyperbolic conic section

From Fig.2.4 looking at the base of the cone C we see that � r x r . Alternatively rearranging the defininginequality for C we obtain

x2 y2 r2 �1 � z2 � h2 �

the range of x is maximum when z � 0 � y i.e. on the x axis, when the inequality becomes x2 r2 � � r x r . From either point of view � r x r

xy

z

Figure 2.3: hyperbolic section of a (full) cone

The plane x constant cuts the cone surface in an hyperbola, see Fig.2.3. lets denote the enclosed area by H.

� ��C

f�x � y � z � d

�x � y � z � � � � r

r�

Hf

�x � y � z � d

�y � z � dx

Moving x, now regarded as constant, to the RHS of the defining inequality of C we obtain

y2 r2 �1 � z � h � 2 x2

In this z � 0 attains maximum range when y � 0; this range is given by

r2 �1 � z � h � 2 x2 � 0 z h

�1 � x

r�

Move z now also held constant to the RHS to obtain

y2 r2 �1 � z � h � 2 � x2

this yields the range of y which is

� � r2�1 � z � h � 2 � x2 y � r2

�1 � z � h � 2 � x2

We have alternative limits

� � �C

f�x � y � z � d

�x � y � z � � � r

� r� h

�1 � x � r �

0 �� r2

�1 � z � h � 2 � x2

� � r2 � 1 � z � h � 2 � x2f

�x � y � z � dydzdx


jbquig-UCD


2.2.2 integration using the new different limits

m � �� C

1d�x � y � z �

� � r

� r� h

�1 � x � r �

0� � r2 � 1 � z � h � 2 � x2

� � r2 � 1 � z � h � 2 � x21 dydzdx

� � r

� r� h

�1 � x � r �

02 � r2

�1 � z � h � 2 � x2 dzdx

� 4 � r

0� h

�1 � x � r �

0

� r2�1 � z � h � 2 � x2 dzdx put u � r � 1 � z

h �� 4h

r� r

0� r

x

�u2 � x2 dudx

� 4hr� r

0

� 12

u�

u2 � x2 � x2

2log�� ux � u2 � x2

x

�� rx dx see subcomputation A below

� 4hr� r

0

r2

�r2 � x2 � x2

2log�� rx � � r

x � 2 � 1�� dx

� 2h � r

0

�r2 � x2 dx � 2h

r� r

0x2 log

�� rx � � rx � 2 � 1

�� dx

� 2h

�πr2

4 � � 2hr

�πr3

12 � see subcomputations B and C below

� πr2h2 � �

πhr2

6 �� πr2h

�12 � 1

6 �� 1

3πr2h

which finishes this tricky calculation, supplied by Dr.K.Hutchinson of UCD Math Dept, except for hissubcomputations A,B and C which follow next.

Subcomputation A/

Put I � � �u2 � x2 du then

I � � �u2 � x2 du note x is constant here

� � � �xsecθ � 2 � x2 d

�xsecθ �

� � x�

sec2 θ � 1d�xsecθ �

� x2 � tanθd�secθ �

� x2 � tanθ�secθ tanθ � dθ

� x2 � tan2 θsecθdθ� � � � �

� x2 � tanθd�secθ � by backtracking two lines

� x2 tanθsecθ � x2 � secθd�tanθ �

c�


2.2. SOLUTION: INTEGRATION ON THE CONE 215

� x2�

sec2 θ � 1secθ � x2 � secθd�tanθ �

� x2 � � ux � 2 � 1 � ux � � x2 � secθd

�tanθ �

� u�

u2 � x2 � x2 � secθd�tanθ �

� u�

u2 � x2 � x2 � secθsec2 θdθ

� u�

u2 � x2 � x2 � secθ � 1 tan2 θ � dθ

� u�

u2 � x2 � x2 � secθdθ � x2 � secθ tan2 θdθ

� u�

u2 � x2 � x2 � secθdθ � I see line (***) above

Thus

I � u�

u2 � x2 � x2 � secθdθ � I

Thus

2I � u�

u2 � x2 � x2 � secθdθ

� u�

u2 � x2 � x2 � secθ�secθ tanθ �

secθ tanθdθ

� u�

u2 � x2 � x2 � secθ tanθ sec2 θsecθ tanθ

dθ

� u�

u2 � x2 � x2 � d�secθ tanθ �

secθ tanθdθ

� u�

u2 � x2 � x2 log�secθ tanθ �

� u�

u2 � x2 � x2 log � secθ �sec2 θ � 1 �

� u�

u2 � x2 � x2 log ux � � u

x � 2 � 1We have finished subcomputation (A) and have proven

I � u2

�u2 � x2 � x2

2log u

x � � u

x � 2 � 1 Subcomputation B/

� r

0

�r2 � x2 dx � � π � 2

0

�r2 � r2 sin2 θd

�r sinθ �

� � π � 20

r cosθ � r cosθdθ

� r2 � π � 20

cos2 θdθ

� r2

2� π � 2

0

�1 cos2θ � dθ

� r2

2

�θ 1

2sin2θ � �� π20


jbquig-UCD


� r2

2� π

2 0 �

� π4

r2

Subcomputation C/Put r

x� secθ � x � r cosθ � θ � arccos

xr

Then

� r

0x2 log

�� rx � � rx � 2 � 1

�� dx

� r3 � π � 20

cos2 θ log �� secθ tanθ �� sinθdθ

� r3 � π � 20

log �� secθ tanθ �� d � � cos3 θ3 �

� r3 log �� secθ tanθ �� cos3 θ3 � �� 0 π

2 r3 � π � 20

cos3 θ3

d� � log �� secθ tanθ ��

� �0 � 0 � r3 � π � 2

0

cos3 θ3

d� � log �� secθ tanθ ��

� r3 � π � 20

cos3 θ3

secθ tanθ sec2 θsecθ tanθ

dθ

� r3 � π � 20

cos3 θ3

secθdθ

� r3 � π � 20

cos3 θ3

secθdθ

� r3

3� π � 2

0cos2 θdθ

� π12

r3

2.3 solution: integration on the solid ball

This problem on integration over the solid ball has been covered in lectures, see � 2.3 for all the details.

2.4 solution: integration on the solid ellipsoid

2.4.1 sketch of the solid ellipsoid

2.4.2 finding the limits of integration

The ellipsoid is a distorted ball, and the limits are variations of those for the ball. We obtain these limits bymanipulation of the inequality

x2

a2 y2

b2 z2

c2 1

of the ellipsoid; study of detailed drawings of the ellipsoid is not so easy.

z attains maximum range when x � 0 � y and that range is

z2

c2 1 � � c z c

c�


2.4. SOLUTION: INTEGRATION ON THE SOLID ELLIPSOID 217

z

x

y

Figure 2.4: the (cutaway) solid ellipsoid

Fix z in this range, move it to RHS and

�xy � is governed by the inequality

x2

a2 y2

b2 1 � z2

c2

Here y attains maximum range when x � 0 and that range is

y2

b2 1 � z2

c2� � b

�1 � z2

c2 y b

�1 � z2

c2

Fix y and z in these ranges, move them to R.H.S. of the original inequality and x is governed by the inequality

x2

a2 1 � z2

c2 � y2

b2� � a

�1 � z2

c2 � y2

b2 x a

�1 � z2

c2 � y2

b2

Finally we have, given any function f�x � y � z � defined over E

� � �E

f�x � y � z � d

�x � y � z � � � c

� c� b � 1 � z2 � c2

� b � 1 � z2 � c2� a � 1 � z2 � c2 � y2 � b2

� a � 1 � z2 � c2 � y2 � b2f


2.4.3 evaluation of the mass (volume) integral

The calculation is parallel to that for the ball but harder, explanation is omitted, see � 2.3.

m � � c

� c� b � 1 � z2 � c2

� b � 1 � z2 � c2� a � 1 � z2 � c2 � y2 � b2

� a � 1 � z2 � c2 � y2 � b21dxdydz

� � c

� c� b � 1 � z2 � c2

� b � 1 � z2 � c22a � 1 � z2 � c2 � y2 � b2 dydz


jbquig-UCD


� 2a � c

� c� A

� A

�A2

b2 � y2

b2 dydz

we put A � b� �

1 � z2 � c2 �

� 2ab� c

� c� A

� A

�A2 � y2 dydz

next move appears already in � 2.3

� 2ab� c

� cπA2 dz

next use the value of A, above

� 2πab � c

� c

�1 � z2

c2 � dz

� 2πab

�z � z3

3c2 � �� c� c

� 2πabc2

�1 � 1

3 �� 4

3πabc

2.5 volume of hyper-hedron T

This is the problem of the volume of the tetrahedron but elevated to 5-dimensional space. The solution is

vol�T �� abcde

5!

Find limits for the hyperhedron in R5; use the method of inequalities, drawing is not possible in R5. Imitatethe argument for the tetrahedron in R3. Obtain

� �� T

1d�v � w � x � y � z �

� � e

0� d

�1 � z � e �

0� c�1 � z � e � y � d �

0� b

�1 � z � e � y � d � x � c �

0� a

�1 � z � e � y � d � x � c � w � b �

01dvdwdxdydz

The integration is an elaboration of that for the tetrahedron and in the end (no details given here)

�� T

1d�v � w � x � y � z � � abcde

1 � 2 � 3 � 4 � 5

c�


Chapter 3

solutions: scalar and vector fields withdifferential operators

Eight solutions were added to this document 13–02–2003.There are four more solutions to be written.Crossreferences are as yet blank.proofreading for typos has yet to be done.Hope to finish it next week.jbquig 13–02–2003

3.1This problem was fully answered in the lectures, see chapter?? � section??.It is a good idea Student should attempt these calculations without theassistance of the lecture notes.

3.2This problem was fully answered in the lectures, see chapter?? � section??.It is a good idea Student should attempt these calculations without theassistance of the lecture notes.

3.3Let

θ � x � y � z �� arctany

x� x � R3 �� x x � 0 � y � 0 �

u � x � y � z �� yi xjx2 y2

� x � R3 �� x �� x2 y2 � 0 �φ � x � y � z �� log � x2 y2 � x � R3 �� x �� x2 y2 � 0 �v � x � y � z �� xi yj

x2 y2 � x � R3 � � x �� x2 y2 � 0 �

Note that θ is the angle in polar co-ordinates in R2 or is the longitude anglein R3.

3.3.1

gradθ

� ∂θ∂x

i ∂θ∂y

j ∂θ∂z

k

� ∂θ∂x

i ∂θ∂y

j

since θ � arctan � y � x � is independent of z and thus ∂θ∂z � 0. We must

compute two partial derivatives. But

∂θ∂x

� ∂∂x

arctan � y � x �� 1

1 �� y � x � 2 �∂∂x

yx

� 11 �� y � x � 2 �

� yx2

� � yx2 y2

and

∂θ∂y

� ∂∂x

arctan � y � x �� 1

1 �� y � x � 2 �∂∂y

yx

� 11 �� y � x � 2 �

1x

� xx2 y2


gradθ

� ∂θ∂x

i ∂θ∂y

j

� � yi xjx2 y2

� u

219

220CHAPTER 3. SOLUTIONS: SCALAR AND VECTOR FIELDS WITH DIFFERENTIAL OPERATORS

3.3.2First

divu

� div � � yx2 y2 i x

x2 y2 j 0k �� ∂

∂x

� yx2 y2 i ∂

∂yx

x2 y2 j ∂0∂z

� 2xy� x2 y2 � 2 i � 2xy

x2 y2 j 0

� 0

Above, computation of two partial derivatives has been left to the readerNext

curlu

� curl � gradθ �� curl � gradθ� 0

by theorem??

Lastly

∇2θ� div � gradθ� div � gradθ �� div � u �

since divu � 0

� 0

3.3.3One must be clever to find a vector potential p i.e. with curlp � u, nostraightforward method has been lectured. Write

u � M � x � y � i N � x � y � j �Note that both M and N depend on x and y only, (i.e. are independent of z)and that P � 0. One might expect to find a vector potential of the sameform, i.e.

p � M � � x � y � i N � � x � y � jBut this is not the correct approach.An approach that works is to seek a vector potential of the form

p � P � x � y � k

i.e. with M � N � 0 and P independent of z. In that case we want

curlp � curl � Pk � � u � � yi xjx2 y2

But curl � Pk � � ∂P

∂yi � ∂P

∂xj . We seek a scalar function P � x � y � with

∂P∂x� � x

x2 y2 and∂P∂y� � y

x2 y2

From computations in � � ?? P � � φ satisfies these two requirements.Finally, the vector potential of u is

p � � φ � x � y � k � � 12

log � x2 y2 � k � log1

� x2 y2k

3.3.4

gradφ

� ∂φ∂x

i ∂φ∂y

j ∂φ∂z

k

� ∂φ∂x

i ∂φ∂y

j

since φ � log � � x2 y2 � is independent of z and thus ∂φ∂z � 0. We must

compute two partial derivatives. But

∂φ∂x

� ∂∂x

log � � x2 y2 �� ∂

∂x� 1

2log � x2 y2 ��

� 12

∂∂x

log � x2 y2 �� 1

21

x2 y2

∂∂x� x2 y2 �

� 12

1x2 y2 � 2x 0 �

� xx2 y2

and similarly

∂φ∂y

� yx2 y2


gradφ

� ∂φ∂x

i ∂φ∂y

j

� xi yjx2 y2

� v

3.3.5First

divv

� div � xx2 y2 i y

x2 y2 j 0k �� ∂

∂xx

x2 y2 i ∂∂y

yx2 y2 j ∂0

∂z

� � x2 y2

� x2 y2 � 2 i x2 � y2

x2 y2 j 0

� 0

Above, computation of two partial derivatives has been left to the readerNext

curlv

� curl � gradφ �� curl � gradφ� 0

by theorem??

Lastly

∇2θ� div � gradφ� div � gradφ �� div � v �� 0

c�


3.4. 221

3.3.6Find a vector potential q of the vector field v; ie. find a vector field q withcurlq � v. We argue as in � � ?? where we found the vector potential of u.Skipping details we need a scalar function Q � x � y � with

∂Q∂x� � y

x2 y2 and∂Q∂y� x

x2 y2

From � � ??, Q � θarctan � y � x � works. Finally

q � Q � x � y � k � arctan � yx� k

3.3.7Explain the physical significance of all this. See the lecture noteschapter?? � ??

3.4Solution to be typed 11/02/2003 jbq

3.5Let

rn � x � y � z � � � x2 y2 z2 � n � 2 for each integern

(i) Compute grad rn .

(ii) Compute ∇2rn .

(iii) Prove that, ∇2 rn � 0 , for two values of n only: find thesevalues.

3.5.1First we compute

∂rn

∂x

� ∂∂x

1� x2 y2 z2 � n � 2

� ∂∂x� x2 y2 z2 �� n � 2

� � n

2� x2 y2 z2 � � � n � 2 � � 1 � 2x 0 0 �

� � nx � x2 y2 z2 � � � n � 2 � � 2

� � nx

� x2 y2 z2 � � n � 2 � � 2

Similarly, with no further work

∂rn

∂y� � ny

� x2 y2 z2 � � n � 2 � � 2

∂rn

∂z� � nz

� x2 y2 z2 � � n � 2 � � 2

Finally

grad � rn �� ∂rn

∂xi ∂rn

∂yj ∂rn

∂zk

� � n � xi yj zk �� x2 y2 z2 � � n � 2 � � 2

3.5.2First we compute

∂2rn

∂x2

� ∂∂x

� ∂rn

∂x�

� ∂∂x

� � nx

� x2 y2 z2 � � n � 2 � � 2�

� ∂∂x

� � nx � x2 y2 z2 � � � n � 2 � � 2 �� n � x2 y2 z2 � � � n � 2 � � 2

� � n � x � � n 2

2� � x2 y2 z2 �� n � 2 � � 2 � 1 � 2x 0 0 �

� � n � x2 y2 z2 � � � n � 2 � � 2

n � n 2 � x2 � x2 y2 z2 � � � n � 4 � � 2

� � n� x2 y2 z2 � �� n 2 � x2

� x2 y2 z2 � � n � 4 � � 2

Similarly, with no further work

∂2rn

∂y2� � n

� x2 y2 z2 � �� n 2 � y2

� x2 y2 z2 � � n � 4 � � 2

∂2rn

∂z2 � � n� x2 y2 z2 � �� n 2 � z2

� x2 y2 z2 � � n � 4 � � 2

Finally

∇2 rn

� ∂2rn

∂x2 ∂2rn

∂y2 ∂2rn

∂z2

� n� 3 � x2 y2 z2 � � n 2 � � x2 y2z2 �

� x2 y2 z2 � � n � 4 � � 2

� n � n � 1 � � x2 y2z2 �� x2 y2 z2 � � n � 4 � � 2

� n � n � 1 � 1� x2 y2 z2 � � n � 2 � � 2

3.5.3

∇2rn � 0� n � n � 1 � � 0� n � 0 or n � 1

3.6Let f and g be scalar fields and v and w be vector fields. Prove

3.6.1

grad � f g �� ∂ f g

∂xi ∂ f g

∂yj ∂ f g

∂zk

� � ∂ f∂x

g f∂g∂x

� i � ∂ f∂y

g f∂g∂y

� j � ∂ f∂z

g f∂g∂z

� k

� � ∂ f

∂xi ∂ f

∂yj ∂ f

∂zk � g f � ∂g

∂xi ∂g

∂yj ∂g

∂zk �

� grad � f � � g f � grad � g �


jbquig-UCD


3.6.2

div � f v �� div � f � Mi Nj Pk � �� div � � f M � i �� f N � j �� f P � k � �� ∂ f M

∂x ∂ f N

∂y ∂ f P

∂z

� � ∂ f∂x

M f∂M∂x

� � ∂ f∂y

N f∂N∂y

� � ∂ f∂z

P f∂P∂z

�� f � ∂M

∂x ∂N

∂y ∂P

∂z� � ∂ f

∂xM ∂ f

∂yN ∂ f

∂zP �

� f divv �� ∂ f∂x

i ∂ f∂y

j ∂ f∂z

k � Mi Nj Pk �� f divv �� grad f � v �

3.6.3Let v � Mi Nj Pk and w � M � i N � j P � k. Then

div � v � w �� div

��

i j kM N PM � N � P �

�� div � � NP � � PN � � i � PM � � MP � � j � MN � � NM � � k �� ∂

∂x� NP � � PN � � i ∂

∂y� PM � � MP � � j ∂

∂z� MN � � NM � � k

� ∂N

∂xP � N

∂P �∂x� ∂P

∂xN � � P

∂x

∂N � ∂P

∂yM � P

∂M �∂y� ∂M

∂yP � � M

∂y

∂P � ∂M∂z

N � M∂N �∂z� ∂N

∂zM � � N

∂z∂M �

� � ∂P∂y� ∂N

∂z� M �

� ∂M

∂z� ∂P

∂x� N �

� ∂N

∂x� ∂M

∂y� P �

�M � ∂P �

∂y� ∂N �

∂z�

�N � ∂M �

∂z� ∂P �

∂x�

�P � ∂N �

∂x� ∂M �

∂y�

� ��

∂P∂y� ∂N

∂z∂M∂z� ∂P

∂x∂N∂x� ∂M

∂y

��

�� M �N �P �

� ��

�� M

NP

� ��

∂P �∂y� ∂N �

∂z∂M �∂z� ∂P �

∂x∂N �∂x� ∂M �

∂y

� ��

� � curlv � w � � � v � curlw �

3.6.4

curl � f v �� curl � � f M � i �� f N � j �� f P � k �� ∂ � f P �

∂y� ∂ � f N �

∂z� i

� ∂ � f M �

∂z� ∂ � f P �

∂x� j

� ∂ � f N �

∂x� ∂ � f M �

∂y� k

� � ∂ f∂y

P f∂P∂y� ∂ f

∂zN � f

∂N∂z

� i

� ∂ f

∂zM f

∂M∂z� ∂ f

∂xP � f

∂P∂x

� j

� ∂ f

∂xN f

∂N∂x� ∂ f

∂yM � f

∂M∂y

� k

� � ∂ f∂y

P � ∂ f∂z

N � i � ∂ f∂z

M � ∂ f∂x

P � j � ∂ f∂x

N � ∂ f∂y

M � k

f � � ∂P

∂y� ∂N

∂z� i � ∂M

∂z� ∂P

∂x� j � ∂N

∂x� ∂M

∂y� k �

�

��

∂ f∂x∂ f∂y∂ f∂z

� ��

�� MNP

� f curl

�� MNP

�� grad f � v f curlv

3.7Let

f � x2 y2 z2

2 � x2 y2 � and F � � yi xj

be a scalar and a vector field respectively. Prove that

curl � f F � � � xzi � yzj �� x2 y2 � kx2 y2

curl � f F �� 1

2curl � � x2 y2 z2 � � yi xj

x2 y2 �use the result of previous question

� 12

� grad � x2 y2 z2 � � � yi xjx2 y2 � x2 y2 z2 � curl

� yi xjx2 y2 �

� 12

� grad � x2 y2 z2 � � � yi xjx2 y2

� x2 y2 z2 � curlu �see q3, gradθ � u

� 12

� grad � x2 y2 z2 � � � yi xjx2 y2 � x2 y2 z2 � curl � gradθ �

see theorem ??, curl � grad � 0

� 12

� grad � x2 y2 z2 � � � yi xjx2 y2 � x2 y2 z2 � � 0 �

� 12

grad � x2 y2 z2 � � � yi xjx2 y2

� 12� 2xi 2yj 2zk � � � yi xj

x2 y2

� 1x2 y2 � xi yj zk � � � � yi xj �

c�


3.8. 223

� 1x2 y2

��

i j kx y z� y x 0

�� 1

x2 y2 � � zxi � zyj �� x2 y2 � k �� zxi zyj �� x2 y2 � k

x2 y2






jbquig-UCD


c�


Chapter 4

solutions:- convolution, laplacetransform, differential equations

4.1 answer

225

226CHAPTER 4. SOLUTIONS:- CONVOLUTION, LAPLACE TRANSFORM, DIFFERENTIAL EQUATIONS

4.2 answer

c�


Chapter 5

solutions:-change of variable th and thejacobean

This problem set has been gleaned from an earlier document without much editing. There is repetition and nodiagrams but still much useful material. Will not be further edited this academic year, jbquig 03–02–2000.

5.1 answer

5.1.1

h :�0 � ∞ � � �

0 � π � � � � π � π � �

�� rφθ

��

�� xyz

�� r sinφcosθ

r sinφsinθr cosφ

�� R3 � Q

h � 1 : R3 � Q �

�� xyz

��

�� r�x � y � z �

θ�x � y � z �

φ�x � y � z �

�� 0 � ∞ � � �

0 � π � � � � π � π � � R3

wherer

�x � y � z � � �

x2 y2 z2


�y � x �� arccos

�x � �

x2 y2 � � arcsin�y � �

x2 y2 �

φ�x � y � z � � arctan

� �x2 y2 � z �� arccos

�z � � x

2 y2 z2 � � arcsin� �

x2 y2 � � x2 y2 z2 �

Note that θ and φ are not defined every where on R3. They are both unambiguously defined on R3 � Q, andthat’s only reason the half plane Q is mentioned (a minor technical point).

5.1.2

First one must compute the 3 � 3 Dh�r� θ � φ � matrix known as the differential of h.

Dh�r� θ � φ � �

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

� ��

�� sinφcosθ � r sinφsinθ r cosφcosθsinφsinθ r sinφcosθ r cosφcosθ

cosφ 0 � r sinφ

��Since the columns of Dh are mutually � (check it) by a result proven during lectures (see notes)

�DetDh � 2 ��Dh1 �� 2 � ��Dh2 �� 2 � ��Dh3 �� 2 � 12 � �

r sinφ � 2 � r2 Thus we have our Jacobian � detDh � � r2 sinφ. However adirect computation yields detDh � � r2 sinφ. The minus here tells us that h transforms FRHR into FLHR.Draw a picture and see this!

227

228 CHAPTER 5. SOLUTIONS:-CHANGE OF VARIABLE TH AND THE JACOBEAN

5.1.3

Write m and I resp for the volume and moment of inertia of B resp. Applying the Spherical Polar Transformh which carries

�0 � a � � � � π � π � � �

0 � π � bijectively onto B � Q (B � Q is of measure zero so can be ignored)and using the Jacobian and the C.O.V. theorem.

m � ��B

1d�x � y � z � � ��

0 � a � � � � π � π � � � 0 � π � 1 � r2 sinφd�r� θ � φ �

I � � ��B

�x2 y2 � d

�x � y � z � � ��

0 � a � � � � π � π � � � 0 � π � r2 sin2 φ � r2 sinφd�r� θ � φ �

Lets work out m, apply Fubini’s theorem

m � � a

0� π

0� π

� πr2 sinφdθdφdr

� 2π � a

0� π

0r2 sinφdφdr

� 2 � 2π � a

0r2 dr

� 2 � 2π � a3 � 3� 4πa3 � 3

Next lets work out I , apply Fubini’s theorem

I � � π

0� a

0� π

� πr2 sin2 � r2 sinφφdθdr dφ

� � π

0� a

0� π

� πr4 sin3 φdθdr dφ

� 2π � π

0� a

0r4 sin3 φdr dφ

� 2π ��a5 � 5 � � π

0sin3 dφ

� 2π ��a5 � 5 �� 1 � � π

φ 0

�1 � cos2 θ � d cosθ

� 2π ��a5 � 5 �� 1 � � � 1

1

�1 � u2 � du

� 2π ��a5 � 5 �� 1

� 1

�1 � u2 � du

� 2π ��a5 � 5 ��

u � u3 � 3 � � 1� 1� 2π �

�a5 � 5 ��

4 � 3 �� 4π �

�a3 � 3 ��

2a2 � 5 �

From this we deduce that the radius of gyration is�

I � m � �2 � 5a

5.1.4

Denote the moment of the whole ball B about the xy-plane in not interesting being zero by symmetry. Insteadwe work with the upper hemiball H. On H the latidude angle runs only from 0 to π � 2, see the altered φ limitin the following integrals.

M � � π � 20

� a

0� π

� πr cosφ � r2 sinφdθdr dφ

� 2π � π � 20

� a

0r3 sinφcosφdr dφ

c�


5.2. ANSWER 229

� �a4 � 4 �� 2π � π � 2

0sinφcosφdφ

� �1 � 2 � �

a4 � 4 �� 2π � π � 20

sin2φdφ

� � � 1 � 2 � �1 � 2 � �

a4 � 4 �� 2π cos2φ � π � 20� � � 1 � 2 � �

1 � 2 � �a4 � 4 �� 2π �

� � 2 ��

2πa3 � 3 � �3a � 8 � �

From this we deduce that the height z of the c.o.g. of the Hemiball above its base is z � M � m � 3a � 8.

5.2 answer

This question is a variant on the preceding. Draw the solid ellipsoid E its like a rugby ball only worse, arugby ball has two different radii, this thing has three. The key idea is to find a transform k which carries arectangloid onto E, so we can use the COV theorem Here it is

k :�0 � ∞ � � �

0 � π � � � � π � π � �

�� rφθ

��

�� xyz

�� rasinφcosθ

rbsinφsinθrccosφ

�� R3 � Q

Note that k carries�0 � 1 � � � � π � π � � �

0 � π � bijectively onto E � Q (the portion E � Q is of measure zero, canbe ignored)

5.2.1

As in the solution to Q1 we obtain Dk

Dk�r� θ � φ � �

�� asinφcosθ � rasinφsinθ racosφcosθbsinφsinθ rbsinφcosθ rccosφcosθ

ccosφ 0 � rcsinφ

��To obtain the Jacobian we divide a � b and c out of rows 1 � 2 and 3 resp. and r out of columns 2 and 3 then

� det�Dk � � � abc � r2 det �

�� sinφcosθ � sinφsinθ cosφcosθsinφsinθ sinφcosθ cosφcosθ

cosφ 0 � sinφ

�� abc � r2 sinφ

The computation of the determinant of the latter matrix being a variant on the computation of the Jacobian ofthe Spherical Polar Transform see Ans Q1 above. A direct computation of detDk shows that there is a “ � ”in it so again FRHR is changed to FLHR by k.

5.2.2

Write m and I for the volume and moment of inertia of E. Then using the COV th with Jacobian as above

m � � � � �0 � 1 � � � � π � π � � � 0 � π � 1 � � det Dk � d �

r� θ � φ � � I � �� 0 � 1 � � � � π � π � � � 0 � π � �

x2 y2 � � � det Dk � d �r� θ � φ �

Thus using Fubini’s theorem

m � � 1

0� π

� π� π

01 � abcr2 sinφdr dθdφ

� abc � 1

0� π

� π� π

01 � r2 sinφdr dθdφ

� abc ��4π � 13 � 3 � done as in Q1

� �4πabc � 3 �


jbquig-UCD


Again using Fubini’s theorem

I � � π

0� π

� π� 1

0

� �ar sinφcosθ � 2 �

br sinφsinθ � 2 �� abcr2 sinφdr dθdφ

� � π

0� π

� π� 1

0abcr4 sin3 φ

�a2 cos2 θ b2 sin2 θ � dr dθdφ

� abc � π

0� π

� π

�r5

5 � � 10 sin3 φ�a2 cos2 θ b2 sin2 θ � dθdφ

� �abc � 5 � � π

0� π

� πsin3 φ

�a2 cos2 θ b2 sin2 θ � dθdφ

Now� π� π cos2 θdθ � π and

� π� π sin2 θdθ � π so

I � �abc � 5 � �

a2 b2 � π � π

0sin3 φdφ

� �abcπ � 5 � �

a2 b2 �� 4 � 3 � see soln Q1 for latter integral

� �4πabc � 3 � � �

a2 b2 � � 5 �

� m � �a2 b2 � � 5

From this we deduce that the radius of gyration of E about the z � axis is

�Im

��

a2 b2

5 � 1 � 25.3 answer

First consider the line L in the xz-plane with intercepts x � a on the x-axis and z � h on the z-axis. This linehas equation x � a z � h � 1 . A cone surface Σ � R3 can be created by rotating the line L about the z-axis.This equation of Σ is obtained by the standard method; replace x (the distance of the point xi yj from thez-axis in R2) by

�x2 y2 (the distance of the point xi yj zk from the z-axis in R3). Thus the equation of

the cone surface Σ is �x2 y2

a z

h� 1 � z � h 1 � �

x2 y2

a more accurately this is the implicit equation of Σ. Now we can give a concise mathematical definition of thesolid cone C in the question, C is the set of points in R3 lying under the surface Σ but above the xy-planez � 0;thus

C ��

x : 0 z h 1 � �x2 y2

a �The student should sketch L � Σ � C � R3.

Next we want to introduce the Cylindrical Polar Transform which is simply a three dimensional versionof the Polar Transform in R2. Recall the latter

h :�0 � ∞ � � � � π � π � �

�rθ � ��

�xy � �

�r cosθr sinθ � R2 � S

Here S � � xi yj : y � 0 � x 0�

being the negative half of the x-axis, where θ is not defined.The three dimensional version the Cylindrical Polar transform is

h :�0 � ∞ � � � � π � π � � R �

�� rθZ

��

�� x�r� θ � Z �

y�r� θ � Z �

z�r� θ � Z �

�� r cosθ

r sinθZ

�� R3 � Q

c�


5.3. ANSWER 231

see ??Q2 for definition of Q.Let T be the triangle in rZ-space

T ��

rZ � : 0 r a

�1 � z � h � � 0 Z h �

Sketch T in rZ-space, or you won’t understand this. Eventually we must integrate over T using Fubini’stheorem, and right here we have the limits of integration which will be needed. Let W be the wedge inr � θ � Z-space

W � T � � � π � π � � �� rθZ

�� : � π � θ � π � 0 Z h�1 � r � a � � 0 Z h

� ��then h carries W bijectively onto C � Q ( C � Q is of measure zero and can be ignored).

The next task is to compute the differential Dh�r� θ � Z � and the Jacobian of � det

�Dh � �

Dh�r � θ � Z � �

�� ∂x∂r

∂x∂θ

∂x∂Z

∂y∂r

∂y∂θ

∂y∂Z

∂z∂r

∂z∂θ

∂z∂Z

�� cosθ � r sinθ 0

sinθ r cosθ 00 0 1

��The computation of the determinant is easy as it reduces to the determinant of the top-left 2 � 2 matrix.

� det�Dh � � � � det

�cosθ � r sinθsinθ r cosθ � � � r

Thus we have all the machinery to use the C.O.V. theorem for integration over the solid cone C.We are assuming that the solid cone C is of uniform density 1, denote by m the mass=volume of C, by M

the moment about the xy-plane and by I the moment of inertia of C about the z-axis, then

m � �� C

1d�x � y � z � M � � ��

Czd

�x � y � z � I � � ��

Cx2 y2 d

�x � y � z �

Applying the COV theorem we finally compute all three integrals

m � � � �C

1d�x � y � z �

� � � �W T � � � π � π � 1 � r d

�r� θ � φ � by COV theorem, Jacobian � r

� � �T� π

� πr dθd

�r� Z � by Fubini’s theorem

� 2π � �T

r d�r� Z �

� 2π � h

0� a

�1 � Z � h �

0r dr dZ by Fubini’s th. see limits of T above

� 2π � h

0� r2

2� � a � 1 � Z � h �

0 dZ

� 2π � h

0

�a

�1 � Z � h � � 2 � 2dZ

� πa2 � h

0

�1 � Z � h � 2 dZ

� πa2 � � h � � �1 � Z � h � 3 � 3 � � h0

� πa2h � 3


jbquig-UCD


M � �� C

zd�x � y � z �

� �� W T � � � π � π � Z � r d

�r� θ � φ � by COV theorem Jacobian � r

� ��T� π

� πZr dθd

�r� Z � by Fubini’s theorem.

� 2π � �T

Zr d�r� Z �

� 2π � h

0� a

�1 � Z � h �

0Zr dr dZ by Fubini’s th., for limits of T see above

� 2π � h

0� r2

2� � a � 1 � Z � h �

0 Z dZ

� πa2 � h

0Z

� �1 � Z � h � � 2 dZ

� πa2 � h

0

�Z � 2Z2 � h Z3 � h2 � dZ

� πa2 � �h2 � 2 � � 2

�h3 � 3 � � h �

h4 � 4 � � h2 � �� πa2h2 � 1 � 2 � 2 � 3 1 � 4 �� πa2h2 � 12� �

πa2h � 3 �� h � 4 �

From this we see that the height of the center of gravity of the solid cone C above its base is

z � M � m � h � 4

I� � ��

C

�x2 y2 � d

�x � y � z �

� � ��W T � � � π � π � r2 � r d

�r� θ � φ � by COV th, Jacobian � r

� � �T� π

� πr3 dθd

�r� Z � by Fubini’s th.

� 2π � �T

r3 d�r� Z �

� 2π � h

0� a

�1 � Z � h �

0r3 dr dZ as twice before

� 2π � h

0� r4

4� � a � 1 � Z � h �

0 dZ

� 2π � h

0

�a

�1 � Z � h � � 4 � 4dZ

� �πa4 � 2 �� h

0

�1 � Z � h � 4 dZ

� �πa4 � 2 �� h � �

1 � Z � h � 5 � 5 � h0� πa4h � 10� �

πa2h � 3 �� 3a2 � 10 �

From this we deduce that the radius of gyration of the solid cone about it’s axis of symmetry is�

I � m � a�

3 � 10

c�


5.4. ANSWER 233

5.4 answer

It is worth while spending a little time to understand the solid torus T. Consider the xz-plane which contains

the point

�b0 � . Consider next the circle C with this point as center and radius a � 0 � a � b. The equation

of C is�x � b � 2 z2 � a2, the disc D is the area bounded by the circle C and has equation (inequality)�

x � b � 2 z2 a2. Consider the toroidal surface Σ obtained by rotating C about the z-axis, we obtain theequation of Σ from that of C by replacing x by

�x2 y2. Similarly we obtain the inequality which describes

the solid torus T from that for the disc D. Thus

Σ �� x :� �

x2 y2 � b � 2 z2 � a2 � � T �� x :� �

x2 y2 � b � 2 z2 a2 �To understand the torus the student should sketch C � D � Σ and T � R3.

Next we tackle the problem of parameterization of the solid torus. This is too difficult to be done in onestep, however when its finished we expect it to be somewhat like the familiar Spherical Polar parameterizationof the solid ball. First

� � π � π � � φ ��

�xz � �

�acosφasinφ �

parameterizes a circle of radius a center 0 in xz-space. Next we obtain a parameterization of C by making a

small adjustment to force the center to lie at

�xz � �

�b0 � .

� � π � π � � φ ��

�xz � �

�b acosφ

asinφ �Next by introducing the angle of longitude and using it to split the x-coordinate up we obtain a parameteri-zation of the toroidal surface Σ.

� � π � π � � � � π � π � ��

θφ � ��

�� xyz

��

b acosφ � cosθ�b acosφ � sinθ

asinφ

�� Σ

Finally by introduce a varying radius we parameterize the solid torus T.

�0 � a � � � � π � π � � � � π � π � �

�� rθφ

��

�� xyz

��

b r cosφ � cosθ�b r cosφ � sinθ

r sinφ

�� T

Discussion of h � 1 omitted

5.4.1

We compute Dh�r� θ � φ � , the 3 � 3 matrix known as the differential of h.

Dh�r� θ � φ � �

��

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

��

Thus

Dh�r � θ � φ � �

�� cosφcosθ � �b r cosφ � sinθ � r sinφcosθ

cosφsinθ�b r cosφ � cosθ � r sinφcosθ

sinφ 0 r cosφ

��Since the columns of Dh are mutually � (check it) by a result proven during lectures (see notes)

�DetDh � 2 ��Dh1 �� 2 � ��Dh2 �� 2 � ��Dh3 �� 2 � 12 � �

b r cosφ � 2 � r2 Thus we have our Jacobian � detDh � � r�b r cosφ � .


jbquig-UCD


5.4.2

Assuming T has density ρ � 1, write m for volume = mass of T, and I for the moment of inertia of T aboutthe z-axis, then

m � � ��T

1d�x � y � z � � M � ��

T

�x2 y2 � d

�x � y � z �

Using the Toroidal Polar transformation and its Jacobian and the C.O.V. theorem we can now compute theseintegrals.

m � �� T

1d�x � y � z �

� �� 0 � a � � � � π � π � � � � π � π � 1 � r

�b r cosφ � d

�r� θ � φ �

� � a

0� π

� π� π

� π1 � r

�b r cosφ � dθdφdr

� 2π � a

0� π

� πr

�b r cosφ � dφdr � π

� πcos � 0 being over a full period

� 2π � a

0� π

� πbr dφdr

� 2π � 2π � a

0br dr

� 2π � 2π�br2 � 2 � � a0

� 2π � 2π�ba2 � 2 �

� �2πb ��

πa2 �

I � � ��T

�x2 y2 � d

�x � y � z �

� � �� 0 � a � � � � π � π � � � � π � π �

�b r cosφ � 2 � r

�b r cosφ � d

�r� θ � φ �

� � a

0� π

� π� π

� πr

�b r cosφ � 3 dθdφdr using Fubini’s theorem

� 2π � a

0� π

� πr

�b r cosφ � 3 dφdr

� 2π � a

0� π

� π

�b3r 3b2r2 cosφ 3br3 cos2 φ r4 cos3 φ � dφdr

Now the student will verify that

� π

� πcosφdφ � 0 � � π

� πcos2 φdφ � � π

� π

�1 � 2 � �

1 cos2φ � dφ � � π

� π

�1 � 2 � �

1 � dφ � π � � π

� πcos3 φdφ � 0

Thus

I � 2π � a

0

�2πb3r 3b2r2 � 0 3br3 � π r4 � 0 � dr

� 2π2 � a

0

�2b3r 0 3br3 0 � dr � 2π2 � a

0

�2b3r 3br3 � dr

� 2π2 �b3r2 3br4 � 4 � � a0

� 2π2 �b3a2 3ba4 � 4 � � a0

� �2πb � �

πa2 � �b2 3a2 � 4 �

We can now compute the radius of gyration

r ��

M � m � � b2 3a2 � 4

c�


5.5. ANSWER 235

5.4.3

The subset S � T is not so easy to figure out. S is given by � � �x2 y2 � b � z � �

x2 y2 � b � . Considerthe corresponding region S

�in the disc D which replacing

�x2 y2 by x is given by � �

x � b � z �x � b � .

Thus S� � D is the segment between two lines z � �

�x � b � passing through the center of the disc D. Draw

S� � D in the xz-plane. You can now draw S � T as S

�rotated about the z-axis.

5.4.4

To integrate over S we observe that S is parameterized by the Toroidal Polar transform but with the angle φonly running from � π � 4 to π � 4. Thus

m � � ��S

1d�x � y � z � � � ��

0 � a � � � � π � π � � � � π � 4 � π � 4 � 1 � r�b r cosφ � d

�r� θ � φ �

AlsoM � � ��

S

�x2 y2 � d

�x � y � z � � � ��

0 � a � � � � π � π � � � � π � 4 � π � 4 � r �b r cosφ � 2 d

�r� θ � φ �

I will not carry out the computations which are merely variants on those for T above.

5.5 answer

5.5.1

The region B is given by three inequalities, lets look at them one at a time, the key to the problem is to spotfrom the outset that Spherical Polars are involved.

The first inequality x2 y2 z2 1 is equivalent to r 1. The second inequality � x y x can bewritten � 1 y � x 1 i.e. � 1 tanθ 1 i.e. � π � 4 θ π � 4. The third inequality x2 y2 z2 can bewritten

�x2 y2 � � z2 1 i.e. tan2 φ 1. Since 0 � φ � π we have φ �

0 � π � 4 � � �3π � 4 � π � . The student can

now presumably draw the region B, which consists of two portions of the unit ball. To obtain the first portion,take that segment of the ball which lies between longitude � π � 4 but above the latitude cone φ � π � 4. Thesecond portion is similar but lies below the latitude cone φ � 3π � 4.

5.5.2

To compute the volume m of B we must evaluate the integral m � � � �B

1d�x � y � z � . We observe that it

follows from the above remarks that the Spherical Polar transform h carries the region

�0 � 1 � � � � π � 4 � π � 4 � � � �

0 � π � 4 � � �3π � 4 � π � � � �

0 � 1 � � � � π � 4 � π � 4 � � �0 � π � 4 � � �

0 � 1 � � � � π � 4 � π � 4 � � �3π � 4 � π

which is the union of two rectangloids in r� θ � φ-space bijectively onto B. Thus applying the COV theorem,using the Spherical Polar transform and its Jacobian r2 sinφ we obtain

m � � ��B

1d�x � y � z �

� � �� 0 � 1 � � � � π � 4 � π � 4 � � � 0 � π � 4 � r2 sinφd

�r� θ � φ � ��

0 � 1 � � � � π � 4 � π � 4 � � � 3π � 4 � π � r2 sinφd�r� θ � φ �

The computations are similar to those carried out above in answers ?1 and ?2, and so are left to the student.

5.5.3

The trick here is to notice that there is a transformation k carrying the (good) region B in xyz-space bijectivelyonto the (bad) region C in XYZ-space. A clue to what k might be is that k should carry the solid ball of radius1 x2 y2 z2 � 1 onto the solid ellipsoid

�X2 � 22 � �

Y2 � 32 � �Z2 � 52 � � 1 Thus k should expand the


jbquig-UCD


x-direction by a factor of 2 the y-direction by a factor of 3 and the z � direction by a factor of 5, so here is thetransformation k which is in fact linear i.e. a matrix mapping

k : R3 �

�� xyz

��

�� X�x � y � z �

Y�x � y � z �

Z�x � y � z �

�� 2 0 0

0 3 00 0 5

�� xyz

�� 2x

3y5z

�� R3

Note that k being invertible is a bijection from R3 to R3. Finally we check that k converts the inequalitieswhich define C into those which define B, thus proving that the transformation carries B onto C. thus wecheck �

X2 � 22 � �Y 2 � 32 � �

Z2 � 52 � 1 � x2 y2 z2 1� 3X � 2Y � 3X � � x � y � x�X2 � 4 � �

y2 � 9 � �Z2 � 25 � � x2 y2 � z2

by substituting for X � Y � Z on the LHS. The region C can now be sketched being the same as B but suitabledeformed by k as described above.

5.5.4

We use the COV theorem. Note that Dk � k, you can see this by hacking out the nine p.d.s or from theprinciple that Dk is the linearization of k but k is already linear. Thus the Jacobian � det Dk � � � detk � �� 2 � 3 � 5 � � 30. Finally the volume of C is

m � �� C

1d�X � Y � Z �

� �� B

1 � 30d�x � y � z �

� 30 � ��B

1d�x � y � z �

etc.,etc.,etc., see part (ii).

5.6 answer

5.6.1

The equation x2 y2 � 1 in R3 describes a cylindrical surface with crossection a circle of radius 1 and withaxis of symmetry the z-axis. Thus the inequality x2 y2 1 describes the solid cylinder with crossection adisc of radius 1 and with axis of symmetry again the z axis.

The equation z � arctan�y � x � may be converted using cylindrical polar coordinates (see Q3 above) to the

form z � θ (as you twirl so do you rise) and so describes the SCREW surface. Now we can visualize theregion K and the student should sketch it

K � � x : x2 y2 1 and arctan�y � x � z 2π

�First construct a solid cylinder of radius 1 with base at z � 0 and top at z � 2π, next imagine a screw surfaceinside it which goes from bottom to top in exactly one twist, then K is the space above the screw but beneaththe ceiling. Imagine a circular staircase in a cylindrical well, just like the one beside the theatre, store coalbeneath , K is the air space left above the stairs and beneath the ceiling. Well now we have the general idea,draw it , next the mathematics.

5.6.2

Consider the Cylindrical polar transform

h :�0 � ∞ � � � � π � π � � R �

�� rθZ

��

�� x�r� θ � Z �

y�r� θ � Z �

z�r � θ � Z �

�� r cosθ

r sinθZ

�� R3 � S

c�


5.6. ANSWER 237

the full details are in Q3 above, in particular the Jacobian � det Dh � � r. Consider the triangle T in θ � Z-space.

T � � �θZ � : θ Z 2π � 0 θ 2π �

which the student should sketch. Next consider the wedge W in rθZ-space

W � �0 � 1 � � T � � �� r

θZ

�� : 0 r 1 � θ Z 2π � 0 θ 2π

The student should sketch W . The Cylindrical Polar transform carries W bijectively onto K. Thus using theCOV theorem, the Jacobian being r,

m � �� K

1d�x � y � z �

� �� W � 0 � 1 � � T

1 � r d�r� θ � Z �

� � 1

0�

T1 � r d

�θ � Z � dr

� � 1

0� 2π

0� 2π

θr dZ dθdr

� � 1

0� 2π

0

�2π � θ � r dθdr

� � 1

0� �

2πθ � θ2 � 2 � � 2πθ 0 � r dr

� � 1

0� �

2π2 � r dr

� � �2π2 � r2 � 2 � 10

� π2

5.6.3

The idea is to find a transformation k from xyz-space to XYZ-space which carries K bijectively to L. Thetransformation k is in fact the matrix mapping

k : R3 �

�� xyz

��

�� X�x � y � z �

Y�x � y � z �

Z�x � y � z �

�� 2 0 1

0 3 10 0 1

�� xyz

�� 2x z

3y zz

�� R3

To prove that k carries the region K to the region L check that the two defining inequalities for L are carriedto the defining inequalities for K by k, substitute for X � Y � Z on RHS below.

x2 y2 1 � � X � Z2

� 2 � Y � Z3

� 2 1

arctan yx z 2π � arctan � 2

�Y � Z �

3�X � Z � � Z 2π

5.6.4

To obtain the volume of L apply the COV th

m � � ��L

1d�x � y � z �

� � ��K

1 � detDk � d �X � Y � Z �


jbquig-UCD


� � ��K

6d�X � Y � Z �

� 6 �� K

1d�X � Y � Z �

� 6π2 by part (ii)

5.7 answer

5.7.1

The region V is similar to the region K sketched in the previous question except that the radius of the cylinderis now a and the region V is now above rather than below the screw surface.

Next lets discuss the region W which is similar to V but now the enclosing cylinder has been replaced byan enclosing cone x2 y2 � z2 with apex at 0 axis of symmetry the z-axis and apexial angle π � 4. i.e. thecone where the latitude angle φ � π � 4. The student should now be able to sketch these regions.

5.7.2

Let h be the Cylindrical Polar transform as in the last question. Then h carries the regions A and B respectivelyin rθZ-space bijectively onto the regions V and W in xyz-space where

A � � �� rθZ

� �� : 0 r a � 0 Z θ � 0 θ 2π�

B �� rθZ

� �� : 0 θ 2π � 0 Z θ � 0 r Z�

The student should sketch these regions A and B in rθZ-space, its tricky to see that h carries B to W

5.7.3

Now we use the COV theorem to calculate the two volumes

m�V � � ��

V1d

�x � y � z �

� ��A

1 � � det Dh � d �r� θ � Z �

� ��A

r d�r� θ � Z �

� � a

0� 2π

0� θ

0r dZ dθdr

� � a

0� 2π

0rθdθdr

� � a

0rθ2 � 2 � 2π

θ 0 dr

� � a

02π2r dr

� 2π2r2 � 2 � a0 dr� �

πa2 �� π �

c�


5.8. ANSWER 239

Now the volume of W

m�W � � ��

W1d

�x � y � z �

� �� B

1 � � det Dh � d �r� θ � Z �

� �� B

r d�r� θ � Z �

� � 2π

0� θ

0� Z

0r dr dZ dθ

� � 2π

0� θ

0Z2 � 2dZ dθ

� � 2π

0θ3 � 6dθ

� θ4 � 24 � 2π0

� 16π4 � 24� 2π4 � 3

5.8 answer

In this question one meets the standard solid ball Bn �1 � of radius 1 in general n space and computes its

measure or hypervolume. Of course in dimensions 1 � 2 and 3 respectively, this object is the interval � � 1 � 1 �the disc and the usual three dimensional ball, respectively and the measures are length 2, area π and volume�4 � 3 � π; can one continue this sequence up into higher dimensions 4,5,6 etc.? Read on. Once one has dealt

with the standard ball of radius 1 the ball of radius a and the general solid ellipsoidal ball are easily dealtwith.

5.8.1

Consider the transformation h from Rn with coordinates x1 � x2 � � � x j � � � xn to Rn with coordinates y1 � y2 � � � y j � � � yn

which expands the j-th direction by the factor a j � 1 j n.

h : Rn �

��

x1

x2

��

x j ��

xn

��

��

��

y1

y2

��

yi ��

yn

��

�

��

a1x1

a2x2

��

aixi

��

anxn

�� Rn

In fact h is a linear mapping being given by the n � n matrix Diag�a1 � a2 � � � � � a j � � � � � an � with a j down the

diagonal and all off diagonal entries zero. Thus Dh � h as we have seen in previous questions and the Jaco-bian � detDh � � ∏n

1 a j. It is clear that h maps the standard n-ball Bn �1 � bijectively onto the n-dimensional

ellipsoid En �a1 � a2 � � � � � a j � � � � � an

� . Now we are ready to use the COV th to study the volume of the solidellipsoid

µ�En �

a1 � a2 � � � � � a j � � � � � an� � � �

En�a1 � a2 � � � � � a j � � � � � an �

1d�y �

� �Bn�1 �

1 � � detDh � d �x �


jbquig-UCD


� �Bn�1 �

1 � n

∏1

a j d�x �

� n

∏1

a j �Bn�1 �

1d�x �

� n

∏1

a j µBn �1 �

Note that all the above integrals are in n-space.Since Bn �

a � can be looked on as an ellipsoid with all n semi-axes equal to a, we obtain µ�Bn �

a � � . Letsstate both of our main results

µ�En �

a1 � a2 � � � � � a j � � � � � an� �� n

∏1

a j µ�Bn �

1 � � � µ�Ba �

n � � � �an � µ

�Bn �

1 � �

5.8.2

This is the key step as it will be used to obtain the volume of the standard n-ball by induction.

µ�Bn �

a � � � �Bn�a �

1d�x1 � x2 � � � � � xn � 2 � xn � 1 � xn �

� � �x1 � 2 � � x2 � 2 � � � � � � xn � 2 � 2 � � xn � 1 � 2 � xn � 2 � 1

1d�x1 � x2 � � � � � xn � 2 � xn � 1 � xn �

� � �xn � 1 � 2 � � xn � 2 � 1

� �x1 � 2 � � x2 � 2 � � � � � � xn � 2 � 2 � 1 �

�xn � 1 � 2 �

�xn � 2

1d�x1 � x2 � � � � � xn � 2 � d

�xn � 1 � xn �

� �B2 � 1 �

�Bn � 2 � � 1 �

�xn � 1 � 2 �

�xn � 2 �

1d�x1 � x2 � � � � � xn � 2 � d

�xn � 1 � xn �

� �B2�1 �

µ�Bn � 2 � � 1 � �

xn � 1 � 2 � �xn � 2 � � d

�xn � 1 � xn �

� �B2 � 1 �

�1 � �

xn � 1 � 2 � �xn � 2 � � n � 2 � � 2µ

�Bn � 2 �

1 � � d�xn � 1 � xn �

� µ�Bn � 2 �

1 � � �B2�1 �

�1 � x2 � y2 � � n � 2 � � 2 d

�x � y �

� µ�Bn � 2 �

1 � � �� 0 � 1 � � � 0 � 2π �

�1 � r2 � � n � 2 � � 2 � r d

�r� θ �

� µ�Bn � 2 �

1 � � � 1

0� 2π

0

�1 � r2 � � n � 2 � � 2 � r dθdr

� µ�Bn � 2 �

1 � �� 2π � 1

0

�1 � r2 � � n � 2 � � 2 � r dr

� µ�Bn � 2 �

1 � �� 1 � π � 1

r 0

�1 � r2 � � n � 2 � � 2 d

�1 � r2 �

� µ�Bn � 2 �

1 � �� 1 � π�2 � n � �

1 � r2 � n � 2 � 1r 0� �

2 � n � πµ�Bn � 2 �

1 � �

To summarize we have proven

µ�Bn �

1 � ��

2n � µ

�Bn � 2 �

1 � �

5.8.3

We can now compute µ�Bn �

1 � � separately for even and odd n.µ

�B1 �

1 � �� µ� � � 1 � 1 � � � 2,

c�


5.8. ANSWER 241

µ�B3 �

1 � �� 2 ��2 � 3 � π � �

4 � 3 � π,µ

�B5 �

1 � �� 4π � 3 ��

2 � 5 � π � �8 � 15 � π2,

etc. for odd n. Next the even nµ

�B2 �

1 � �� πµ

�B4 �

1 � �� π ��2 � 4 � π � �

1 � 2 � π2,µ

�B6 �

1 � �� 1 � 2 � π2 � �

2 � 6 � π � �1 � 6 � π3,

etc.. If you don’t believe it recall we used a computer program and already have the hypervolume for n=4,5.You may compare the results by both methods. From part (i) µBn �

a � � anµBn �1 � and µ

�En �

a1 � a2 � � � � � a j � � � � � an� � �

�∏n

1 a j � µ�Bn �

1 � � .

5.8.4

The student should first convince him/herself that the n dimensional Spherical Polar transform h indeedcarries the given rectangloid from rθ1θ2

� � � θn � 1-space bijectively onto the ball Bn �1 � � Rn. Next comes the

question of the n � n differential matrix Dh. The columns of this matrix are obtained by partial differentiationof the original column

x �

��

r cosθ1

r sinθ1 cosθ2

r sinθ1 sinθ2 cosθ3

r sinθ1 sinθ2 sinθ3 cosθ4��

r sinθ1 sinθ2� � � sinθn � 2 cosθn � 1

r sinθ1 sinθ2� � � sinθn � 2 sinθn � 1

��

w.r.t. r � θ1 � θ2 � � � � � θn � 1 in order. These n-columns will be found to be mutually orthogonal and so the Jacobian� detDh � of h is, as for the 3 � 3 case, simply the product of the lengths of these columns being

1 � r � r sinθ1� r sinθ1 sinθ2

� � � r sinθ1 sinθ2� � � sinθn � 2

thus � detDh � � rn � 1 sinn � 2 θ1 sinn � 3 θ2 sinn � 4 θ3� � � sin2 θn � 3 sinθn � 2

Now we can apply the COV theorem to compute the measure or hypervolume

µ�Bn �

1 � � � �Bn�1 �

1d�x �

� � �0 � 1 � � � 0 � π � � � 0 � π � � � � � � � 0 � π � � � 0 � 2π �

1 � � det Dh � d �r� θ1 � θ2 � � � � � θn � 2 � θn � 1

�

� � 1

0� π

0� π

0

� � � � π

0� 2π

0rn � 1 sinn � 2 θ1 sinn � 3 θ2 sinn � 4 θ3

� � � sin2 θn � 3 sinθn � 2 dr dθ1 dθ2� � � � dθn � 1 dθn � 1

Because the domain of integration is a rectangloid, and because the integrand function is separable (ie aproduct of simple functions each of one variable only ) it is routine to break the latter integral up into aproduct of simple integrals each in dimension 1.

µ�Bn �

1 � � � � � 1

0rn � 1 dr �� π

0sinn � 2 θ1

� dθ1� � � π

0sinn � 3 θ2

� dθ2� � � � � π

0sinθn � 2

� dθn � 2� � � 2π

01dθn � 1

Next evaluate the first and last of these n integrals which is easy. Also note that in the remaining n � 2integrals we may change the variable of integration to some thing brief say ω

µ�Bn �

1 � � � 2π�

rn

n � � n � 2

∏j 1

� π

0sin j ωdω


jbquig-UCD


c�


Chapter 6

Answers surface integration, divergencetheorem of Gauss

This answer set has been gleaned from an earlier document without much editing. There is repetition and nodiagrams but still much useful material. Will not be further edited this academic year, jbquig 14–02–2000.

6.1 answer

From the above we see that the total buoyancy force on the body exerted by liquid pressure acting on the its

surface is ��Σ

�ρgzn � k � dσ and after rearranging the integrand can be written ��

Σ

�ρgzk � � ndσ which is

now the flux of the vector field 0i 0j ρgzk across the surface Σ. We compute the divergence

div�0i 0j ρgzk � � ∂0

∂x ∂0

∂y ∂ρgz

∂z� 0 0 ρg� ρg

Applying the divergence thoerem we obtain that the total buoyancy force on the Body B is

��Σ

�ρgzk �� ndσ � ��

Bρgd

�x � y � z �� ρg � � �

B1d

�x � y � z � � ρg

�volumeB �

Now density by volume by gravitational acceleration is weight, but the weight of what? The weight of thevolume B filled with a material of densit ρ i.e. filled with liquid, thus we have proven the Principle ofArchimedes, thatBUOYANCY = WEIGHT OF LIQUID DISPLACED

6.2 answer

Note Q2 in the book was wrong (or too hard), the corrected version is given here. There is an easier questioninvolving flux across a toroidal surface below S4-Q3, I suggest you attempt it first.

6.2.1

The inequality�x � b � 2 z2 a2 describes the disc D in the xz-plane with center at the point

�xz � �

�b0 �

and radius a � 0. The equations z � ��x � b � yield two lines passing through the center of the disc D making

243

244 CHAPTER 6. ANSWERS SURFACE INTEGRATION, DIVERGENCE THEOREM OF GAUSS

angles � π � 4 with the x-axis. The inequalities � �x � b � z �

x � b � describes one quadrant formed by thesetwo lines. Thus

A � � �xz � :

�x � b � 2 z2 a2 and � �

x � b � z �x � b � � � R2

is that segment of the disc D lying as described between these two lines crossing at the center This area A isdescribed using toroidal polar coordinates r� θ � φ by

0 r a � θ � 0 and � π � 4 φ π � 4

Now the student can easily draw A � R2.The solid torus T is obtained by rotating the disc D about the z-axis. B � T � R3 is obtained by rotating

the subset A of the disc. Hopefully the student can now sketch B � T � R3. One notes that in toroidal polarcoords B is given by

0 r a � � π θ π � � π � 4 φ π � 4

Next we turn to discussion of the curve Γ which forms the boundary of the area A � R2. This curve hasthree pieces Γ � L1

�L2

�C where L1 and L2 are parts of the lines where φ � � π � 4 and C is a portion of

the circle r � a. The curve Γ when rotated sweeps out the boundary surface Σ of the solid region B, whichthus consists of three subsurfaces; Σ � Σ1

�Σ2

�S. The line segment L1 sweeps out the surface Σ1 which is

thus part of the cone φ � � π � 4. Similarly Σ2 is swept out by L2 and is part of the cone φ � π � 4. Finally S isswept out by C and is part of the Toroidal surface r � a.

Hopefully now the student can sketch

Σ � Σ1�

Σ2�

S � B � T � R3

6.2.2

We will evaluate the integral I directly by using the parametrization formula for flux of a vector field acrossa surface.

I � ��Σ Σ1 � Σ2 � S

x� �

x2 y2 � b � dy � dz y� �

x2 y2 � b � dz � dx z�

x2 y2 dx � dy

� ��Σ Σ1 � Σ2 � S

v � ndσ

� ��Σ1

v � ndσ ��Σ2

v � ndσ � �S

v � ndσ

where the vector field is

v � Mi Nj Pk � x� �

x2 y2 � b � i y� �

x2 y2 � b � j z�

x2 y2 k

The next step, as we must integrate across all three surfaces is to find a parametrization of each; this sends usoff on a long deviation. Recall the toroidal polar transform for the solid torus T

h :�0 � a � � � � π � π � � � � π � π � �

�� rθφ

��

�� xyz

��


r sinφ

�� T � R3

Lets give the reverse formulae for r� θ � φ in terms of x � y � z which will soon be useful, these are obtained bysimple manipulation of the above (do it) or using trigonometry from a good diagram of T (do it).

r �� x2 y2 � b � 2 z2 � 1 � 2

φ � arctanz�

x2 y2 � b

c�


6.2. ANSWER 245

θ � arctanyx

Now the surface S is part of the toroidal surface where r � a and so its parameterization k is obtained from hessentially by fixing r � a, and is

k :� � π � π � � � � π � 4 � π � 4 � �

�θφ � ��

�� xyz

��


asinφ

�� S � R3

The surface Σ1 is part of the cone surface where φ � � π � 4 and so its parameterization l is obtained from hessentially by fixing φ � � π � 4, and is

l :�0 � a � � � � π � π � �

�rθ � ��

�� xyz

��

b r cos� � π � 4 � � cosθ�

b r cos� � π � 4 � � sinθ

r sin� � π � 4 �

��

b r � � 2 � cosθ�b r � � 2 � sinθ� r � � 2

�� Σ1 � R3

The parameterization m of Σ2 is obtained similarly, this time φ is fixed at π � 2, and is

m :�0 � a � � � � π � π � �

�rθ � ��

�� xyz

��

b r cos�π � 4 � � cosθ�

b r cos�π � 4 � � sinθ

r sin�π � 4 �

��

b r � � 2 � cosθ�b r � � 2 � sinθ

r � � 2

�� Σ1 � R3

We return to the main computation.

I � ��Σ1

v � ndσ � �Σ2

v � ndσ � �S

v � ndσ

We will work out these three integrals one at a time, the last one first.

��S

v � ndσ use the parametrization formula

� �� π � π � � � � π � 4 � π � 4 � ��

M N P∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

�� d �θ � dφ �

� �� π � π � � � � π � 4 � π � 4 �

��x

� �x2 y2 � b � y

� �x2 y2 � b � z

�x2 y2

∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

�� d �θ � dφ �

� �� π � π � � � � π � 4 � π � 4 � ��

�b acosφ � cosθ

�acosφ � �

b acosφ � sinθ�acosφ � �

asinφ � �b acosφ ��

b acosφ � sinθ�b cosφ � cosθ 0� asinφcosθ � asinφsinθ acosφ

�� d �θ � dφ �

� �� π � π � � � � π � 4 � π � 4 � � a � �

b acosφ � � � � b acosφ � � � a � �� cosφcosθ cosφsinθ sinφ� sinθ cosθ 0� sinφcosθ � sinφsinθ cosφ

�� d �θ � φ �

� �� π � π � � � � π � 4 � π � 4 � a2 �

b acosφ � 2�� cosφcosθ cosφsinθ sinφ� sinθ cosθ 0� sinφcosθ � sinφsinθ cosφ

�� d�θ � φ � rows orthogonal so get det

� �� π � π � � � � π � 4 � π � 4 � a2 �

b acosφ � 2 � 1d�θ � φ � apply Fubini’s theorem

� � π � 4� π � 4 � π

� πa2 �

b acosφ � 2 dθdφ

� 2πa2 � π � 4� π � 4 �

b acosφ � 2 dφ

� 2πa2 � π � 4� π � 4 b2 2bacosφ a2 cos2 φdφ


jbquig-UCD


� 2πa2b2 � π � 4� π � 4 1dφ 4πa3b � π � 4

� π � 4 cosφdφ 2πa4 � π � 4� π � 4 cos2 φdφ

� π2a2b2 4 � 2πa3b πa4 � π � 4� π � 4 � �

�1 cos2φ � dφ

� π2a2b2 4 � 2πa3b πa4 � π � 4� π � 4 1dφ πa4 � π � 4

� π � 4 cos2φdφ

� π2a2b2 4 � 2πa3b �1 � 2 � π2a4 πa4

�12

sin2φ � � π � 4� π � 4

� π2a2b2 4 � 2πa3b �1 � 2 � π2a4 πa4

We have one of the three integrals

��S

v � ndσ � π2a2b2 4 � 2πa3b 12

π2a4 πa4

Next consider � �Σ1

v � ndσ . We have developed a parameterization of Σ1 and so can evaluate this in-

tegral in the usual way, the student should do it, but here is a quick trick. The Surface Σ1 is given byφ � � π � 4(constant), and so also where tanφ � tan

� � π � 4 � � � 1(const) By second year analysis grad�tanφ �

is a normal vector field to this surface where

grad�tanφ � � grad

�tanarctan

z�x2 y2 � b

�

� gradz�

x2 y2 � b

�

��

∂∂x z

� �x2 y2 � 1 � 2 � b � � 1

∂∂y z

� �x2 y2 � 1 � 2 � b � � 1

∂∂z z

� �x2 y2 � 1 � 2 � b � � 1

� ��

�� z

� � 1 � � �x2 y2 � 1 � 2 � b � � 2x

�x2 y2 � � 1 � 2

z� � 1 � � �

x2 y2 � 1 � 2 � b � � 2y�x2 y2 � � 1 � 2

� �x2 y2 � 1 � 2 � b � � 1

��We get a simpler normal field N to Σ1 if we multiply the latter by the scalar function� �

x2 y2 � 1 � 2 � b � 2 �x2 y2 � 1 � 2.

N �� zx� zy� �

x2 y2 � 1 � 2 � b � �x2 y2 � 1 � 2

��Next

v � N � � �� x� �

x2 y2 � b �y

� �x2 y2 � b �

z�

x2 y2

��

�� zx� zy� �x2 y2 � 1 � 2 � b � �

x2 y2 � 1 � 2�� 0

The unit normal field n differs from N only by multiplication by a scalar function and so also v � n � 0.Thus � �

Σ1

v � ndσ � � �Σ1

0dσ � 0

Similarly ��Σ1

v � ndσ � 0. Finally

I� ��

Σv � ndσ

c�


6.2. ANSWER 247

� ��Σ1

v � ndσ ��Σ2

v � ndσ � �S

v � ndσ

� 0 0 ��S

v � ndσ

� ��S

v � ndσ

� π2a2b2 4 � 2πa3b 12

π2a4 πa4

We finally have our result

I � � �Σ

v � ndσ � π � 1 π2 � a4 4 � 2πa3b π2a2b2

6.2.3

We compute the surface integral I again by converting it to a volume integral using the divergence Theoremof Gauss, followed by the COV th using Toroidal Polar Coordinates, followed by Fubini’s theorem.

I � ��Σ

x� �

x2 y2 � b � dy � dz y� �

x2 y2 � b � dz � dx z�

x2 y2 dx � dy � � �Σ

v � ndσ

where the vector field is

v � Mi Nj Pk � x� �

x2 y2 � b � i y� �

x2 y2 � b � j z�

x2 y2 k

Now

∂M∂x

� ∂∂x

x� �

x2 y2 � 1 � 2 � b �

� � �x2 y2 � 1 � 2 � b � x2 �

x2 y2 � � 1 � 2Similarly ∂N

∂y� � �

x2 y2 � 1 � 2 � b � y2 �x2 y2 � � 1 � 2 . Also

∂P∂z

� ∂∂z

z�x2 y2 � 1 � 2

� �x2 y2 � 1 � 2

Thus

divv � ∂M∂x

∂N∂y

∂P∂z

� 2� �

x2 y2 � 1 � 2 � b � �x2 y2 � �

x2 y2 � � 1 � 2 �x2 y2 � 1 � 2

� 2� �

x2 y2 � 1 � 2 � b � 2�x2 y2 � 1 � 2

Now we are ready to compute the integral I using the divergence theorem,

I � ��Σ

v � ndσ next use divergence th,

� �� B


� �� B

� 2 � �x2 y2 � 1 � 2 � b � 2

�x2 y2 � 1 � 2 � d

�x � y � z � next COV with T.P.Tform. h

� �� 0 � a � � � � π � π � � � � π � 4 � π � 4 � �

4r cosφ 2b � � det Dhd�r� θ � φ �

� �� 0 � a � � � � π � π � � � � π � 4 � π � 4 � �

4r cosφ 2b � � r�b r cosφ � d

�r� θ � φ �


jbquig-UCD


� � � � �0 � a � � � � π � π � � � � π � 4 � π � 4 � r

�4r cosφ 2b � �

b r cosφ � d�r� θ � φ �

� � � � �0 � a � � � � π � π � � � � π � 4 � π � 4 � r

�2b 4r cosφ � �

b r cosφ � d�r� θ � φ � Fubini’s th. next

� � π � 4� π � 4 � a

0� π

� πr


b r cosφ � dθdr dφ

� 2π � π � 4� π � 4 � a

0r


b r cosφ � dr dφ

� 2π � π � 4� π � 4 � a

0

�2b2r 6r2bcosφ 4r3 cos2 φ � dr dφ

� 2π � π � 4� π � 4 �

2b2a2 2a3bcosφ a4 cos2 φ � dφ

� 4a2b2π � π � 4� π � 4 dφ 4a3bπ � π � 4

� π � 4 cosφdφ 2a4π � π � 4� π � 4 cos2 φdφ

� 4a2b2π�π � 2 � 4a3bπ

� � 2 � a4π � π � 4� π � 4 �

1 cos2φ � dφ

� 4a2b2π�π � 2 � 4a3bπ

� � 2 � a4π � π � 4� π � 4 1dφ a4π � π � 4

� π � 4 cos2φdφ

� 4a2b2π�π � 2 � 4a3bπ

� � 2 � a4π�π � 2 � a4π

�1 � 2 � sin2φ � π � 4

� π � 4� 4a2b2π

�π � 2 � 4a3bπ

� � 2 � a4π�π � 2 � a4π

�1 � 2 � �

2 �� 2a2b2π2 4 � 2a3bπ a4π2 � 2 a4π

Check that got the same answer as in (ii).

I � � �Σ

v � ndσ � π � 1 π2 � a4 4 � 2πa3b 2π2a2b2

6.3 answer

Detailed remarks about the solid torus and the toroidal surface have been made elsewhere (eg. S3-Q4 also seelecture notes). The student should review the standard drawing of the disc D spun around the z-axis to yieldthe solid toros T, and its bounding toroidal surface Σ. Then the student should review the implicit equationsof Σ and the implicit inequality which describes T. Most importantly the should remind him/herself of theparamaterization of Σ and of the Toroidal Polar transform which parameterizes T, we will meet these below.

In the problem we are asked to compute the flux

I � � �Σ

v � ndσ � � �Σ


across the surface Σ of the vector field

v � Mi Nj Pk � x � bx�x2 y2 i y � by�

x2 y2 j zk

A direct method evaluation of I uses the parameterization mapping

h :� � π � π � � � � π � π � �

�θφ � ��

�� xyz

��


asinφ

�� Σ � R3

of the surface Σ and the the PARAMETERIZATION FORMULA for evaluation of a vector flux integralacross a surface, which in the present example is

I � � �Σ

v � ndσ

c�


6.3. ANSWER 249

� � �Σ


� � � �� π � π � � � � π � π �

��M N P∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

�� d �θ � dφ �

Now we must slug out all nine entries. Lets start with the first row

M � x � 1 � b � �x2 y2 �

� �b acosφ � cosθ

�1 � b � �

b acosφ � ��

b acosφ � cosθ� �

acosφ � � �b acosφ � �

� acosφcosθ

By similar methods we obtain N � acosφsinθ and of course P � z � asinφ . In addition hacking outthe six pds we obtain

I � � � �� π � π � � � � π � π �

�� acosφcosθ acosφsinθ asinφ� �b acosφ � sinθ

�b cosφ � cosθ 0� asinφcosθ � asinφsinθ acosφ

�� d �θ � dφ �

� � � �� π � π � � � � π � π � a �

�b acosφ � � a

�� cosφcosθ cosφsinθ sinφ� sinθ cosθ 0� sinφcosθ � sinφsinθ cosφ

�� d �θ � φ �

where we have divided constants out of all three rows. The remaining determinant has mutually perpendicularrows and so is � �� row 1 �� row 2 �� row 3 �� Thus

I � � � �� π � π � � � � π � π � a �

�b acosφ �� a

�� 1 � 1 � 1d

�θ � φ �

� � � �� π � π � � � � π � π � a2 �

b acosφ � d�θ � φ �

� a2 � π

� π� π

� π

�b acosφ � dθdφ by Fubini’s Theorem

� a2 � π

� π� π

� πbdθdφ since � π

� πcos � 0

� a2b � π

� π� π

� π1dθdφ

� 4π2a2b

Next we turn to the evaluation of the integral I using the DIVERGENCE theorem of GAUSS, which tells us.

� �Σ

v � ndσ � � � �T


or in more detail

� �Σ

M dy � dz N dz � dx Pdz � dx � �� T

�∂M∂x

∂N∂y

∂P∂z � d

�x � y � z �

We must slug out

divv � ∂M∂x

∂N∂y

∂P∂z

� ∂∂x� x � bx � x2 y2 � � 1 � 2 � ∂

∂y� y � by � x2 y2 � � 1 � 2 � ∂

∂z

�z �

� � 1 � b � x2 y2 � � 1 � 2 bx2 � x2 y2 � � 3 � 2 � � 1 � b � x2 y2 � � 1 � 2 by2 � x2 y2 � � 3 � 2 � 1

� 3 � 2b � x2 y2 � � 1 � 2 b � x2 y2 � � x2 y2 � � 3 � 2� 3 � b � x2 y2 � � 1 � 2


jbquig-UCD


Thus, going back to the main computation

I � ��T

� 3 � b � x2 y2 � � 1 � 2 � d�x � y � z �

To evaluate the latter scalar integral over the solid region T we of course appeal to the COV theorem and usethe Toroidal Polar transform (see full details elsewhere, S3-Q4 or lecture notes)

h :�0 � a � � � � π � π � � � � π � π � �

�� rθφ

��

�� xyz

��


r sinφ

�� T

with its Jacobian det Dh � r�b r cosφ � . All these facts about the T.P.Tform should be know by heart;

applying them we obtain

I � � � �T

� 3 � b � x2 y2 � � 1 � 2 � d�x � y � z �

� � � � �0 � a � � � � π � π � � � � π � π � � 3 � b � x2 y2 � � 1 � 2 � � detDhd

�r� θ � φ �

� � � � �0 � a � � � � π � π � � � � π � π � � 3 � b � �

b r cosφ � 2 � � 1 � 2 � � r �b r cosφ � d

�r� θ � φ �

� � � � �0 � a � � � � π � π � � � � π � π � � 3 � b

�b r cosφ � � 1 � � r �

b r cosφ � d�r� θ � φ �

� � � � �0 � a � � � � π � π � � � � π � π �

�3r

�b r cosφ � � br � d

�r � θ � φ �

� � a

0� π

� π� π

� π

�2rb r cosφ � dθdφdr � by Fubini’s th

� 2π � a

0� π

� π

�2rb r cosφ � dφdr

� 2π � a

0� π

� π

�2rb � dφdr since � π

� πcos � 0

� 4π2 � a

02rbdr

� 4π2a2b

Thus, we have obtained the same result by two methods

I � � �Σ x � bx�

x2 y2 dy � dz y � by�x2 y2 dz � dx zdx � dy � 4π2a2b

6.4 answer

6.4.1

To be able to draw the surface Σ one must study the equation

x2 y2 � z2 � 0 z 3 � � 2

which describes it.The various expressions in this question would be immediately meaningful if converted

into spherical polars r� θ � φ . The equation x2 y2 � z2 may be writtenx2 y2

z2� 1 or tan2 φ � 1 or

φ � π � 4 or � 3pi � 4, which in the end yields the full latitude cone with apexial angle π � 4. The inequality 0 z 3 � � 2 may be written 0 z2 9 � 2 and since x2 y2 � z2 it may in turn be written r2 � x2 y2 z2 9.Thus the surface Σ is the cone of latitude φ � π � 4 running out from the origin above the xy-plane andterminating on the sphere where r � 3.

From all this, after a glance at the inequalities which describe the solid D, we conclude that D is the regionbounded by the spherical surface r 3 and the cone φ � π � 4. The student can now easily draw Σ � D � R3.

c�


6.4. ANSWER 251

6.4.2

In fact the boundary of the solid region D consists of two surfaces, the first being Σ which we have justdiscussed in detail and the second being a piece S of the sphere r � 3, in fact

S �� x : r � 3 � 0 φ π � 4�

Writingv � Mi Nj Pk � xzi yzj � �

x2 y2 � k

the divergence theorem tells us that

��Σ � S

v � ndσ � � ��D


� � ��D

�∂M∂x

∂N∂y

∂P∂z � d

�x � y � z �

� � ��D

�z z 0 � d

�x � y � z �

� � ��D

2zd�x � y � z �

Thus � �Σ

v � ndσ ��S

v � ndσ � �� D

2zd�x � y � z �

But we can easily compute the second integral here, using the parameterization

h :� � π � π � � �

0 � π � 4 � ��

θφ � ��

�� xyz

�� 3sinφcosθ

3sinφsinθ3cosφ

�� S � R3

of the surface S.

� �S

v � ndσ � � � �� π � π � � � 0 � π � 4 � ��

M N P∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

�� d�θ � φ �

� � � �� π � π � � � 0 � π � 4 � �� 9sinφcosφcosθ 9sinφcosφsinθ � 9sin2 φ� 3sinφsinθ 3sinφcosθ 0

3cosφcosθ 3cosφsinθ � 3sinφ

�� d�θ � φ � row1 is a multiple of row2

� � � �� π � π � � � 0 � π � 4 � 0 d

�θ � φ �

� 0

Substituting this zero value above we have the required equality

I � � �Σ

xzdy � dz yzdz � dx � �x2 y2 � dx � dy � � ��

D2zd

�x � y � z �

Next we will compute I as required by evaluation of the volume integral on the RHS, we use of course theCOV theorem, clearly the spherical polar transform ( Jacobian � r2 sinφ ) h carries

�0 � 3 � � � � π � π � � �

0 � π � 4 �onto D.

I � � ��D

2zd�x � y � z �

� � �� 0 � 3 � � � � π � π � � � 0 � π � 4 � 2z � det Dhd

�r� θ � φ �

� � �� 0 � 3 � � � � π � π � � � 0 � π � 4 � 2r cosφ � r2 sinφd

�r� θ � φ �


jbquig-UCD


�

� � 3

0� π � 4

0� π

� π2r3 sinφcosφdθdφdr

� 2π � 3

0� π � 4

02r3 sinφcosφdφdr

� 2π � 3

0� π � 4

0r3 sin2φdφdr

� π � 3

0r3

�� cos2φ�� π � 40

�� dr

� π � 3

0r3 � � 1 � � 0 � dr

� � π � 3

0r3 dr

� � 81π � 4

6.4.3

We compute the surface integral I again, this time by parameterization of Σ a conical surface on which thelatitude angle is fixed at φ � π � 4, so the spherical polar transform (with φ fixed) will parameterize Σ. Thisis a useful trick but its not new; the standard parameterization of the sphere is obtained the same way butholding r fixed. The parameterization of Σ is

h :�0 � 3 � � � � π � π � �

�rθ � ��

�� xyz

�� r sinπ � 4cosθ

r sinπ � 4sinθr cosπ � 4

��

1� 2 �� r cosθ

r sinθr

�� Σ � R3

Using the evaluation by parameterization theorem for flux integrals we obtain

I � � �Σ

Mdy � dz Ndz � dx Pdx � dy

� � �Σ

xzdy � dz yzdz � dx � �x2 y2 � dx � dy

� � � �0 � 3 � � � � π � π �

�� M N P∂x∂r

∂y∂r

∂z∂r

∂x∂θ

∂y∂θ

∂z∂θ

�� d�r� θ �

� � � �0 � 3 � � � � π � π �

�� xz yz � �x2 y2 �

cosθ sinθ 0r sinθ � r cosθ 1

�� d�r� θ �

� � � �0 � 3 � � � � π � π �

��1 � 2 � r2 cosθ

�1 � 2 � r2 sinθ � �

1 � 2 � r2�1 � � 2 � cosθ

�1 � � 2 � sinθ

�1 � � 2 �

�1 � � 2 � r sinθ � �

1 � � 2 � r cosθ 0

�� d�r� θ �

��

14 � ��

0 � 3 � � � � π � π � r3�� cosθ sinθ � 1cosθ sinθ 1sinθ � cosθ 0

�� d�r� θ �

��

14 � ��

0 � 3 � � � � π � π � r3 �2 � d

�r� θ �

� �2 �

�14 � � 3

0� π

� πr3 dθdr

� �2π � �

2 ��

14 � � 3

0r3 dr

c�


6.5. ANSWER 253

� �2π � �

2 ��

14 � �

34

4 �� 81π � 4

Check that this agrees with the previous computation of I except for � 1, this carelessness is tied up withFRHR and the direction of the surface normal etc.etc. but enough � � � .

6.5 answer

6.5.1

In S4-Q3 we obtained the parametrization

h :� � π � π � � � � π � π � �

�θφ � ��

�� xyz

��


asinφ

�� Σ � R3

of the toroidal surface Σ

6.5.2

Using the notation v � Mi Nj Pk we compute the integral I using the evaluation by parametrizationformula.

I � ��Σ

v�� v �� 2 � ndσ

� ��Σ

�M�� v �� 2 dy � dz N�� v �� 2 dz � dx P�� v �� 2 dx � dy � dσ

� �� π � π � � � � π � π �

��M � �� v �� 2 N � �� v �� 2 P �� v �� 2

∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

�� d �θ � φ �

� �� π � π � � � � π � π �

�1�� v �� 2 � �� M N P� �


�� d �θ � φ �

� �� π � π � � � � π � π �

�1�� v �� 2 � �� acosφcosθ acosφsinθ asinφ� �


�� d �θ � φ �

� �� π � π � � � � π � π � a2 �

b acosφ ��

1�� v �� 2 � �� cosφcosθ cosφsinθ sinφ� sinθ cosθ 0� sinφcosθ � sinφsinθ cosφ

�� d �θ � φ �

� �� π � π � � � � π � π � a2 �

b acosφ ��

1�� v �� 2 � �1 � d

�θ � φ � rows mutually orthogonal, det � 1

We must deviate from our main argument to compute �� v �� 2 at points x Σ in terms of the parameters θ � φ.

�� v �� 2 � M2 N2 P2

� x � bx�x2 y2 2 y � by�

x2 y2 2 z2

� �x2 y2 � 1 � b�

x2 y2 2 z2


jbquig-UCD


� �b acosφ � 2

�1 � b�

b acosφ � � 2 a2 sin2 φ

� � �b acosφ � � b � 2 a2 sin2 φ

� �acosφ � 2 a2 sin2 φ

� a2

Returning to our main computation

I � � �Σ

v�� v �� 2 � ndσ

� � � �� π � π � � � � π � π � a2 �

b acosφ ��

1�� v �� 2 � d�θ � φ �

� � � �� π � π � � � � π � π � a2 �

b acosφ ��

1a2 � d

�θ � φ �

� � � �� π � π � � � � π � π �

�b acosφ � d

�θ � φ � next use Fubini’s th.

� � π

� π� π

� π

�b acosφ � dθdφ

� � π

� π� π

� πbdθdφ since � π

� πcos � 0

� � π

� π2πbdφ

� 4π2b

Summing up, we obtain as required a result independent of a

I � � �Σ

v�� v �� 2 � ndσ � 4π2b

6.5.3

The task in this part of the qustion is to study and compute the volume integral.

J � � � �T


�x � y � z �

The horrendous divv�� v �� 2 is worked out below and is indeed infinite on the circle B

x2 y2 � b2 � z � 0 ), which runs round the center of the torus; the student may look ahead at this point ifhe/she so wishes. However one can avoid the onerous divergence by using the theorem of Gauss to turn Jinto a surface integral.

J � �� T


�x � y � z � now apply th. of Gauss

� ��Σ

v�� v �� 2 � ndσ� I� 4π2b

But wait a minute, the answer is supposed to be zero. To instruct the reader I have introduced a subtle error.Given a C 1 vector field w defined over a domain D with boundary the closed surface S, the divergence

theorem of Gauss tells us that � ��D

divwd�x � y � z � � � �

Sw � ndσ

c�


6.5. ANSWER 255

Note that, amongst other things, the hypotheses demand that the vector field w is over ALL OF the solidregion D.

In our example the fieldv�� v �� 2 is not well defined over all of the solid region T but is infinite where the

denominator is zero, i.e. where v is zero, ie where�

x2 y2 � b and z � 0 i.e. on the circle C mentionedabove which passes round the center of the torus T. I now give the the correct argument.

Let ε � 0 be small and Σ�be the toroidal surface whose implicit equation is

� �x2 y2 � b � 2 z2 � ε2

Thus Σ�is like Σ but with the radius a replaced by the very small ε. Let T

�denote the solid region between the

two toroidal surfaces Σ�and Σ. The vector field

v�� v �� 2 is well defined over the solid T�

and Gauss’ theorem

applies

�� T

�div

v�� v �� 2 d�x � y � z � � � �

Σ � Σ�

v�� v �� 2 � ndσ

� � �Σ

v�� v �� 2 � ndσ � � �S

igma� v�� v �� 2 � ndσ

� 4π2b � 4π2b see (ii) above� 0

The reason for the minus sign is that the outward normal vector from the solid T�

is the inward normal forthe surface S. Now let ε tend to zero so that the solid T

�tends in the limit to the solid T.

J � �� T


�x � y � z � � lim

ε � 0��

T�div

v�� v �� 2 d�x � y � z � � lim

ε � 00 � 0

The heroic student who perhaps mistrusts sophistry will now with boundless energy,

(a) hack out divv�� v �� 2 and

(b) by direct computation prove that indeed

limε � 0

�� T

�div

v�� v �� 2 d�x � y � z � � 0

(a)

�� v �� 2 � x � bx�x2 y2 2 y � by�

x2 y2 2 z2

� �x2 y2 � 1 � b�

x2 y2 2 z2

� � �x2 y2 � b � 2 z2

Thus

v�� v �� 2 � � x � bx�x2 y2

�i � y � by�

x2 y2

�j zk � � �

x2 y2 � b � 2 z2 � � 1

�� M

NP

��

��

� x � bx�x2 y2 � � 1 � 2 � � � �

x2 y2 � b � 2 z2 � � 1

� y � by�x2 y2 � � 1 � 2 � � � �

x2 y2 � b � 2 z2 � � 1

z � �x2 y2 � 1 � 2 � b � 2 z2 � � 1

� ��


jbquig-UCD


Since divv�� v �� 2 � ∂M

∂x ∂N

∂y ∂P

∂zwe next must hack out these three partial derivatives;

∂M∂x

� ∂∂x� x � bx

�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1

� � 1 � b�x2 y2 � � 1 � 2 bx2 �

x2 y2 � � 3 � 2 � � � �x2 y2 � 1 � 2 � b � 2 z2 � � 1

� x � bx�x2 y2 � � 1 � 2 � � � 1 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 2

� �2 � � �

x2 y2 � 1 � 2 � b � x�x2 y2 � � 1 � 2

� � 1 � b�x2 y2 � � 1 � 2 bx2 �

x2 y2 � � 3 � 2 � � � �x2 y2 � 1 � 2 � b � 2 z2 � � 1

� 2x2 � � �x2 y2 � 1 � 2 � b � 2 z2 � � 2 � �

x2 y2 � 1 � 2 � b � 2 �x2 y2 � � 1

We obtain a similar expression for∂N∂y

and adding

∂M∂x

∂N∂y

� � 2 � 2b�x2 y2 � � 1 � 2 b

�x2 y2 � �

x2 y2 � � 3 � 2 � � � �x2 y2 � 1 � 2 � b � 2 z2 � � 1

� 2�x2 y2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 2 � �x2 y2 � 1 � 2 � b � 2 �

x2 y2 � � 1

� � 2 � b�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1

� 2 � � �x2 y2 � 1 � 2 � b � 2 z2 � � 2 � �

x2 y2 � 1 � 2 � b � 2

Next we work out

∂P∂z

� ∂∂z� z � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1 ��

x2 y2 � 1 � 2 � b � 2 z2 � � 1 � � 1 � 2z2 � �x2 y2 � 1 � 2 � b � 2 z2 � � 2

Thus

divv�� v �� 2 � ∂M

∂x ∂N

∂y ∂P

∂z

� � 2 � b�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1

� 2 � � �x2 y2 � 1 � 2 � b � 2 z2 � � 2 � �

x2 y2 � 1 � 2 � b � 2

� �x2 y2 � 1 � 2 � b � 2 z2 � � 1 � 2z2 � �

x2 y2 � 1 � 2 � b � 2 z2 � � 2

� � 3 � b�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1 � 2 � � �x2 y2 � 1 � 2 � b � 2 z2 � � 1

� � 1 � b�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1

To sum up we have proven

divv � � 1 � b�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1

(b)

c�


6.6. ANSWER 257

�� T

�div

v�� v �� 2 d�x � y � z � apply COV th. T.P.Tform. h

� �� ε � a � � � � π � π � � � � π � π � div

v�� v �� 2 � � det Dh � d �r� θ � φ �

In this latter formula we substitute the Jacobian � det Dh � � r�b r cosφ � and also the above formula for

divv in terms of toroidal polar coordinates r � θ � φ.

�� T

�div

v�� v �� 2 d�x � y � z �

� �� ε � a � � � � π � π � � � � π � π � � 1 � b

�x2 y2 � � 1 � 2 � � � �

x2 y2 � 1 � 2 � b � 2 z2 � � 1 � r�b r cosφ � d

�r� θ � φ �

� �� ε � a � � � � π � π � � � � π � π �

�1 � b � �

b r cosφ � � � a2 � � 1 � � r �b r cosφ � d

�r� θ � φ �

� �� ε � a � � � � π � π � � � � π � π � a � 2r2 cosφd

�r� θ � φ � apply Fubini’s theorem

� a � 2 � π

� π� a

ε� π

� πr2 cosφdθ � dr dφ

� 2πa � 2 � π

� π� a

εr2 cosφdr dφ

� 23

πa � 2 �a3 � ε3 � � π

� πcosφdφ

� 23

πa � 2 �a3 � ε3 � �

0 �

� 0

Thus �� T

�div

v�� v �� 2 d�x � y � z � � 0 and for a second time we have proven

�� T


�x � y � z � � lim

ε � 0��

T�div

v�� v �� 2 d�x � y � z � � lim

ε � 00 � 0

6.6 answer

6.6.1

We are asked to compute divv where the vector field is closely related to the inverse square law field but hasa different denominator.

divv � div�xi yj zk � �

x2 y2 z2 � � 1

� ∂∂x

� x �x2 y2 z2 � � 1 � ∂

∂y� y �

x2 y2 z2 � � 1 � ∂∂z

� z �x2 y2 z2 � � 1 �

Lets work out the first of these three partial derivatives

∂∂x

� x �x2 y2 z2 � � 1 � � �

x2 y2 z2 � � 1 x� � 1 � �

x2 y2 z2 � � 2 �2x �

� �x2 y2 z2 � � 1 � 2x2 �

x2 y2 z2 � � 2

� � x2 y2 z2�x2 y2 z2 � 2

Similarly

∂∂y

� y �x2 y2 z2 � � 1 � � x2 � y2 z2

�x2 y2 z2 � 2 � ∂

∂z� z �

x2 y2 z2 � � 1 � � x2 y2 � z2�x2 y2 z2 � 2


jbquig-UCD


Going back to our main computation, add these three together to get

divv � 1x2 y2 z2

6.6.2

We will first compute I as the flux of the vector field v across the ellipsoidal surface E. Such a parameteriza-tion can be obtained from the parameterization of the solid ellipsoid, (see S3-Q2 for full details) or again byaltering appropriately the parameterization of the spherical surface. Here it is

h :� � π � π � � �

0 � π � �

�� rθφ

��

�� xyz

�� asinφcosθ

bsinφsinθccosφ

�� E

Applying the EVALUATION by PARAMETERIZATION FORMULA of flux of a vector field across a sur-face we obtain

I& � � �E � v � n � dσ

� � �E


� � �E


� � � �� π � π � � � 0 � π � ��

M N P∂x∂θ

∂y∂θ

∂z∂θ

∂x∂φ

∂y∂φ

∂z∂φ

�� d �θ � φ �

� � � �� π � π � � � 0 � π � �� x �

�x2 y2 z2 � y � �

x2 y2 z2 � z � �x2 y2 z2 �� asinφsinθ bsinφcosθ 0


�� d �θ � φ �

� � � �� π � π � � � 0 � π � �

x2 y2 z2 � � 1�� x y z� asinφsinθ bsinφcosθ 0


�� d �θ � φ �

� � � �� π � π � � � 0 � π � �

x2 y2 z2 � � 1�� asinφcosθ bsinφsinθ ccosφ� asinφsinθ bsinφcosθ 0


�� d �θ � φ �

� � � �� π � π � � � 0 � π � abc

�x2 y2 z2 � � 1


�� d �θ � φ �

� � � �� π � π � � � 0 � π � abc

�x2 y2 z2 � � 1 sinφd

�θ � φ �

we have seen this det be f ore and know itis sinφ

� � � �� π � π � � � 0 � π � � abcsinφ

x2 y2 z2 � d�θ � φ �

� � � �� π � π � � � 0 � π � � abcsinφ

a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � d�θ � φ �

To sum up we have so far proven that

I � � �E

� v � n � dσ � �� π � π � � � 0 � π � � abcsinφ

a2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ � d�θ � φ �

Next we apply the DIVERGENCE theorem of GAUSS to reduce the surface integral I to a volume integralover the solid ellipsoidal ball E whose boundary surface is of course the original E

I � ��E

� v � n � dσ � ��E


c�


6.7. ANSWER 259

� � ��E

divvd�x � y � z � � � � �

E

�∂M∂x

∂N∂y

∂P∂z � d

�x � y � z �

� � ��E� 1x2 y2 z2 � d

�x � y � z � from (i) above

� � �x2 � a2 � y2 � b2 � 1 � c � 1 � x2 � a2 � y2 � b2

� c � 1 � x2 � a2 � y2 � b2

� 1x2 y2 z2 � dz d

�x � y � � by Fubini’s theorem

� � �F � c � 1 � x2 � a2 � y2 � b2

� c � 1 � x2 � a2 � y2 � b2

� 1x2 y2 z2 � dz d

�x � y �

� � �F 1�

x2 y2arctan

z�x2 y2 �� c � 1 � x2 � a2 � y2 � b2

� c � 1 � x2 � a2 � y2 � b2d

�x � y �

� � �F

21�

x2 y2arctan c

�1 � x2 � a2 � y2 � b2

�x2 y2 d

�x � y �

Thus we have two expression for the original flux integral I, equating them we obtain as required

� 2π

0� π

0

abcsinφa2 sin2 φcos2 θ b2 sin2 φsin2 θ c2 cos2 φ

dθdφ

� � �F

2�x2 y2

arctan c�

1 � x2 � a2 � y2 � b2�

x2 y2 d�x � y �

6.7 answer

MISSING


jbquig-UCD

third year engineering math3602 mathematics (integral

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