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Sreenivasa Institute of Technology and Management Studies
Engineering Mathematics-III
II B.TECH. I SEMESTER
Course Code: 18SAH211 Regulation:R18
UNIT-I
NUMERICAL INTEGRATION
Motivation:
Most such integrals cannot be evaluated explicitly. Many others it is often faster to integrate them numerically rather than evaluating them exactly using a complicated antiderivative of f(x)
Example:
The solution of this integral equation with Matlab is 1/2*2^(1/2)*pi^(1/2)*FresnelS(2^(1/2)/pi^(1/2)*x)
we cannot find this solution analytically by techniques in calculus.
2
dxx )sin( 2
3
3
4
5
Graphical Representation of
Integral
Integral = area
under the curve
Use of a grid to
approximate an integral
6
Use of strips to
approximate an
integral
7
Numerical Integration
Net force
against a
skyscraper
Cross-sectional area
and volume flowrate
in a river
Survey of land
area of an
irregular lot
Methods of Numerical Integration
Trapezoidal Rule’s
1/3 Simpson’s method
3/8 Simpson’s method
Applied in two dimensional domain
8
9
10
11
12
Simpson’s 1/3- Rule is given by
Note: While applying the Simpson’s 1/3 rule, the number
of sub-intervals (n) should be taken as multiple of 2.
13
13
Simpson’s 3/8- Rule is given by
Note: While applying the Simpson’s 3/8 rule, the number
of sub-intervals (n) should be taken as multiple of 3.
14
15
16
17
18
19
19
Numerical solution of
ordinary differential
equations
20
Introduction
21
Taylor’s Series Method
22
23
24
Picard’s Method
25
26
27
28
29
Euler’s Method
30
31
32
Runge-Kutta Formula
33
UNIT-II
Multiple Integrals
Double Integration
34
Evaluation of Double Integration
35
36
37
38
39
40
Triple Integration
41
42
43
UNIT-III
Partial Differential Equations
44
45
46
47
48
49
UNIT-IV
Vector Differentiation
Elementary Vector Analysis
Definition (Scalar and vector)
Vector is a directed quantity, one with both
magnitude and direction.
For instance acceleration, velocity, force
Scalar is a quantity that has magnitude but not
direction.
For instance mass, volume, distance
50
We represent a vector as an arrow from the
origin O to a point A.
The length of the arrow is the magnitude of
the vector written as or .
O
A
or O
A
OAa
aOA
51
Basic Vector System
Unit vectors , ,
•Perpendicular to each other
•In the positive directions
of the axes
•have magnitude (length) 1
52
Magnitude of vectors
Let P = (x, y, z). Vector is defined by
with magnitude (length)
OP = = + +p x i y j z k
= [ ]x, y, z
OP = p
OP = = + +p x y z2 2 2
53
Calculation of Vectors
Vector Equation
Two vectors are equal if and only if the
corresponding components are equals
332211
321321
, ,
Then
. and Let
babababa
kbjbibbkajaiaa
====
++=++=
54
Addition and Subtraction of Vectors
Multiplication of Vectors by Scalars
kbajbaibaba )()()( 332211 ++=
kbjbibb )()()(
thenscalar, a is If
321
++=
55
55
Example 1
Given 5 3 and 4 3 2 . Findp i j k q i j k= + - = - +
a p q) +
) b p q-
) 2 10 d q p-
c p) Magnitude of vector
56
Vector Products
1 2 3 1 2 3~ ~ ~ ~ ~ ~~ ~
If and , a a i a j a k b b i b j b k= + + = + +
1 1 2 2 3 3~ ~a b a b a b a b = + +
1) Scalar Product (Dot product)
2) Vector Product (Cross product)
~ ~~
1 2 3~ ~
1 2 3
2 3 3 2 1 3 3 1 1 2 2 1~ ~~
i j k
a b a a a
b b b
a b a b i a b a b j a b a b k
=
= - - - + -
~~~~ and between angle theis ,cos||||.or bababa =
57
The area of parallelogram
The volume of tetrahedrone
The volume of parallelepiped
a
ba bxA =
a b
c
321
321
321
6
1
ccc
bbb
aaa
=6
1=V a . b cx
a b
c321
321
321
ccc
bbb
aaa
==V a . b cx
58
Differentiation of Two Vectors
If both and are vectors, then
)(~
uA )(~
uB
58
~
~~
~~~
~
~~
~~~
~~
~~
~
~
)()
..).()
)()
)()
Bdu
Ad
du
BdABA
du
dd
Bdu
Ad
du
BdABA
du
dc
du
Bd
du
AdBA
du
db
du
AdcAc
du
da
+=
+=
+=+
=
Del Operator Or Nabla (Symbol )
• Operator is called vector differential operator,
defined as
59
.~~~
+
+
= k
zj
yi
x
60
Grad (Gradient of Scalar Functions)
• If x,y,z is a scalar function of three
variables and is differentiable, the gradient
of is defined as
. grad~~~k
zj
yi
x
+
+
==
function vector a is *
functionscalar a is *
61
Example 1
zxyyzx
xyzzx
zyxyz
zxyyzx
zxyyzx
222
232
223
2232
2232
23z
2y
2x
hence ,Given
(1,3,2).Pat grad determine , If
+=
+=
+=
+=
=+=
Solution
61
62
.723284
.))2()3)(1(2)2)(3()1(3(
))2)(3)(1(2)2()1(())2()3()2)(3)(1(2(
have we(1,3,2),PAt
.)23(
)2()2(
zyx
Therefore,
~~~
~
222
~
232
~
223
~
222
~
232
~
223
~~~
kji
k
ji
kzxyyzx
jxyzzxizyxyz
kji
++=
++
+++=
=
++
+++=
+
+
=
63
Exercise 2
(1,2,3).Ppoint at grad determine
, If 323
=
+=
zxyyzx
63
Solution
.110111126(1,2,3),PAt
Grad
z
y
x
then,Given
~~~
323
kji
zxyyzx
++==
==
=
=
=
+=
63
64
Grad Properties
If A and B are two scalars, then
)()()()2
)()1
ABBAAB
BABA
+=
+=+
64
65
Divergence of a Vector
..
).(
.
as defined
is of divergence the, If
~~
~~~~~~
~~
~~~~~
z
a
y
a
x
aAAdiv
kajaiakz
jy
ix
AAdiv
AkajaiaA
zyx
zyx
zyx
+
+
==
++
+
+
=
=
++=
65
66
Example 1
.13
)3)(2(2)3)(1()2)(1(2
(1,2,3),point At
.22
.
(1,2,3).point at determine
, If
~
~~
~
~
2
~~
2
~
=
+-=
+-=
+
+
==
+-=
Adiv
yzxzxy
z
a
y
a
x
aAAdiv
Adiv
kyzjxyziyxA
zyx
Answer
67
Exercise 2
.114
(3,2,1),point At
.
(3,2,1).point at determine
, If
~
~~
~
~
3
~
2
~
23
~
=
=
=
+
+
==
-+=
Adiv
z
a
y
a
x
aAAdiv
Adiv
kyzjzxyiyxA
zyxAnswer
68
Remarks
.called is vector ,0 If
function.scalar a is but function, vector a is
~~
~~
vectorsolenoidAAdiv
AdivA
=
69
Curl of a Vector
.
)(
by defined is of curl the,If
~~~
~~
~~~~~~
~~
~~~~~
zyx
zyx
zyx
aaa
zyx
kji
AAcurl
kajaiakz
jy
ix
AAcurl
AkajaiaA
==
++
+
+
=
=
++=
70
Example 1
.)2,3,1(at determine
,)()( If
~
~
2
~
22
~
224
~
-
-++-=
Acurl
kyzxjyxizxyA
Solution
.)42()22(
)()(
)()(
)()(
~
3
~
2
~
2
~
22422
~
2242
~
222
222224
~~~
~~
kyxjzxxyzizx
kzxyy
yxx
jzxyz
yzxx
iyxz
yzxy
yzxyxzxy
zyx
kji
AAcurl
-++---=
-
-+
+
-
--
-
+
--
=
-+-
==
70
71
Exercise 2
.10682
))3(4)1(2(
))2()1(2)2)(3)(1(2()2()1(
(1,3,-2),At
~~~
~
3
~
2
~
2
~
kji
k
jiAcurl
--=
-+
-+-----=
.)3,2,1(point at determine
,)()( If
~
~
22
~
22
~
223
~
Acurl
kyzxjzxizyxyA -++-=
72
Answer
.261215 (1,2,3),At
.)232(
)22()2(
~~~~
~
22
~
22
~
22
~
kjiAcurl
kyzxyx
jzyxyzizzxAcurl
++-=
+-+
+----=
Remark
function. vector a also is
andfunction vector a is
~
~
Acurl
A
73
UNIT-V
Vector Integration
Polar Coordinate for Plane (r, θ)
ddrrdS
ry
rx
=
=
=
sin
cos
x
ds
y
d
74
Polar Coordinate for Cylinder (, , z)
dzdddV
dzddS
zz
y
x
=
=
=
=
=
sin
cos
x
y
z
dv
z
ds
75
Polar Coordinate for Sphere (r, ,
dddrrdV
ddrdS
rz
ry
rx
sin
sin
cos
sinsin
cossin
2
2
=
=
=
=
=
y
x
r
z
76
Example 1 (Volume Integral)
.9 and
4,0by bounded space a is and
22 where Calculate
22
~~~~~
=+
==
++=
yx
zzV
kyjziFdVFV
x
z
y
4 -
3 3
77
Solution
Since it is about a cylinder, it is easier if we use
cylindrical polar coordinates, where
.40,20,30 where
,,sin,cos
====
z
dzdddVzzyx
78
Line Integral
Ordinary integral f (x) dx, we integrate along
the x-axis. But for line integral, the integration
is along a curve.
f (s) ds = f (x, y, z) ds
A
O
B
~~rdr+
~r
79
Scalar Field, V Integral
If there exists a scalar field V along a curve C,
then the line integral of V along C is defined by
.where~~~~
~
kdzjdyidxrd
rdVc
++=
80
Example 1
(3,2,1).B to(0,0,0)A from
along findthen
,,2,3
by given is curve a and z If
~
32
2
==
===
=
CrdV
uzuyux
CxyV
c
81
Solution
.1
,1,22,33 (3,2,1),BAt
.0
,0,02,03 (0,0,0),AAt
.343
And,
.12)()2)(3(
zGiven
32
32
~
2
~~
~~~~
8322
2
=
====
=
====
++=
++=
==
=
u
uuu
u
uuu
kduujduuidu
kdzjdyidxrd
uuuu
xyV
82
.11
36
5
244
11
36
5
244
364836
)343)(12(
~~~
~
1
0
11
~
1
0
10
~
1
0
9
1
0 ~
101
0~
91
0~
8
1
0 ~
2
~~
8
~
kji
kujuiu
kduujduuiduu
kduujuduiduurdVu
u
B
A
++=
+
+=
++=
++=
=
=
83
83
Exercise 2
.11
768144
5
384
(4,3,2).B to(0,0,0)A from
curve thealong calculate
,2,3,4
by given is curve theand If
~~~~
~
23
22
kjirdV
C rdV
uzuyux
CyzxV
B
A
c
++=
==
===
=
Answer
84
Vector Field, Integral
Let a vector field
and
The scalar product is written as
.
)).((. ~~~~~~~~
dzFdyFdxF
kdzjdyidxkFjFiFrdF
zyx
zyx
++=
++++=
~F
~~~~kFjFiFF zyx ++=
.~~~~kdzjdyidxrd ++=
~~. rdF
85
. .
bygiven is Bpoint another A topoint a from
curve thealong of integral line then the
, curve thealong is field vector a If
~~
~
~
++=c
zc
yc
xc
dzFdyFdxFrdF
CF
CF
86
Example 1
.y2y
if,2,4curve thealong
(4,2,1)B to(0,0,0)A from .Calculate
~~~
2
~
32
~~
kzjxzixF
tztytx
rdFc
-+=
===
==
87
Solution
.344
And
.4432
)()2(2)()4()2()4(
2yGiven
~
2
~~
~~~~
~
5
~
4
~
4
~
32
~
3
~
22
~~~
2
~
kdttjdttidt
kdzjdyidxrd
ktjtit
kttjttitt
kyzjxzixF
++=
++=
-+=
-+=
-+=
88
.)1216128(
1216128
)3)(4()4)(4()4)(32(
)344)(4432(.
Then
754
754
2544
~
2
~~~
5
~
4
~
4
~~
dtttt
dttdttdtt
dttttdttdtt
kdttjdttidtktjtitrdF
-+=
-+=
-++=
++-+=
.1
,1,22,44 (4,2,1),Bat and,
.0
,0,02,04 (0,0,0),AAt
32
32
=
====
=
====
t
ttt
t
ttt
89
.30
2326
2
3
3
8
5
128
2
3
3
8
5
128
)1216128(.
1
0
865
1
0
754
~~
=
-+=
-+=
-+= =
=
ttt
dttttrdFt
t
B
A
90
Surface Integral
Scalar Field, V Integral
If scalar field V exists on surface S, surface
integral V of S is defined by
=S S
dSnVSVd~~
where
S
Sn
=
~
91
Example 1
Scalar field V = x y z defeated on the surface
S : x2 + y2 = 4 between z = 0 and z = 3 in the
first octant.
Evaluate SSVd~
Solution
Given S : x2 + y2 = 4 , so grad S is
~~~~~22 jyixk
z
Sj
y
Si
x
SS +=
+
+
=
92
Also,
4422)2()2( 2222 ==+=+= yxyxS
Therefore,
)(2
1
4
22
~~
~~
~jyix
jyix
S
Sn +=
+
=
=
Then,
+
=
S SdSjyixxyzdSnV )(
2
1
~~~
+= dSjzxyiyzx )(2
1
~
2
~
2
93
Surface S : x2 + y2 = 4 is bounded by z = 0 and z = 3
that is a cylinder with z-axis as a cylinder axes and
radius,
So, we will use polar coordinate of cylinder to find
the surface integral.
.24 ==
x
z
y
2
2
3
O
94
Polar Coordinate for Cylinder
cos 2cos
sin 2sin
ρ
x
y
z z
dS d dz
= =
= =
=
=
where 2
0
30 z(1st octant) and
95
Using polar coordinate of cylinder,
cossin8)()sin2)(cos2(
sincos8)sin2()cos2(
222
222
zzzxy
zzyzx
==
==
From
=+=S
SS
SVddSjzxyiyzxdSnV~~
2
~
2
~)(
2
1
96
96
= =+=
Sz
dzdjzizSVd 2
0
3
0 ~
2
~
2
~)2)(cossin8sincos8(
2
1
3
2 2 2 22
0 ~ ~ 0
2 22
0 ~ ~
1 18 cos sin sin cos
2 2
9 98 cos sin sin cos
2 2
z i z j d
i j d
= +
= +
2 22
0 ~ ~
3 3 2
~ ~0
~ ~
98 cos sin sin cos
2
cos sin sin cos36
3( sin ) 3(cos )
12( )
i j d
i j
i j
= +
= +
-
= +
Therefore,
97
Green’s Theorem
If c is a closed curve in counter-clockwise on
plane-xy, and given two functions P(x, y) and
Q(x, y),
where S is the area of c.
+=
-
cSdyQdxPdydx
y
P
x
Q)(
98
Example 1
2 2
2 2
Prove Green's Theorem for
[( ) ( 2 ) ]
which has been evaluated by boundary that defined as
0, 0 4 in the first quarter.
cx y dx x y dy
x y and x y
+ + +
= = + =
y
2
x 2
C3
C2
C1 O
x2 + y2 = 22
Solution
99
1 1
2 2
2 2
1 2 3
1
2 2
22
0
2
3
0
Given [( ) ( 2 ) ] where
and 2 . We defined curve
as , .
i) For : 0, 0 0 2
( ) ( ) ( 2 )
1 8.
3 3
c
c c
x y dx x y dy
P x y Q x y c
c c and c
c y dy and x
Pdx Qdy x y dx x y dy
x dx
x
+ + +
= + = +
= =
+ = + + +
=
= =
100
2 2
2ii) For : 4 , in the first quarter from (2,0) to (0,2).
This curve actually a part of a circle.
Therefore, it's more easier if we integrate by using polar
coordinate of plane,
2cos , 2sin , 0
c x y
x y
+ =
= =2
2sin , 2cos .dx d dy d
= - =
101
.448
sin42sin2cos8
)cossin82cos22sin8(
)cossin8cos4sin8(
)]cos2))(sin2(2cos2((
)sin2)()sin2()cos2(([
)2()()(
2
2
2
2
22
0
2
0
0
2
22
0
22
-=++-=
+++=
+++-=
++-=
++
-+=
+++=+
d
d
d
d
dyyxdxyxQdyPdxcc
102
3 3
3
2 2
0
2
02
2
iii) : 0, 0, 0 2
( ) ( ) ( 2 )
2
4.
8 16( ) ( 4) 4 .
3 3
c c
c
For c x dx y
Pdx Qdy x y dx x y dy
y dy
y
Pdx Qdy
= =
+ = + + +
=
=
= -
+ = + - - = -
103
b) Now, we evaluate
where 1 2 .
Again,because this is a part of the circle,
we shall integrate by using polar coordinate of plane,
cos , sin
where
S
Q Pdxdy
x y
Q Pand y
x y
x r y r
-
= =
= =
0 r 2, 0 .2
and dxdy dS r dr d
= =
104
.3
16
cos3
162
sin3
162
sin3
2
2
1
)sin21(
)21(
2
2
2
2
0
0
2
00
32
0
2
0
-=
+=
-=
-=
-=
-=
-
=
=
= =
d
drr
ddrrr
dydxydydxy
P
x
Q
r
SS
105
Therefore,
( )
16.
3
Green's Theorem has been proved.
C S
Q PPdx Qdy dx dy
x y
LHS RHS
+ = -
= -
=
106
Divergence Theorem (Gauss’ Theorem)
If S is a closed surface including region V
in vector field
..~~~ =
SVSdFdVFdiv
~F
~
yx zff f
div Fx y z
= + +
107
Example 1
2
~ ~ ~~
2 2
Prove Gauss' Theorem for vector field,
2 in the region bounded by
planes 0, 4, 0, 0 4
in the first octant.
F x i j z k
z z x y and x y
= + +
= = = = + =
108
Solution
x
z
y
2
2
4
O
S3
S4
S2
S1
S5
109
1
2
3
4
2 2
5
~
~ ~
For this problem, the region of integration is bounded
by 5 planes :
: 0
: 4
: 0
: 0
: 4
To prove Gauss' Theorem, we evaluate both
. ,
The answer should be the same.
V
S
S z
S z
S y
S x
S x y
div F dV
and F d S
=
=
=
=
+ =
110
2
~ ~ ~ ~~
2
~
~
1) We evaluate . Given 2 .
So,
( ) (2) ( )
1 2 .
Also, (1 2 ) .
The region is a part of the cylinder. So, we integrate by using
polar c
V
V V
div F dV F x i j z k
div F x zx y z
z
div F dV z dV
= + +
= + +
= +
= +
oordinate of cylinder ,
; sin ;
where 0 2, 0 , 0 4.2
x = cos y z z
dV d d dz
z
= =
=
111
2
2
2
2
2
2
2 4
0 0 0
22 4
00 0
2
0 0
2 2
00
0
0
~
Therefore,
(1 2 ) (1 2 )
[ ]
(20 )
[10 ]
(40)
40
20 .
20 .
V z
V
z dV z dzd d
z z d d
d d
d
d
div F dV
= = =
= =
= =
=
=
+ = +
= +
=
=
=
=
=
=
112
1
~ ~~ ~
1~ ~
~ ~ ~~
~ ~ ~ ~~
~ ~
2) Now, we evaluate . . .
i) : 0, ,
2 0
. ( 2 ).( ) 0
. 0.
S S
S
F d S F n dS
S z n k dS rdrd
F x i j k
F n x i j k
F n dS
=
= = - =
= + +
= + - =
=
113
2
2
2~ ~
2
~ ~ ~ ~ ~~ ~
~ ~ ~ ~ ~~
2
2
0 0~ ~
ii) : 4, ,
2 (4) 2 16
. ( 2 16 ).( ) 16.
Therefore for , 0 r 2, 02
. 16
16 .
S r
S z n k dS rdrd
F x i j k x i j k
F n x i j k k
S
F n dS rdrd
= =
= = =
= + + = + +
= + + =
=
=
=
114
3
3~ ~
2
~ ~ ~~
2
~ ~ ~ ~~ ~
3
2 4
0 0~ ~
iii) : 0, ,
2
. ( 2 ).( )
2.
Therefore for S , 0 2, 0 4
. ( 2)
16.
S x z
S y n j dS dxdz
F x i j z k
F n x i j z k j
x z
F n dS dzdx= =
= = - =
= + +
= + + -
= -
= -
=
= -
115
4
4~ ~
2 2
~ ~ ~ ~~ ~
2
~ ~ ~ ~~
~ ~
iv) : 0, ,
0 2 2
. (2 ).( ) 0.
. 0.S
S x n i dS dydz
F i j z k j z k
F n j z k i
F n dS
= = - =
= + + = +
= + - =
=
116
2 2
5
5 5~ ~
~5 ~
~5
~ ~
5
v) : 4,
2 2 4
2 2
4
1( ).
2
By using polar coordinate of cylinder :
cos , sin ,
where for :
2, 0 , 0 4, 22
S x y dS d dz
S x i y j and S
x i y jS
nS
x i y j
x y z z
S
z dS d dz
+ = =
= + =
+
= =
= +
= = =
= =
117
.416
)2)(sin)(cos2(.
).sin(cos2
2.kerana ;sin2cos2
)sin()cos(2
1
2
1
2
1
2
1).2(.
5
2
0
4
0
2
~~
2
2
2
2
~~~
2
~~~~
+=
=
+=
+=
=+=
+=
+=
+++=
= =
S zdzddSnF
yx
jyixkzjixnF
118
1 2 3 4 5~ ~ ~ ~ ~ ~~ ~ ~ ~ ~ ~
~ ~
Finally,
. . . . . .
0 16 16 0 16 4
20 .
. 20 .
Gauss' Theorem has been proved.
S S S S S S
S
F d S F d S F d S F d S F d S F d S
F d S
LHS RHS
= + + + +
= + - + + +
=
=
=
119
Stokes’ Theorem
If is a vector field on an open surface S and
boundary of surface S is a closed curve c,
therefore
=S c
rdFSdFcurl~~~~
~F
~ ~~
~ ~
x y z
i j k
curl F Fx y z
f f f
= =
120
Example 1
Surface S is the combination of
2 2
~ ~ ~~
i) part of the cylinder 9 0
and 4 0.
ii) half of the circle with radius 3 at 4, and
iii) 0
, prove Stokes' Theorem
for this case.
a x y between z
z for y
a z
plane y
If F z i xy j xz k
+ = =
=
=
=
= + +
121
Solution
2 2
1
2
3
We can divide surface S as
S : x y 9 0 z 4 y 0
S : z 4, half of the circle with radius 3
S : y 0
for and+ =
=
=
z
y x
3
4
O
S3
C2
S2
C1
S1
3
122
We can also mark the pieces of curve C as
C1 : Perimeter of a half circle with radius 3.
C2 : Straight line from (-3,0,0) to (3,0,0).
Let say, we choose to evaluate first.
Given
~ ~Scurl F d S
~~~~kxzjxyizF ++=
123
So,
~~
~
~~
~~~
~
)1(
)()(
)()()()(
kyjz
kzy
xyx
jxzx
zz
ixyz
xzy
xzxyzzyx
kji
Fcurl
+-=
-
+
-
+
-
=
=
124
By integrating each part of the surface,
2 2
1
1~ ~
2 2
1
2 2
( ) : 9,
2 2
(2 ) (2 )
2 6
i For surface S x y
S x i y j
and S x y
x y
+ =
= +
= +
= + =
125
)(3
1
6
22
~~
~~
1
1
~jyix
jyix
S
Sn +=
+
=
=
and
).1(3
1
3
1
3
1)1(
~~~~~~
zy
jyixkyjznFcurl
-=
+
+-=
Then ,
126
By using polar coordinate of cylinder ( because
is a part of the cylinder), 9: 22
1 =+ yxS
cos , sin ,
3, 0 0 4.
x y z z
dS d dz
where
dan z
= = =
=
=
127
Therefore,
~ ~
1(1 )
3
1sin 1
3
sin (1 ) ; 3
curl F n y z
z
z because
= -
= -
= - =
Also, dzddS 3=
128
1 1~ ~~ ~
4
0 0
4
00
4
0
3 sin (1 )
3 (1 ) cos
3 (1 )(1 ( 1))
24
S S
z
curl F d S curl F n dS
z d dz
z dz
z dz
= =
=
= -
= - -
= - - -
= -
129
(ii) For surface , normal vector unit to the
surface is
By using polar coordinate of plane ,
4:2 =zS
.~~kn =
ddrrdSdanzry === 4,sin
0 r 3 and 0 .where
130
2 2
~ ~ ~ ~~
~ ~~ ~
3
0 0
32
0 0
(1 )
sin
( sin )( )
sin
18
S S
r
r
curl F n z j y k k
y r
curl F d S curl F n dS
r rdrd
r d dr
= =
= =
= - +
= =
=
=
=
=
131
(iii) For surface S3 : y = 0, normal vector unit
to the surface is
dS = dxdz
The integration limits :
.~~jn -=
3 3 0 4x and z-
So,
1
)())1((~~~~~
-=
-+-=
z
jkyjznFcurl
132
3 3
1 2 3
~ ~~ ~
3 4
3 0
~ ~ ~ ~~ ~ ~ ~
Then,
. .
( 1)
24.
. . . .
24 18 24
18.
S S
x z
S S S S
curl F d S curl F n dS
z dzdx
curl F d S curl F d S curl F d S curl F d S
=- =
=
= -
=
=
= + +
= - + +
=
133
~ ~
1
1
Now, we evaluate . for each pieces of the curve C.
i) is a half of the circle.
Therefore, integration for will be more easier if we use
polar coordinate for plane with radius
CF d r
C
C
3, that is
3cos , 3sin dan z 0
where 0 .
r
x y
=
= = =
134
~ ~ ~~
~
~
~ ~~
~ ~
(3cos )(3sin )
9sin cos
and
3sin 3cos .
F z i xy j xz k
j
j
dr dx i dy j dz k
d i d j
= + +
=
=
= + +
= - +
135
1
2
~ ~
2
0~ ~
3
0
From here,
. 27sin cos .
. 27sin cos
9cos
18.
C
F d r d
F d r d
=
=
= -
=
136
2
2
~ ~ ~~
~
~ ~
ii) Curve is a straight line defined as
, 0 z 0, where 3 3.
Therefore,
0.
. 0.C
C
x t y and t
F z i xy j xz k
F d r
= = = -
= + +
=
=
137
1 2~ ~ ~ ~ ~ ~
~ ~ ~~
. . .
18 0
18.
We already show that
. .
Stokes' Theorem has been proved.
C C C
S C
F d r F d r F d r
curl F d S F d r
= +
= +
=
=