VC.07

Triple Integrals

Example 1: Triple Integrals to Compute Volume

I R

Recall that in previous chapters we could find the length of an interval Iby computing dx or the area of a region R by computing dA.

R

It follows that we can compute the volume of a 3-dimensional region Rby calculating 1dV.For this cube, the calculation is not exciting:

1 x 1,0 y 2,1 z 3

3 2 1

1 0 11dx dy dz

3 2x 1

x 11 0

x dy dz

3 y 2

y 01

2y dz

z 3

z 14z

8

Example 2: Triple Integrals to Compute Volume

We can spice things up a bit. We can findthe volume of the region abovethexy-plane, the xz-plane, the yz-plane, andbelow the plane given by x y 4z 8:

If x y 4z 8, thenz . So z-t8 x y4

8 x

op

is andz-bottomy4 is .0

88 x

x y

0

8

00

4d1 yd dxz

If x y 4z 8, thenwhen z 0, y . S

8 x8 xo y-top is and y-bottom s .0 i

y

x

Finally, integrate x from x = to x 0 = 8.

Example 2: Triple Integrals to Compute Volume

We can spice things up a bit. We can findthe volume of the region abovethexy-plane, the xz-plane, the yz-plane, andbelow the plane given by x y 4z 8:

88 x

x y

0

8

00

4d1 yd dxz

0

8

0

8 x

dy dx1 (8 x y)4

y 8 x

y

8

0 0

2y1 8y xy4 2 dx

28

0

1 (8 x)8(8 x) x(8 x)4 2 dx

643

pyramid base1V A h31 32 23643

g g

Example 3: Changing the Order of Integration

RRepeat Example 2 by computing dy dx dz whereR istheregion

abovethexy-plane, the xz-plane, the yz-plane, andbelow the plane given by x y 4z 8

If x y 4z 8, theny . So y-t8 x 4z8 x 4

op is and y-bottom is .z 0

8 8 x 4z

0

4z2

0 0dzxdy1 d

If x y 4z 8, thenwhen y 0, x . S

8 4z8 4z 0o x-top is and x-bottom is .

xFinally, integrate z from z = to z 0 = 2.

z

643

Example 4: 3D-Change of VariablesxyzCompute the volume of the parallelepiped describedby the region R

between the planes z 3x andz 3x 2,y x and y x 4,and y 2xand y 2x 3.

u z 3x,andleturunfrom0to 2v y x,andlet v runfrom0to 4w y 2x,andlet wrunfrom0to3

xyz uvw

xyzR R

1dx dy dz V (u,v,w) dudv dw

First, pick a clever change of variables:

Our integral is much more manageable now:

3 4 2

xyz0 0 0

V (u,v,w) dudv dw

xyz

yx zu u u

y yxV (u,v,w) v v vyx z

w w w

The Volume Conversion Factor:

xyz uvw

xyzR R

1dx dy dz V (u,v,w) dudv dw

1 2 3

31 2x

31 2

yz

31 2

Let T(u,v,w) be a transformation from uvw-space to xyz-space.That is, T(u,v,w) T (u,v,w),T (u,v,w),T (u,v,w) (x(u,v,w),y(u,v,w),z(u,v,w)).

ThenV (u,v,

TT Tw w w

w) TT Tv

TT Tu u u

v v

xy xyzNote: Just likeA (u,v),weneed V (u,v,w) tobepositive.Hence, the absolutevaluebars in the formula above.

xyz

yx zu u u

yx zvOr V (u,v,w

yv v

xw

)

zw w

Example 4: 3D-Change of VariablesxyzCompute the volume of the parallelepiped describedby the region R

between the planes z 3x andz 3x 2,y x and y x 4,and y 2xand y 2x 3.

u z 3xv y xw y 2x

w vx 32v wy 3

z u w v

xyz

0 0 11 2V (u,v,w) 13 3

1 1 13 3

1 23 31 1 1

3 3

13

1Use .3

3 4 2

xyz0 0 0

V (u,v,w) dudv dw3 4 2

0 0 0

1 dudv dw3

8

Example 4: 3D-Change of VariablesxyzCompute the volume of the parallelepiped describedby the region R

between the planes z 3x andz 3x 2,y x and y x 4,and y 2xand y 2x 3.

parallelepiped

parallelepiped

Check :For a parallelepiped generated by three intersecting vectors X, Y and Z,V (X Y) Z

4 8Parallelepiped is generatedby vectors (1,1,3), , , 4 ,and(0,0,2).3 3

V (1,1,3)

g

4 8, , 4 (0,0,2) 83 3

g

No need to study this, it's just nice to verify this works...

Example 5: A Mathematica-Assisted Change of Variables

2

Find the volume of the "football" whose outer skin is described by the pa 0,0,4s) 4 s cos(t),rametric equation sin(t),0 for2 s

F(s2and0 t

t2

, ) (.

First,fill in the football: 20,0,4s) r 4 s cos(t),sin(t),0

2 sF(r

20 t 20

,s

r

,t) (

1

Change of variables:

2

2

x(r,s,t) r cos(t)y(r,s,t) r sin(t)z(r,

4

s

s

,t) 4s4 s

2 2 1

rst0 2 0

V (r,s,t) dr dsdt

Example 5: A Mathematica-Assisted Change of Variables

2Find the volume of the "football" whose outer skin is described by the param 0,0,4s) (4 s ) cos(t),sinetric equation F(s,t (t),0 for2 s 2and0 t

(2 .

)

xyz

yx zr r r

y yxV (r,s,t) s s syx z

t t t

2 2 1

rst0 2 0

V (r,s,t) dr dsdt

2 44(16r 8rs rs ) (Mathematica)

2048 (Mathematica)15

Example 5: A Mathematica-Assisted Change of Variables

2Find the volume of the "football" whose outer skin is described by the param 0,0,4s) (4 s ) cos(t),sinetric equation F(s,t (t),0 for2 s 2and0 t

(2 .

)

228

8

x4 dx4

204815

Check using solids of revolution:

Example 6: Beyond Volume Calculations

R

R

It is easy to see that dV computes the volume of a solid.But it's

harder to interpret f(x,y,z)dV since f(x,y,z) lives in 4 dimensions.

We need an example to keep referring back to to give us some

intuition.

1 x 1,0 y 2,1 z 3

32 4 g

cm

A cube of varying density has its density at each point (x,y,z) described by f(x,y,z) x y .Findthemassof the cube.

3 2 12 4

1 0 1x y dx dy dz

128 grams15

3D Integrals:

xyz uvw

xyzR R

f(x,y,z)dx dy dz f(x,y,z) V (u,v,w) dudv dw

xyz

yx zu

yx zw

V (u,v,u u

w) yx

w

v

w

zv v

xyzR

xyz

In my opinion, a good way to think about f(x,y,z)dx dy dz is

asacalculation of the mass of R where f(x,y,z) is the density of the solid at any given point (x,y,z).

Example 7: A 3D-Change of Variables with an Integrand

xyz

xyzR

Compute 9y dx dy dz whereR is the parallelepiped that is between

the planes z 3x andz 3x 2, y x and y x 4,and y 2x andy 2x 3.

0 u 20 v 40 w 3

3 4 2

xyz0 0 0

9y(u,v,w) V (u,v,w) dudv dw

u z 3xv y xw y 2x

w vx 32v wy 3

z u w v

xyz1V (u,v,w) 3

xyzR9y dx dy dz

All from Example4:

3 4 2

0 0 0

2v w 19 dudv dw3 3

3 4 2

0 0 02v w dudv dw 132