Self-Help Work Sheets C11: Triple Integration Problems for Fun and Practice

SHWS C11: TRIPLE INTEGRATION

29

Self-Help Work Sheets C11: Triple Integration

These problems are intended to give you more practice on some of the skills the chapter on Triple Integration has sought to develop. They do not cover everything so a careful review of the Chapter and your class notes is also in order.

Problems for Fun and Practice

1. For each of the following solids give a description in rectangular coordinates in the order speci?ed:

(a)

S in

is bounded above terms of range on

by z,

x2 y2 th9en+y,3t6he+n

z2 x1.6

=

1 and below by z

=

2.

Description:

(b)

S

is

bounded

above

by

x2 16

+

y2 36

+

z2 16

= 1 and below by

yz 6+4

= 1. Description:

in terms of range on z, then x, then y.

2. Each of the following iterated integrals cannot be easily done in the order given. Convince yourself that this is true and then convert each one to an equivalent iterated integral that can be done and evaluate it.

(a) !2!1!1 sinh "z2# dzdyd x

00y

!2 !4 !2

(b)

yzex3dxdydz

00z

3. Convert each of the following to an equivalent triple integ4al in cylindrical coordinates and evaluate.

!5 (a)

!25-x2!6 d zd yd x $

0 0 0 x2 + y2

(b)

!2

!4-x 2

x2+y2

!2

zdzdydx

$

00

0

x2 + y2

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4. Convert each of the following to an equivalent triple integral in spherical coordinates and evaluate.

(a)

!1

!1-x

2 1-!x

2-

y2

dzdydx

1 + x2 + y2 + z2

00

0

!3

!9-x

2 9-!x

2-

y2

(b)

xzdzdydx

00

0

5. Convert to cylindrical coordinates and evaluate the integral

!!!$

(a)

x2 + y2d V where S is the solid in the ?rst octant

S

bounded by the coordinate plane, the plane z = 4, and the cylinder x2 + y2 =

(b)

!

!

!

"x2

+

y

2

#

3 2

dV

25. where S is the solid bounded above by

S

the low

paraboloid z = by the x y-plane,

21an"dx

2 + y2#, laterally

beby

the cylinder x2 + y2 = 4.

6. Convert to spherical coordinates and evaluate the integral.

(a)

!

!

!

"x2

+

y2

+

z

2

#

3 2

dV

where S is the solid in the ?rst oc-

S

tant bounded by the sphere x2 +

$y$2x+2 +z

2

y

= 25, 2, and

the the

cone cone

z z

= =

2 x2 + y2.

!!! $

(b)

sin x2 + y2 + z2d V where S is the solid bounded

S

above by the sphere x2 + y2 +

zz2==$4x92

and below + y2.

by

the

cone

7. Find the coordinates of the center of gravity of the solid S with indicated mass density = (x, y, z). (Choose whichever system of coordinates would be best.)

(a) S: 0 x 1, 0 y 1, 0 z 1; = 3 - x - y - z.

SHWS C11: TRIPLE INTEGRATION

31

(b)

S:

x2 + y2

a2,

b a

$ x

2

+

y2

z

b

for constants b

>

0,

a

>

0 and

= x2 + y2 + z2.

8. Find the centroid for each of the following solids S:

(a) S: x2 + y2 1, x 0, y 0, 0 z x y (b) S: 9 x 2 + y2 + z2, z 0

9. Find the designated moment for each solid S with density = (x, y, z).

(a) Mxy ? the moment relative to the x y-plane where S: 0 x 1, 0 y 1 - x, 0 z 1 = 3xy

(b) Mxz ? the moment relative to the x z-plane where S: 0 x 1, 0 y 1, 0 z x y = 2 (x + y)

(c) Myz ? the moment relative to the yz-plane where S: bounded in the ?rst octant by x + z = 1, x = y, and the coordinate planes = 5y

10. Find the moment of inertia about the z-axis for the solid S bounded by 0 x 1, 0 y 1 - x, 0 z 5 with density = (x, y, z) = 3.

11. Find the moment of inertia about the y-axis of the solid S in the ?rst octant bounded by x2 + z2 = 1, y = x, y = 0, z = 0 with density = 2z.

12. Find the moment of inertia about the central axes of a homogeneous right circular cylindrical shell with total mass m, inner radius a, outer radius b and height h.

Proposed Solutions/Answers

1. (a)

(x/3)2+(y/6)2+(z/4)2=1 the plane z=2

To ?nd the range of x and y we

substitute the value of z into the

equation will give

x2 th9e

y2 z2 l+arg3e6st+ran1g6e

= for

1 x

that and

y. From inspecting the ?gure we

see that this value is z = 2.

32

For

z

=

2,

we

have

x2 9

+

y2 36

+

(2)2 16

=

1

or

x2 9

+

y2 36

=

3 4

.

From

this

we

obtain:

% 2z 4 1- --63%34x-x932

x2 y2

9 -%36

y6

3 4

-

x2 9

(b)

the plane (y/6)+(z/4)=1

To ?nd the scope of x and y

we notice from the ?gure that

(x/4)2+(y/36)2+(z/4)2=1 the largest range of x and y

iyn thze 6+4

solid = 1.

occur on Thus we

the plane solve this

for z and substitute this into

the equation Substitution

x2 y2 z2 16 + 36 + 16 this value of z

= 1. into

Solving

y 6

+

x2 y2 z2

16 + 36 + 16

z 4

=

1 for z

we get

z

=

1,

we

get

x2 16

+

y2 36

= +

146&&11--6y '26y

16

' .

=

1

or

x2 16

+

y2 18

+

y 3

=

1

or

x2 16

+

(y

- 3)2 18

=

1 2

.

From

this

we

see

that

y

will

be

maximized when x = 0. Thus we obtain the desired description.

4

& 1

y -(6

'

z

% 41

-

x2 16

-

y2 36(

-22

1

-

(y

- 3)2 9

x

22

1

-

(y

- 9

3)2

0 y 6.

SHWS C11: TRIPLE INTEGRATION

33

2. (a)

!2 !1 !1

&'

sinh z2 d zd yd x

00y

!2 !1 !z

&'

=

sinh z2 dydzd x

000

!2 !1

&'

= z sinh z2 dzd x

00

=

!2

1 2

(cosh

(1)

-

cosh

(0))

d

x

0

= cosh (1) - cosh (0)

=

e1

+ e-1 2

-1

The region indicated by the integral is

bounded by z = y, y = 0, z = 1, x = 0,

and x = 2 which is indicated by the ?gure

above. The dif?culty with integrating the

ogrriagteinsailnthri"pzl2e#i,nwteegrnaleeisd

that to a zdz

easily rather

intethan

just d z. Note that if we switch the dz and

d y, we might get a z where we need it.

(b)

!2 !4 !2 yzex3dxdydz

00z

!2 !4 !x

=

yzex3dzdydx

000

=

!2 !4

y

x2 2

ex3dyd

x

00

=

!2

0

) y2 *4 20

x2 2

ex

3

d

x

!2 = 4 x2ex3dx

0

=

4 3

ex

3

*2

0

=

4 3

"e8

-

1#

The region described by the integral is bounded by y = 0, y = 4, z = 0, z = x, and x = 2. A picture of the region is indicated above. In the original integral, if we try to integrate ex3d x we have a problems. We can easily integrate x2ex3, so this suggests

switching d x and dz.

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