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 onTriple Integration has sought to develop. They do not cover everything so a careful reviewof the Chapter and your class notes is also in order.

Problems for Fun and Practice1. For each of the following solids give a description in rectangular coordinates in the

order speciÞed:

(a) S is bounded above byx2

9+ y2

36+ z2

16= 1 and below by z = 2. Description:

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

(b) S is bounded above byx2

16+ y

2

36+ z

2

16= 1 and below by

y6+ z

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 iteratedintegral that can be done and evaluate it.

(a)2!

0

1!0

1!y

sinh"z2# dzdydx

(b)2!

0

4!0

2!z

yzex3dxdydz

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

(a)5!

0

√25−x2!0

6!0

dzdydx$x2 + y2

(b)2!

0

√4−x2!

0

x2+y2

2!0

zdzdydx$x2 + y2

30

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

(a)1!

0

√1−x2!

0

√1−x2−y2!

0

dzdydx1+ x2 + y2 + z2

(b)3!

0

√9−x2!

0

√9−x2−y2!

0

xzdzdydx

5. Convert to cylindrical coordinates and evaluate the integral

(a)! !S

! $x2 + y2dV where S is the solid in the Þrst octant

bounded by the coordinate plane, theplane z = 4, and the cylinder x2 + y2 =25.

(b)! !S

! "x2 + y2# 3

2 dV where S is the solid bounded above by

the paraboloid z = 12"x2 + y2#, be-

low by the xy-plane, and laterally bythe cylinder x2 + y2 = 4.

6. Convert to spherical coordinates and evaluate the integral.

(a)! !S

! "x2 + y2 + z2# 3

2 dV where S is the solid in the Þrst oc-tant bounded by the sphere x2 +y2 + z2 = 25, the cone z =$x2 + y2, and the cone z =

2$x2 + y2.

(b)! !S

!sin$x2 + y2 + z2dV where S is the solid bounded

above by the sphere x2 + y2 +z2 = 49 and below by the conez = $x2 + y2.

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

(a) S: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1; δ = 3− x − y − z.

SHWS C11: TRIPLE INTEGRATION 31

(b) S: x2 + y2 ≤ a2,ba$x2 + y2 ≤ z ≤ b for constants b > 0, a > 0 and

δ = 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 ≤ xy(b) S: 9 ≤ x2 + y2 + z2, z ≥ 0

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

(a) Mxy � the moment relative to the xy-plane whereS: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x , 0 ≤ z ≤ 1δ = 3xy

(b) Mxz � the moment relative to the xz-plane whereS: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ xyδ = 2 (x + y)

(c) Myz � the moment relative to the yz-plane whereS: bounded in the Þrst octant by x + z = 1, x = y,and the coordinate planesδ = 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 δ = δ (x, y, z) = 3.

11. Find the moment of inertia about the y-axis of the solid S in the Þrst octant boundedby x2 + z2 = 1, y = x , y = 0, z = 0 with density δ = 2z.

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

Proposed Solutions/Answers

1. (a)(x/3) +(y/6) +(z/4) =12 2 2

the plane z=2

To Þnd the range of x and y wesubstitute the value of z into the

equationx2

9+ y2

36+ z2

16= 1 that

will give the largest range for x andy. From inspecting the Þgure wesee that this value is z = 2.

32

For z = 2, we havex2

9+ y

2

36+ (2)

2

16= 1 or

x2

9+ y

2

36= 3

4. From this we obtain:

2 ≤ z ≤ 4%

1− x2

9− y2

36

−6%

34− x

2

9≤ y ≤ 6

%34− x

2

9−3 ≤ x ≤ 3

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

(x/4) +(y/36) +(z/4) =12 2 2

To Þnd the scope of x and ywe notice from the Þgure thatthe largest range of x and yin the solid occur on the planey6+ z

4= 1. Thus we solve this

for z and substitute this into

the equationx2

16+ y

2

36+ z

2

16= 1. Solving

y6+ z

4= 1 for z we get z = 4

&1− y

6

'.

Substitution this value of z intox2

16+ y

2

36+ z

2

16= 1, we get

x2

16+ y

2

36+ 16

&1− y6

'2

16 =

1 orx2

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− y6

'≤ z ≤ 4

%1− x

2

16− y2

36

−2√

2

(1− (y − 3)2

9≤ x ≤ 2

√2

(1− (y − 3)2

90 ≤ y ≤ 6.

SHWS C11: TRIPLE INTEGRATION 33

2. (a)

2!0

1!0

1!y

sinh&z2'dzdydx

=2!

0

1!0

z!0

sinh&z2'dydzdx

=2!

0

1!0

z sinh&z2'dzdx

=2!

0

12(cosh (1)− cosh (0)) dx

= cosh (1)− cosh (0)

= e1 + e−1

2− 1

The region indicated by the integral isbounded by z = y, y = 0, z = 1, x = 0,and x = 2 which is indicated by the Þgureabove. The difÞculty with integrating theoriginal triple integral is that to easily inte-grate sinh

"z2#, we need a zdz rather than

just dz. Note that if we switch the dz anddy, we might get a z where we need it.

(b)

2!0

4!0

2!z

yzex3dxdydz

=2!

0

4!0

x!0

yzex3dzdydx

=2!

0

4!0

yx2

2ex

3dydx

=2!

0

)y2

2

*4

0

x2

2ex

3dx

= 42!

0

x2ex3dx

= 43ex

3*2

0= 4

3"e8 − 1

#

The region described by the integral isbounded by y = 0, y = 4, z = 0, z = x ,and x = 2. A picture of the region is indi-cated above. In the original integral, if wetry to integrate ex3dx we have a problems.We can easily integrate x2ex3, so this suggestsswitching dx and dz.

34

3. (a)

5!0

√25−x2!0

6!0

dzdydx$x2 + y2

=π/2!0

5!0

6!0

rdzdrdθr

= 15π

The region described by the integral is the set of all point (x, y, z) satisfyingthe three inequalities 0 ≤ z ≤ 6, 0 ≤ y ≤ √25− x2, and 0 ≤ x ≤ 5. It is 1

4 ofa cylinder as shown in the Þgure above.

(b)

2!0

√4−x2!

0

x2+y2

2!0

zdzdydx$x2 + y2

=π/2!0

2!0

r2/2!0

zrdzdrdθr

= 2π5

The region or solid is in the Þrst octant between the paraboloid and the cylinderas indicated in the picture above. It is the set of all points (x, y, z) satisfying

the three inequalities 0 ≤ z ≤ x2 + y2

2, 0 ≤ y ≤ √4− x2, and 0 ≤ x ≤ 2.

SHWS C11: TRIPLE INTEGRATION 35

4. (a)

1!0

√1−x2!

0

√1−x2−y2!

0

dzdydx1+ x2 + y2 + z2

=π/2!0

π/2!0

1!0

ρ2 sinφdρdφdθ1+ ρ2

=π/2!0

π/2!0

1!0

sinφ+

1− 11+ ρ2

*dρdφdθ

=&

1− π4

' π/2!0

(cosφ]π/20 dθ

=&

1− π4

' π2= π

2− π

2

8

The solid is18

of the spherewith radius 1 centered at theorigin as indicated in the Þg-ure.

(b)

3!0

√9−x2!

0

√9−x2−y2!

0

xzdzdydx

=π/2!0

π/2!0

3!0

(ρ sinφ cos θ) (ρ cosφ) ρ2 sinφdρdφdθ

=π/2!0

π/2!0

3!0

ρ4 cos θ sin2 φ cosφdρdφdθ

= 35

5

π/2!0

cos θ

,sin3 φ

3

-φ=π/2φ=0

dθ = 34

5

The solid is18

of thesphere with radius 3centered at the originas depicted in the Þg-ure in part a.x = ρ sinφ cos θz = ρ cosφ

36

5. (a) ! !S

! .x2 + y2dV

=2π!

0

5!0

4!0

r · rdzdrdθ

= 43(5)3 2π

x +y =252 2

4

(b) ! !S

! &x2 + y2

' 32 dV

=2π!

0

2!0

r2/2!0

r3 · rdzdrdθ

=2π!

0

2!0

r6

2drdθ

=2π!

0

r7

14

*2

0dθ =

)26

7

/(2π)

z=(x +y )/22 2

x +y =42 2

z goes upto 2

SHWS C11: TRIPLE INTEGRATION 37

6. (a) ! !S

! &x2 + y2 + z2

' 32 dV

=π/2!0

π/4!tan−1

&12

'5!

0

ρ3 · ρ2 sinφdρdφdθ0 12 3dV

=)

56

6

/ π/2!0

cosφ]π/4tan−1

&12

' dθ

=)

56

6

/,4−√10

2√

5

4&π2

'

part of sphere x +y +z = 252 2 2

inside cone z = +

outside cone z = +

To Þnd the range of φ we use the

fact that tanφ =$x2 + y2

z. For the

inside z = 2$x2 + y2, so tanφ =$

x2 + y2

2$x2 + y2

= 12

. For the outside z =$x2 + y2, so tanφ =

$x2 + y2$x2 + y2

= 1.

Aside:5φ2

1If tanφ = 1

2, then cosφ = 2√

5.

(b) ! !S

!sin.x2 + y2 + z2dV

=2π!

0

π/4!0

7!0

(sin ρ) ρ2 sinφdρdφdθ

= 2π

,√2

2− 1

4(−47 cos 7+ 14 sin 7− 2)

the sphere x +y +z = 492 2 2

the cone z = +

Using tanφ =$x2 + y2

zwe

get tanφ =$x2 + y2$x2 + y2

= 1 or

φ = π

4.

38

7. (a) First Þnd the mass (which we denote by m):

m =1!

0

1!0

1!0

(3− x − y − z) dxdydz = 32

Then

Mxy =!S

zδdV =1!

0

1!0

1!0

&3z − xz − yz − z2

'dxdydz = 2

3

Mxz =!S

yδdV =1!

0

1!0

1!0

&3y − xy − y2 − zy

'dxdydz = 2

3

Myz =!S

xδdV =1!

0

1!0

1!0

&3x − x2 − yx − zx

'dxdydz = 2

3

The coordinates of the center of gravity are

(x, y, z) =)Myzm,Mxzm,Mxym

/=)

49,

49,

49

/.

(b)

h

a The solid is the cone with height h and coverradius a. To compute the integral in sphericalcoordinates we need φ.

tanφ = ah

so φ = tan−1&ah

'.

also

z = ρ cosφ = h gives ρ = hcosφ

.

SHWS C11: TRIPLE INTEGRATION 39

mass =! !S

!δdV

=2π!

0

tan−1( ah )!0

hcosφ!0

ρ2 · ρ2 sinφdρdφdθ

=2π!

0

tan−1( ah )!0

h5

5sinφ

cos5 φdφdθ

=2π!

0

h5

5

)1

cos4 φ

*tan−1( ah )

0dθ

= 2πh5

5

,"a2 + h2#2h4 − 1

4

a2 h2+

φ

a

h

cosφ = h√a2 + h2

Also

!S

xδdV =2π!

0

tan−1( ah )!0

hcosφ!0

ρ sinφ cos θ · ρ2 · ρ2 sinφdρdφdθ = 0

!S

yδdV =2π!

0

tan−1( ah )!0

hcosφ!0

ρ sinφ sin θρ2 · ρ2 sinφdρdφdθ = 0

!S

zδdV =2π!

0

tan−1( ah )!0

hcosφ!0

ρ cosφρ2·ρ2 sinφdρdφdθ = πh6

3

,"a2 + h2#2h4 − 1

4

Therefore (x, y, z) =)

0, 0,56h/

.

8. The centroid of a solid S has coordinates (x, y, z) where

x =5SxdV

volume of S, y =

5SydV

volume of S, z =

5SzdV

volume of S.

40

(a)cover is z = xy

x +y = 12 2

volume of S =1!

0

√1−y2!

0

xy!0

dzdxdy

=1!

0

√1−y2!

0

xydxdy

=1!

0

√1−y2!

0

y"1− y2#

2dy =

)y2

4− y4

8

*1

0= 1

8

!S

xdV =1!

0

√1−y2!

0

xy!0

xdzdxdy =1!

0

√1−y2!

0

x2ydxdy =1!

0

"1− y2# 3

2 y3

dy

= −16

&1− y2

' 52 2

5

*1

0= 1

15

!S

ydV =1!

0

√1−y2!

0

xy!0

ydzdxdy =1!

0

x2y2

2

*√1−y2

0dy = 1

2

)y3

3− y5

5

*1

0= 1

15

!S

zdV =1!

0

√1−y2!

0

xy!0

zdzdxdy =1!

0

√1−y2!

0

x2y2dxdy = 16

1!0

y2&

1− y2' 3

2 dy

SHWS C11: TRIPLE INTEGRATION 41

= 16

π/2!0

sin2 θ cos3 θ cos θdθ

= 16

π/2!0

sin2 θ cos4 θdθ

= 16

π/2!0

(1− cos 2θ) (1+ cos 2θ)2

8dθ

= 16

π/2!0

"1− cos2 2θ + cos 2θ − cos3 2θ

#8

dθ

= 16

& π32

'

Aside:

1 ! y2

1y

θ

y = sin θ$1− y2 = cos θ

Therefore the centroid is)

815,

815,π

24

/.

(b)3

4

S

The volume of a sphere of radius r is43πr3. Using this the volume of S is

12

)43π (4)3 − 4

3π (3)3

/= 2

3π (64− 27)

= 74π3

!S

xdV =4!

3

π/2!0

2π!0

ρ sinφ cos θρ2 sinφdθdφdρ = 0

!S

ydV =4!

3

π/2!0

2π!0

ρ sinφ sin θρ2 sinφdθdφdρ = 0

42

!S

zdV =2π!

0

π/2!0

4!3

ρ cosφρ2 sinφdρdφdθ = 44 − 34

4

2π!0

)− cos2 φ

2

*π20dθ

= π"44 − 34#

4= 175π

4

The centroid is)

0, 0,525296

/.

9. (a)

S

z = 1

y = 1 - x

δ = 3xy

Mxy =!S

zδdV =1!

0

1−x!0

1!0

3xyzdzdydx

=1!

0

1−x!0

32xydydx = 3

4

1!0

&1− x2

'dx = 1

4

(b) δ = 2 (x + y)

Mxz =!S

yδdV

=1!

0

1!0

xy!0

2y (x + y) dzdxdy

S z = xy

x = 1 y = 1

=1!

0

1!0

2xy2 (x + y) dxdy =1!

0

)23y2 + y3

/dy = 2

9+ 1

4= 17

36

SHWS C11: TRIPLE INTEGRATION 43

(c)

x + z = 1

x = y

S

δ = 5y

Myz =!S

xδdV =1!

0

x!0

1−x!0

5xydzdydx

=1!

0

x!0

5xy (1− x) dydx =1!

0

52x3 (1− x) dx

= 52

1!0

&x3 − x4

'dx = 5

2

)14− 1

5

/= 1

8

10.

y = 1 - x

z = 5

S

δ = 3

Iz =!S

&x2 + y2

'δdV = 3

1!0

1−x!0

5!0

&x2 + y2

'dzdydx

= 151!

0

1−x!0

&x2 + y2

'dydx = 15

1!0

,x2 − x3 + (1− x)

3

3

4dx

= 15

,x3

3− x

4

4− (1− x)

4

12

-1

0

= 15)

212

/= 5

2

11.

x + y = 12 2

x = yS

δ = 2z

Iy =!S

2z&x2 + z2

'δdV

=1!

0

x!0

√1−x2!

0

&2zx2 + 2z2

'dzdydx

=1!

0

x!0

)x2&

1− x2'+ 1

2

&1− x2

'2/dydx

=1!

0

)x3&

1− x2'+ x

2

&1− x2

'2/dx =

1!0

)x2− x

5

2

/dx = 1

6

44

12.radius is b

radius is a

h

S

The mass is m. Since the solid is homoge-

neous δ = mass of Svolume of S

. To Þnd the volume,we can use the formula for the volume of acylinder of radius r and height h which isπr2h.

volume of S = πb2h−πa2h = π&b2 − a2

'h.

Thus δ = mπ"b2 − a2

#h

.

Iz =!S

&x2 + y2

'δdV = m

π"b2 − a2

#h

2π!0

h!0

b!a

r2 · rdrdzdθ

= mπ"b2 − a2

#hb4 − a4

4(h) (2π) = m

"b4 − a4#

2"b2 − a2

# = m"b2 + a2#

2