Calculus one and several variables 10E Salas solutions manual ch17

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU

JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38

866 SECTION 17.1

CHAPTER 17

SECTION 17.1

1.3∑

i=1

3∑j=1

2i−13j+1 =

(3∑

i=1

2i−1

)⎛⎝ 3∑j=1

3j+1

⎞⎠ = (1 + 2 + 4)(9 + 27 + 81) = 819

2. 2 + 22 + 3 + 32 + 4 + 42 + 5 + 52 = 68

3.4∑

i=1

3∑j=1

(i2 + 3i)(j − 2) =

[4∑

i=1

(i2 + 3i)

]⎡⎣ 3∑j=1

(j − 2)

⎤⎦ = (4 + 10 + 18 + 28)(−1 + 0 + 1) = 0

4.22

+23

+24

+25

+26

+27

+42

+43

+44

+45

+46

+47

+62

+63

+64

+65

+66

+67

= 19435

.

5.m∑i=1

Δxi = Δx1 + Δx2 + · · · + Δxm = (x1 − x0) + (x2 − x1) + · · · + (xm − xm−1)

= xm − x0 = a2 − a1

6. (y1 − y0) + (y2 − y1) + · · · + (yn − yn−1) = yn − y0 = b2 − b1

7.m∑i=1

n∑j=1

Δxi Δyj =

(m∑i=1

Δxi

)⎛⎝ n∑j=1

Δyj

⎞⎠ = (a2 − a1) (b2 − b1)

8.n∑

j=1

q∑k=1

Δyj Δzk =

⎛⎝ n∑

j=1

Δyj

⎞⎠( q∑

k=1

Δzk

)= (b2 − b1) (c2 − c1)

9.m∑i=1

(xi + xi−1)Δxi =m∑i=1

(xi + xi−1)(xi − xi−1) =m∑i=1

(xi2 − x2

i−1)

= xm2 − x0

2 = a22 − a1

2

10.n∑

j=1

12(yj2 + yjyj−1 + yj−1

2)Δyj =12

n∑j=1

(yj3 − yj−13) =

12(b23 − b1

3)

11.m∑i=1

n∑j=1

(xi + xi−1)ΔxiΔyj =

(m∑i=1

(xi + xi−1)Δxi

)( n∑j=1

Δyj

)

(Exercise 9)∧

=(a2

2 − a12)(b2 − b1)

12.m∑i=1

n∑j=1

(yi + yj−1)ΔxiΔyj =

(m∑i=1

Δxi

)⎡⎣ n∑j=1

(yj2 − yj−12)

⎤⎦ = (a2 − a1)(b22 − b1

2)



SECTION 17.2 867

13.m∑i=1

n∑j=1

(2Δxi − 3Δyj) = 2

(m∑i=1

Δxi

)( n∑j=1

1)− 3

(m∑i=1

1

)( n∑j=1

Δyj

)

= 2n(a2 − a1) − 3m(b2 − b1)

14.m∑i=1

n∑j=1

(3Δxi − 2Δyj) = 3m∑i=1

n∑j=1

Δxi − 2m∑i=1

n∑j=1

Δyj = 3n(a2 − a1) − 2m(b2 − b1).

15.m∑i=1

n∑j=1

q∑k=1

Δxi Δyj Δzk =

(m∑i=1

Δxi

)⎛⎝ n∑j=1

Δyj

⎞⎠( q∑

k=1

Δzk

)

= (a2 − a1)(b2 − b1)(c2 − c1)

16.m∑i=1

n∑j=1

q∑k=1

(xi + xi−1)ΔxiΔyjΔzk =

[m∑i=1

(xi2 − xi−1

2)

]⎛⎝ n∑j=1

Δyj

⎞⎠( q∑

k=1

Δzk

)

= (a22 − a1

2)(b2 − b1)(c2 − c1)

17.n∑

i=1

n∑j=1

n∑k=1

δijkaijk = a111 + a222 + · · · + annn =n∑

p=1

appp

18. Start withm∑i=1

n∑j=1

aij . Take all the aij (there are only a finite number of them) and order them

in any order you chose. Call the first one b1, the second b2, and so on. Thenm∑i=1

n∑j=1

aij =r∑

p=1

bp where r = m× n.

SECTION 17.2

1. Lf (P ) = 2 14 , Uf (P ) = 5 3

4 2. Lf (P ) = 3, Uf (P ) = 5

3. (a) Lf (P ) =m∑i=1

n∑j=1

(xi−1 + 2yj−1) Δxi Δ yj , Uf (P ) =m∑i=1

n∑j=1

(xi + 2yj) Δxi Δyj

(b) Lf (P ) ≤m∑i=1

n∑j=1

[xi−1 + xi

2+ 2(yj−1 + yj

2

)]Δxi Δyj ≤ Uf (P ).

The middle expression can be writtenm∑i=1

n∑j=1

12(xi

2 − x2i−1

)Δyj +

m∑i=1

n∑j=1

(yj

2 − y2j−1

)Δxi.

The first double sum reduces tom∑i=1

n∑j=1

12(xi

2 − x2i−1

)Δyj =

12

(m∑i=1

(xi

2 − x2i−1

))( n∑j=1

Δyj

)=

12

(4 − 0) (1 − 0) = 2.

In like manner the second double sum also reduces to 2. Thus, I = 4; the volume of the prism

bounded above by the plane z = x + 2y and below by R.



868 SECTION 17.2

4. Lf (P ) = −7/16, Uf (P ) = 7/16 5. Lf (P ) = −7/24, Uf (P ) = 7/24

6. (a) Lf (p) =m∑i=1

n∑j=1

(xi−1 − yj)ΔxiΔyj , Uf (P ) =m∑i=1

n∑j=1

(xi − yj−1)ΔxiΔyj

(b) Lf (P ) ≤m∑i=1

n∑j=1

(xi + xi−1

2− yj + yj−1

2

)ΔxiΔyj ≤ Uf (P )


n∑j=1

12(xi

2 − xi−12)Δyj −

m∑i=1

n∑j=1

12(yj2 − yj−1

2)Δxi.

The first sum reduces to

12

(m∑i=1

(xi2 − xi−1

2)

)⎛⎝ n∑j=1

Δyj

⎞⎠ =

12(1 − 0)(1 − 0) =

12.

In like manner the second sum also reduces to 12 . Thus I = 1

2 − 12 = 0.

7. (a) Lf (P ) =m∑i=1

n∑j=1

(4xi−1 yj−1) Δxi Δyj , Uf (P ) =m∑i=1

n∑j=1

(4xi yj) Δxi Δyj

(b) Lf (P ) ≤m∑i=1

n∑j=1

(xi + xi−1) (yj + yj−1) Δx1 Δyj ≤ Uf (P ).


n∑j=1

(xi

2 − x2i−1

) (yj

2 − y2j−1

)=

(m∑i=1

xi2 − x2

i−1

)( n∑j=1

yj2 − y2

j−1

)

by (17.1.5)∧

=(b2 − 02

) (d2 − 02

)= b2d2.

It follows that I = b2d2.

8. (a) Lf (P ) =m∑i=1

n∑j=1

3(xi−12 + yj−1

2)ΔxiΔyj , Uf (P ) =m∑i=1

n∑j=1

3(xi2 + yj

2)ΔxiΔyj

(b) Lf (P ) ≤m∑i=1

n∑j=1

[(xi

2 + xixi−1 + xi−12) + (yj2 + yjyj−1 + yj−1

2)]ΔxiΔyj ≤ Uf (P )

Since in general (A2 + AB + B2)(A−B) = A3 −B3, the middle expression can be written

m∑i=1

n∑j=1

(xi3 − xi−1

3)Δyj +m∑i=1

n∑j=1

(yj3 − yj−13)Δxi,

which reduces to(m∑i=1

xi3 − xi−1

3

)⎛⎝ n∑j=1

Δyj

⎞⎠+

(m∑i=1

Δxi

)⎛⎝ n∑j=1

yj3 − yj−1

3

⎞⎠ .

This can be evaluated as b3d + bd3 = bd(b2 + d2). It follows that I = bd(b2 + d2).



SECTION 17.2 869

9. (a) Lf (P ) =m∑i=1

n∑j=1

3(x2i−1 − yj

2)

Δxi Δyj , Uf (P ) =m∑i=1

n∑j=1

3(xi

2 − y2j−1

)Δxi Δyj

(b) Lf (P ) ≤m∑i=1

n∑j=1

[ (xi

2 + xixi−1 + x2i−1

)−(yj

2 + yjyj−1 + y2j−1

) ]Δxi Δyj ≤ Uf (P ).

Since in general(A2 + AB + B2

)(A−B) = A3 −B3, the middle expression can be written

m∑i=1

n∑j=1

(xi

3 − x3i−1

)Δyj −

m∑i=1

n∑j=1

(yj

3 − y3j−1

)Δxi,

which reduces to(m∑i=1

xi3 − x3

i−1

)⎛⎝ n∑j=1

Δyj

⎞⎠−

(m∑i=1

Δxi

)⎛⎝ n∑j=1

yj3 − y3

j−1

⎞⎠ .

This can be evaluated as b3d− bd3 = bd(b2 − d2

). It follows that I = bd (b2 − d2).

10. On each subrectangle, the minimum and the maximum of f are equal, so f is constant on each

subrectangle and therefore (since f is continuous) on the entire rectangle R. Then∫∫R

f(x, y) dxdy = f(a, c)(b− a)(d− c).

11.∫∫Ω

dxdy =∫ b

a

φ(x) dx

12. Suppose that there is a point (x0, y0) on the boundary of Ω at which f is not zero. As (x, y)

tends to (x0, y0) through that part of R which is outside Ω, f(x, y), being zero, tends to zero.

Since f(x0, y0) is not zero, f(x, y) does not tend to f(x0, y0). Thus the extended function f can not

be continuous at (x0, y0).

13. Suppose f(x0, y0) �= 0. Assume f(x0, y0) > 0. Since f is continuous, there exists a disc Ωε with radius

ε centered at (x0, y0) such that f(x, y) > 0 on Ωε. Let R be a rectangle contained in Ωε. Then∫∫R

f(x, y) dxdy > 0, which contradicts the hypothesis.

14.∫∫R

(x + 2y) dx dy = 4; area(R) = (2)(1) = 2, so average value =42

= 2

15. By Exercise 7, Section 17.2,∫∫R

4xy dxdy = 2232 = 36. Thus

favg =1

area (R)

∫∫R

4xy dxdy =16

(36) = 6

16.∫∫R

(x2 + y2) dxdy =bd(b2 + d2)

3; area(R) = bd, so average value =

b2 + d2

3



870 SECTION 17.3

17. By Theorem 16.2.10, there exists a point (x1, y1) ∈ Dr such that∫∫Dr

f(x, y) dxdy = f(x1, y1)∫∫R

dxdy = f(x1, y1)πr2 =⇒ f(x1, y1) =1

πr2

∫∫Dr

f(x, y) dxdy

As r → 0, (x1, y1) → (x0, y0) and f(x1, y1) → f(x0, y0) since f is continuous.

The result follows.

18. 0 ≤ sin(x + y) ≤ 1 for all (x, y) ∈ R. Thus, 0 ≤∫∫R

sin(x + y) dxdy ≤∫∫R

dxdy = 1

19. z =√

4 − x2 − y2 on Ω : x2 + y2 ≤ 4, x ≥ 0, y ≥ 0;∫∫Ω

√4 − x2 − y2 dxdy is the volume V of one

quarter of a hemisphere; V = 43π.

20. 8 − 4√x2 + y2 on Ω is a cone with height h = 8 and radius r = 2; V =

32π3

21. z = 6 − 2x− 3y ⇒ x

3+

y

2+

z

6= 1; the solid is the tetrahedron bounded by the coordinate planes

and the plane:x

3+

y

2+

z

6= 1; V = 1

6 (3)(2)(6) = 6

22. (a) Lf (P ) ∼= 35.4603; Uf (P ) ∼= 36.5403 (c)∫∫R

(3y2 − 2x

)dxdy = 36

SECTION 17.3

1.∫ 1

0

∫ 3

0

x2 dy dx =∫ 1

0

3x2 dx = 1

2.∫ 3

0

∫ 1

0

ex+y dx dy =∫ 3

0

(e1+y − ey) dy =[e1+y − ey

]30

= e4 − e3 − e + 1

3.∫ 1

0

∫ 3

0

xy2 dy dx =∫ 1

0

x

[13y3

]30

dx =∫ 1

0

9x dx =92

4.∫ 1

0

∫ x

0

x3y dy dx =∫ 1

0

x3x2

2dx =

112

5.∫ 1

0

∫ x

0

xy3 dy dx =∫ 1

0

x

[14y4

]x0

dx =∫ 1

0

14x5 dx =

124

6.∫ 1

0

∫ x

0

x2y2 dy dx =∫ 1

0

x2x3

3dx =

118

7.∫ π/2

0

∫ π/2

0

sin (x + y) dy dx =∫ π/2

0

[− cos (x + y)]π/20 dx =∫ π/2

0

[cosx− cos

(x +

π

2

)]dx = 2



SECTION 17.3 871

8.∫ π/2

0

∫ π/2

0

cos(x + y) dx dy =∫ π/2

0

[sin(π

2+ y)− sin y

]dy =

[cos y − cos

(π2

+ y)]π/2

0= 0

9.∫ π/2

0

∫ π/2

0

(1 + xy) dy dx =∫ π/2

0

[y +

12xy2

]π/20

dx =∫ π/2

0

(12π +

18π2x

)dx =

14π2 +

164

π4

10.∫ 1

−1

∫ √1−y2

−√

1−y2(x + 3y3) dx dy =

∫ 1

−1

6y3√

1 − y2 dy = 0 (integrand is odd)

11.∫ 1

0

∫ y

y2

√xy dx dy =

∫ 1

0

√y

[23x3/2

]yy2

dy =∫ 1

0

23

(y2 − y7/2

)dy =

227

12.∫ 1

0

∫ y2

0

yex dx dy =∫ 1

0

y(ey2 − 1) dy =

[ey

2

2− y2

2

]10

=12(e− 2)

13.∫ 2

−2

∫ 4− 12y

2

12y

2

(4 − y2

)dx dy =

∫ 2

−2

(4 − y2

) [(4 − 1

2y2

)−(

12y2

)]dy

= 2∫ 2

0

(16 − 8y2 + y4

)dy =

51215

14. I =∫ 1

0

∫ x2

x3(x4 + y2) dy dx =

∫ 1

0

[x4y +

y3

3

]x2

x3

dx =∫ 1

0

(4x6

3− x7 − x9

3

)dx

=[4x7

21− x8

8− x10

30

]10

=9

280

15. 0 by symmetry (integrand odd in y, Ω symmetric about x-axis)

16.∫ 1

0

∫ 2y

0

e−y2/2 dx dy =∫ 1

0

2ye−y2/2 dy =[−2e−y2/2

]10

= 2(

1 − 1√e

)

17.∫ 2

0

∫ x/2

0

ex2dy dx =

∫ 2

0

12xex

2dx =

[14ex

2]20

=14(e4 − 1

)

18.∫ 0

−1

∫ x4

x3(x + y) dy dx +

∫ 1

0

∫ x3

x4(x + y) dydx =

∫ 0

−1

[xy +

y2

2

]x4

x3

dx +∫ 1

0

[xy +

y2

2

]x3

x4

dx

=∫ 0

−1

(x5 +

x8

2− x4 − x6

2

)dx +

∫ 1

0

(x4 +

x6

2− x5 − x8

2

)dx

=[x6

6+

x9

18− x5

5− x7

14

]0−1

+[x5

5+

x7

14− x6

6− x9

18

]10

= −13



872 SECTION 17.3

19.

∫ 1

0

∫ y1/4

y1/2f(x, y) dx dy

20.

∫ 1

0

∫ 1

√x

f(x, y) dydx

21.

∫ 0

−1

∫ 1

−x

f(x, y) dy dx +∫ 1

0

∫ 1

x

f(x, y) dy dx

22.

∫ 1/2

3√

1/2

∫ y1/3

1/2

f(x, y) dxdy +∫ 1

1/2

∫ y1/3

y

f(x, y) dxdy

23.

∫ 2

1

∫ y

1

f(x, y) dx dy +∫ 4

2

∫ y

y/2

f(x, y) dx dy

+∫ 8

4

∫ 4

y/2

f(x, y) dx dy

24.

∫ −1

−3

∫ 3

−y

f(x, y) dx dy +∫ 1

−1

∫ 3

1

f(x, y) dx dy

+∫ 9

1

∫ 3

√y

f(x, y) dx dy

25.∫ 4

−2

∫ 12x+2

1/4x2dy dx =

∫ 4

−2

[12x + 2 − 1

4x2

]dx = 9

26.∫ 3

0

∫ 4y−y2

y

dx dy =∫ 3

0

(3y − y2) dy =92

27.∫ 1/4

0

∫ y

2y3/2dx dy =

∫ 1/4

0

[y − 2y3/2

]dy =

1160



SECTION 17.3 873

28.∫ 3

2

∫ 5−x

6/x

dy dx =∫ 3

2

(5 − x− 6/x) dx =52

+ 6 ln23

29. ∫ 1

0

∫ y2

0

sin(y3 + 1

2

)dx dy =

∫ 1

0

y2 sin(y3 + 1

2

)dy

=[−2

3cos(y3 + 1

2

)]10

=23

(cos

12− cos 1

)

30.∫ 1

−1

∫ 0

x2−1

x2 dy dx =∫ 1

−1

x2(1 − x2) dx

=415

31. ∫ ln 2

0

∫ 2

exe−x dy dx =

∫ ln 2

0

e−x (2 − ex) dx

=[−2e−x − x

]ln 2

0= 1 − ln 2

32. ∫ 1

0

∫ √y

0

x3√x4 + y2

dx dy =∫ 1

0

12(√

2 − 1)y dy

=14(√

2 − 1)

33.∫ 2

1

∫ 2/y

y−1

dx dy =∫ 2

1

[2y− (y − 1)

]dy = ln 4 − 1

2

34.∫ 1

0

∫ 1−x

0

(x + y) dy dx =∫ 1

0

[x(1 − x) +

(1 − x)2

2

]dx =

13



874 SECTION 17.3

35.∫ 2

0

∫ 3− 32x

0

(4 − 2x− 4

3y

)dy dx =

∫ 3

0

∫ 2− 23y

0

(4 − 2x− 4

3y

)dx dy = 4

36.∫ 1

0

∫ 1

0

(2x + 3y) dy dx =∫ 1

0

(2x +32) dx =

52

37.∫ 2

0

∫ 1− 12x

0

x3y dy dx =∫ 2

0

∫ 2−2y

0

x3y dx dy =215

38.∫ 1

−1

∫ √1−x2

−√

1−x2(x2 + y2) dy dx = 2

∫ 1

−1

[x2√

1 − x2 +13(1 − x2)3/2

]dx =

π

2

39.∫ 2

0

∫ √2x−x2

−√

2x−x2(2x + 1) dy dx =

∫ 1

−1

∫ 1+√

1−y2

1−√

1−y2(2x + 1) dx dy

=∫ 1

−1

[x2 + x

]1+√1−y2

1−√

1−y2

dy

= 6∫ 1

−1

√1 − y2 dy = 6

(π2

)= 3π

40.∫ 1

−1

∫ 1+√

1−x2

1−√

1−x2(4 − y2 − 1

4x2) dy dx =

∫ 1

−1

[6√

1 − x2 − 12x2√

1 − x2 − 23(1 − x2)3/2

]dx =

4316

π.

41.∫ 1

0

∫ 1−x

0

(x2 + y2) dy dx =∫ 1

0

(2x2 − 4

3x3 − x +

13

)dx =

16

42.∫ 1

−1

∫ √1−x2

−√

1−x2(1 − x) dy dx =

∫ 1

−1

(2√

1 − x2 − 2x√

1 − x2) dx = π

43.∫ 1

0

∫ x

x2(x2 + 3y2) dy dx =

∫ 1

0

(2x3 − x4 − x6

)dx =

1170

44.∫ 2

1

∫ 5−y

2y−1

(1 + xy) dx dy =∫ 2

1

(6 + 9y − 3y2 − 32y3) dy =

558

45.∫ a

0

∫ √a2−x2

0

√a2 − x2 dy dx =

∫ a

0

(a2 − x2) dx =23a3

46.∫ a

0

∫ b(1−x/a)

0

c(1 − x

a− y

b

)dy dx =

∫ a

0

bc

2

(1 − x

a

)2dx =

abc

6

47.∫ 1

0

∫ 1

y

ey/xdx dy =∫ 1

0

∫ x

0

ey/xdy dx =∫ 1

0

[xey/x

]x0dx =

∫ 1

0

x(e− 1) dx =12

(e− 1)

48.∫ 1

0

∫ cos−1 y

0

esinx dx dy =∫ π/2

0

∫ cosx

0

esinx dy dx =∫ π/2

0

cosxesinx dx = e− 1



SECTION 17.3 875

49.∫ 1

0

∫ 1

x

x2ey4dy dx =

∫ 1

0

∫ y

0

x2ey4dx dy =

∫ 1

0

[13x3ey

4]y0

dy =13

∫ 1

0

y3ey4dy =

112

(e− 1)

50.∫ 1

0

∫ y

0

ey2dx dy =

∫ 1

0

yey2dy =

12(e− 1)

51. favg =18

∫ 1

−1

∫ 4

0

x2y dy dx =18

∫ 1

−1

8x2 dx =∫ 1

−1

x2 dx =23

52. area of Ω = 14π (quarter circle of radius 1)

Average value =4π

∫ 1

0

∫ √1−x2

0

xy dy dx =2π

∫ 1

0

x(1 − x2) dx =12π

53. favg =1

(ln 2)2

∫ 2 ln 2

ln 2

∫ 2 ln 2

ln 2

1xy

dy dx =1

(ln 2)2

∫ 2 ln 2

ln 2

1x

ln 2 dx = 1

54. area of Ω = 1 (parallelogram)

Average value =∫ 1

0

∫ x+1

x−1

ex+y dy dx =∫ 1

0

(e2x+1 − e2x−1

)dx =

12(e3 − 2e + e−1).

55.∫∫R

f(x)g(y) dxdy =∫ d

c

∫ b

a

f(x)g(y) dx dy =∫ d

c

(∫ b

a

f(x)g(y) dx)

dy

=∫ d

c

g(y)(∫ b

a

f(x) dx)

dy =(∫ b

a

f(x) dx) (∫ d

c

g(y) dy)

56. Symmetry about the origin [ we want Ω to contain (−x,−y) whenever it contains (x, y)].

57. Note that Ω = { (x, y) : 0 ≤ x ≤ y, 0 ≤ y ≤ 1}.Set Ω′ = { (x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}.∫∫

Ω

f(x)f(y) dxdy =∫ 1

0

∫ y

0

f(x)f(y) dx dy

=∫ 1

0

∫ x

0

f(y)f(x) dy dx

x and y are dummy variables

=∫ 1

0

∫ x

0

f(x)f(y) dy dx =∫∫Ω′

f(x)f(y) dxdy.

Note that Ω and Ω′ don’t overlap and their union is the unit square

R = { (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.



876 SECTION 17.3

If∫ 1

0

f(x) dx = 0, then

0 =(∫ 1

0

f(x) dx) (∫ 1

0

f(y) dy)

=∫∫R

f(x)f(y) dxdy

by Exercise 55

=∫∫Ω

f(x)f(y) dxdy +∫∫Ω′

f(x)f(y) dxdy

= 2∫∫Ω

f(x)f(y) dxdy

and therefore∫∫Ω

f(x)f(y) dxdy = 0.

58.∫ x

0

∫ b

a

∂f

∂x(t, y) dy dt =

∫ b

a

∫ x

0

∂f

∂x(t, y) dt dy

=∫ b

a

[f(x, y) − f(0, y)] dy

=∫ b

a

f(x, y) dy −∫ b

a

f(0, y) dy.

Thus ∫ b

a

f(x, y) dy =∫ x

0

∫ b

a

∂f

∂x(t, y) dy dt +

∫ b

a

f(0, y) dy

and

d

dx

[∫ b

a

f(x, y) dy]

=d

dx

[∫ x

0

∫ b

a

∂f

∂x(t, y) dy dt

]+

d

dx

[∫ b

a

f(0, y) dy]

=d

dx

[∫ x

0

H(t) dt]

+ 0 = H(x) =∫ b

a

∂f

∂x(x, y) dy.

59. Let M be the maximum value of | f(x, y) | on Ω.∫ φ2(x+h)

φ1(x+h)

=∫ φ1(x)

φ1(x+h)

+∫ φ2(x)

φ1(x)

+∫ φ2(x+h)

φ2(x)

|F (x + h) − F (x) | =∣∣∣∣∫ φ2(x+h)

φ1(x+h)

f(x, y) dy −∫ φ2(x)

φ1(x)

f(x, y) dy∣∣∣∣

=∣∣∣∣∫ φ1(x)

φ1(x+h)

f(x, y) dy +∫ φ2(x+h)

φ2(x)

f(x, y) dy∣∣∣∣

≤∣∣∣∣∫ φ1(x)

φ1(x+h)

f(x, y) dy∣∣∣∣ +∣∣∣∣∫ φ2(x+h)

φ2(x)

f(x, y) dy∣∣∣∣

≤∣∣φ1(x) − φ1 (x + h)

∣∣M + |φ2 (x + h) − φ2(x)∣∣M.

The expression on the right tends to 0 as h tends to 0 since φ1 and φ2 are continuous.



SECTION 17.4 877

60. (a)∫ 3

−1

∫ 5

2

xe−xy dy dx ∼= −25.9893 (b)∫ 7

3

∫ 4

1

xy

x2 + y2dy dx ∼= 4.5720

61. (a)∫ 2

1

∫ 1+√x−1

x2−2x+2

1 dy dx = 13 (b)

∫ 2

1

∫ 1+√y−1

y2−2y+2

1 dx dy = 13

62. (a)∫ 1

0

∫ 3−y

2y

√2x + y dx dy ∼= 2.8133 (b)

∫ 2

0

∫ x/2

0

√2x + y dy dx+

∫ 3

2

∫ 3−x

0

√2x + y dy dx

∼= 2.8133

SECTION 17.4

1.∫ π/2

0

∫ sin θ

0

r cos θ dr dθ =∫ π/2

0

12

sin2 θ cos θ dθ =[16

sin3 θ

]π/20

=16

2.∫ π/4

0

∫ cos 2θ

0

r dr dθ =∫ π/4

0

cos2 2θ2

dθ =π

16

3.∫ π/2

0

∫ 3 sin θ

0

r2 dr dθ =∫ π/2

0

9 sin3 θ dθ = 9∫ π/2

0

(1 − cos2 θ) sin θ dθ = 9[− cos θ +

13

cos3 θ]π/20

= 6

4.∫ 2π/3

−π/3

∫ 2 cos θ

0

r sin θ dr dθ =∫ 2π/3

−π/3

2 cos2 θ sin θ dθ =16

5. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1∫∫Γ

(cos r2

)r drdθ =

∫ 2π

0

∫ 1

0

(cos r2

)r dr dθ = 2π

∫ 1

0

r cos r2 dr = π sin 1

(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2∫∫Γ

(cos r2

)r drdθ =

∫ 2π

0

∫ 2

1

(cos r2

)r dr dθ = 2π

∫ 2

1

r cos r2 dr = π(sin 4 − sin 1)

6. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1.∫∫Γ

(sin r)r dr dθ =∫ 2π

0

∫ 1

0

(sin r)r dr dθ = 2π∫ 1

0

r sin r dr = 2π(sin 1 − cos 1)

(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2.∫∫Γ

(sin r)r dr dθ =∫ 2π

0

∫ 2

1

(sin r)r dr dθ = 2π∫ 2

1

r sin r dr = 2π[cos 1 − 2 cos 2 + sin 2 − sin 1]

7. (a) Γ : 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 1∫∫Γ

(r cos θ + r sin θ)r drdθ =∫ π/2

0

∫ 1

0

r2(cos θ + sin θ) dr dθ

=( ∫ π/2

0

(cos θ + sin θ) dθ)(∫ 1

0

r2 dr

)= 2(

13

)=

23



878 SECTION 17.4

(b) Γ : 0 ≤ θ ≤ π/2, 1 ≤ r ≤ 2∫∫Γ

(r cos θ + r sin θ)r drdθ =∫ π/2

0

∫ 2

1

r2(cos θ + sin θ) dr dθ

=(∫ π/2

0

(cos θ + sin θ) dθ)(∫ 2

1

r2 dr

)= 2(

73

)=

143

8. Γ : 0 ≤ θ ≤ π3 , 0 ≤ r ≤ 1

cos θ∫∫Γ

√x2 + y2 dx dy =

∫ π/3

0

∫ 1/ cos θ

0

r · r dr dθ =∫ π/3

0

dθ

3 cos3 θ=

13

∫ π/3

0

sec3 θ dθ

=13

[12

sec θ tan θ +12

ln | sec θ + tan θ|]π/30

=13

√3 +

16

ln(2 +√

3)

9.∫ π/2

−π/2

∫ 1

0

r2 dr dθ =13π 10.

∫ π/2

0

∫ 2

0

r2 dr dθ =π

2

∫ 2

0

r2 dr =43π

11.∫ 1

1/2

∫ √1−x2

0

dy dx =∫ π/3

0

∫ 1

12 sec θ

r dr dθ =∫ π/3

0

(12− 1

8sec2 θ

)dθ =

16π −

√3

8

12.∫ π/3

0

∫ 1/2 sec θ

0

r4 cos θ sin θ dr dθ +∫ π/2

π/3

∫ 1

0

r4 cos θ sin θ dr dθ

=∫ π/3

0

sin θ

5(32) cos4 θdθ +

∫ π/2

π/3

15

cos θ sin θ dθ =7

480+

140

=19480

13.∫ 1

0

∫ √1−x2

0

sin√x2 + y2 dy dx =

∫ π/2

0

∫ 1

0

sin(r) r dr dθ =∫ π/2

0

(sin 1 − cos 1) dθ =π

2(sin 1 − cos 1)

14.∫ 2π

0

∫ 1

0

e−r2r dr dθ = 2π

∫ 1

0

re−r2dr = π(1 − 1

e)

15.∫ 2

0

∫ √2x−x2

0

x dy dx =∫ π/2

0

∫ 2 cos θ

0

r cos θ r dr dθ =83

∫ π/2

0

cos4 θ dθ =83· 34· 12· π

2=

π

2

(See Exercise 62, Section 8.3)∧

16. The region is the inside of the circle (x− 1/2)2 + y2 = 1/4, which has polar equation r = cos θ.

So the integral becomes ∫ π/2

−π/2

∫ cos θ

0

r3 dr dθ =∫ π/2

−π/2

14

cos4 θ dθ =3π32

17. A =∫ π/3

0

∫ 3 sin 3θ

0

r dr dθ =92

∫ π/3

0

sin2 3θ dθ =94

∫ π/3

0

(1 − 6 cos θ) dθ =3π4



SECTION 17.4 879

18. A =∫ 2π

0

∫ 2(1−cos θ)

0

r dr dθ =∫ 2π

0

2(1 − cos θ)2 dθ = 6π

19. First we find the points of intersection:

r = 4 cos θ = 2 =⇒ cos θ =12

=⇒ θ = ±π

3.

A =∫ π/3

−π/3

∫ 4 cos θ

2

r dr dθ =∫ π/3

−π/3

(8 cos2 θ − 2) dθ =∫ π/3

−π/3

(2 + 4 cos 2θ) dθ =4π3

+ 2√

3

20. A =∫ 2π

0

∫ 1+2cos θ

0

r dr dθ =∫ 2π

0

12(1 + 2 cos θ)2 dθ = 3π

21. A = 4∫ π/4

0

∫ 2√

cos 2θ

0

r dr dθ = 8∫ π/4

0

cos 2θ dθ = 4

22. A =∫ π/3

−π/3

∫ 3 cos θ

1+cos θ

r dr dθ =∫ π/3

−π/3

12[9 cos2 θ − (1 + cos θ)2

]dθ =

[3θ2

+ sin 2θ − sin θ

]π/3−π/3

= π

23.∫ 2π

0

∫ b

0

(r2 sin θ + br

)dr dθ =

∫ 2π

0

[13r3 sin θ +

b

2r2

]b0

dθ

= b3∫ 2π

0

(13

sin θ +12

)dθ = b3π

24. V =∫ 2π

0

∫ 1

0

(1 − r2)r dr dθ = 2π∫ 1

0

(r − r3) dr =π

2

25. 8∫ π/2

0

∫ 2

0

r

2

√12 − 3r2 dr dθ = 8

∫ π/2

0

[− 1

18(12 − 3r2

)3/2]20

dθ

= 8∫ π/2

0

43

√3 dθ =

163

√3π

26. V =∫ 2π

0

∫ √5

0

6√

5 − r2 r dr dθ = 2π∫ √

5

0

r6√

5 − r2 dr =307

(5)1/6π

27.∫ 2π

0

∫ 1

0

r√

4 − r2 dr dθ =∫ 2π

0

[−1

3(4 − r2

)3/2]10

dθ

=∫ 2π

0

(83−√

3)dθ =

23(8 − 3

√3 )π

28. V =∫ π/2

−π/2

∫ cos θ

0

(1 − r2)r dr dθ =∫ π/2

−π/2

(cos2 θ

2− cos4 θ

4

)dθ =

5π32



880 SECTION 17.5

29.∫ π/2

−π/2

∫ 2 cos θ

0

2r2 cos θ dr dθ =∫ π/2

−π/2

[23r3 cos θ

]2 cos θ

0

dθ

=∫ π/2

−π/2

163

cos4 θ dθ =323

∫ π/2

0

cos4 θ dθ =323

(316

π

)= 2π

Ex. 46, Sect. 8.3∧

30.∫ π/2

−π/2

∫ 2a cos θ

0

r2 dr dθ =∫ π/2

−π/2

8a3

3cos3 θdθ =

329a3

31.b

a

∫ π

0

∫ a sin θ

0

r√a2 − r2 dr dθ =

b

a

∫ π

0

[−1

3(a2 − r2

)3/2]a sin θ

0

dθ

=13a2b

∫ π

0

(1 − cos3 θ

)dθ =

13πa2b

32. (a) V = 2∫ 2π

0

∫ r

0

√R2 − s2 s ds dθ = 4π

∫ r

0

S√R2 − S2 dS =

4π3

[R3 − (R2 − r2)3/2

].

(b)43πR3 − V =

43π(R2 − r2)3/2

33. A = 2∫ π/4

0

∫ 2 cos 2θ

0

r dr dθ = 2∫ π/4

0

12 (2 cos 2θ)2 dθ = 2

∫ π/4

0

(1 + cos 4θ) dθ =π

2

34.∫∫

Ω

ex2+y2

dxdy =∫ 2π

0

∫ 4

2

rer2dr dθ = π

(e16 − e4

)

SECTION 17.5

1. M =∫ 1

−1

∫ 1

0

x2 dy dx =23

xM M =∫ 1

−1

∫ 1

0

x3 dy dx = 0 =⇒ xM = 0

yM M =∫ 1

−1

∫ 1

0

x2y dy dx =∫ 1

−1

12x2 dx =

13

=⇒ yM =1/32/3

=12

2. M =∫ 1

0

∫ √x

0

(x + y) dy dx =∫ 1

0

(x3/2 +

x

2

)dx =

1320

xMM =∫ 1

0

∫ √x

0

x(x + y) dy dx =∫ 1

0

(x5/2 +

x2

2

)dx =

1942

=⇒ xM =190273

yMM =∫ 1

0

∫ √x

0

y(x + y) dy dx =∫ 1

0

(x2

2+

x3/2

3

)dx =

310

=⇒ yM =613



SECTION 17.5 881

3. M =∫ 1

0

∫ 1

x2xy dy dx =

12

∫ 1

0

(x− x5) dx =16

xM M =∫ 1

0

∫ 1

x2x2y dy dx =

12

∫ 1

0

(x2 − x6) dx =221

=⇒ xM =2/211/6

=47

yM M =∫ 1

0

∫ 1

x2xy2 dy dx =

13

∫ 1

0

(x− x7) dx =18

=⇒ yM =1/81/6

=34

4. M =∫ π

0

∫ sinx

0

y dy dx =∫ π

0

sin2 x

2dx =

π

4

xMM =∫ π

0

∫ sinx

0

xy dy dx =∫ π

0

xsin2 x

2dx =

π2

8=⇒ xM =

π

2

yMM =∫ π

0

∫ sinx

0

y2 dy dx =∫ π

0

sin3 x

3dx =

49

=⇒ yM =169π

5. M =∫ 8

0

∫ x1/3

0

y2 dy dx =13

∫ 8

0

x dx =323

xM M =∫ 8

0

∫ x1/3

0

xy2 dy dx =13

∫ 8

0

x2 dx =5129

=⇒ xM =512/932/3

=163

yM M =∫ 8

0

∫ x1/3

0

y3 dy dx =14

∫ 8

0

x4/3 dx =967

=⇒ yM =96/732/3

=97

6. M =∫ a

0

∫ √a2−x2

0

xy dy dx =∫ a

0

x

2(a2 − x2) dx =

a4

8

xMM =∫ a

0

∫ √a2−x2

0

x2y dy dx =∫ a

0

x2

2(a2 − x2) dx =

a5

15=⇒ xM =

815

a

yMM =∫ a

0

∫ √a2−x2

0

xy2 dy dx =∫ a

0

x

3(a2 − x2)3/2 dx =

a5

15=⇒ yM =

815

a

7. M =∫ 1

0

∫ 3x

2x

xy dy dx =52

∫ 1

0

x3 dx =58

xM M =∫ 1

0

∫ 3x

2x

x2y dy dx =52

∫ 1

0

x4 dx =12

=⇒ xM =1/25/8

=45

yM M =∫ 1

0

∫ 3x

2x

xy2 dy dx =193

∫ 1

0

x4 dx =1915

=⇒ yM =19/155/8

=15275

8. M =∫ 2

0

∫ 3− 32x

0

(x + y) dy dx =∫ 2

0

[x(3 − 3

2x) +

12(3 − 3

2x)2]dx = 5

xMM =∫ 2

0

∫ 3− 32x

0

x(x + y) dy dx =∫ 2

0

[x2(3 − 3

2x) +

x

2(3 − 3

2x)2]dx =

72

=⇒ xM =710

yMM =∫ 2

0

∫ 3− 32x

0

y(x + y) dy dx =∫ 2

0

[x

2(3 − 3

2x)2 +

(3 − 32x)3

3

]dx = 6 =⇒ yM =

65



882 SECTION 17.5

9. M =∫ 2π

0

∫ 1+cos θ

0

r2 dr dθ =13

∫ 2π

0

(1 + 3 cos θ + 3 cos2 θ + cos3 θ) dθ =5π3

xM M =∫ 2π

0

∫ 1+cos θ

0

r3 cos θ dr dθ =14

∫ 2π

0

(1 + cos θ)4 cos θ dθ

=14

∫ 2π

0

[cos θ + 4 cos2 θ + 6 cos3 θ + 4 cos4 θ + cos5 θ

]dθ

=7π4

Therefore, xM =7π/45π/3

=2120

.

yM M =∫ 2π

0

∫ 1+cos θ

0

r3 sin θ dr dθ =14

∫ 2π

0

(1 + cos θ)4 sin θ dθ =14

[15(1 + cos θ)5

]2π0

= 0

Therefore, yM = 0.

10. M =∫∫Ω

y dx dy =∫ 5π/6

π/6

∫ 2 sin θ

1

r sin θ r dr dθ =∫ 5π/6

π/6

(83

sin4 θ − 13

sin θ

)dθ =

3√

3 + 8π12

xM = 0 by symmetry

yM M =∫ 5π/6

π/6

∫ 2 sin θ

1

r2 sin2 θ r dr dθ =∫ 5π/6

π/6

(4 sin6 θ − 1

4sin2 θ

)dθ =

11√

3 + 12π16

Therefore, yM =33√

3 + 36π12√

3 + 32π

11. Ω : −L/2 ≤ x ≤ L/2, −W/2 ≤ y ≤ W/2

Ix =∫∫Ω

M

LWy2 dxdy =

4MLW

∫ W/2

0

∫ L/2

0

y2dx dy =112

MW 2

symmetry∧

Iy =∫∫Ω

M

LWx2 dxdy =

112

ML2, Iz =∫∫Ω

M

LW

(x2 + y2

)dxdy =

112

M(L2 + W 2

)

Kx =√Ix/M =

W√

36

, Ky =√Iy/M =

L√

36

Kz =√Iz/M =

√3√L2 + W 2

6

12. λ(x, y) = k

(x +

L

2

)M =

∫ W/2

−W/2

∫ L/2

−L/2

k

(x +

L

2

)dx dy =

kWL2

2

Ix =∫ W/2

−W/2

∫ L/2

−L/2

k

(x +

L

2

)y2 dx dy =

(∫ W/2

−W/2

y2 dy

)(∫ L/2

−L/2

k

(x +

L

2

)dx

)

=(W 3

12

)(M

W

)=

112

MW 2



SECTION 17.5 883

Iy =∫ W/2

−W/2

∫ L/2

−L/2

k

(x +

L

2

)x2 dx dy =

kL4W

24=

112

ML2

Iz =∫ W/2

−W/2

∫ L/2

−L/2

k

(x +

L

2

)(x2 + y2) dx dy = Ix + Iy =

112

M(L2 + W 2).

13. M =∫∫Ω

k

(x +

L

2

)dxdy =

∫∫Ω

12kLdxdy =

12kL( area of Ω) =

12kL2W

symmetry∧

xMM =∫∫Ω

x

[k

(x +

L

2

)]dxdy =

∫∫Ω

(kx2 +

12Lx

)dxdy

=∫∫Ω

kx2 dxdy = 4k∫ W/2

0

∫ L/2

0

x2 dx dy =112

kWL3

symmetry∧

symmetry∧

= 16

(12kL

2W)L = 1

6 ML; xM = 16 L

yMM =∫∫Ω

y

[k

(x +

L

2

)]dxdy = 0; yM = 0

by symmetry∧

14. Iz =∫∫Ω

λ(x, y)[x2 + y2] dx dy =∫∫Ω

λ(x, y)x2 dx dy +∫∫Ω

λ(x, y)y2 dx dy = Ix + Iy.

Since Iz = Ix + Iy, we have MKz2 = MKx

2 + MKy2 therefore Kz

2 = Kx2 + Ky

2.

15. Ix =∫∫Ω

4MπR2

y2 dxdy =4MπR2

∫ π/2

0

∫ R

0

r3 sin2 θ dr dθ

=4MπR2

(∫ π/2

0

sin2 θ dθ

)(∫ R

0

r3 dr

)=

4MπR2

(π4

)(14R4

)=

14MR2

Iy = 14MR2, Iz = 1

2MR2

Kx = Ky = 12R, Kz = R/

√2

16. Iz = IM + d2M. Rotation doesn’t change d, doesn’t change M , and doesn’t change IM .



884 SECTION 17.5

17. IM , the moment of inertia about the vertical line through the center of mass, is∫∫Ω

M

πR2

(x2 + y2

)dxdy

where Ω is the disc of radius R centered at the origin. Therefore

IM =M

πR2

∫ 2π

0

∫ R

0

r3 dr dθ =12MR2.

We need I0 = 12 MR2 + d2M where d is the distance from the center of the disc to the origin. Solving

this equation for d, we have d =√I0 − 1

2 MR2

/√M.

18. Ix =∫ b

a

∫ f(x)

0

λy2 dy dx =λ

3

∫ b

a

[f(x)]3 dx

Iy =∫ b

a

∫ f(x)

0

λx2 dy dx = λ

∫ b

a

x2f(x) dx.

19. Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b

Ix =∫∫Ω

4Mπab

y2 dxdy =4Mπab

∫ a

0

∫ ba

√a2−x2

0

y2 dy dx =14Mb2

Iy =∫∫Ω

4Mπab

x2 dxdy =4Mπab

∫ a

0

∫ ba

√a2−x2

0

x2 dy dx =14Ma2

Iz = 14M(a2 + b2

)

20. Ix =∫ 1

0

∫ √x

0

(x + y)y2 dy dx =∫ 1

0

(x5/2

3+

x2

4

)dx =

528

Iy =∫ 1

0

∫ √x

0

(x + y)x2 dy dx =∫ 1

0

(x7/2 +

x3

2

)dx =

2572

; Iz = Ix + Iy.

21. Ix =∫ 1

0

∫ 1

x2xy3 dy dx =

14

∫ 1

0

(x− x9) dx =110

Iy =∫ 1

0

∫ 1

x2x3y dy dx =

12

∫ 1

0

(x3 − x7) dx =116

Iz =∫ 1

0

∫ 1

x2xy(x2 + y2) dy dx = Ix + Iy =

1380

22. Ix =∫ 8

0

∫ 3√x

0

y2 · y2 dy dx =∫ 8

0

x5/3

5dx =

965

Iy =∫ 8

0

∫ 3√x

0

y2 · x2 dy dx =∫ 8

0

x3

3dx =

10243

; Iz = Ix + Iy



SECTION 17.5 885

23. Ix =∫ 2π

0

∫ 1+cos θ

0

r4 sin2 θ dr dθ =15

∫ 2π

0

(1 + cos θ)5 sin2 θ dθ =33π40

Iy =∫ 2π

0

∫ 1+cos θ

0

r4 cos2 θ dr dθ =15

∫ 2π

0

(1 + cos θ)5 cos2 θ dθ =93π40

Iz =∫ 2π

0

∫ 1+cos θ

0

r4 dr dθ = Ix + Iy =63π20

24. xM =x1M1 + x2M2

M1 + M2, yM =

y1M1 + y2M2

M1 + M2

25. Ω : r21 ≤ x2 + y2 ≤ r2

2, A = π(r22 − r2

1

)(a) Place the diameter on the x-axis.

Ix =∫∫Ω

M

Ay2 dxdy =

M

A

∫ 2π

0

∫ r2

r1

(r2 sin2 θ

)r dr dθ =

14M(r22 + r2

1

)(b) 1

4M(r22 + r2

1

)+ Mr2

1 = 14M(r22 + 5r2

1

)(parallel axis theorem)

(c) 14M(r22 + r2

1

)+ Mr2

2 = 14M(5r2

2 + r21

)26. Set r1 = r2 = r in the proceeding problem. Then the required moments of inertia are

(a) 12Mr2 (b) 3

2Mr2.

27. Ω : r21 ≤ x2 + y2 ≤ r2

2, A = π(r22 − r2

1

)I =∫∫Ω

M

A

(x2 + y2

)dxdy =

M

A

∫ 2π

0

∫ r2

r1

r3 dr dθ =12M(r2

2 + r21)

28. Let l be the x-axis and let the plane of the plate be the xy-plane. Then

I − IM =∫∫Ω

λ(x, y)y2 dx dy −∫∫Ω

λ(x, y)(y − yM )2 dx dy

=∫∫Ω

λ(x, y)[2yMy − y2M ] dx dy

= 2yM∫∫Ω

yλ(x, y) dx dy − y2M

∫∫Ω

λ(x, y) dx dy

= 2y2MM − y2

MM = y2MM = d2M.

29. M =∫∫Ω

k(R−√x2 + y2

)dxdy = k

∫ π

0

∫ R

0

(Rr − r2

)dr dθ =

16kπR3

xM = 0 by symmetry

yMM =∫∫Ω

y[k(R−√x2 + y2

)]dxdy = k

∫ π

0

∫ R

0

(Rr2 − r3

)sin θ dr dθ =

16kR4

yM = R/π



886 SECTION 17.5

30. Ix =∫∫Ω

k(R−√x2 + y2)y2 dx dy = k

∫ π

0

∫ R

0

(R− r) r2 sin2 θ r dr dθ =kπR5

40=

3MR2

20.

Iy = k

∫ π

0

∫ R

0

(R− r)r2 cos2 θr dr dθ =kπR5

40=

3MR2

20

Iz = Ix + Iy =3MR2

10.

31. Place P at the origin.

M =∫∫Ω

k√x2 + y2 dxdy

= k

∫ π

0

∫ 2R sin θ

0

r2 dr dθ =329kR3

xM = 0 by symmetry

yMM =∫∫Ω

y(k√x2 + y2

)dxdy = k

∫ π

0

∫ 2R sin θ

0

r3 sin θ dr dθ =6415

kR4

yM = 6R/5

Answer: the center of mass lies on the diameter through P at a distance 6R/5 from P .

32. Putting the right angle at the origin, we have λ(x, y) = k(x2 + y2).

M =∫ b

0

∫ h−hb x

0

k(x2 + y2) dy dx =112

kbh(b2 + h2)

xMM =∫ b

0

∫ h−hb x

0

kx(x2 + y2) dy dx =kb2h(3b2 + h2)

60=⇒ xM =

b(3b2 + h2)5(b2 + h2)

yMM =∫ b

0

∫ h−hb x

0

ky(x2 + y2) dy dx =kbh2(b2 + 3h2)

60=⇒ yM =

h(b2 + 3h2)5(b2 + h2)

33. Suppose Ω, a basic region of area A, is broken up into n basic regions Ω1, · · · , Ωn with areas

A1, · · · , An. Then

xA =∫∫Ω

x dxdy =n∑

i=1

⎛⎝∫∫

Ωi

x dxdy

⎞⎠ =

n∑i=1

xi Ai = x1 A1 + · · · + xn An.

The second formula can be derived in a similar manner.

34. (a) M =∫ 2

0

∫ 1

x/2

(x + y) dy dx =43

xM M =∫ 2

0

∫ 1

x/2

x(x + y) dy dx =76; xM =

78

yM M =∫ 2

0

∫ 1

x/2

y(x + y) dy dx = 1; yM =34

(b) Ix =∫ 2

0

∫ 1

x/2

y2(x + y) dy dx =45

Iy =∫ 2

0

∫ 1

x/2

x2(x + y) dy dx =43



SECTION 17.6 887

SECTION 17.6

1. They are equal; they both give the volume of T .

2. (a) Lf (P ) =m∑i=1

n∑j=1

q∑k=1

xi−1yj−1zk−1ΔxiΔyjΔzk, Uf (P ) =m∑i=1

n∑j=1

q∑k=1

xiyjzkΔxiΔyjΔzk

(b) xi−1yj−1zk−1 ≤(xi + xi−1

2

)(yj + yj−1

2

)(zk + zk−1

2

)≤ xiyjzk

xi−1yj−1zk−1ΔxiΔyjΔzk ≤ 18(xi

2 − xi−12) (

yj2 − yj−1

2) (

zk2 − zk−1

2)≤ xiyjzkΔxiΔyjΔzk

Lf (P ) ≤ 18

m∑i=1

n∑j=1

q∑k=1

(xi

2 − xi−12) (

yj2 − yj−1

2) (

zk2 − zk−1

2)≤ Uf (P ).

The middle term can be written

18

(m∑i=1

xi2 − xi−1

2

)⎛⎝ n∑j=1

yj2 − yj−1

2

⎞⎠( q∑

k=1

zk2 − zk−1

2

)=

18(1)(1)(1) =

18.

Therefore I =18.

3.∫∫∫Π

αdx dy dz = α

∫∫∫Π

dx dy dz = α (volume of Π) = α(a2 − a1)(b2 − b1)(c2 − c1)

4. Since the volume is 1, the average value is∫∫∫Ω

xyz dx dy dz =18.

5. Let P1 = {x0, · · · , xm}, P2 = {y0, · · · , yn}, P3 = {z0, · · · , zq} be partitions of [ 0, a ], [ 0, b ], [ 0, c ]

respectively and let P = P1 × P2 × P3. Note that

xi−1yj−1 ≤(xi + xi−1

2

)(yj + yj−1

2

)≤ xiyj

and therefore

xi−1yj−1 Δxi Δyj Δzk ≤ 14

(xi

2 − x2i−1

) (yj

2 − y2j−1

)Δzk ≤ xiyj Δxi ΔyjΔzk.

It follows that

Lf (P ) ≤ 14

m∑i=1

n∑j=1

q∑k=1

(xi

2 − x2i−1

) (yj

2 − y2j−1

)Δzk ≤ Uf (P ).


14

(m∑i=1

xi2 − x2

i−1

)⎛⎝ n∑j=1

yj2 − y2

j−1

⎞⎠( q∑

k=1

Δzk

)=

14a2b2c.

6. Ix = Ixy + Ixz, Iy = Ixy + Iyz, Iz = Ixz + Iyz



888 SECTION 17.6

7. x1 = a, y1 = b, z1 = c; x0 = A, y0 = B, z0 = C

x1V1 + xV = x0V0 =⇒ a2bc + (ABC − abc)x = A2BC

=⇒ x =A2BC − a2bc

ABC − abc

similarly

y =AB2C − ab2c

ABC − abc, z =

ABC2 − abc2

ABC − abc

8. Encase T in a box Π. A partition P of Π breaks up II into little boxes Πijk. Since f is

nonnegative on Π, all the mijk are nonnegative. Therefore

0 ≤ Lf (P ) ≤∫∫∫T

f(x, y, z) dx dy dz.

9. M =∫∫∫Π

Kz dxdydz

Let P1 = {x0, · · · , xm}, P2 = {y0, · · · , yn}, P3 = {z0, · · · , zq} be partitions of [ 0, a ] and let

P = P1 × P2 × P3. Note that

zk−1 ≤ 12 (zk + zk−1) ≤ zk

and therefore

Kzk−1 Δxi Δyj Δzk ≤ 12K Δxi Δyj

(zk

2 − z2k−1

)≤ Kzk Δxi ΔyjΔzk.

It follows that

Lf (P ) ≤ 12K

m∑i=1

n∑j=1

q∑k=1

Δxi Δyj(zk

2 − z2k−1

)≤ Uf (P ).


12K

(m∑i=1

Δxi

)⎛⎝ n∑j=1

Δyj

⎞⎠ ( q∑

k=1

zk2 − z2

k−1

)=

12K(a) (a) (a2) =

12Ka4.

M = 12Ka4 where K is the constant of proportionality for the density function.

10. xMM =∫∫∫Π

Kzxdx dy dz =Ka5

4=⇒ xM =

12a

yMM =∫∫∫Π

Kzy dx dy dz =Ka5

4=⇒ yM =

12a

zMM =∫∫∫Π

Kz2 dx dy dz =Ka5

3=⇒ zM =

23a.



SECTION 17.7 889

11. Iz =∫∫∫Π

Kz(x2 + y2

)dxdydz

=∫∫∫Π

Kx2z dxdydz

︸︷︷︸I1

+∫∫∫Π

Ky2z dxdydz

︸︷︷︸I2

.

We will calculate I1 using the partitions we used in doing Exercise 9. Note that

x2i−1zk−1 ≤

(xi

2 + xixi−1 + x2i−1

3

)(zk + zk−1

2

)≤ xi

2zk

and therefore

Kx2i−1zk−1 Δxi Δyj Δzk ≤ 1

6 K(xi

3 − x3i−1

)Δyj(zk

2 − z2k−1

)≤ Kxi

2zk2 Δxi Δyj Δzk.

It follows that

Lf (P ) ≤ 16K

m∑i=1

n∑j=1

q∑k=1

(xi

3 − x3i−1

)Δyj(zk

2 − z2k−1

)≤ Uf (P ).


16K

(m∑i=1

xi3 − x3

i−1

)⎛⎝ n∑j=1

Δyj

⎞⎠( q∑

k=1

zk2 − z2

k−1

)=

16Ka3 (a)(a2) =

16Ka6.

Similarly I2 = 16 Ka6 and therefore Iz = 1

3 Ka6 = 23

(12 Ka4

)a2 = 2

3 Ma2.

by Exercise 9∧

12. (a) Lf (P ) ∼= 56.4803 Uf (P ) ∼= 57.5603 (c)∫∫R

(3y2 − 2x

)dxdy = 57

SECTION 17.7

1.∫ a

0

∫ b

0

∫ c

0

dx dy dz =∫ a

0

∫ b

0

c dy dz =∫ a

0

bc dz = abc

2.∫ 1

0

∫ x

0

∫ y

0

y dz dy dx =∫ 1

0

∫ x

0

y2 dy dx =∫ 1

0

x3

3dx =

112

.

3.∫ 1

0

∫ 2y

1

∫ x

0

(x + 2z) dz dx dy =∫ 1

0

∫ 2y

1

[xz + z2

]x0dx dy =

∫ 1

0

∫ 2y

1

2x2 dx dy

=∫ 1

0

[23x3

]2y1

dy =∫ 1

0

(163

y3 − 23

)dy =

23

4.∫ 1

0

∫ 1+x

1−x

∫ xy

0

4z dz dy dx =∫ 1

0

∫ 1+x

1−x

2x2y2 dy dx =∫ 1

0

2x2

3[(1 + x)3 − (1 − x)3

]dx =

119

5.∫ 2

0

∫ 1

−1

∫ 3

0

(z − xy) dz dy dx =∫ 2

0

∫ 1

−1

[12z2 − xyz

]31

dy dx

=∫ 2

0

∫ 1

−1

(4 − 2xy) dy dx =∫ 2

0

[2y − xy2

]1−1

dx =∫ 2

0

8 dy = 16



890 SECTION 17.7

6.∫ 2

0

∫ 1

−1

∫ 3

1

(z − xy) dy dx dz =∫ 2

0

∫ 1

−1

(2z − 4x) dx dz =∫ 2

0

4z dz = 8

7.∫ π/2

0

∫ 1

0

∫ √1−x2

0

x cos z dy dx dz =∫ π/2

0

∫ 1

0

[xy cos z]√

1−x2

0 dx dz

=∫ π/2

0

∫ 1

0

x√

1 − x2 cos z dx dz =∫ π/2

0

[−1

3(1 − x2)3/2 cos z

]10

dz =13

∫ π/2

0

cos z dz =13

8.∫ 2

−1

∫ y−2

1

∫ e2

e

x + y

zdz dx dy =

∫ 2

−1

∫ y−2

1

(x + y) dx dy =∫ 2

−1

[(y − 2)2 − 1

2+ y(y − 3)

]dy =

32

9.∫ 2

1

∫ y2

y

∫ lnx

0

yez dz dx dy =∫ 2

1

∫ y2

y

[yez]lnx0 dx dy

=∫ 2

1

∫ y2

y

y(x− 1) dx dy =∫ 2

1

[12x2y − xy

]y2

y

dy =∫ 2

1

(12y5 − 3

2y3 + y2

)dy =

4724

10.∫ π/2

0

∫ π/2

0

∫ 1

0

ez cosx sin y dz dy dx =∫ π/2

0

∫ π/2

0

(e− 1) cosx sin y dy dx

=∫ π/2

0

(e− 1) cosx dx = e− 1

11.∫∫∫Π

f(x)g(y)h(z) dxdydz =∫ c2

c1

[∫ b2

b1

(∫ a2

a1

f(x)g(y)h(z) dx)dy

]dz

=∫ c2

c1

[∫ b2

b1

g(y)h(z)(∫ a2

a1

f(x) dx)dy

]dz

=∫ c2

c1

[h(z)

(∫ a2

a1

f(x) dx)(∫ b2

b1

g(y) dy)dz

]

=(∫ a2

a1

f(x) dx) (∫ b2

b1

g(y) dy

)(∫ c2

c1

h(z) dz)

12.(∫ 1

0

x3 dx

)(∫ 2

0

y2 dy

)(∫ 3

0

z dz

)=(

14

)(83

)(92

)= 3

13.(∫ 1

0

x2 dx

)(∫ 2

0

y2 dy

)(∫ 3

0

z2 dz

)=(

13

)(83

)(273

)= 8

14. M =∫∫∫Π

kxyz dx dy dz = k

(∫ a

0

x dx

)(∫ b

0

y dy

)(∫ c

0

z dz

)=

18ka2b2c2

15. xMM =∫∫∫Π

kx2yz dxdydz = k

(∫ a

0

x2 dx

) (∫ b

0

y dy

) (∫ c

0

z dz

)

= k(

13 a

3) (

12 b

2) (

12 c

2)

= 112 ka

3 b2 c2.

By Exercise 14, M = 18 ka

2 b2 c2. Therefore x = 23 a. Similarly, y = 2

3 b and z = 23 c.



SECTION 17.7 891

16. (a)

I =∫∫∫Π

kxyz[(x− a)2 + (y − b)2

]dx dy dz

= k

(∫ a

0

x(x− a)2 dx)(∫ b

0

y dy

)(∫ c

0

z dz

)+ k

(∫ a

0

x dx

)(∫ b

0

y(y − b)2 dy)(∫ c

0

z dz

)

=148

ka2b2c2(a2 + b2).

Since M = 18ka

2b2c2, I = 16M(a2 + b2).

(b) IM − 118

(a2 + b2) by parallel axis theorem

17.

18. V =∫∫∫T

dx dy dz =∫ 1

0

∫ 1

0

∫ 1−y

0

dz dy dx =12

19. center of mass is the centroid

x = 12 by symmetry

yV =∫∫∫T

y dxdydz =∫ 1

0

∫ 1

0

∫ 1−y

0

y dz dy dx =∫ 1

0

∫ 1

0

(y − y2

)dy dx

=∫ 1

0

[12y2 − 1

3y3

]10

dx =∫ 1

0

16dx =

16

zV =∫∫∫T

z dxdydz =∫ 1

0

∫ 1

0

∫ 1−y

0

z dz dy dx =∫ 1

0

∫ 1

0

12(1 − y)2 dy dx

=12

∫ 1

0

∫ 1

0

(1 − 2y + y2

)dy dx =

12

∫ 1

0

[y − y2 1

3y3

]10

dx =12

∫ 1

0

13dx =

16

V = 12 (by Exercise 18 ); y = 1

3 , z = 13

20. Ix =∫∫∫T

M

V(y2 + z2) dx dy dz =

13M

Iy =∫∫∫T

M

V(x2 + z2) dx dy dz =

12M Iz =

∫∫∫T

M

V(x2 + y2) dx dy dz =

12M



892 SECTION 17.7

21.∫ r

−r

∫ φ(x)

−φ(x)

∫ ψ(x,y)

−ψ(x,y)

k(r −√x2 + y2 + z2

)dz dy dx with φ(x) =

√r2 − x2,

ψ(x, y) =√r2 − (x2 + y2) , k the constant of proportionality

22.∫ 1

−1

∫ √1−x2

−√

1−x2

∫ 1

√x2+y2

k√x2 + y2 + z2 dz dy dx

23.∫ 1

0

∫ √x−x2

−√x−x2

∫ 1−y2

−2x−3y−10

dz dy dx

24.∫ √

2

−√

2

∫ √1−x2/2

−√

1−x2/2

∫ 4−x2−y2

2+y2dz dy dx

25.∫ 1

−1

∫ 2√

2−2x2

−2√

2−2x2

∫ 4−x2−y2/4

3x2+y2/4

k

(z − 3x2 − 1

4y2

)dz dy dx

26.∫ √

2

−√

2

∫ √2−y2

−√

2−y2

∫ 4−z2

z2+2y2k√x2 + y2 dx dz dy

27.∫∫T

∫ (x2z + y

)dx dy dz =

∫ 2

0

∫ 3

1

∫ 1

0

(x2z + y

)dx dy dz =

∫ 2

0

∫ 3

1

[13x3z + xy

]10

dy dz

=∫ 2

0

∫ 3

1

(13z + y

)dy dz =

∫ 2

0

[13zy +

12y2

]31

dz =∫ 2

0

(23z + 4

)dz =

283

28.∫ 1

0

∫ y

0

∫ x+y

0

2yex dz dx dy =∫ 1

0

∫ y

0

2y(x + y)ex dx dy =∫ 1

0

(4y2ey − 2yey + 2y − 2y2) dy = 4e− 293

29.∫∫T

∫x2y2z2 dx dy dz =

∫ 0

−1

∫ y+1

0

∫ 1

0

x2y2z2 dx dz dy +∫ 1

0

∫ 1−y

0

∫ 1

0

x2y2z2 dx dz dy

=∫ 0

−1

∫ y+1

0

[13x3y2z2

]10

dz dy +∫ 1

0

∫ 1−y

0

[13x3y2z2

]10

dz dy

=13

∫ 0

−1

∫ y+1

0

y2z2 dz dy +13

∫ 1

0

∫ 1−y

0

[y2z2]10dz dy

=13

∫ 0

−1

[13y2z3

]y+1

0

dy +13

∫ 1

0

[13y2z3

]1−y

0

dy

=19

∫ 0

−1

(y5 + 3y4 + 3y3 + y2

)dy +

19

∫ 1

0

(y2 − 3y3 + 3y4 − y5

)dy =

1270



SECTION 17.7 893

30.∫ 2

0

∫ √4−x2

0

∫ √4−x2−y2

0

xy dz dy dx =∫ 2

0

∫ √4−x2

0

xy√

4 − x2 − y2 dy dx

=∫ π/2

0

∫ 2

0

r2 cos θ sin θ√

4 − r2r dr dθ

=12

∫ 2

0

r3√

4 − r2 dr =14

∫ 4

0

(4√u− u3/2) du =

3215

31.∫∫∫T

y2 dx dy dz =∫ 3

0

∫ 2−2x/3

0

∫ 6−2x−3y

0

y2 dz dy dx =∫ 3

0

∫ 2−2x/3

0

[y2z]6−2x−3y

0dy dx

=∫ 3

0

∫ 2−2x/3

0

(6y2 − 2xy2 − 3y3) dy dx

=∫ 3

0

[2y3 − 2

3xy3 − 3

4y4

]2−2x/3

0

dx

=14

∫ 3

0

(2 − 2

3x

)dx =

125

32.∫ 1

0

∫ 1−x2

0

∫ √1−y

0

y2 dz dy dx =∫ 1

0

∫ 1−x2

0

y2√

1 − y dy dx

=∫ 1

0

[−2

3x3 +

45x5 − 2

7x7 +

16105

]dx =

112

33. V =∫ 2

0

∫ x+2

x2

∫ x

0

dz dy dx =∫ 2

0

∫ x+2

x2x dy dx =

∫ 2

0

(x2 + 2x− x3

)dx =

83

xV =∫ 2

0

∫ x+2

x2

∫ x

0

x dz dy dx =∫ 2

0

∫ x+2

x2x2 dy dx =

∫ 2

0

(x3 + 2x2 − x4

)dx =

4415

yV =∫ 2

0

∫ x+2

x2

∫ x

0

y dz dy dx =∫ 2

0

∫ x+2

x2xy dy dx =

∫ 2

0

12(x3 + 4x2 + 4x− x5

)dx = 6

zV =∫ 2

0

∫ x+2

x2

∫ x

0

z dz dy dx =∫ 2

0

∫ x+2

x2

12x2 dy dx =

∫ 2

0

12(x3 + 2x2 − x4

)dx =

2215

x =1110

, y =94, z =

1120

34. (a) M =∫ 1

0

∫ 1

0

∫ 1

0

kz dx dy dz =12k

(b) M =∫ 1

0

∫ 1

0

∫ 1

0

k(x2 + y2 + z2) dz dy dx = k

35. V =∫ 2

−1

∫ 3

0

∫ 4−x2

2−x

dz dy dx = 272 ; (x, y, z) =

(12 ,

32 ,

125

)



894 SECTION 17.7

36.∫∫∫T

(x− x) dx dy dz =∫∫∫T

x dx dy dz − x

∫∫∫T

dx dy dz = xV − xV = 0.

Similarly, the other two integrals are zero.

37. V =∫ a

0

∫ φ(x)

0

∫ ψ(x,y)

0

dz dy dx =16abc with φ(x) = b

(1 − x

a

), ψ(x, y) = c

(1 − x

a− y

b

)(x, y, z) =

(14 a,

14 b,

14 c)

38. Iz =∫∫∫T

M

V(x2 + y2) dx dy dz =

130

(M

V

)=

15M

39. Π: 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c

(a) Iz =∫ a

0

∫ b

0

∫ c

0

M

abc

(x2 + y2

)dz dy dx =

13M(a2 + b2

)(b) IM = Iz − d2M = 1

3 M(a2 + b2

)− 1

4

(a2 + b2

)M = 1

12M(a2 + b2

)∧

parallel axis theorem (17.5.7)

(c) I = IM + d2M = 112 M

(a2 + b2

)+ 1

4 a2M = 1

3 Ma2 + 112 Mb2

∧parallel axis theorem (17.5.7)

40. V =∫ 2

1

∫ 2

1

∫ 1+x+y

−2

dz dy dx =∫ 2

1

∫ 2

1

(3 + x + y) dy dx = 6

xV =∫ 2

1

∫ 2

1

∫ 1+x+y

−2

x dz dy dx =10912

=⇒ x =10972

= y by symmetry

zV =∫ 2

1

∫ 2

1

∫ 1+x+y

−2

z dz dy dx =7312

=⇒ z =7372

.

41. M =∫ 1

0

∫ 1

0

∫ y

0

k(x2 + y2 + z2

)dz dy dx =

∫ 1

0

∫ 1

0

k

(x2y + y3 +

13y3

)dy dx

=∫ 1

0

k

(12x2 +

13

)dx =

12k

(xM , yM , zM ) =(

712 ,

3445 ,

3790

)

42. T is symmetric (a) about the yz-plane, (b) about the xz-plane, (c) about the xy-plane,

(d) about the origin.



SECTION 17.7 895

43. (a) 0 by symmetry

(b)∫∫∫T

(a1 x + a2 y + a3z + a4) dxdydz =∫∫∫T

a4 dxdydz = a4 (volume of ball) = 43 πa4

by symmetry∧

44.∫ 2

0

∫ 2

0

∫ 4−y2

2−y

x2y2 dz dy dx =35245

45. V = 8∫ a

0

∫ √a2−x2

0

∫ √a2−x2−y2

0

dz dy dx = 8∫ a

0

∫ √a2−x2

0

√a2 − x2 − y2 dy dx

polar coordinates∧

= 8∫ π/2

0

∫ a

0

√a2 − r2 r dr dθ

= −4∫ π/2

0

[23(a2 − r2)3/2

]a0

dθ

=83

∫ π/2

0

dθ =43π a3

46. 8∫ a

0

∫ b√

1−x2/a2

0

∫ c√

1−x2/a2−y2/b2

0

dz dy dx =43πabc.

47. M =∫ 2

−2

∫ √4−x2/2

−√

4−x2/2

∫ 4−y2

x2+3y2k|x| dz dy dx = 4

∫ 2

0

∫ √4−x2/2

0

∫ 4−y2

x2+3y2kx dz dy dx

= 4 k∫ 2

0

∫ √4−x2/2

0

(4x− x3 − 4xy2

)dy dx =

43k

∫ 2

0

x(4 − x2

)3/2dx =

12815

k

48. using polar coordinates V = 2∫ 2π

0

∫ 1

0

(r − r3) dr dθ = π

49. M =∫ 2

−1

∫ 3

0

∫ 4−x2

2−x

k(1 + y) dz dy dx =1354

k; (xM , yM , zM ) =(

12,

95,

125

)

50. (a) V =∫ 2

0

∫ 2−z

0

∫ 9−x2

0

dy dx dz

(b) V =∫ 2

0

∫ 2−x

0

∫ 9−x2

0

dy dz dx

(c) V =∫ 5

0

∫ 2

0

∫ 2−x

0

dz dx dy +∫ 9

5

∫ √9−y

0

∫ 2−x

0

dz dx dy



896 SECTION 17.8

51. (a) V =∫ 6

0

∫ 3

z/2

∫ 6−x

x

dy dx dz

(b) V =∫ 3

0

∫ 2x

0

∫ 6−x

x

dy dz dx

(c) V =∫ 6

0

∫ 3

z/2

∫ y

z/2

dx dy dz +∫ 6

0

∫ (12−z)/2

3

∫ 6−y

z/2

dx dy dz

52. (a) V =∫∫Ωxy

2√

4 − y dx dy (b) V =∫∫Ωxy

(∫ √4−y

−√4−y

dz

)dx dy

(c) V =∫ 4

−4

∫ 4

|x|

∫ √4−y

−√4−y

dz dy dx (d) V =∫ 4

0

∫ y

−y

∫ √4−y

−√4−y

dz dx dy

53. (a) V =∫∫Ωxy

2y dydz (b) V =∫∫Ωxy

( ∫ y

−y

dx

)dydz

(c) V =∫ 4

0

∫ √4−y

−√4−y

∫ y

−y

dx dz dy (d) V =∫ 2

−2

∫ 4−z2

0

∫ y

−y

dx dy dz

54. (a) V =∫∫Ωxy

(4 − z2 − |x|) dx dz (b) V =∫∫Ωxy

(∫ 4−z2

|x|dy

)dx dz

(c) V =∫ 2

−2

∫ 4−z2

z2−4

∫ 4−z2

|x|dy dx dz

(d) V =∫ 0

−2

∫ √4+x

−√

4+x

∫ 4−z2

|x|dy dz dx +

∫ 2

0

∫ √4−x

−√

4−x

∫ 4−z2

|x|dy dz dx

55. (a)∫ 4

2

∫ 5

3

∫ 2

1

lnxy

zdz dy dx ∼= 6.80703 (b)

∫ 4

0

∫ 2

1

∫ 3

0

x√yz dz dy dx =

16√

33

(4√

2 − 2)

56. Let f(x, z) = 36 − 9x2 − 4z2 and g(x, z) = 1 − 12 x− 1

3 z. Then

V =∫ 2

0

∫ 3−(3/2)x

0

∫ f(x,z)

g(x,z)

1 dy dz dx = 71

SECTION 17.8

1. r2 + z2 = 9 2. r = 2 3. z = 2r

4. r cos θ = 4z 5. 4r2 = z2 6. r2 sin2 θ + z2 = 8



SECTION 17.8 897

7.∫ π/2

0

∫ 2

0

∫ 4−r2

0

r dz dr dθ

=∫ π/2

0

∫ 2

0

(4r − r2

)dr dθ

=∫ π/2

0

4 dθ = 2π0

1

2

0 1 2

0

2

4

8.∫ π/4

0

∫ 1

0

∫ √1−r2

0

r dz dr dθ =π

12

9.∫ 2π

0

∫ 2

0

∫ r2

0

r dz dr dθ

=∫ 2π

0

∫ 2

0

r3 dr dθ

=∫ 2π

0

4 dθ = 8π-2

0

2

20

0

2

4

-

10.∫ 3

0

∫ 2π

0

∫ 3

r

r dz dθ dr = 9π

11.∫ 1

0

∫ √1−x2

0

∫ √4−(x2+y2)

0

dz dy dx =∫ π/2

0

∫ 1

0

∫ √4−r2

0

r dz dr dθ

=∫ π/2

0

∫ 1

0

r√

4 − r2 dr dθ

=∫ π/2

0

(83−√

3)

dθ =16

(8 − 3

√3)π

12.∫ π

0

∫ 1

0

∫ 1

r

z3r dz dr dθ =∫ π

0

∫ 1

0

14(1 − r4) r dr dθ =

π

12



898 SECTION 17.8

13.∫ 3

0

∫ √9−y2

0

∫ √9−x2−y2

0

1√x2 + y2

dz dx dy =∫ π/2

0

∫ 3

0

∫ √9−r2

0

1r· r dz dr dθ

=∫ π/2

0

∫ 3

0

√9 − r2 dr dθ

=∫ π/2

0

[r

2

√9 − r2 +

92

sin−1 r

3

]30

dθ

=9π4

∫ π/2

0

dθ =98π2

14.∫ π/2

0

∫ 1

0

∫ √1−r2

0

zr dz dr dθ =∫ π/2

0

∫ 1

0

r

2(1 − r2) dr dθ =

π

16

15.∫ 1

0

∫ √1−x2

0

∫ 2

0

sin(x2 + y2) dz dy dx =∫ π/2

0

∫ 1

0

∫ 2

0

sin(r2)r dz dr dθ

=∫ π/2

0

∫ 1

0

2r sin(r2) dr dθ = 12π(1 − cos 1) ∼= 0.7221

16.∫ 2π

0

∫ 1

0

∫ 2−r2

r2r2dz dr dθ = 4π

∫ 1

0

(1 − r2)r2 dr =8π15

17. (0, 1, 2) → (1, 12π, 2) 18. (0, 1,−2) → (1, 1

2π,−2)

19. (0,−1, 2) → (1, 32π, 2) 20. (0, 0, 0) → (0, arbitrary, 0)

21. V =∫ π/2

−π/2

∫ 2a cos θ

0

∫ r

0

r dz dr dθ =∫ π/2

−π/2

∫ 2a cos θ

0

r2 dr dθ

=∫ π/2

−π/2

83a3 cos3 θ dθ =

329

a3

22. V =∫ π/2

−π/2

∫ 2a cos θ

0

∫ r2/a

0

r dz dr dθ =32πa3

23. V =∫ π/2

−π/2

∫ a cos θ

0

∫ a−r

0

r dz dr dθ =∫ π/2

−π/2

∫ a cos θ

0

r(a− r) dr dθ

=∫ π/2

−π/2

a3

(12

cos2 θ − 13

cos3 θ)

dθ =136

a3(9π − 16)

24. V =∫ π/2

−π/2

∫ 2 cos θ

0

∫ 2+ 12 r cos θ

0

r dz dr dθ =52π



SECTION 17.8 899

25. V =∫ π/2

−π/2

∫ cos θ

0

∫ r cos θ

r2r dz dr dθ =

∫ π/2

−π/2

∫ cos θ

0

(r2 cos θ − r3

)dr dθ

=∫ π/2

−π/2

112

cos4 θ dθ =132

π

26. V =∫ 2π

0

∫ 3

0

∫ √25−r2

r+1

r dz dr dθ =413π

27. V =∫ 2π

0

∫ 1/2

0

∫ √1−r2

r√

3

r dz dr dθ =∫ 2π

0

∫ 1/2

0

(r√

1 − r2 − r2√

3)dr dθ =

13π(2 −

√3)

28. V =∫ 2π

0

∫ a

0

∫ √a2+r2

√2r

r dz dr dθ =2π3a3(

√2 − 1)

29. V =∫ 2π

0

∫ 3

1

∫ √9−r2

0

r dz dr dθ =∫ 2π

0

∫ 3

1

r√

9 − r2 dr dθ = 323 π

√2

30. V =∫ 2π

0

∫ 2

1

∫ 12

√36−r2

0

r dz dr dθ =13π(35

√35 − 128

√2).

31. Set the lower base of the cylinder on the xy-plane so that the axis of the cylinder coincides with the

z-axis. Assume that the density varies directly as the distance from the lower base.

M =∫ 2π

0

∫ R

0

∫ h

0

kzr dz dr dθ =12kπR2h2

32. xM = yM = 0 by symmetry

zMM =∫ 2π

0

∫ R

0

∫ h

0

kz2r dz dr dθ =13kπR2h3

M =12kπR2h2, zM =

23h

The center of mass lies on the axis of the cylinder at a distance 23h from the base of zero mass

density.

33. I = Iz = k

∫ 2π

0

∫ R

0

∫ h

0

zr3 dr dθ dz

= 14 kπR

4h2 = 12

(12 kπR

2h2)R2 = 1

2 MR2

∧from Exercise 31



900 SECTION 17.9

34. (a) I =M

πR2h

∫ 2π

0

∫ R

0

∫ h

0

r3 dz dr dθ =12MR2

(b) I =M

πR2h

∫ 2π

0

∫ R

0

∫ h

0

(r2 sin2 θ + z2)r dz dr dθ =14MR2 +

13Mh2

(c) I = 14MR2 + 1

3Mh2 −M( 12h)2 = 1

4MR2 + 112Mh2

35. Inverting the cone and placing the vertex at the origin, we have

V =∫ h

0

∫ 2π

0

∫ (R/h)z

0

r dr dθ dz =13πR2h.

36. xM = yM = 0 by symmetry

zMM =∫ h

0

∫ 2π

0

∫ (R/h)z

0

(M

V

)zr dr dθ dz =

(M

V

)πR2h2

4=⇒ zM =

πR2h2

4V=

34h

On the axis of the cone at a distance 34h from the vertex.

37. I =M

V

∫ h

0

∫ 2π

0

∫ (R/h)z

0

r3 dr dθ dz =310

MR2

38. I =M

V

∫ h

0

∫ 2π

0

∫ (R/h)z

0

z2 r dr dθ dz =35Mh2.

39. V =∫ 2π

0

∫ 1

0

∫ 1−r2

0

r dz dr dθ =12π 40. M =

∫ 2π

0

∫ 1

0

∫ 1−r2

0

kzr dz dr dθ =16πk

41. M =∫ 2π

0

∫ 1

0

∫ 1−r2

0

k(r2 + z2

)r dz dr dθ =

14kπ

SECTION 17.9

1.(√

3, 14 π, cos−1

[13

√3])

2.(

12

√6, 1

2

√2,

√2)

3. ( 34 ,

34

√3, 3

2

√3 ) 4. (2

√10, 2

3π, cos−1[ 310

√10])

5. ρ =√

22 + 22 + (2√

6/3)2 =4√

63

6.(

8, −π

4,

5π6

)

φ = cos−1

(2√

6/34√

6/3

)= cos−1(1/2) =

π

3

θ = tan−1(1) =π

4

(ρ, θ, φ) =

(4√

63

,π

4,π

3

)



SECTION 17.9 901

7. x = ρ sinφ cos θ = 3 sin 0 cos(π/2) = 0

z = ρ cosφ = 3 cos 0 = 3

y = ρ sinφ sin θ = 3 sin 0 sin(π/2) = 0

(x, y, z) = (0, 0, 3)

8. (a) (5, 12π, arccos 4

5 )

(b) (5, 32π, arccos 4

5 )

9. The circular cylinder x2 + y2 = 1; the radius of the cylinder is 1 and the axis is the z-axis.

10. The xy-plane.

11. The lower nappe of the circular cone z2 = x2 + y2.

12. Vertical plane which bisects the first and third quadrants of the xy-plane.

13. Horizontal plane one unit above the xy-plane.

14. sphere x2 + y2 + (z − 12 )2 = 1

4 of radius 12 and center (0, 0, 1

2 )

15. Sphere of radius 2 centered at the origin:∫ 2π

0

∫ π

0

∫ 2

0

ρ2 sinφdρ dφ dθ =83

∫ 2π

0

∫ π

0

sinφdφ dθ =163

∫ 2π

0

dθ =32π3

16. That part of the sphere of radius 1 that lies in the first quadrant between the x, z-plane and the plane

y = x ∫ π/4

0

∫ π/2

0

∫ 1

0

ρ2 sinφdρ dφ dθ =π

12

17. The first quadrant portion of the sphere that lies between the x, y-plane and the plane z = 32

√3.

∫ π/2

π/6

∫ π/2

0

∫ 3

0

ρ2 sinφdρ dθ dφ = 9∫ π/2

π/6

∫ π/2

0

sinφdθ dφ

= 92 π

∫ π/2

π/6

sinφdφ

= 92 π [− cosφ]π/2π/6 = 9

4π√

3

18. A cone of radius 1 and height 1;∫ π/4

0

∫ 2π

0

∫ secφ

0

ρ2 sinφdρ dθ dφ =13π

19.∫ 1

0

∫ √1−x2

0

∫ √2−x2−y2

√x2+y2

dz dy dx =∫ π/4

0

∫ π/2

0

∫ √2

0

ρ2 sinφdρ dθ dφ

= 23

√2∫ π/4

0

∫ π/2

0

sinφdθ dφ

=√

23 π

∫ π/4

0

sinφdφ =√

26

π (2 −√

2)



902 SECTION 17.9

20.∫ π/4

0

∫ π/2

0

∫ 2

0

ρ4 sinφdρ dθ dφ =16π5

∫ π/4

0

sinφdφ =8π5

(2 −√

2)

21.∫ 3

0

∫ √9−y2

0

∫ √9−x2−y2

0

z√x2 + y2 + x2 dz dx dy

=∫ π/2

0

∫ π/2

0

∫ 3

0

ρ cosφ · ρ · ρ2 sinφdρ dθ dφ

=∫ π/2

0

12

sin 2φdφ

∫ π/2

0

dθ

∫ 3

0

ρ4 dρ =[− 1

4cos 2φ

]π/20

(π2

) [15ρ5

]30

=243π20

22.∫ π/2

0

∫ π/2

0

∫ 1

0

1ρ2

ρ2 sinφdρ dθ dφ =π

2

23. V =∫ 2π

0

∫ π

0

∫ R

0

ρ2 sinφdρ dφ dθ =43πR3

24. r = ρ sinφ, θ = θ, z = ρ cosφ

25. V =∫ α

0

∫ π

0

∫ R

0

ρ2 sinφdρ dφ dθ =23αR3

26. M =∫ 2π

0

∫ π

0

∫ R

0

k(R− ρ)ρ2 sinφdρ dφ dθ =13kπR4

27. M =∫ 2π

0

∫ tan−1(r/h)

0

∫ h secφ

0

kρ3 sinφdρ dφ dθ

=∫ 2π

0

∫ tan−1(r/h)

0

kh4

4tanφ sec3 φdφ dθ

=kh4

4

∫ 2π

0

13[sec3 φ

]tan−1(r/h)

0dθ =

kh4

4

∫ 2π

0

13

⎡⎣(√

r2 + h2

h

)3

− 1

⎤⎦ dθ

=16kπh(r2 + h2

)3/2 − h3

28. V =∫ 2π

0

∫ tan−1 r/h

0

∫ h secφ

0

ρ2 sinφdρ dφ dθ =2π3

∫ tan−1(r/h)

0

h3 tanφ sec2 φdφ

=π

3h3 tan2(tan−1

( rh

)=

13πr2h

29. center ball at origin; density =M

V=

3M4πR3

(a) I =3M

4πR3

∫ 2π

0

∫ π

0

∫ R

0

ρ4 sin3 φdρ dφ dθ =25MR2

(b) I = 25 MR2 + R2M = 7

5 MR2 (parallel axis theorem)



SECTION 17.9 903

30. The center of mass is the centroid; V =23πR3

zV =∫ 2π

0

∫ π/2

0

∫ R

0

(ρ cosφ)ρ2 sinφdρ dφ dθ =14πR4

z =38R; (x, y, z) =

(0, 0,

38R

)

31. center balls at origin; density =M

V=

3M4π(R2

3 −R13)

(a) I =3M

4π(R2

3 −R13) ∫ 2π

0

∫ π

0

∫ R2

R1

ρ4 sin3 φdρ dφ dθ =25M

(R2

5 −R15

R23 −R1

3

)This result can be derived from Exercise 29 without further integration. View the solid as a ball

of mass M2 from which is cut out a core of mass M1.

M2 =M

VV2 =

3M4π(R2

3 −R13) (4

3πR2

3

)=

MR23

R23 −R1

3 ; similarly M1 =MR1

3

R23 −R1

3 .

Then

I = I2 − I1 = 25 M2R2

2 − 25 M1R1

2 =25

(MR2

3

R23 −R1

3

)R2

2 − 25

(MR1

3

R23 −R1

3

)R1

2

=25M

(R2

5 −R15

R23 −R1

3

).

(b) Outer radius R and inner radius R1 gives

moment of inertia =25M

(R5 −R1

5

R3 −R13

). [part (a) ]

As R1 → R,

R5 −R15

R3 −R13 =

R4 + R3R1 + R2R12 + RR1

3 + R14

R2 + RR1 + R12 −→ 5R4

3R2=

53R2.

Thus the moment of inertia of spherical shell of radius R is

25M

(53R2

)=

23MR2.

(c) I = 23MR2 + R2M = 5

3 MR2 (parallel axis theorem)

32. (a) The center of mass is the centroid; using the result of Exercise 30,

x = y = 0

z =z2V2 − z1V1

V=

38R2

43πR2

3 − 38R1

43πR1

3

43π(R2

3 −R13)

=3(R2

2 + R12)(R2 + R1)

8(R22 + R2R1 + R1

2)

(b) Setting R1 = R2 = R in (a), we get x = y = 0, z = 12R

33. V =∫ 2π

0

∫ α

0

∫ a

0

ρ2 sinφdρ dφ dθ =23π (1 − cosα) a3



904 SECTION 17.9

34.∫ 2π

0

∫ π/4

0

∫ 1

0

eρ3ρ2 sinφdρ dφ dθ =

13π(e− 1)

(2 −

√2)

35. (a) Substituting x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cosφ

into x2 + y2 + (z −R)2 = R2

we have ρ2 sin2 φ + (ρ cosφ−R)2 = R2,

which simplifies to ρ = 2R cosφ.

(b) 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, R secφ ≤ ρ ≤ 2R cosφ

36. (a) M =∫ 2π

0

∫ π/2

0

∫ 2R cosφ

0

kρ3 sinφdρ dφ dθ =85kπR4

(b) M =∫ 2π

0

∫ π/2

0

∫ 2R cosφ

0

kρ3 sin2 φdρ dφ dθ =14kπ2R4

(c) M =∫ 2π

0

∫ π/2

0

∫ 2R cosφ

0

kρ3 cos2 θ sin2 φdρ dφ dθ =18kπ2R4

37. V =∫ 2π

0

∫ π/4

0

∫ 2

0

ρ2 sinφdρ dφ dθ +∫ 2π

0

∫ π/2

π/4

∫ 2√

2 cosφ

0

ρ2 sinφdρ dφ dθ

=13

(16 − 6

√2)π

38. V =∫ 2π

0

∫ π

0

∫ 1−cosφ

0

ρ2 sinφdρ dφ dθ =83π

39. Encase T in a spherical wedge W . W has spherical coordinates in a box Π that contains S. Define

f to be zero outside of T . Then

F (ρ, θ, φ) = f (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ)

is zero outside of S and

∫∫T

∫f(x, y, z) dxdydz =

∫∫W

∫f(x, y, z) dxdydz

=∫∫

Π

∫F (ρ, θ, φ) ρ2 sinφdρdθdφ

=∫∫S

∫F (ρ, θ, φ) ρ2 sinφdρdθdφ.



SECTION 17.9 905

40. Break up T into little basic solids T1, . . . , TN . Choose a point (x∗i , y

∗i , z

∗i ) from each Ti and view

all the mass as concentrated there. Now Ti attracts m with a force

Fi∼= −Gmλ(x∗

i , y∗i , z

∗i )(Volume of Ti)ri3

ri

where ri is the vector from (x∗i , y

∗i , z

∗i ) to (a, b, c). We therefore have

Fi∼= Gmλ(x∗

i , y∗i , z

∗i )[(x

∗i − a)i + (y∗i − b)j + (z∗i − c)k]

[(x∗i − a)2 + (y∗i − b)2 + (z∗i − c)2]3/2

(Volume of Ti).

The sum of these approximations is a Riemann sum for the triple integral given and tends to thattriple integral as the maximum diameter of the Ti tends to zero.

41. T is the set of all (x, y, z) with spherical coordinates (ρ, θ, φ) in the set

S : 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, R secφ ≤ ρ ≤ 2R cosφ.

T has volume V = 23 πR

3. By symmetry the i, j components of force are zero and

F =

⎧⎨⎩3GmM

2πR3

∫∫T

∫z

(x2 + y2 + z2)3/2dxdydz

⎫⎬⎭k

=

⎧⎨⎩3GmM

2πR3

∫∫S

∫ (ρ cosφρ3

)ρ2 sinφdρdθdφ

⎫⎬⎭k

=

{3GmM

2πR3

∫ 2π

0

∫ π/4

0

∫ 2R cosφ

R secφ

cosφ sinφdρ dφ dθ

}k

=GmM

R2

(√2 − 1

)k.

42. With the coordinate system shown in the figure, T

is the set of all points (x, y, z) with cylindrical coor-dinates (r, θ, z) in the set

S : 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, α ≤ z ≤ α + h.

The gravitational force is

F =

⎡⎣∫∫

T

∫ (GmM

V

)z

(x2 + y2 + z2)3/2dxdydz

⎤⎦k

=

⎡⎣GmM

πR2h

∫∫S

zr

(r2 + z2)3/2dr dθ dz

⎤⎦k

=[GmM

πR2h

∫ 2π

0

∫ R

0

∫ α+h

α

zr

(r2 + z2)3/2dzdrdθ

]k

=2GmM

R2h

(√R2 + α2 −

√R2 + (α + h)2 + h

)k



906 SECTION 17.10

SECTION 17.10

1. ad− bc 2. 1 3. 2(v2 − u2

)4. u ln v − u 5. −3u2v2 6. 1 +

1uv

7. abc 8. 2 9. ρ2 sinφ

10. |J(r, θ, z)| =

∣∣∣∣∣∣∣∣cos θ sin θ 0

−r sin θ r cos θ 0

0 0 1

∣∣∣∣∣∣∣∣ = r.

11. J(ρ, θ, φ) =

∣∣∣∣∣∣∣∣sin φ cos θ sin φ sin θ cos θ

−ρ sin φ sin θ ρ sin φ cos θ 0

ρ cos φ cos θ ρ cos φ sin θ −ρ sin φ

∣∣∣∣∣∣∣∣ = −ρ2 sin φ; |J(ρ, θ, φ)| = ρ2 sin φ.

12. (a) dx− by = (ad− bc)u0 (b) cx− ay = (bc− ad)v0

13. Set u = x + y, v = x− y. Then

x =u + v

2, y =

u− v

2and J (u, v) = −1

2.

Ω is the set of all (x, y) with uv-coordinates in

Γ : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2.

Then∫ ∫Ω

(x2 − y2

)dxdy =

∫ ∫Γ

12uv dudv =

12

∫ 1

0

∫ 2

0

uv dv du

=12

(∫ 1

0

u du

)(∫ 2

0

v dv

)=

12

(12

)(2) =

12.

14. Using the changes of variables from Exercise 13,∫∫Ω

4xy dx dy =∫ 1

0

∫ 2

0

4(u2 − v2

4

)12dv du =

12

∫ 1

0

∫ 2

0

(u2 − v2

)dv du = −1

15.12

∫ 1

0

∫ 2

0

u cos (πv) dv du =12

(∫ 1

0

u du

)(∫ 2

0

cos (πv) dv)

=12

(12

)(0) = 0

16. Set u = x− y, v = x + 2y. Then

x =2u + v

3, y =

v − u

3, and J(u, v) =

13

Ω is the set of all (x, y) with uv-coordinates in the set

Γ : 0 ≤ u ≤ π, 0 ≤ v ≤ π/2.



SECTION 17.10 907

Therefore∫∫Ω

(x + y) dxdy =∫∫Γ

19(u + 2v) du dv =

19

∫ π

0

∫ π/2

0

(u + 2v) dv du =118

π3.

17. Set u = x− y, v = x + 2y. Then

x =2u + v

3, y =

v − u

3, and J(u, v) =

13.


Γ : 0 ≤ u ≤ π, 0 ≤ v ≤ π/2.

Therefore∫∫Ω

sin (x− y) cos (x + 2y) dxdy =∫∫Γ

13

sinu cos v dudv =13

∫ π

0

∫ π/2

0

sinu cos v dv du

=13

(∫ π

0

sinu du

)(∫ π/2

0

cos v dv)

=13(2)(1) =

23.

18. Using the change of variables from Exercise 16,∫∫Ω

sin 3x dx dy =∫ π

0

∫ π/2

0

sin(2u + v)13du dv = 0.

19. Set u = xy, v = y. Thenx = u/v, y = v and J (u, v) = 1/v.

xy = 1, xy = 4 =⇒ u = 1, u = 4

y = x, y = 4x =⇒ u/v = v, 4u/v = v =⇒ v2 = u, v2 = 4u


Γ : 1 ≤ u ≤ 4,√u ≤ v ≤ 2

√u.

(a) A =∫∫Γ

1vdudv =

∫ 4

1

∫ 2√u

√u

1vdv du =

∫1

4

ln 2 du = 3 ln 2

(b) xA =∫ 4

1

∫ 2√u

√u

u

v2dv du =

73

; x =7

9 ln 2

yA =∫ 4

1

∫ 2√u

√u

dv du =143

; y =14

9 ln 2

20. J(r, θ) = abr, Γ : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π

A =∫∫

Γ

abr dr dθ = ab

∫ 2π

0

∫ 1

0

r dr dθ = πab

21. Set u = x + y, v = 3x− 2y. Then

x =2u + v

5, y =

3u− v

5and J (u, v) = −1

5.



908 SECTION 17.10

With Γ : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2

M =∫ 1

0

∫ 2

0

15λ dv du =

25λ where λ is the density.

Then

Ix =∫ 1

0

∫ 2

0

(3u− v

5

)2 15λ dv du =

8λ375

=475

(25λ

)=

475

M,

Iy =∫ 1

0

∫ 2

0

(2u + v

5

)2 15λ dv du =

28λ375

=1475

(25λ

)=

1475

M,

Iz = Ix + Iy =1875

M.

22. x =u + v

2, y =

v − u

2, J(u, v) =

12

Γ : −2 ≤ u ≤ 2, −4 ≤ v ≤ −u2

A =∫∫Γ

12du dv =

12

∫ 2

−2

∫ −u2

−4

dv du =163

23. Set u = x− 2y, v = 2x + y. Then

x =u + 2v

5, y =

v − 2u5

and J (u, v) =15.

Γ is the region between the parabola v = u2 − 1 and the line v = 2u + 2. A sketch of the curves showsthat

Γ : −1 ≤ u ≤ 3, u2 − 1 ≤ v ≤ 2u + 2.Then

A =15

(area of Γ) =15

∫ 3

−1

[(2u + 2) −

(u2 − 1

) ]du =

3215

.

24. xA =12

∫ 2

−2

∫ −u2

−4

u + v

2dv du = −32

5yA =

12

∫ 2

−2

∫ −u2

−4

v − u

2dv du = −32

5

A =163

=⇒ x = y = −65

25. The choice θ = π/6 reduces the equation to 13u2 + 5v2 = 1. This is an ellipse in the uv-planewith area πab = π/

√65. Since J(u, v) = 1, the area of Ω is also π/

√65.

26.∫∫Sa

e−(x−y)2

1 + (x + y)2dx dy =

12

∫∫Γ

e−u2

1 + v2du dv

where Γ is the square in the uv-plane with vertices (−2a, 0), (0, −2a), (2a, 0), (0, 2a).Γ contains the square −a ≤ u ≤ a, −a ≤ v ≤ a and is contained in the square−2a ≤ u ≤ 2a, −2a ≤ v ≤ 2a. Therefore

12

∫ a

−a

∫ a

−a

e−u2

1 + v2du dv ≤ 1

2

∫∫Γ

e−u2

1 + v2dudv ≤ 1

2

∫ 2a

−2a

∫ 2a

−2a

e−u2

1 + v2du dv.



SECTION 17.10 909

The two extremes can be written

12

(∫ a

−a

e−u2du

) (∫ a

−a

11 + v2

dv

)and

12

(∫ 2a

−2a

e−u2du

) (∫ 2a

−2a

11 + v2

dv

).

As a → ∞ both expressions tend to 12 (

√π) (π) = 1

2π3/2. It follows that∫ ∞

−∞

∫ ∞

−∞

e−(x−y)2

1 + (x + y)2dx dy =

12π3/2.

27. J = abcρ2 sinφ; V =∫ 2π

0

∫ π

0

∫ 1

0

abcρ2 sinφdρ dφ dθ =43πabc

28. x = y = 0

zV =∫ 2π

0

∫ π/2

0

∫ 1

0

(cρ cosφ)abcρ2 sinφdρ dφ dθ =πabc2

4=⇒ z =

38c .

29. V =23πabc, λ =

M

V=

3M2πabc

Ix =3M

2πabc

∫ 2π

0

∫ π/2

0

∫ 1

0

(b2ρ2 sin2 φ sin2 θ + c2ρ2 cos2 φ

)abcρ2 sinφdρ dφ dθ

= 15 M(b2 + c2

)Iy = 1

5 M(a2 + c2

), Iz = 1

5 M(a2 + b2

)30. I =

∫ 2π

0

∫ 1

0

∫ π

0

ρ2(abcρ2 sinφ) dφ dρ dθ =45πabc

PROJECT 17.10

1. (a) θ = tan−1

[(aybx

)1/α], r =

[(xa

)2/α+(yb

)2/α]α/2(b) ar1(cos θ1)α = ar2(cos θ2)α

br1(sin θ1)α = br2(sin θ2)α

r1 > 0, 0 < θ < 12π

⎫⎪⎪⎬⎪⎪⎭ =⇒ r1 = r2, θ1 = θ2

2. J = abαr cosα−1 θ sinα−1 θ

3. (a)

x

y

a

a



910 REVIEW EXERCISES

(b) x = ar cos3 θ, y = ar sin3 θ; x23 + y

23 = a

23 =⇒ r = 1 and x = a cos3 θ, y = a sin3 θ

A =∫ 0

π2

y(θ)x′(θ) dθ =∫ 0

π2

a sin3 θ(3a cos2 θ[− sin θ]) dθ

= 3a2

∫ π2

0

sin4 θ cos2 θ dθ = 3a2

∫ π2

0

(sin4 θ − sin6 θ) dθ

= 3a2

[3 · 14 · 2

π

2− 5 · 3 · 1

6 · 4 · 2π

2

](See Exercise 62(b) in 8.3)

=3a2π

32

(c) Entire area enclosed: 4 · 3a2π

32=

3a2π

8

4. (a)

a = 3, b = 2 a = 2, b = 3

(b) From Problem 2, Jacobian J = 8abr cos7 θ sin7 θ

A =∫ π

2

0

∫ 1

0

8abr cos7 θ sin7 θ dθ = 4ab∫ π

2

0

cos7 θ sin7 θ dθ =ab

70

REVIEW EXERCISES

1.∫ 1

0

∫ √y

y

xy2 dx dy =∫ 1

0

[12x2y2]√y

ydy =

∫ 1

0

(12y3 − 1

2y4

)dy =

[18y4 − 1

10y5]10

=140

2.∫ 1

0

∫ y

−y

ex+y dx dy =∫ 1

0

[ex+y]y−y

dy =∫ 1

0

(e2y − 1) dy =[12e2y − y

]10

=e2

2− 3

2

3.∫ 1

0

∫ 3x

x

2yex3dy dx =

∫ 1

0

[y2ex

3]3xx

dx =∫ 1

0

(9x2ex3 − x2ex

3) dx =

[3ex

3 − 13ex

3]10

=83e− 8

3

4.∫ 2

1

∫ lnx

0

xey dy dx =∫ 2

1

[xey]lnx

0dx =

∫ 2

1

x(x− 1) dx =[13x3 − 1

2x2]21

=56

5.∫ π/4

0

∫ 2 sin θ

0

r cos θ dr dθ =∫ π/4

0

[12r2 cos θ

]2 sin θ

0dθ =

∫ π/4

0

2 sin2 θ cos θ dθ =[23

sin3 θ]π/40

=√

26

6.∫ 2

−1

∫ 4

0

∫ 1

0

xyz dx dy dz =∫ 2

−1

∫ 4

0

[12x2yz]10dy dz =

∫ 2

−1

∫ 4

0

12yz dy dz =

∫ 2

−1

4z dz = 6



REVIEW EXERCISES 911

7.∫ 2

0

∫ 2−3x

0

∫ x+y

0

x dz dy dx =∫ 2

0

∫ 2−3x

0

[xz]x+y

0dy dx =

∫ 2

0

∫ 2−3x

0

(x2 + xy) dy dx

=∫ 2

0

[x2y +

12xy2

]2−3x

0

dx =∫ 2

0

(32x3 − 4x2 + 2x

)dx

=[38x4 − 4

3x3 + x2

]20

= −23

8. ∫ π2

0

∫ π2

z

∫ sin z

0

3x2 sin y dx dy dz =∫ π

2

0

∫ π2

z

[x3 sin y

]sin z

0dy dz =

∫ π2

0

∫ π2

z

sin3 z sin y dy dz

=∫ π

2

0

sin3 z cos z dz =14

sin4 z]π

2

0=

14

9.∫ 0

−π2

∫ 2 sin θ

0

∫ r2

0

r2 cos θ dz dr dθ =∫ 0

−π2

∫ 2 sin θ

0

r4 cos θ dr dθ =∫ 0

−π2

325

sin5 θ cos θ dθ = −1615

10.∫ π

2

−π6

∫ π2

0

∫ 1

0

ρ3 sinϕ cosϕdρ dθ dϕ =∫ π

2

−π6

∫ π2

0

14

sinϕ cosϕdθ dϕ

=∫ π

2

−π6

π

16sin 2ϕdϕ = −

[ π32

cos 2ϕ]π

2

−π6

=3π64

11.∫ 1

0

∫ 1

y

ex2dx dy =

∫ 1

0

∫ x

0

ex2dy dx =

∫ 1

0

ex2y∣∣∣x0dx =

∫ 1

0

xex2dx =

12ex

2∣∣∣10

=e− 1

2

12.∫ 2

0

∫ 1

x2

cos y2 dy dx =∫ 1

0

∫ 2y

0

cos y2 dx dy =∫ 1

0

x cos y2∣∣∣2y0

dy =∫ 1

0

2y cos y2dy = sin y2∣∣∣10

= sin 1

13.∫ 1

0

∫ √1−y2

0

1√1 − y2

dxdy =∫ 1

0

dy = 1

14.∫ 1

0

∫ 1−x

0

y cos(x + y) dy dx =∫ 1

0

∫ 1−y

0

y cos(x + y) dx dy

=∫ 1

0

y sin(x + y)]1−y

0dy =

∫ 1

0

y(sin 1 − sin y) dy

=[12y2 sin 1 + y cos y − sin y

]10

= cos 1 − 12

sin 1

15.∫ 1

0

∫ √1−x2

0

xy dy dx =∫ 1

0

[12xy2]√1−x2

0dx =

∫ 1

0

(12x− 1

2x3

)dx =

18

16.∫ √

3

−√

3

∫ 4−y2

y2/3

(x− y) dx dy =∫ √

3

−√

3

[x2

2− xy)

]4−y2

y2/3dy =

∫ √3

−√

3

(49y4 +

43y3 − 4y2 − 4y + 8

)dy =

48√

35

17.∫ 2

0

∫ 3x−x2

x

(x2 − xy) dy dx =∫ 2

0

[x2y − 1

2xy2]3x−x2

xdx =

∫ 2

0

(2x4 − 2x3 − 1

2x5

)dx = − 8

15




18.∫ 2

0

∫ 2−x

0

x(x− 1)exy dy dx =∫ 2

0

[(x− 1)exy

]2−x

0dx =

∫ 2

0

(x− 1)e2x−x2dx−

∫ 2

0

(x− 1) dx

= −[12e2x−x2

]20−[12x2 − x

]20

= 0

19.∫ 2

0

∫ y

0

∫ √4−y2

0

2xyz dz dx dy =∫ 2

0

∫ 2

x

xyz2∣∣∣√4−y2

0dx dy =

∫ 2

0

∫ 2

0

xy(4 − y2) dx dy

=∫ 2

0

[12y3(4 − y2)

]dy =

83

20.∫∫∫T

z dxdydz = 2∫ 1

0

∫ x

0

∫ 1−x

0

z dz dy dx =∫ 1

0

∫ x

0

(1 − x)2 dy dx =∫ 1

0

(1 − x)2x dx =112

21.∫ 2

0

∫ √4−x2

0

∫ √4−x2−y2

0

xy dz dy dx =∫ 2

0

∫ √4−x2

0

[xyz]√4−x2−y2

0dy dx

=∫ 2

0

[− 1

3x√

4 − x2 − y2]√4−x2

0dx

=∫ 2

0

13x(4 − x2)

32 dx =

3215

22.∫∫∫T

(x2 + 2z) dx dy dz =∫ 2

−2

∫ 4

x2

∫ 4−y

0

(x2 + 2z) dz dy dx

=∫ 2

−2

∫ 4

x2(4x2 − x2y + 16 − 8y + y2) dy dx

=∫ 2

−2

(16x6 − 8x2 +

643

)dx =

210

21

23.∫ 2

0

∫ √4−y2

0

e√

x2+y2dx dy =

∫ π/2

0

∫ 2

0

err dr dθ =π

2

∫ 2

0

rerdr =π

2

[rer − er

]20

=π

2(e2 + 1)

24.∫ 1

−1

∫ √1−x2

0

arctan (y/x) dy dx =∫ π/2

0

∫ 1

0

rθ dr dθ +∫ π

π/2

∫ 1

0

(θ − π)r dr dθ

=∫ π/2

0

θ

2dθ +

∫ π

π/2

12(θ − π) dθ = 0

25. V =∫ 3

0

∫ 2π

0

(9 − r2)r dr dθ = 2π∫ 3

0

(9 − r2)r dr =81π2

26.∫ 1

0

∫ √x

x2(2 − x2 − y2) dy dx = −

∫ 1

0

(2x1/2 − x5/2 − 1

3x3/2 − 2x2 + x4 + 1

3x6)dx =

52105

27. V =∫ 1

0

∫ 1−x

0

(x2 + y2) dy dx =∫ 1

0

[x2 − x3 +

13(1 − x)3

]dx =

[13x3 − 1

4x4 − 1

12(1 − x)4

]10

=16




28. V =∫ 3

0

∫ π/2

0

r2 sin θ dr dθ =∫ π/2

0

9 sin θ dθ = 9

29. M =∫ π/2

−π/2

∫ cosx

0

y dy dx =∫ π/2

−π/2

12

cos2 x dx =π

4

xMM =∫ π/2

−π/2

∫ cosx

0

xy dy dx = 0 by symmetry

yMM =∫ π/2

−π/2

∫ cosx

0

y2 dy dx =∫ π/2

−π/2

13

cos3 x dx =49

The center of mass is: (0, 169π )

30. M =∫ 1

0

∫ y

y22x dx dy =

∫ 1

0

(y2 − y4) dy =215

xMM =∫ 1

0

∫ y

y22x2 dx dy =

∫ 1

0

(23y3 − 2

3y6

)dy =

114

yMM =∫ 1

0

∫ y

y22xy dx dy =

∫ 1

0

(y3 − y5) dy =112

The center of mass is: (15/28, 5/8)

31. M =∫ π/2

0

∫ R

r

u3 du dθ =π

8(R4 − r4); (polar coordinates [u, θ])

By symmetry, x = y.

xMM =∫ π/2

0

∫ R

r

u4 cos θ du dθ =15(R5 − r5); xM =

8(R5 − r5)5π(R4 − r4)

32. M =∫ π

0

∫ 2(1+cos θ)

0

r2 dr dθ =∫ π

0

83(1 + cos θ)3 dθ =

203π

xMM =∫ π

0

∫ 2(1+cos θ)

0

r3 cos θ dr dθ =∫ π

0

4(1 + cos θ)4 cos θ dθ = 14π

yMM =∫ π

0

∫ 2(1+cos θ)

0

r3 sin θ dr dθ =∫ π

0

4(1 + cos θ)4 sin θ dθ = −45(1 + cos θ)5|π0 =

1285

The center of mass is: ( 2110 ,

9625π )

33. Introduce a coordinate system as shown in the figure.

(a) A = 12bh; by symmetry, x = 0

y A =∫ 0

−b/2

∫ 2hb (x+ b

2 )

0

y dy dx +∫ b

2

0

∫ − 2hb (x− b

2 )

0

y dy dx

=bh2

6=⇒ y =

h

3 x

y

−b 2 b 2

h

(b) I =∫ 0

−b/2

∫ 2hb (x+ b

2 )

0

λy2 dy dx +∫ b/2

0

∫ − 2hb (x− b

2 )

0

λy2 dy dx =λbh3

12= 1

6Mh2




(c) I = 2∫ b/2

0

∫ − 2hb (x− b

2 )

0

λx2 dx dy =148

λhb3 = 124Mb2

34. Let λ =k√

x2 + y2

(a) M =∫ π

0

∫ R

r

k dr dθ = kπ(R− r)

By symmetry, xM = 0; yMM =∫ π

0

∫ R

r

kr sin θ dr dθ = k(R2 − r2)

The center of mass is: (0, R+rπ )

(b) Ix =∫ π

0

∫ R

r

kr2 sin2 θ dr dθ =k

3(R3 − r3)

∫ π

0

sin2 θ dθ =kπ

6(R3 − r3)

(c) Iy =∫ π

0

∫ R

r

kr2 cos2 θ dr dθ =k

3(R3 − r3)

∫ π

0

cos2 θ dθ =kπ

6(R3 − r3)

35. V =∫ 2

0

∫ x

0

∫ 2x+2y+1

0

dz dy dx =∫ 2

0

∫ x

0

(2x + 2y + 1) dy dx = 10

36. V =∫ 1

0

∫ x

x2

∫ 4(x2+y2)

−1

dz dy dx =∫ 1

0

∫ x

x2(4x2 + 4y2 + 1) dy dx

=∫ 1

0

(x +

163x3 − x2 − 4x4 − 4

3x6

)dx =

107210

37. The curve of intersection of the two surfaces is the circle: x2 + y2 = 4, x = 3

V =∫ 2

−2

∫ √4−x2

−√

4−x2

∫ 12−x2−2y2

2x2+y2dz dy dx =

∫ 2

−2

∫ √4−x2

−√

4−x23(4 − x2 − y2

)dy dx

= 3∫ 2π

0

∫ 2

0

(4 − r2

)r dr dθ

= 3∫ 2π

0

[2r2 − 1

4 r4]20dθ = 12

∫ 2π

0

= 24π

38. V =∫ 1

0

∫ √1−y2

0

∫ (2−y−z)/2

0

dx dz dy =∫ 1

0

∫ π/2

0

2 − r cos θ − r sin θ

2r dθ dr

=∫ 1

0

12(π − 2r)r dr =

3π − 412

39. V =∫ 2π

0

∫ π/3

0

∫ 2

secφ

ρ2 sinφdρ dφ dθ =∫ 2π

0

∫ π/3

0

[13 ρ

3]2secφ

dφ dθ

=13

∫ 2π

0

∫ π/3

0

(8 − sec3φ

)sinφdφ dθ

=13

∫ 2π

0

[− 8 cosφ− 1

2 sec2 φ]π/30

dθ

=13

(52

)(2π) =

5π3




40. V =∫ 4

0

∫ (12−3x)/4

0

∫ 16−x2

0

dz dy dx=∫ 4

0

∫ 12−3x4

0

(16 − x2) dy dx=34

∫ 4

0

(64 − 16x− 4x2 + x3) dx = 80

41. V =∫ 2π

0

∫ π/2

π/4

∫ 1

0

ρ2 sinφdρ dφ dθ =∫ 2π

0

∫ π/2

π/4

13

sinφdφ dθ =√

2π3

42. V =∫ 2π

0

∫ π/4

0

∫ 1

0

ρ2 sinφdρ dφ dθ =2 −

√2

3π

43. (a) V =∫ 1

0

∫ x

0

∫ √1−x2

0

dz dy dx +∫ 1

0

∫ y

0

∫ √1−y2

0

dz dx dy

= 2∫ 1

0

∫ x

0

√1 − x2 dy dx = 2

∫ 1

0

x√

1 − x2dx =23

By symmetry, x = y.

xV =∫ 1

0

∫ x

0

∫ √1−x2

0

x dz dy dx +∫ 1

0

∫ y

0

∫ √1−y2

0

x dz dx dy

For the first integral:

∫ 1

0

∫ x

0

∫ √1−x2

0

x dz dy dx =∫ 1

0

∫ x

0

x√

1 − x2 dy dx

=∫ 1

0

x2√

1 − x2 dx =∫ π/2

0

sin2 u cos2 u du =π

16

x = sinu∧

For the second integral:∫ 1

0

∫ y

0

∫ √1−y2

0

x dz dx dy =∫ 1

0

∫ y

0

x√

1 − y2dx dy =∫ 1

0

12y2√

1 − y2 dy =π

32

Thus, xV =3π32

=⇒ x = y =9π64

Now calculate z:

z V =∫ 1

0

∫ x

0

∫ √1−x2

0

z dz dy dx +∫ 1

0

∫ y

0

∫ √1−y2

0

z dz dx dy;

∫ 1

0

∫ x

0

∫ √1−x2

0

z dz dy dx =∫ 1

0

∫ x

0

12(1 − x2)dy dx =

12

∫ 1

0

(x− x3

)dx =

18

and similarly,∫ 1

0

∫ y

0

∫ √1−y2

0

z dz dy dx =18.

Therefore, zV =14

=⇒ z =38




(b) Iz =∫ 1

0

∫ x

0

∫ √1−x2

0

λ(√

x2 + y2)2

dz dy dx +∫ 1

0

∫ y

0

∫ √1−y2

0

λ(√

x2 + y2)2

dz dx dy;

∫ 1

0

∫ x

0

∫ √1−x2

0

λ(√

x2 + y2)2

dz dy dx =∫ π/4

0

∫ sec θ

0

∫ r sin θ

0

λr3dz dr dθ =320

λ

and∫ 1

0

∫ y

0

∫ √1−y2

0

λ(√x2 + y2)2dz dx dy =

320

λ =⇒ Iz =310

λ

44. (a) V =∫ 2π

0

∫ 1

0

(r − r2)r dr dθ = 2π∫ 1

0

(r2 − r3) dr =π

6By symmetry, x = y = 0

zV =∫ 2π

0

∫ 1

0

∫ r

r2zr dz dr dθ =

π

12and hence z = 1

2

(b) Iz = K

∫ 2π

0

∫ 1

0

∫ r

r2r2 dz dr dθ = 2πK

∫ 1

0

r2(r − r2)dr =πK

10

Here, K is the density of the mass.

45. Denote polar coordinates by [u, θ].

(a) M =∫ 2π

0

∫ r

0

∫ h

0

u3 dz du d θ = 2πh∫ r

0

u3 du =πhr4

2

(b) By symmetry, xM = yM = 0

(c) zMM =∫ 2π

0

∫ r

0

∫ h

0

u3z dz du dθ =πh2r4

4=⇒ zM = h/2

46. λ =√x2 + y2

(a) M =∫ 2π

0

∫ π/2

0

∫ r

0

ρ3 sin2 φdρ dφ dθ =r4π

2

∫ π2

0

sin2 φdφ =r4π2

8


zMM =∫ 2π

0

∫ π/2

0

∫ r

0

ρ4 sin2 φ cosφdρ dφ dθ =2r5π

15=⇒ zM =

16r15π

47. (a) M =∫ 1

0

∫ 2π

0

∫ 1

r

r2 dz dθ dr = 2π∫ 1

0

∫ 1

r

r2 dz dr = 2π∫ 1

0

r2(1 − r) dr =π

6


zMM =∫ 1

0

∫ 2π

0

∫ 1

r

r2z dz dθ dr = π

∫ 1

0

r2(1 − r2) dr =2π15

=⇒ zM =45

(c) Iz =∫ 1

0

∫ 2π

0

∫ 1

r

r4 dz dθ dr =π

15

48. J(u, v) =

∣∣∣∣∣ 2u 2v

−2v 2u

∣∣∣∣∣ = 4u2 + 4v2




49. J(u, v) =

∣∣∣∣∣ eu cos v eu sin v

−eu sin v eu cos v

∣∣∣∣∣ = e2u

50. Ju, v, w) =

∣∣∣∣∣∣∣∣2u 2w vw

2w 2v uw

2v 2u uv

∣∣∣∣∣∣∣∣ = (vw − u2)(4uw − 4v2)

51. Set x =v − u

2, y =

v + u

2=⇒ u = y − x, v = y + x, 1 ≤ v ≤ 2, J = −1

2at x = 0, y = u, y = v =⇒ u = v

at y = 0, −x = u, x = v =⇒ u = −v∫∫Ω

cos(y − x

y + x

)dx dy =

∫ 2

1

∫ v

−v

12

cos(uv

)du dv =

∫ 2

i

v sin 1 dv =32

sin 1

52. By the hint, J =

∣∣∣∣∣∣∣∣∣∣∣

cos θu

sin θ

u2r

−r cos θu2

−r sin θ

u20

−r sin θ

u

r cos θu

0

∣∣∣∣∣∣∣∣∣∣∣= −2r3

u3

∫∫∫T

dx dy dz =∫ 2π

0

∫ 2

1

∫ 2

1

2r3

u3dr du dθ =

45π8

Calculus one and several variables 10E Salas solutions manual ch17

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Transcript of Calculus one and several variables 10E Salas solutions manual ch17