Chapter 15 Solutions Calculus

7/30/2019 Chapter 15 Solutions Calculus

1/37

620

CHAPTER 15

Multiple Integrals

EXERCISE SET 15.1

1.

10

20

(x + 3)dydx =

10

(2x + 6)dx = 7 2.

31

11

(2x 4y)dydx =31

4x dx = 16

3.

42

10

x2ydxdy =

42

1

3y dy = 2 4.

02

21

(x2 + y2)dxdy =

02

(3 + 3y2)dy = 14

5.

ln 30

ln 20

ex+ydydx =

ln 30

exdx = 2

6.

20

10

y sin xdydx =

20

1

2sin x dx = (1 cos2)/2

7.

01

52

dxdy =

01

3 dy = 3 8.

64

73

dydx =

64

10dx = 20

9.

10

10

x

(xy + 1)2dydx =

10

1 1

x + 1

dx = 1 ln 2

10.

/2

21

x cos xydydx =

/2

(sin2x sin x)dx = 2

11.

ln 20

10

xy ey2xdydx =

ln 20

1

2(ex 1)dx = (1 ln2)/2

12.

43

21

1

(x + y)2dydx =

43

1

x + 1 1

x + 2

dx = ln(25/24)

13.

11

22

4xy3dydx =

11

0 dx = 0

14.

10

10

xyx2 + y2 + 1

dydx =

10

[x(x2 + 2)1/2 x(x2 + 1)1/2]dx = (3

3 4

2 + 1)/3

15.

10

32

x

1 x2 dydx =10

x(1 x2)1/2dx = 1/3

16.

/20

/30

(x sin y y sin x)dydx =/20

x

2

2

18sin x

dx = 2/144

17. (a) xk = k/2 1/4, k = 1, 2, 3, 4; yl = l/2 1/4, l = 1, 2, 3, 4, R

f(x, y) dxdy 4

k=1

4l=1

f(xk, yl )Akl =

4k=1

4l=1

[(k/21/4)2+(l/21/4)](1/2)2 = 37/4

(b)

20

20

(x2 + y) dxdy = 28/3; the error is |37/4 28/3| = 1/12


2/37

Exercise Set 15.1 621

18. (a) xk = k/2 1/4, k = 1, 2, 3, 4; yl = l/2 1/4, l = 1, 2, 3, 4, R

f(x, y) dxdy 4

k=1

4l=1

f(xk, yl )Akl =

4k=1

4l=1

[(k/2 1/4) 2(l/2 1/4)](1/2)2 = 4

(b)

20

20

(x 2y) dxdy = 4; the error is zero

19. V =

5

3

2

1

(2x + y)dydx =

5

3

(2x + 3/2)dx = 19

20. V =

31

20

(3x3 + 3x2y)dydx =

31

(6x3 + 6x2)dx = 172

21. V =

20

30

x2dydx =

20

3x2dx = 8

22. V = 3

0

4

0

5(1

x/3)dydx =

3

0

5(4

4x/3)dx = 30

23. (a)

(1, 0, 4)

(2, 5, 0)

x

y

z (b)(0, 0, 5)

(3, 4, 0)

(0, 4, 3)

z

x

y

24. (a)

(1, 1, 0)

(0, 0, 2)

x

y

z (b)

(2, 2, 0)

(2, 2, 8)

x

y

z

25.

1/20

0

x cos(xy)cos2 xdydx =

1/20

cos2 x sin(xy)0

dx

=

1/20

cos2 x sin xdx = 13

cos3 x1/20

=1

3


3/37

622 Chapter 15

26. (a)

y

x

z

5

3

(0, 2, 2)

(5, 3, 0)

(b) V =

50

20

y d y d x +

50

32

(2y + 6) dydx

= 10 + 5 = 15

27. fave =2

/20

10

y sin xydxdy =2

/20

cos xy

x=1x=0

dy =

2

/20

(1 cos y) dy = 1 2

28. average =1

3

30

10

x(x2 + y)1/2dxdy =

30

1

9[(1 + y)3/2 y3/2]dy = 2(31 9

3)/45

29. Tave =1

2 1

0 2

0 10 8x2 2y2

dydx =

1

2 1

0

44

3 16x2

dx =

14

3

30. fave =1

A(R)

ba

dc

k dydx =1

A(R)(b a)(d c)k = k

31. 1.381737122 32. 2.230985141

33.

R

f(x, y)dA =

ba

dc

g(x)h(y)dy

dx =

ba

g(x)

dc

h(y)dy

dx

=

ba

g(x)dx

dc

h(y)dy

34. The integral of tan x (an odd function) over the interval [1, 1] is zero.

35. The first integral equals 1/2, the second equals 1/2. No, because the integrand is not continuous.

EXERCISE SET 15.2

1.

10

xx2

xy2dydx =

10

1

3(x4 x7)dx = 1/40

2. 3/2

1

3y

y

ydxdy = 3/2

1

(3y

2y2)dy = 7/24

3.

30

9y20

ydxdy =

30

y

9 y2 dy = 9

4.

11/4

xx2

x/y dydx =

11/4

xx2

x1/2y1/2dydx =11/4

2(x x3/2)dx = 13/80


4/37


5.

2

x30

sin(y/x)dydx =

2

[x cos(x2) + x]dx = /2

6.

11

x2x2

(x2 y)dydx =11

2x4dx = 4/5 7.

/2

x20

1

xcos(y/x)dydx =

/2

sin x dx = 1

8. 1

0

x

0

ex2

dydx = 1

0

xex2

dx = (e

1)/2 9.

1

0

x

0

yx2 y2 dydx =

1

0

1

3

x3dx = 1/12

10.

21

y20

ex/y2

dxdy =

21

(e 1)y2dy = 7(e 1)/3

11. (a)

20

x20

xydydx =

20

1

2x5dx =

16

3

(b)

31

(y+7)/2(y5)/2

xydxdy =

31

(3y2 + 3y)dy = 38

12. (a) 1

0x

x2

(x + y)dydx = 1

0

(x3/2 + x/2 x3 x4/2)dx = 3/10

(b)

11

1x21x2

xdydx +

11

1x21x2

y d y d x =

11

2x

1 x2 dx + 0 = 0

13. (a)

84

x16/x

x2dydx =

84

(x3 16x)dx = 576

(b)

42

816/y

x2 dxdy +

84

8y

x2 dxdy =

84

512

3 4096

3y3

dy +

84

512 y33

dy

=640

3+

1088

3= 576

14. (a)21

y0

xy2dxdy =21

12

y4dy = 31/10

(b)

10

21

xy2 dydx +

21

2x

xy2 dydx =

10

7x/3 dx +

21

8x x43

dx = 7/6 + 29/15 = 31/10

15. (a)

11

1x21x2

(3x 2y)dydx =11

6x

1 x2 dx = 0

(b)

11

1y2

1y2(3x 2y) dxdy =

114y

1 y2 dy = 0

16. (a)5025x2

5x y d y d x =50 (5x x

2

)dx = 125/6

(b)

50

25y25y

y dxdy =

50

y

25 y2 5 + y

dy = 125/6

17.

40

y0

x(1 + y2)1/2dxdy =40

1

2y(1 + y2)1/2dy = (

17 1)/2


5/37

624 Chapter 15

18.

0

x0

x cos y d y d x =

0

x sin x dx =

19.

20

6yy2

xydxdy =

20

1

2(36y 12y2 + y3 y5)dy = 50/3

20. /4

0

1/2

sin y

xdxdy = /4

0

1

4cos2y dy = 1/8

21.

10

xx3

(x 1)dy dx =10

(x4 + x3 + x2 x)dx = 7/60

22.

1/20

2xx

x2dydx +

11/2

1/xx

x2dydx =

1/20

x3dx +

11/2

(x x3)dx = 1/8

23. (a)

-1.5 -1-2 -0.5 0.5 1 1.5

x

y

1

2

3

4

(b) x = (1.8414, 0.1586), (1.1462, 3.1462)

(c)

R

x dA 1.14621.8414

x+2ex

xdydx =

1.14621.8414

x(x + 2 ex) dx 0.4044

(d)

R

x dA 3.14620.1586

ln yy2

xdxdy =

3.14620.1586

ln2 y

2 (y 2)

2

2

dy 0.4044

24. (a)

1 2 3

5

15

25

x

y

R

(b) (1, 3), (3, 27)

(c) 3

1

4x3x4

34x+4x2xdydx =

3

1

x[(4x3

x4)

(3

4x + 4x2)] dx =

224

15

25. A =

/40

cos xsinx

dydx =

/40

(cos x sin x)dx =

2 1

26. A =

14

y23y4

dxdy =

14

(y2 3y + 4)dy = 125/6


6/37


27. A =

33

9y21y2/9

dxdy =

33

8(1 y2/9)dy = 32

28. A =

10

cosh xsinhx

dydx =

10

(cosh x sinh x)dx = 1 e1

29. 4063x/20

(3

3x/4

y/2) dydx = 40

[(3

3x/4)(6

3x/2)

(6

3x/2)2/4] dx = 12

30.

20

4x20

4 x2 dydx =

20

(4 x2) dx = 16/3

31. V =

33

9x29x2

(3 x)dydx =33

(6

9 x2 2x

9 x2)dx = 27

32. V =

10

xx2

(x2 + 3y2)dydx =

10

(2x3 x4 x6)dx = 11/70

33. V =30

20

(9x2 + y2)dydx =30

(18x2 + 8/3)dx = 170

34. V =

11

1y2

(1 x)dxdy =11

(1/2 y2 + y4/2)dy = 8/15

35. V =

3/23/2

94x294x2

(y + 3)dydx =

3/23/2

6

9 4x2 dx = 27/2

36. V =

30

3y2/3

(9 x2)dxdy =30

(18 3y2 + y6/81)dy = 216/7

37. V = 8

50

25x20

25 x2dydx = 8

50

(25 x2)dx = 2000/3

38. V = 2

20

1(y1)20

(x2 + y2)dxdy = 2

20

1

3[1 (y 1)2]3/2 + y2[1 (y 1)2]1/2

dy,

let y 1 = sin to get V = 2/2/2

1

3cos3 + (1 + sin )2 cos

cos d which eventually yields

V = 3/2

39. V = 4 1

01x2

0

(1

x2

y2)dydx =

8

3

1

0

(1

x2)3/2dx = /2

40. V =

20

4x20

(x2 + y2)dydx =

20

x2

4 x2 + 13

(4 x2)3/2

dx = 2

41.

20

2y2

f(x, y)dxdy 42.

80

x/20

f(x, y)dydx 43.

e21

2lnx

f(x, y)dydx


7/37

626 Chapter 15

44.

10

eey

f(x, y)dxdy 45.

/20

sinx0

f(x, y)dydx 46.

10

xx2

f(x, y)dydx

47.

40

y/40

ey2

dxdy =

40

1

4yey

2

dy = (1 e16)/8

48. 1

0

2x

0

cos(x2)dydx = 1

0

2x cos(x2)dx = sin1

49.

20

x20

ex3

dydx =

20

x2ex3

dx = (e8 1)/3

50.

ln 30

3ey

xdxdy =1

2

ln 30

(9 e2y)dy = 12

(9ln3 4)

51.

20

y20

sin(y3)dxdy =

20

y2 sin(y3)dy = (1 cos8)/3

52. 1

0 e

e

x

xdydx = 1

0

(ex xex)dx = e/2 1

53. (a)

40

2x

sin y3 dydx; the inner integral is non-elementary.

20

y20

sin

y3

dxdy =

20

y2 sin

y3

dy = 13

cos

y3 2

0

= 0

(b)

10

/2sin1 y

sec2(cos x)dxdy ; the inner integral is non-elementary.

/20

sin x0

sec2(cos x)dydx =

/20

sec2(cos x)sin x dx = tan 1

54. V = 4

2

0

4x20

(x2 + y2) dy dx = 4

2

0

x2

4 x2 + 13

(4 x2)3/2

dx (x = 2sin )

=

/20

64

3+

64

3sin2 128

3sin4

d =

64

3

2+

64

3

4 128

3

2

1 32 4 = 8

55. The region is symmetric with respect to the y-axis, and the integrand is an odd function of x,hence the answer is zero.

56. This is the volume in the first octant under the surface z =

1 x2 y2, so 1/8 of the volume ofthe sphere of radius 1, thus

6.

57. Area of triangle is 1/2, so f = 210

1x

1

1 + x2 dydx = 210

11 + x2

x

1 + x2

dx =

2 ln 2

58. Area =

20

(3x x2 x) dx = 4/3, so

f =3

4

20

3xx2x

(x2 xy)dydx = 34

20

(2x3 + 2x4 x5/2)dx = 34

8

15= 2

5


8/37


59. Tave =1

A(R)

R

(5xy + x2) dA. The diamond has corners (2, 0), (0,4) and thus has area

A(R) = 41

22(4) = 16m2. Since 5xy is an odd function of x (as well as y),

R

5xydA = 0. Since

x2 is an even function of both x and y,

Tave =4

16 R

x,y>0

x2 dA =1

4

2

0

42x

0

x2 dydx =1

4

2

0

(4

2x)x2 dx =1

44

3x3

1

2x4

2

0

=2

3

C

60. The area of the lens is R2 = 4 and the average thickness Tave is

Tave =4

4

20

4x20

1 (x2 + y2)/4 dydx = 1

20

1

6(4 x2)3/2dx (x = 2cos )

=8

3

0

sin4 d =8

3

1 32 4

2=

1

2in

61. y = sin x and y = x/2 intersect at x = 0 and x = a = 1.895494, so

V = a

0 sinx

x/2 1 + x + y dy dx = 0.676089

EXERCISE SET 15.3

1.

/20

sin 0

r cos dr d =

/20

1

2sin2 cos d = 1/6

2.

0

1+cos 0

rdrd =

0

1

2(1 + cos )2d = 3/4

3. /2

0

a sin

0

r2drd = /2

0

a3

3

sin3 d =2

9

a3

4.

/60

cos30

rdrd =

/60

1

2cos2 3 d = /24

5.

0

1sin 0

r2 cos drd =

0

1

3(1 sin )3 cos d = 0

6.

/20

cos 0

r3drd =

/20

1

4cos4 d = 3/64

7. A = 2

0

1cos

0

rdrd = 2

0

1

2

(1

cos )2d = 3/2

8. A = 4

/20

sin 20

rdrd = 2

/20

sin2 2 d = /2

9. A =

/2/4

1sin2

rdrd =

/2/4

1

2(1 sin2 2)d = /16


9/37

628 Chapter 15

10. A = 2

/30

2sec

r d r d =

/30

(4 sec2 )d = 4/3

3

11. A = 2

/2/6

4sin 2

rdrd =

/2/6

(16 sin2 4)d = 4/3 + 2

3

12. A = 2

/2

1

1+cos

rdrd =

/2

(

2cos

cos2 )d = 2

/4

13. V = 8

/20

31

r

9 r2 drd = 1283

2

/20

d =64

3

2

14. V = 2

/20

2sin 0

r2drd =16

3

/20

sin3 d = 32/9

15. V = 2

/20

cos 0

(1 r2)rdrd = 12

/20

(2cos2 cos4 )d = 5/32

16. V = 4 /2

0

3

1

drd = 8 /2

0

d = 4

17. V =

/20

3 sin 0

r2 sin drd = 9

/20

sin4 d = 27/16

18. V = 4

/20

22cos

r

4 r2 drd +/2

20

r

4 r2 drd

=32

3

/20

sin3 d +32

3

/2

d =64

9+

16

3

19. 2

0

1

0

er2

rdrd =1

2

(1

e1)

2

0

d = (1

e1)

20.

/20

30

r

9 r2 drd = 9/20

d = 9/2

21.

/40

20

1

1 + r2rdrd =

1

2ln 5

/40

d =

8ln 5

22.

/2/4

2cos 0

2r2 sin drd =16

3

/2/4

cos3 sin d = 1/3

23. /2

0

1

0

r3drd =1

4

/2

0

d = /8

24.

20

20

er2

rdrd =1

2(1 e4)

20

d = (1 e4)

25.

/20

2cos 0

r2drd =8

3

/20

cos3 d = 16/9


10/37


26.

/20

10

cos(r2)rdrd =1

2sin1

/20

d =

4sin1

27.

/20

a0

r

(1 + r2)3/2drd =

2

1 1/

1 + a2

28. /4

0

sec tan

0

r2drd =1

3

/4

0

sec3 tan3 d = 2(

2 + 1)/45

29.

/40

20

r1 + r2

drd =

4(

5 1)

30.

/2tan1(3/4)

53csc

r d r d =1

2

/2tan1(3/4)

(25 9csc2 )d

=25

2

2 tan1(3/4)

6 = 25

2tan1(4/3) 6

31. V =

20

a0

hrdrd =

20

ha2

2d = a2h

32. (a) V = 8

/20

a0

c

a(a2 r2)1/2 rdrd = 4c

3a(a2 r2)3/2

a0

=4

3a2c

(b) V 43

(6378.1370)26356.5231 1,083,168,200,000 km3

33. V = 2

/20

a sin 0

c

a(a2 r2)1/2r d r d = 2

3a2c

/20

(1 cos3 )d = (3 4)a2c/9

34. A = 4

/40

a2cos20

rdrd = 4a2/40

cos2 d = 2a2

35. A =

/4

/6

4sin

8cos2

r d r d +

/2

/4

4sin

0

rdrd

=

/4/6

(8sin2 4cos2)d +/2/4

8sin2 d = 4/3 + 2

3 2

36. A =

0

2a sin 0

rdrd = 2a20

sin2 d = a2 12

a2 sin2

37. (a) I2 =

+0

ex2

dx

+0

ey2

dy

=

+0

+0

ex2

dx

ey

2

dy

=+0

+0

ex2

ey2

dxdy =+0

+0

e(x2

+y2

)dxdy

(b) I2 =

/20

+0

er2

rdrd =1

2

/20

d = /4 (c) I =

/2

38. (a) 1.173108605 (b)

0

10

rer4

drd =

10

rer4

dr 1.173108605


11/37

630 Chapter 15

39. V =

20

R0

D(r)rdrd =

20

R0

kerrdrd = 2k(1 + r)erR0

= 2k[1 (R + 1)eR]

40.

tan1(2)tan1(1/3)

20

r3 cos2 drd = 4

tan1(2)tan1(1/3)

cos2 d =1

5+ 2[tan1(2) tan1(1/3)] = 1

5+

2

EXERCISE SET 15.4

1. (a) z

xy

(b)

x

y

z

(c)

x

y

z

2. (a)

y

x

z (b) z

y

x

(c) z

yx


12/37


3. (a) x = u, y = v, z =5

2+

3

2u 2v (b) x = u, y = v, z = u2

4. (a) x = u, y = v, z =v

1 + u2(b) x = u, y = v, z =

1

3v2 5

3

5. (a) x = 5cos u, y = 5sin u, z = v; 0 u 2, 0 v 1(b) x = 2 cos u, y = v, z = 2sin u; 0 u 2, 1 v 3

6. (a) x = u, y = 1 u, z = v;1 v 1 (b) x = u, y = 5 + 2v, z = v; 0 u 3

7. x = u, y = sin u cos v, z = sin u sin v 8. x = u, y = eu cos v, z = eu sin v

9. x = r cos , y = r sin , z =1

1 + r210. x = r cos , y = r sin , z = er

2

11. x = r cos , y = r sin , z = 2r2 cos sin

12. x = r cos , y = r sin , z = r2(cos2 sin2 )

13. x = r cos , y = r sin , z =

9

r2; r

5

14. x = r cos , y = r sin , z = r; r 3 15. x = 12

cos , y =1

2 sin , z =

3

2

16. x = 3cos , y = 3sin , z = 3cot 17. z = x 2y; a plane

18. y = x2 + z2, 0 y 4; part of a circular paraboloid

19. (x/3)2 + (y/2)2 = 1; 2 z 4; part of an elliptic cylinder

20. z = x2 + y2; 0 z 4; part of a circular paraboloid

21. (x/3)

2

+ (y/4)

2

= z2

; 0 z 1; part of an elliptic cone22. x2 + (y/2)2 + (z/3)2 = 1; an ellipsoid

23. (a) x = r cos , y = r sin , z = r, 0 r 2; x = u, y = v, z = u2 + v2; 0 u2 + v2 4

24. (a) I: x = r cos , y = r sin , z = r2, 0 r 2; II: x = u, y = v, z = u2 + v2; u2 + v2 2

25. (a) 0 u 3, 0 v (b) 0 u 4,/2 v /2

26. (a) 0 u 6, v 0 (b) 0 u 5, /2 v 3/2

27. (a) 0 /2, 0 2 (b) 0 , 0

28. (a) /2 , 0 2 (b) 0 /2, 0 /2

29. u = 1, v = 2, ru rv = 2i 4j + k; 2x + 4y z = 5

30. u = 1, v = 2, ru rv = 4i 2j + 8k; 2x + y 4z = 6

31. u = 0, v = 1, ru rv = 6k; z = 0 32. ru rv = 2i j 3k; 2x y 3z = 4


13/37

632 Chapter 15

33. ru rv = (

2/2)i (2/2)j + (1/2)k; x y +

2

2z =

2

8

34. ru rv = 2i ln 2k; 2x (ln2)z = 0

35. z =

9 y2, zx = 0, zy = y/

9 y2, z2x + z2y + 1 = 9/(9 y2),

S =2033

39 y2 dydx =

2

0 3 dx = 6

36. z = 8 2x 2y, z2x + z2y + 1 = 4 + 4 + 1 = 9, S =40

4x0

3 dydx =

40

3(4 x)dx = 24

37. z2 = 4x2 + 4y2, 2zzx = 8x so zx = 4x/z, similarly zy = 4y/z thus

z2x + z2y + 1 = (16x

2 + 16y2)/z2 + 1 = 5, S =

10

xx2

5 dydx =

5

10

(x x2)dx =

5/6

38. z2 = x2 + y2, zx = x/z, zy = y/z, z2x + z

2y + 1 = (z

2 + y2)/z2 + 1 = 2,

S =R

2 dA = 2 /2

0

2cos 0 2 rdrd = 42

/20 cos

2

d = 2

39. zx = 2x, zy = 2y, z2x + z2y + 1 = 4x2 + 4y2 + 1,

S =

R

4x2 + 4y2 + 1 dA =

20

10

r

4r2 + 1 drd

=1

12(5

5 1)20

d = (5

5 1)/6

40. zx = 2, zy = 2y, z2x + z2y + 1 = 5 + 4y

2,

S =10

y0

5 + 4y

2

dxdy =10 y

5 + 4y

2

dy = (27 55)/12

41. r/u = cos vi + sin vj + 2uk, r/v = u sin vi + u cos vj,

r/u r/v = u4u2 + 1; S =20

21

u

4u2 + 1 du dv = (17

17 5

5)/6

42. r/u = cos vi + sin vj + k, r/v = u sin vi + u cos vj,

r/u r/v = 2u; S =/20

2v0

2 u du dv =

2

123

43. zx = y, zy = x, z2x + z

2y + 1 = x

2 + y2 + 1,

S =

R

x2 + y2 + 1 dA =

/60

30

r

r2 + 1 drd =1

3(10

101)

/60

d = (10

101)/18


14/37


15/37

634 Chapter 15

54. (a) x = v cos u, y = v sin u, z = f(v), for example (b) x = v cos u, y = v sin u, z = 1/v2

(c) z

yx

55. (x/a)2 + (y/b)2 + (z/c)2 = cos2 v(cos2 u + sin2 u) + sin2 v = 1, ellipsoid

56. (x/a)2 + (y/b)2 (z/c)2 = cos2 u cosh2 v + sin2 u cosh2 v sinh2 v = 1, hyperboloid of one sheet

57. (x/a)2 + (y/b)2 (z/c)2 = sinh2 v + cosh2 v(sinh2 u cosh2 u) = 1, hyperboloid of two sheets

EXERCISE SET 15.5

1.

11

20

10

(x2 + y2 + z2)dxdydz =

11

20

(1/3 + y2 + z2)dydz =

11

(10/3 + 2z2)dz = 8

2.

1/21/3

0

10

zx sin xy dz dy dx =

1/21/3

0

1

2x sin xydydx =

1/21/3

1

2(1 cos x)dx = 1

12+

3 24

3.

20

y21

z1

yz dx dz dy =

20

y21

(yz2 + yz)dz dy =

20

1

3y7 +

1

2y5 1

6y

dy =

47

3

4.

/4

0

1

0

x2

0

x cos y dz dxdy =

/4

0

1

0

x3 cos ydxdy =

/4

0

14

cos y dy = 2/8

5.

30

9z20

x0

xydydxdz =

30

9z20

1

2x3dxdz =

30

1

8(81 18z2 + z4)dz = 81/5

6.

31

x2x

ln z0

xey dy dz dx =

31

x2x

(xz x)dz dx =31

1

2x5 3

2x3 + x2

dx = 118/3

7.

20

4x20

3x2y25+x2+y2

x dz dydx =

20

4x20

[2x(4 x2) 2xy2]dydx

=

2

0

43

x(4 x2)3/2dx = 128/15

8.

21

2z

3y0

y

x2 + y2dxdydz =

21

2z

3dydz =

21

3(2 z)dz = /6


16/37


9.

0

10

/60

xy sin yz dz dy dx =

0

10

x[1 cos(y/6)]dydx =0

(1 3/)x dx = ( 3)/2

10.

11

1x20

y0

y dz dydx =

11

1x20

y2dydx =

11

1

3(1 x2)3dx = 32/105

11.20

x0

2x20

xyz dz dy dx =20

x0

1

2 xy(2 x2)2dydx =20

1

4 x3(2 x2)2dx = 1/6

12.

/2/6

/2y

xy0

cos(z/y)dz dx dy =

/2/6

/2y

y sin xdxdy =

/2/6

y cos y dy = (5 6

3)/12

13.

30

21

12

x + z2

ydzdydx 9.425

14. 8

10

1x20

1x2y20

ex2y2z2 dz dy dx 2.381

15. V =

40

(4x)/20

(123x6y)/40

dz dy dx =

40

(4x)/20

1

4(12 3x 6y)dydx

=

40

3

16(4 x)2dx = 4

16. V =

10

1x0

y0

dz dy dx =

10

1x0

y d y d x =

10

2

3(1 x)3/2dx = 4/15

17. V = 2 2

0 4

x2 4y

0

dz dy dx = 2 2

0 4

x2(4 y)dydx = 2

2

0 8 4x2 +

1

2x4 dx = 256/15

18. V =

10

y0

1y20

dz dx dy =

10

y0

1 y2 dxdy =

10

y

1 y2 dy = 1/3

19. The projection of the curve of intersection onto the xy-plane is x2 + y2 = 1,

V = 4

10

1x20

43y24x2+y2

dz dy dx

20. The projection of the curve of intersection onto the xy-plane is 2x2 + y2 = 4,

V = 420

42x20

8x2y23x2+y2

dz dy dx

21. V = 2

33

9x2/30

x+30

dz dy dx 22. V = 8

10

1x20

1x20

dz dy dx


17/37

636 Chapter 15

23. (a)

(0, 0, 1)

(1, 0, 0)(0, 1, 0)

z

y

x

(b)

(0, 9, 9)

(3, 9, 0)

z

x

y

(c)

(0, 0, 1)

(1, 2, 0)

x

y

z

24. (a)

(3, 9, 0)

(0, 0, 2)

x

y

z

(b)(0, 0, 2)

(0, 2, 0)

(2, 0, 0)

x

y

z

(c)

(2, 2, 0)

(0, 0, 4)

x

y

z

25. V = 1

0 1x

0 1xy

0

dz dy dx = 1/6, fave = 6 1

0 1x

0 1xy

0

(x + y + z) dz dy dx =3

4

26. The integrand is an odd function of each of x, y, and z, so the answer is zero.

27. The volume V =3

2, and thus

rave =

2

3

G

x2 + y2 + z2 dV =

2

3

1/21/

2

12x212x2

67x2y25x2+5y2

x2 + y2 + z2dzdydx 3.291


18/37


28. V = 1, dave =1

V

10

10

10

(x z)2 + (y z)2 + z2 dxdydz 0.771

29. (a)

a0

b(1x/a)0

c(1x/ay/b)0

dz dy dx,

b0

a(1y/b)0

c(1x/ay/b)0

dz dx dy,

c0

a(1z/c)0

b(1x/az/c)0

dydxdz,

a0

c(1x/a)0

b(1x/az/c)0

dy dz dx,

c

0

b(1z/c)

0

a(1y/bz/c)

0

dxdy dz,

b

0

c(1y/b)

0

a(1y/bz/c)

0

dx dz dy

(b) Use the first integral in Part (a) to geta0

b(1x/a)0

c

1 xa y

b

dydx =

a0

1

2bc

1 xa

2dx =

1

6abc

30. V = 8

a0

b1x2/a20

c1x2/a2y2/b20

dz dy dx

31. (a) 2

04x2

0

5

0

f(x,y,z) dzdydx

(b)

90

3x0

3xy

f(x, y, z) dzdydx (c)

20

4x20

8yy

f(x,y,z) dzdydx

32. (a)

30

9x20

9x2y20

f(x, y, z)dz dy dx

(b)

40

x/20

20

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

20

4x20

4yx2

f(x,y,z)dz dy dx

33. (a) At any point outside the closed sphere {x2 + y2 + z2 1} the integrand is negative, so tomaximize the integral it suffices to include all points inside the sphere; hence the maximumvalue is taken on the region G = {x2 + y2 + z2 1}.

(b) 4.934802202

(c)

20

0

10

(1 2)ddd = 2

2

34.

ba

dc

k

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

ba

dc

f(x)g(y)

k

h(z)dz

dydx

=

ba

f(x)

dc

g(y)dy

dx

k

h(z)dz

= b

a

f(x)dxd

c

g(y)dy

k

h(z)dz35. (a)

11

x dx

10

y2dy

/20

sin z dz

= (0)(1/3)(1) = 0

(b)

10

e2xdx

ln 30

eydy

ln 20

ezdz

= [(e2 1)/2](2)(1/2) = (e2 1)/2


19/37

638 Chapter 15

EXERCISE SET 15.6

1. Let a be the unknown coordinate of the fulcrum; then the total moment about the fulcrum is5(0 a) + 10(5 a) + 20(10 a) = 0 for equilibrium, so 250 35a = 0, a = 50/7. The fulcrumshould be placed 50/7 units to the right of m1.

2. At equilibrium, 10(0 4) + 3(2 4) + 4(3 4) + m(6 4) = 0, m = 25

3. A = 1, x =

1

0

1

0

xdydx =1

2, y =

1

0

1

0

y d y d x =1

2

4. A = 2, x =1

2

G

xdydx, and the region of integration is symmetric with respect to the x-axes

and the integrand is an odd function of x, so x = 0. Likewise, y = 0.

5. A = 1/2,

R

x dA =

10

x0

xdydx = 1/3,

R

y dA =

10

x0

y d y d x = 1/6;

centroid (2/3, 1/3)

6. A =

10

x20

dydx = 1/3,

R

x dA =

10

x20

xdydx = 1/4,

R

y dA =

10

x20

y d y d x = 1/10; centroid (3/4, 3/10)

7. A =

10

2x2x

dydx = 7/6,

R

x dA =

10

2x2x

xdydx = 5/12,

R

y dA =

10

2x2x

y d y d x = 19/15; centroid (5/14, 38/35)

8. A =

4,

R

x dA =

10

1x20

xdydx =1

3, x =

4

3, y =

4

3by symmetry

9. x = 0 from the symmetry of the region,

A =1

2(b2 a2),

R

y dA =

0

ba

r2 sin drd =2

3(b3 a3); centroid x = 0, y = 4(b

3 a3)3(b2 a2) .

10. y = 0 from the symmetry of the region, A = a2/2,

R

x dA = /2

/2 a

0

r2 cos drd = 2a3/3; centroid 4a3

, 0

11. M=

R

(x, y)dA =

10

10

|x + y 1| dxdy

=

10

1x0

(1 x y) dy +11x

(x + y 1) dy

dx =1

3


20/37


x = 3

10

10

x(x, y) dydx = 3

10

1x0

x(1 x y) dy +11x

x(x + y 1) dy

dx =1

2

By symmetry, y =1

2as well; center of gravity (1/2, 1/2)

12. x =1

M

G

x(x, y) dA, and the integrand is an odd function of x while the region is symmetric

with respect to the y-axis, thus x = 0; likewise y = 0.

13. M =

10

x0

(x + y)dydx = 13/20, Mx =

10

x0

(x + y)y d y d x = 3/10,

My =

10

x0

(x + y)xdydx = 19/42, x = My/M = 190/273, y = Mx/M = 6/13;

the mass is 13/20 and the center of gravity is at (190/273, 6/13).

14. M =

0

sin x0

y d y d x = /4, x = /2 from the symmetry of the density and the region,

Mx =0sinx0 y

2

dydx = 4/9, y = Mx/M =

16

9 ; mass /4, center of gravity

2 ,

16

9

.

15. M =

/20

a0

r3 sin cos drd = a4/8, x = y from the symmetry of the density and the

region, My =

/20

a0

r4 sin cos2 drd = a5/15, x = 8a/15; mass a4/8, center of gravity

(8a/15, 8a/15).

16. M =

0

10

r3drd = /4, x = 0 from the symmetry of density and region,

Mx =

0

1

0

r4 sin drd = 2/5, y =8

5

; mass /4, center of gravity 0,8

5.

17. V = 1, x =

10

10

10

x dz dydx =1

2, similarly y = z =

1

2; centroid

1

2,

1

2,

1

2

18. symmetry,

G

z dz dy dx =

20

20

10

rz dr d dz = 2, centroid = (0, 0, 1)

19. x = y = z from the symmetry of the region, V = 1/6,

x =1

V

10

1x0

1xy0

x dz dydx = (6)(1/24) = 1/4; centroid (1/4, 1/4, 1/4)

20. The solid is described by 1 y 1, 0 z 1 y2, 0 x 1 z;

V =

11

1y20

1z0

dx dz dy =4

5, x =

1

V

11

1y20

1z0

x dxdz dy =5

14, y = 0 by symmetry,

z =1

V

11

1y20

1z0

z dx dz dy =2

7; the centroid is

5

14, 0,

2

7

.


21/37

640 Chapter 15

21. x = 1/2 and y = 0 from the symmetry of the region,

V =

10

11

1y2

dz dy dx = 4/3, z =1

V

G

z dV = (3/4)(4/5) = 3/5; centroid (1/2, 0, 3/5)

22. x = y from the symmetry of the region,

V = 2

0

2

0

xy

0

dz dy dx = 4, x =1

VG

x dV = (1/4)(16/3) = 4/3,

z =1

V

G

z dV = (1/4)(32/9) = 8/9; centroid (4/3, 4/3, 8/9)

23. x = y = z from the symmetry of the region, V = a3/6,

x =1

V

a0

a2x20

a2x2y20

x dz dydx =1

V

a0

a2x20

x

a2 x2 y2 dydx

=1

V

/20

a0

r2

a2 r2 cos drd = 6a3

(a4/16) = 3a/8; centroid (3a/8, 3a/8, 3a/8)

24. x = y = 0 from the symmetry of the region, V = 2a3/3

z =1

V

aa

a2x2a2x2

a2x2y20

z dz dy dx =1

V

aa

a2x2a2x2

1

2(a2 x2 y2)dydx

=1

V

20

a0

1

2(a2 r2)rdrd = 3

2a3(a4/4) = 3a/8; centroid (0, 0, 3a/8)

25. M =

a0

a0

a0

(a x)dz dy dx = a4/2, y = z = a/2 from the symmetry of density and

region, x =1

M a

0

a

0

a

0

x(a x)dz dy dx = (2/a4)(a5/6) = a/3;

mass a4/2, center of gravity (a/3, a/2, a/2)

26. M =

aa

a2x2a2x2

h0

(h z)dz dy dx = 12

a2h2, x = y = 0 from the symmetry of density

and region, z =1

M

G

z(h z)dV = 2a2h2

(a2h3/6) = h/3;

mass a2h2/2, center of gravity (0, 0, h/3)

27. M = 1

1 1

0

1y2

0

yz dz dy dx = 1/6, x = 0 by the symmetry of density and region,

y =1

M

G

y2z dV = (6)(8/105) = 16/35, z =1

M

G

yz2dV = (6)(1/12) = 1/2;

mass 1/6, center of gravity (0, 16/35, 1/2)


22/37


28. M =

30

9x20

10

xz dz dy dx = 81/8, x =1

M

G

x2z dV = (8/81)(81/5) = 8/5,

y =1

M

G

xyz dV = (8/81)(243/8) = 3, z =1

M

G

xz2dV = (8/81)(27/4) = 2/3;

mass 81/8, center of gravity (8/5, 3, 2/3)

29. (a) M =10

10

k(x2 + y2)dy dx = 2k/3, x = y from the symmetry of density and region,

x =1

M

R

kx(x2 + y2)dA =3

2k(5k/12) = 5/8; center of gravity (5/8, 5/8)

(b) y = 1/2 from the symmetry of density and region,

M =

10

10

kxdydx = k/2, x =1

M

R

kx2dA = (2/k)(k/3) = 2/3,

center of gravity (2/3, 1/2)

30. (a) x = y = z from the symmetry of density and region,

M =10

10

10

k(x2

+ y2

+ z2

)dz dy dx = k,

x =1

M

G

kx(x2 + y2 + z2)dV = (1/k)(7k/12) = 7/12; center of gravity (7/12, 7/12, 7/12)

(b) x = y = z from the symmetry of density and region,

M =

10

10

10

k(x + y + z)dz dy dx = 3k/2,

x =1

M

G

kx(x + y + z)dV =2

3k(5k/6) = 5/9; center of gravity (5/9, 5/9, 5/9)

31. V =G

dV =0

sinx0

1/(1+x2+y2)0

dz dy dx = 0.666633,

x =1

V

G

xdV = 1.177406, y =1

V

G

ydV = 0.353554, z =1

V

G

zdV = 0.231557

32. (b) Use polar coordinates for x and y to get

V =

G

dV =

20

a0

1/(1+r2)0

r dz dr d = ln(1 + a2),

z =1

V G

zdV =a2

2(1 + a2) ln(1 + a2)

Thus lima0+

z =1

2; lima+

z = 0.

lima0+

z =1

2; lima+

z = 0

(c) Solve z = 1/4 for a to obtain a 1.980291.


23/37

642 Chapter 15

33. Let x = r cos , y = r sin , and dA = rdrd in formulas (11) and (12).

34. x = 0 from the symmetry of the region, A =

20

a(1+sin )0

rdrd = 3a2/2,

y =1

A

20

a(1+sin )0

r2 sin drd =2

3a2(5a3/4) = 5a/6; centroid (0, 5a/6)

35. x = y from the symmetry of the region, A =/20

sin 20

rdrd = /8,

x =1

A

/20

sin20

r2 cos drd = (8/)(16/105) =128

105; centroid

128

105,

128

105

36. x = 3/2 and y = 1 from the symmetry of the region,R

x dA = xA = (3/2)(6) = 9,

R

y dA = yA = (1)(6) = 6

37. x = 0 from the symmetry of the region, a2/2 is the area of the semicircle, 2y is the distancetraveled by the centroid to generate the sphere so 4a3/3 = (a2/2)(2y), y = 4a/(3)

38. (a) V =

1

2a2

2

a +

4a

3

=

1

3(3 + 4)a3

(b) the distance between the centroid and the line is

2

2

a +

4a

3

so

V =

1

2a2

2

2

2

a +

4a

3

=

1

6

2(3 + 4)a3

39. x = k so V = (ab)(2k) = 22abk

40. y = 4 from the symmetry of the region,

A =

22

8x2x2

dydx = 64/3 so V = (64/3)[2(4)] = 512/3

41. The region generates a cone of volume1

3ab2 when it is revolved about the x-axis, the area of the

region is1

2ab so

1

3ab2 =

1

2ab

(2y), y = b/3. A cone of volume

1

3a2b is generated when the

region is revolved about the y-axis so1

3a2b =

1

2ab

(2x), x = a/3. The centroid is (a/3, b/3).

42. Ix = a

0 b

0

y2 d y d x =1

3ab3, Iy =

a

0 b

0

x2 d y d x =1

3a3b,

Iz =

a0

b0

(x2 + y2) d y d x =1

3ab(a2 + b2)

43. Ix =

20

a0

r3 sin2 d r d = a4/4; Iy =

20

a0

r3 cos2 d r d = a4/4 = Ix;

Iz = Ix + Iy = a4/2


24/37


EXERCISE SET 15.7

1.

20

10

1r20

zr dz drd =

20

10

1

2(1 r2)rdrd =

20

1

8d = /4

2.

/20

cos 0

r20

r sin dz drd =

/20

cos 0

r3 sin drd =

/20

1

4cos4 sin d = 1/20

3.

/2

0

/2

0

1

0

3 sin cos ddd =

/2

0

/2

0

14

sin cos dd =

/2

0

18

d = /16

4.

20

/40

a sec 0

2 sin ddd =

20

/40

1

3a3 sec3 sin dd =

20

1

6a3d = a3/3

5. V =

20

30

9r2

r dz drd =

20

30

r(9 r2)drd =20

81

4d = 81/2

6. V = 2

20

20

9r20

r dz drd = 2

20

20

r

9 r2drd

=

2

3 (27 55) 2

0 d = 4(27 55)/37. r2 + z2 = 20 intersects z = r2 in a circle of radius 2; the volume consists of two portions, one inside

the cylinder r =

20 and one outside that cylinder:

V =

20

20

r220r2

r dz drd +

20

202

20r220r2

r dz drd

=

20

20

r

r2 +

20 r2

drd +

20

202

2r

20 r2 drd

=4

3(10

5 13)

20

d +128

3

20

d =152

3 +

80

3

5

8. z = hr/a intersects z = h in a circle of radius a,

V =

20

a0

hhr/a

r dz drd =

20

a0

h

a(ar r2)drd =

20

1

6a2h d = a2h/3

9. V =

20

/30

40

2 sin ddd =

20

/30

64

3sin dd =

32

3

20

d = 64/3

10. V =

20

/40

21

2 sin ddd =

20

/40

7

3sin dd =

7

6(2

2)

20

d = 7(2

2)/3

11. In spherical coordinates the sphere and the plane z = a are = 2a and = a sec , respectively.They intersect at = /3,

V =

2

0

/3

0

a sec

0

2 sin ddd +

2

0

/2

/3

2a

0

2 sin ddd

=

20

/30

1

3a3 sec3 sin dd +

20

/2/3

8

3a3 sin dd

=1

2a320

d +4

3a320

d = 11a3/3


25/37

644 Chapter 15

12. V =

20

/2/4

30

2 sin ddd =

20

/2/4

9sin dd =9

2

2

20

d = 9

2

13.

/20

a0

a2r20

r3 cos2 dz drd =

/20

a0

(a2r3 r5)cos2 drd

=1

12a6

/2

0

cos2 d = a6/48

14.

0

/20

10

e3

2 sin ddd =1

3(1 e1)

0

/20

sin dd = (1 e1)/3

15.

/20

/40

80

4 cos2 sin ddd = 32(2

2 1)/15

16.

20

0

30

3 sin ddd = 81

17. (a) /2

/3

4

1

2

2

r tan3 1 + z

2dz dr d =

/3

/6

tan3 d4

1

r dr2

2

11 + z

2dz

=

4

3 1

2ln 3

15

2

2ln(

5 2)

=

5

2(8 + 3ln 3)ln(

5 2)

(b)

/2/3

41

22

y tan3 z1 + x2

dxdy dz; the region is a rectangular solid with sides /6, 3, 4.

18.

/20

/40

1

18cos37 cos dd =

2

36

/20

cos37 d =4,294,967,296

755,505,013,725

2 0.008040

19. (a) V = 2

20

a0

a2r20

r dz drd = 4a3/3

(b) V =

20

0

a0

2 sin ddd = 4a3/3

20. (a)

20

4x20

4x2y20

xyz dz dy dx

=

20

4x20

1

2xy(4 x2 y2)dydx = 1

8

20

x(4 x2)2dx = 4/3

(b)

/20

20

4r20

r3z sin cos dz drd

=/20

20

1

2 (4r3

r5

)sin cos drd =

8

3/20 sin cos d = 4/3

(c)

/20

/20

20

5 sin3 cos sin cos ddd

=

/20

/20

32

3sin3 cos sin cos dd =

8

3

/20

sin cos d = 4/3


26/37


21. M =

20

30

3r

(3 z)r dz drd =20

30

1

2r(3 r)2drd = 27

8

20

d = 27/4

22. M =

20

a0

h0

k zrdz dr d =

20

a0

1

2kh2rdrd =

1

4ka2h2

20

d = ka2h2/2

23. M = 2

0

0

a

0

k3 sin ddd = 2

0

0

1

4ka4 sin dd =

1

2ka4

2

0

d = ka4

24. M =

20

0

21

sin ddd =

20

0

3

2sin dd = 3

20

d = 6

25. x = y = 0 from the symmetry of the region,

V =

20

10

2r2r2

r dz drd =

20

10

(r

2 r2 r3)drd = (8

2 7)/6,

z =1

V

20

10

2r2r2

zr dz drd =6

(8

2 7) (7/12) = 7/(16

2 14);

centroid 0, 0, 7162 14

26. x = y = 0 from the symmetry of the region, V = 8/3,

z =1

V

20

20

2r

zr dz drd =3

8(4) = 3/2; centroid (0, 0, 3/2)

27. x = y = z from the symmetry of the region, V = a3/6,

z =1

V

/20

/20

a0

3 cos sin ddd =6

a3(a4/16) = 3a/8;

centroid (3a/8, 3a/8, 3a/8)

28. x = y = 0 from the symmetry of the region, V =20

/30

40

2 sin ddd = 64/3,

z =1

V

20

/30

40

3 cos sin ddd =3

64(48) = 9/4; centroid (0, 0, 9/4)

29. y = 0 from the symmetry of the region, V = 2

/20

2 cos 0

r20

r dz drd = 3/2,

x =2

V

/20

2 cos 0

r20

r2 cos dz drd =4

3() = 4/3,

z =2

V /2

0 2cos

0 r2

0

rz dz dr d =4

3(5/6) = 10/9; centroid (4/3, 0, 10/9)

30. M=

/20

2 cos 0

4r20

zr dz drd =

/20

2cos 0

1

2r(4 r2)2drd

=16

3

/20

(1 sin6 )d = (16/3)(11/32) = 11/6


27/37

646 Chapter 15

31. V =

/20

/3/6

20

2 sin ddd =

/20

/3/6

8

3sin dd =

4

3(

3 1)/20

d

= 2(

3 1)/3

32. M =

20

/40

10

3 sin ddd =

20

/40

1

4sin dd =

1

8(2

2)

20

d = (2

2)/4

33. x = y = 0 from the symmetry of density and region,

M =

20

10

1r20

(r2 + z2)r dz drd = /4,

z =1

M

20

10

1r20

z(r2+z2)r dz drd = (4/)(11/120) = 11/30; center of gravity (0, 0, 11/30)

34. x = y = 0 from the symmetry of density and region, M =

20

10

r0

zr dz drd = /4,

z =1

M 2

0 1

0 r

0

z2r dz drd = (4/)(2/15) = 8/15; center of gravity (0, 0, 8/15)

35. x = y = 0 from the symmetry of density and region,

M =

20

/20

a0

k3 sin ddd = ka4/2,

z =1

M

20

/20

a0

k4 sin cos ddd =2

ka4(ka5/5) = 2a/5; center of gravity (0, 0, 2a/5)

36. x = z = 0 from the symmetry of the region, V = 54/3 16/3 = 38/3,

y =1

V

0

0

32

3 sin2 sin ddd =1

V

0

0

65

4sin2 sin dd

=1V

0

658

sin d =3

38(65/4) = 195/152; centroid (0, 195/152, 0)

37. M =

20

0

R0

0e(/R)32 sin ddd =

20

0

1

3(1 e1)R30 sin dd

=4

3(1 e1)0R3

38. (a) The sphere and cone intersect in a circle of radius 0 sin 0,

V = 2

1 0 sin0

0 20r2

r cot 0

r dz drd = 2

1 0 sin0

0 r20 r2 r2 cot 0 drd

=

21

1

330(1 cos3 0 sin3 0 cot 0)d =

1

330(1 cos3 0 sin2 0 cos 0)(2 1)

=1

330(1 cos 0)(2 1).


28/37


(b) From Part (a), the volume of the solid bounded by = 1, = 2, = 1, = 2, and

= 0 is1

330(1 cos 2)(2 1)

1

330(1 cos 1)(2 1) =

1

330(cos 1 cos 2)(2 1)

so the volume of the spherical wedge between = 1 and = 2 is

V =1

332(cos 1 cos 2)(2 1)

1

331(cos 1 cos 2)(2 1)

=

1

3 (

3

2 3

1)(cos 1 cos 2)(2 1)(c)

d

dcos = sin so from the Mean-Value Theorem cos 2cos 1 = (21)sin where

is between 1 and 2. Similarlyd

d3 = 32 so 3231 = 32(21) where is between

1 and 2. Thus cos 1cos 2 = sin and 3231 = 32 so V = 2 sin .

39. Iz =

20

a0

h0

r2 r dz dr d =

20

a0

h0

r3dz dr d =1

2a4h

40. Iy = 2

0

a

0

h

0

(r2 cos2 + z2)r dz drd = 2

0

a

0

(hr3 cos2 +1

3

h3r)drd

=

20

1

4a4h cos2 +

1

6a2h3

d =

4

a4h +

3a2h3

41. Iz =

20

a2a1

h0

r2 r dz dr d =

20

a2a1

h0

r3dz dr d =1

2h(a42 a41)

42. Iz =

20

0

a0

(2 sin2 ) 2 sin ddd =

20

0

a0

4 sin3 ddd =8

15a5

EXERCISE SET 15.8

1.(x, y)

(u, v)=

1 43 5 = 17 2. (x, y)(u, v) =

1 4v4u 1 = 1 16uv

3.(x, y)

(u, v)=

cos u sin vsin u cos v = cos u cos v + sin u sin v = cos(u v)

4.(x, y)

(u, v)=

2(v2 u2)(u2 + v2)2

4uv(u2 + v2)2

4uv

(u2 + v2)22(v2 u2)(u2 + v2)2

= 4/(u2 + v2)2

5. x =2

9u +

5

9v, y = 1

9u +

2

9v;

(x, y)

(u, v)=

2/9 5/91/9 2/9 = 19

6. x = ln u, y = uv;(x, y)

(u, v)=

1/u 0v u = 1


29/37

648 Chapter 15

7. x =

u + v/

2, y =

v u/2; (x, y)(u, v)

=

1

2

2

u + v

1

2

2

u + v

12

2

v u1

2

2

v u

=

1

4

v2 u2

8. x = u3/2/v1/2, y = v1/2/u1/2;(x, y)

(u, v)

=

3u1/2

2v1/2 u

3/2

2v3/2

v1/22u3/2

12u1/2v1/2

=1

2v

9.(x, y, z)

(u, v, w)=

3 1 01 0 20 1 1

= 5

10.(x, y, z)

(u, v, w)=

1 v u 0

v vw u uw uvvw uw uv

= u2v

11. y = v, x = u/y = u/v,z = w

x = w

u/v;(x, y, z)

(u,v,w)=

1/v u/v2 00 1 0

1/v u/v2 1 = 1/v

12. x = (v + w)/2, y = (u w)/2, z = (u v)/2, (x, y, z)(u, v, w)

=

0 1/2 1/2

1/2 0 1/21/2 1/2 0

= 1

4

13.

x

y

(0, 2)

(1, 0) (1, 0)

(0, 0)

14.

1 2 3

1

2

3

4

x

y

(0, 0) (4, 0)

(3, 4)

15.

-3 3

-3

3

x

y

(2, 0)

(0, 3) 16.

1 2

1

2

x

y

17. x =1

5u +

2

5v, y = 2

5u +

1

5v,

(x, y)

(u, v)=

1

5;

1

5

S

u

vdAuv =

1

5

31

41

u

vdu dv =

3

2ln 3


30/37


18. x =1

2u +

1

2v, y =

1

2u 1

2v,

(x, y)

(u, v)= 1

2;

1

2

S

veuvdAuv =1

2

41

10

veuvdu dv =1

2(e4 e 3)

19. x = u + v, y = u v, (x, y)(u, v)

= 2; the boundary curves of the region S in the uv-plane are

v = 0, v = u, and u = 1 so 2 S

sin u cos vdAuv = 2 1

0

u

0

sin u cos v dv du = 1

1

2

sin2

20. x =

v/u, y =

uv so, from Example 3,(x, y)

(u, v)= 1

2u; the boundary curves of the region S in

the uv-plane are u = 1, u = 3, v = 1, and v = 4 so

S

uv2

1

2u

dAuv =

1

2

41

31

v2du dv = 21

21. x = 3u, y = 4v,(x, y)

(u, v)= 12; S is the region in the uv-plane enclosed by the circle u2 + v2 = 1.

Use polar coordinates to obtainS

12

u2

+ v2

(12) dAuv = 14420

10 r

2

dr d = 96

22. x = 2u, y = v,(x, y)

(u, v)= 2; S is the region in the uv-plane enclosed by the circle u2 + v2 = 1. Use

polar coordinates to obtain

S

e(4u2+4v2)(2) dAuv = 2

20

10

re4r2

dr d = (1 e4)/2

23. Let S be the region in the uv-plane bounded by u2 + v2 = 1, so u = 2x, v = 3y,

x = u/2, y = v/3,(x, y)

(u, v)

= 1/2 00 1/3

= 1/6, use polar coordinates to get1

6

S

sin(u2 + v2)dudv =1

6

/20

10

r sin r2 drd =

24( cos r2)

10

=

24(1 cos1)

24. u = x/a,v = y/b,x = au,y = bv;(x, y)

(u, v)= ab; A = ab

20

10

r d r d = ab

25. x = u/3, y = v/2, z = w,(x, y, z)

(u,v,w)= 1/6; S is the region in uvw-space enclosed by the sphere

u2 + v2 + w2 = 36 so

S

u2

9

1

6dVuvw =

1

54

20

0

60

( sin cos )22 sin d d d

=1

54

20

0

60

4 sin3 cos2 d d d =192

5


31/37

650 Chapter 15

26. Let G1 be the region u2 + v2 + w2 1, with x = au,y = bv,z = cw, (x, y, z)

(u, v, w)= abc; then use

spherical coordinates in uvw-space:

Ix =

G

(y2 + z2)dxdydz = abc

G1

(b2v2 + c2w2) dudv dw

= 2

0

0

1

0

abc(b2 sin2 sin2 + c2 cos2 )4 sin ddd

=

20

abc

15(4b2 sin2 + 2c2)d =

4

15abc(b2 + c2)

27. u = = cot1(x/y), v = r =

x2 + y2

28. u = r =

x2 + y2, v = ( + /2)/ = (1/)tan1(y/x) + 1/2

29. u =3

7x 2

7y, v = 1

7x +

3

7y 30. u = x + 4

3y, v = y

31. Let u = y

4x, v = y + 4x, then x =

1

8

(v

u), y =

1

2

(v + u) so(x, y)

(u, v)

=

1

8

;

1

8

S

u

vdAuv =

1

8

52

20

u

vdu dv =

1

4ln

5

2

32. Let u = y + x, v = y x, then x = 12

(u v), y = 12

(u + v) so(x, y)

(u, v)=

1

2;

12

S

uv dAuv = 12

20

10

uv du dv = 12

33. Let u = x

y, v = x + y, then x =

1

2

(v + u), y =1

2

(v

u) so

(x, y)

(u, v)

=1

2

; the boundary curves of

the region S in the uv-plane are u = 0, v = u, and v = /4; thus

1

2

S

sin u

cos vdAuv =

1

2

/40

v0

sin u

cos vdu dv =

1

2[ln(

2 + 1) /4]

34. Let u = y x, v = y + x, then x = 12

(v u), y = 12

(u + v) so(x, y)

(u, v)= 1

2; the boundary

curves of the region S in the uv-plane are v = u, v = u, v = 1, and v = 4; thus1

2

S

eu/vdAuv =1

2

41

vv

eu/vdu dv =15

4(e e1)

35. Let u = y/x,v = x/y2, then x = 1/(u2v), y = 1/(uv) so(x, y)

(u, v)=

1

u4v3;

S

1

u4v3dAuv =

41

21

1

u4v3du dv = 35/256


32/37


36. Let x = 3u, y = 2v,(x, y)

(u, v)= 6; S is the region in the uv-plane enclosed by the circle u2 + v2 = 1

so

R

(9 x y)dA =S

6(9 3u 2v)dAuv = 620

10

(9 3r cos 2r sin )rdrd = 54

37. x = u, y = w/u,z = v + w/u,(x, y, z)

(u,v,w)

=

1

u

;

S

v2w

udVuvw =

42

10

31

v2w

udu dv dw = 2ln 3

38. u = xy,v = yz,w = xz, 1 u 2, 1 v 3, 1 w 4,

x =

uw/v,y =

uv/w,z =

vw/u,(x, y, z)

(u,v,w)=

1

2

uvw

V =

G

dV =

21

31

41

1

2

uvwdw dv du = 4(

2 1)(

3 1)

39. (b) If x = x(u, v), y = y(u, v) where u = u(x, y), v = v(x, y), then by the chain rule

x

u

u

x+

x

v

v

x=

x

x= 1,

x

u

u

y+

x

v

v

y=

x

y= 0

y

u

u

x+

y

v

v

x=

y

x= 0,

y

u

u

y+

y

v

v

y=

y

y= 1

40. (a)(x, y)

(u, v)=

1 v uv u = u; u = x + y, v = yx + y ,

(u, v)

(x, y)=

1 1

y/(x + y)2 x/(x + y)2

=x

(x + y)2+

y

(x + y)2=

1

x + y=

1

u;

(u, v)

(x, y)

(x, y)

(u, v)= 1

(b)(x, y)

(u, v)=

v u0 2v = 2v2; u = x/y, v = y,

(u, v)

(x, y)=

1/

y x/(2y3/2)0 1/(2

y)

= 12y = 12v2 ; (u, v)(x, y) (x, y)(u, v) = 1(c)

(x, y)

(u, v)=

u vu v

= 2uv; u = x + y, v = x y,

(u, v)(x, y) = 1/(2

x + y) 1/(2

x + y)

1/(2x y) 1/(2x y) = 12x2 y2 = 12uv ; (u, v)(x, y) (x, y)(u, v) = 1

41.(u, v)

(x, y)= 3xy4 = 3v so

(x, y)

(u, v)=

1

3v;

1

3

S

sin u

vdAuv =

1

3

21

2

sin u

vdu dv = 2

3ln 2


33/37

652 Chapter 15

42.(u, v)

(x, y)= 8xy so

(x, y)

(u, v)=

1

8xy; xy

(x, y)(u, v) = xy

1

8xy

=

1

8so

1

8

S

dAuv =1

8

169

41

du dv = 21/8

43.(u, v)

(x, y)= 2(x2 + y2) so (x, y)

(u, v)= 1

2(x2 + y2);

(x4 y4)exy(x, y)(u, v)

= x4 y42(x2 + y2) exy = 12 (x2 y2)exy = 12 veu so1

2

S

veudAuv =1

2

43

31

veudu dv =7

4(e3 e)

44. Set u = x + y + 2z, v = x 2y + z, w = 4x + y + z, then (u, v, w)(x, y, z)

=

1 1 21 2 14 1 1

= 18, and

V =R

dxdy dz =66

22

33

(x, y, z)

(u, v, w) dudv dw = 6(4)(12)

1

18 = 16

45. (a) Let u = x + y, v = y, then the triangle R with vertices (0, 0), (1, 0) and (0, 1) becomes thetriangle in the uv-plane with vertices (0, 0), (1, 0), (1, 1), andR

f(x + y)dA =

10

u0

f(u)(x, y)

(u, v)dvdu =

10

uf(u) du

(b)

10

ueu du = (u 1)eu10

= 1

46. (a) (x,y,z)(r,,z)

=

cos

r sin 0

sin r cos 00 0 1

= r,(x, y, z)(r,,z) = r

(b)(x, y, z)

(,,)=

sin cos cos cos sin sin sin sin cos sin sin cos

cos sin 0

= 2 sin ;(x, y, z)(,,)

= 2 sin

CHAPTER 15 SUPPLEMENTARY EXERCISES

3. (a) R

dA (b) G

dV (c) R

1 +

z

x2

+ z

y 2

dA

4. (a) x = a sin cos , y = a sin sin , z = a cos , 0 2, 0 (b) x = a cos , y = a sin , z = z, 0 2, 0 z h

7.

10

1+1y21

1y2f(x, y) dxdy 8.

20

2xx

f(x, y) dydx +

32

6xx

f(x, y) dy dx


34/37

Chapter 15 Supplementary Exercises 653

9. (a) (1, 2) = (b, d), (2, 1) = (a, c), so a = 2, b = 1, c = 1, d = 2

(b)

R

dA =

10

10

(x, y)

(u, v)dudv =

10

10

3dudv = 3

10. If 0 < x,y < then 0 < sin

xy 1, with equality only on the hyperbola xy = 2/4, so

0 =

0

0

0 dydx <

0

0

sin

xy dy dx <

0

0

1 dydx = 2

11.

11/2

2x cos(x2) dx =1

sin(x2)

11/2

= 1/(

2)

12.

20

x2

2ey

3

x=2yx=y

dy =3

2

20

y2ey3

dy =1

2ey

3

20

=1

2

e8 1

13.

10

22y

exey dxdy 14.

0

x0

sin x

xdydx

15.

6

1

x

y

y = sinx

y = tan (x/2)

16.

0

p/2

p/6

r = a

r= a(1 + cos u)

17. 2

80

y1/30

x2 sin y2dxdy =2

3

80

y sin y2 dy = 13

cos y280

=1

3(1 cos 64) 0.20271

18.

/2

0

2

0

(4 r2)rdrd = 2

19. sin2 = 2sin cos =2xy

x2 + y2, and r = 2a sin is the circle x2 + (y a)2 = a2, so

a0

a+a2x2aa2x2

2xy

x2 + y2dydx =

a0

x

ln

a +

a2 x2 ln

a

a2 x2

dx = a2

20.

/2/4

20

4r2(cos sin ) rdrd = 4cos2/2/4

= 4

21.20

20

16r4

r2 cos2 r dz dr d =20

cos2 d20

r3(16 r4) dr = 32

22.

/20

/20

10

1

1 + 22 sin ddd =

1

4

2

/20

sin d

=

1 4

2

( cos )/20

=

1 4

2


35/37

654 Chapter 15

23. (a)

20

/30

a0

(2 sin2 )2 sin ddd =

20

/30

a0

4 sin3 ddd

(b)

20

3a/20

a2r2r/3

r2 dz rdr d =

20

3a/20

a2r2r/3

r3 dz dr d

(c) 3a/2

3a/2

(3a2/4)x2

(3a2/4)

x2

a2x2y2

x2+y2/

3

(x2 + y2) dz dy dx

24. (a)

40

4xx24xx2

4xx2+y2

dz dy dx

(b)

/2/2

4cos 0

4r cos r2

r dz drd

25.

20

2y/2(y/2)1/3

dxdy =

20

2 y

2y

2

1/3dy =

2y y

2

4 3

2

y2

4/320

=3

2

26. A = 6 /6

0

cos 3

0

r d r d = 3 /6

0

cos2 3 = /4

27. V =

20

a/30

a3r

r dz drd = 2

a/30

r(a

3r) dr =a3

9

28. The intersection of the two surfaces projects onto the yz-plane as 2y2 + z2 = 1, so

V = 4

1/20

12y20

1y2y2+z2

dx dz dy

= 4

1/20

12y20

(1 2y2 z2) dz dy = 41/20

2

3(1 2y2)3/2 dy =

2

4

29. ru rv = 2u2 + 2v2 + 4,

S =

u2+v24

2u2 + 2v2 + 4 dA =

20

20

2

r2 + 2 rdrd =8

3(3

3 1)

30. ru rv =

1 + u2, S =

20

3u0

1 + u2dvdu =

20

3u

1 + u2du = 53/2 1

31. (ru rv)u=1v=2

= 2,4, 1, tangent plane 2x + 4y z = 5

32. u =

3, v = 0, (ru

rv)u=3

v=0

=

18, 0,

3

, tangent plane 6x + z =

9

33. A =

44

2+y2/8y2/4

dxdy =

44

2 y

2

8

dy =

32

3; y = 0 by symmetry;

44

2+y2/8y2/4

xdxdy =

44

2 +

1

4y2 3

128y4

dy =256

15, x =

3

32

256

15=

8

5; centroid

8

5, 0


36/37

Chapter 15 Supplementary Exercises 655

34. A = ab/2, x = 0 by symmetry,aa

b1x2/a20

y d y d x =1

2

aa

b2(1 x2/a2)dx = 2ab2/3, centroid

0,4b

3

35. V =1

3a2h, x = y = 0 by symmetry,

20

a0

hrh/a0 rz dz dr d =

a0 rh

2 1

r

a2

dr = a2

h2

/12, centroid (0, 0, h/4)

36. V =

22

4x2

4y0

dz dy dx =

22

4x2

(4 y)dydx =22

8 4x2 + 1

2x4

dx =256

15,

22

4x2

4y0

y dz dydx =

22

4x2

(4y y2) dydx =22

1

3x6 2x4 + 32

3

dx =

1024

3522

4x2

4y0

z dz dy dx =

22

4x2

1

2(4 y)2dydx =

22

x

6

6+ 2x4 8x2 + 32

3

dx =

2048

105

x = 0 by symmetry, centroid

0,

12

7,

8

7

37. The two quarter-circles with center at the origin and of radius A and 2A lie inside and outsideof the square with corners (0, 0), (A, 0), (A, A), (0, A), so the following inequalities hold:/20

A0

1

(1 + r2)2rdr d

A0

A0

1

(1 + x2 + y2)2dxdy

/20

2A0

1

(1 + r2)2rdrd

The integral on the left can be evaluated asA2

4(1 + A2)and the integral on the right equals

2A2

4(1 + 2A2). Since both of these quantities tend to

4as A +, it follows by sandwiching that

+0

+0

1

(1 + x2 + y2)2dxdy =

4.

38. The centroid of the circle which generates the tube travels a distance

s =

40

sin2 t + cos2 t + 1/16 dt =

17, so V = (1/2)2

17 =

172/4.

39. (a) Let S1 be the set of points (x, y, z) which satisfy the equation x2/3 + y2/3 + z2/3 = a2/3, and

let S2 be the set of points (x,y,z) where x = a(sin cos )3, y = a(sin sin )3, z = a cos3 ,

0 , 0 < 2.If (x, y, z) is a point of S2 then

x2/3 + y2/3 + z2/3 = a2/3[(sin cos )3 + (sin sin )3 + cos3 ] = a2/3

so (x, y, z) belongs to S1.If (x, y, z) is a point of S1 then x

2/3 + y2/3 + z2/3 = a2/3. Let

x1 = x1/3, y1 = y1/3, z1 = z1/3, a1 = a1/3. Then x21 + y21 + z21 = a21, so in spherical coordinatesx1 = a1 sin cos , y1 = a1 sin sin , z1 = a1 cos , with

= tan1

y1x1

= tan1

yx

1/3, = cos1

z1a1

= cos1z

a

1/3. Then

x = x31 = a31(sin cos )

3 = a(sin cos )3, similarly y = a(sin sin )3, z = a cos so (x, y, z)belongs to S2. Thus S1 = S2


37/37

656 Chapter 15

(b) Let a = 1 and r = (cos sin )3i + (sin sin )3j + cos3 k, then

S = 8

/20

/20

r rdd

= 72

/20

/20

sin cos sin4 cos

cos2 + sin2 sin2 cos2 dd 4.4506

(c)

(x, y, z)

(,,) =

sin3 cos3 3 sin2 cos cos3 3 sin3 cos2 sin sin

3

sin3

3 sin2

cos sin3

3 sin3

sin2

cos cos3 3 cos2 sin 0

= 92 cos2 sin2 cos2 sin5 ,

V = 9

20

0

a0

2 cos2 sin2 cos2 sin5 ddd =4

35a3

40. V =4

3a3, d =

3

4a3

a

dV =3

4a3

0

20

a0

3 sin ddd =3

4a32(2)

a4

4=

3

4a

41. (a) (x/a)2+(y/b)2+(z/c)2 = sin2 cos2 +sin2 sin2 +cos2 = sin2 +cos2 = 1, an ellipsoid

(b) r(, ) = 2sin cos , 3sin sin , 4cos ; rr = 26sin2

cos , 4sin

2

sin , 3cos sin ,r r = 2

16 sin4 + 20 sin4 cos2 + 9 sin2 cos2 ,

S =

20

0

2

16sin4 + 20sin4 cos2 + 9 sin2 cos2 dd 111.5457699

Date post:	04-Apr-2018
Category:	Documents
Upload:	aditya-p-banerjee
View:	249 times
Download:	0 times

Chapter 15 Solutions Calculus

Documents