solucion larzon cap13

8/2/2019 solucion larzon cap13

1/41

C H A P T E R 1 3

Multiple Integration

Section 13.1 Iterated Integrals and Area in the Plane . . . . . . . . . . . . . 365

Section 13.2 Double Integrals and Volume . . . . . . . . . . . . . . . . . . . 369

Section 13.3 Change of Variables: Polar Coordinates . . . . . . . . . . . . . 375

Section 13.4 Center of Mass and Moments of Inertia . . . . . . . . . . . . . 379

Section 13.5 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

Section 13.6 Triple Integrals and Applications . . . . . . . . . . . . . . . . . 388

Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates . . . . 393

Section 13.8 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . 397

Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405


2/41

365

C H A P T E R 1 3

Multiple Integration

Section 13.1 Iterated Integrals and Area in the Plane

Solutions to Even-Numbered Exercises

2. x2

x

y

xdy 12

y2

x x2

x

1

2 x4

x

x2

x x

2x2 1 4.

cos y

0

ydx yxcos y

0

y cosy

6. x

x3

x2 3y2dy x2y y3x

x3 x2x x

3 x2x3 x33 x52 x32 x5 x9

8. 1y2

1y2x2 y 2dx 13x3 y2x

1y2

1y2 213 1 y232 y21 y212

21 y 2

31 2y 2

10.

cosx 13 cos3x cosy2

y

cosy 13 cos3y cosy

2

y

sin3x cosydx 2

y

1 cos2x sinx cosydx

12.

1

14x2 163 dx

4x3

3

16

3x

1

1

43 16

3 4

3

16

3 8

1

1

2

2

x2 y2dydx 1

1x2y y

3

3 2

2

dx 1

12x2 83 2x2

8

3dx

14.

4

4

64 x3x2dx 29 64 x3324

4

0 2

912832

2048

92

4

4x2

064 x

3

dydx 4

4y64 x3

x2

0dx

16.

2010y 143 y3 2y3dy 5y2

7y4

6

y4

2 2

0

20 56

3 8

140

3

20

2yy

10 2x2 2y2dxdy 2010x 2x

3

3 2y2x

2y

y

dy 2020y 163 y3 4y3 10y

2

3y3 2y3dy

18. 2

0

2yy2

3y26y

3ydxdy 2

0

3xy2yy2

3y26y

dy 32

0

8y2 4y3dy 383y3 y42

0

16

20. 2

0

2 cos

0

rdrd 2

0

r2

2 2 cos

0

d 2

0

2 cos2d 12 sin 22

0

2

22.

4

0

cos3 sin d cos4

4 4

0

1

41

24

1 316

4

0

cos

0

3r2 sin drd 4

0

r3sin cos

0

d


3/41

24. 3

0

0

x2

1 y2dydx

3

0

x2 arctany

0

dx 3

0

x22dx

2x3

3 3

0

9

2

26. 0

0

xyex2y2dxdy

0

12yex2y2

0

dy 0

1

2yey

2

dy 14ey2

0

1

4

28.

A 3

1

3

1

dxdy 3

1

x3

1

dy 3

1

2 dy 2y3

1

4

2

2

31

3

1

x

y

A 3

1

3

1

dydx 3

1

y3

1

dx 3

1

2 dx 2x3

1

4

30.

3y12

0

1y y1

12 2

12

0

3 dy 1

12 1y 2 1dy

12

0

x5

2

dy 1

12x

11y2

2

dy

x2 5

3

3

5

4

2

41

1

1

x 1y =

y

A 12

0 5

2dxdy

1

1211y2

2dxdy

A 5

2

1x1

0

dydx 5

2

y1x1

0

dx 5

2

1

x 1dx 2x 1

5

2

2

32.

y 2 sin , dy 2 cos d, 4 y 2 2 cos

2 12 sin 22

0

42

0

cos2d 22

0

1 cos 2d

A 2

0

4y 2

0

dxdy 2

0

4 y2dy

x 2 sin , dx 2 cos d, 4 x2 2 cos

2 12 sin 22

0

22

0

1 cos 2d

42

0

cos2d

2

0

4 x2dx

x

2

1 2

1

y x= 4 2

y

A 2

04x2

0

dydx 34.

3

532 16

16

5

35y53 y2

4 8

0

8

0

y23 y2dy

A 8

0

y23

y2dxdy

16 2

532

16

5

x2 25x524

0

4

0

2x x32dx

40

y2x

x32dx

1 1

1

2

3

4

5

6

7

8

2 3 4 5 6 7 8

y

x

(4, 8)

y =x3/2

y = 2x

A 4

0

2x

x32dydx

366 Chapter 13 Multiple Integration


4/41

36.

x

4

6

2 4 6 8

2

2

(3, 3)

(9, 1)

y x=

y =9x

y

9y 12y 21

0

9 lny 12y 23

1

9

21 ln 9

1

0

9 ydy 3

1

9y ydy

1

0

x9

y

dy 3

1

x9y

y

dy

A 1

0

9

y

dxdy 3

1

9y

y

dxdy

9

21 ln 9

12x23

0

9 lnx9

3

9

2 9ln 9 ln 3

3

0

yx

0

dx 9

3

y9x

0

dx 3

0

xdx 9

3

9

xdx

A 3

0

x

0

dydx 9

3

9x

0

dydx 38.

x

2

3

4

1 2 3 4

1

y x= 2

y x=

y

A 2

0

2x

x

dydx 2

0

2x xdx x2

2 2

0

2

1 4 3 2

y2

4 2

0

2y y2

4 4

2

2

0

y

2dy

4

2

2 y2dy

A 2

0

y

y2 dxdy

4

2

2

y2dxdy

40.

x

2

3

4

1 2 3 4

1

y x= 2

y

2

0 x2

0

fx,ydydx

4

0

2

y

fx,ydxdy, y x 2, 0 y 4 42.

1

1

1

2

3

4

1 2 3 4

y

x

4

0 4y

0

fx,ydxdy

2

0

4x

2

0

fx,ydydx, 0 y 4 x2, 0 x 2

44.

1 1 2

2

3

y

x

e2

0

2

1

fx,ydxdy e

e2

ln y

1

fx,ydxdy

2

1

ex

0

fx,ydydx, 0 y ex, 1 x 2 46.

2

x

2

1

23

4

4

y

1

0

arccosy

arccos y

fx,ydxdy

2

2

cosx

0

fx,ydydx, 0 y cosx,

2 x

2

Section 13.1 Iterated Integrals and Area in the Plane 367


5/41

48.

x

2

3

4

1 2 3 4

1

y

2

1

4

2

dxdy 4

2

2

1

dydx 2 50.

1

1

1

1

y

x

2

2

4y2

4y2dxdy 4

2

2

4x2

4x2dydx

2

2

4 x2 4 x2dx 4

52.

2

0

6y

2y

dxdy 2

0

6 3ydy 6y 3y2

2 2

0

6y = x

2 y = 6 x

y

x

11

1

1

2

3

4

5

6

2 3 4 5 6

(4, 2)

4

0

x2

0

dydx 6

4

6x

0

dydx 4

0

x

2dx

6

4

6 xdx 4 2 6

54.

3

0

y2

0

dxdy 3

0

y2dy y3

3 3

0

9

1

2

1

2

3

4

5

3

4

5

1 2 3 4 5 6 7 8 9

y

x

y = x

9

0

3

xdydx

9

0

3 xdx 3x 23x329

0

27 18 9

56. 2

2

4y2

0

dxdy 4

0

4x

4x

dydx 32

3

x

2

1 2 3

1

1

2x y= 4 2

y

58. The first integral arises using vertical representative rectangles. The second integral arises using horizontal

representative rectangles.

1

4cos4

1

2sin4

1

4

4

0

y

y2x sinydxdy

4

0

12y siny 1

8y2 sinydy

1

4cos4

1

2sin4

1

4

x

1

1

2

2

3

4

(2, 4)

y

2

0

2x

x2

x sinydydx

2

0

x cos2x x cosx2dx



6/41

Section 13.2 Double Integrals and Volume

60.

20

xey2y

0

dy 20

yey2

dy 12 ey2

2

0

1

2e4

1

2e0

1

2 1 1

e4 0.4908

20

2x

ey2

dydx 20

y0

ey2

dxdy

62.

40

yx sinxx

0

dx 40

x sinxdx sinx x cosx4

0

sin 4 4 cos 4 1.858

2

0 4

y 2x sinxdxdy

4

0 x

0x sinxdydx

64. 10

2yy

sin x ydxdy sin 2

2

sin 3

3 0.408 66. a

0

ax0

x2 y 2dydx a4

6

68. (a)

(b)

(c) Both orders of integration yield 1.11899.

2

0 2

4y 2

xy

x2 y 2 1dxdy

3

2 2

0

xy

x2 y 2 1dxdy

4

3 164y

0

xy

x2 y 2 1dxdy

y 4 x2

4x 16 4y

x

1

1

2

2

3

4

yy 4 x2x 4 y 2

70. 20

2x

16 x3 y3dydx 6.8520 72. 20

1sin 0

15rdrd452

32

135

8 30.7541

78. False, letfx,y x.

74. A region is vertically simple if it is bounded on the left and right by vertical lines, and bounded on the top and bottom by

functions ofx. A region is horizontally simple if it is bounded on the top and bottom by horizontal lines, and bounded on the

left and right by functions ofy.

76. The integrations might be easier. See Exercise 59-62.

For Exercises 2 and 4, and the midpoints of the squares are

12,1

2, 3

2,

1

2, 5

2,

1

2, 7

2,

1

2, 1

2,

3

2, 3

2,

3

2, 5

2,

3

2, 7

2,

3

2.

x

2

3

4

1 2 3 4

1

yxi y

i 1

2.

40

20

1

2x2ydydx 4

0

x2y2

4 2

0dx 4

0

x2dx x3

3 4

0

64

3 21.3

i1

fxi,yixiyi 1

16

9

16

25

16

49

16

3

16

27

16

75

16

147

16 21

fx,y 1

2x2y

Section 13.2 Double Integrals and Volume 369


7/41

4.

40

ln 3

x 1dx ln 3 lnx 1

4

0 ln 3ln 5 1.768

40

20

1

x 1y 1dydx 4

0

1x 1 lny 12

0dx

8

i1

fxi,yixiyi 4

9

4

15

4

21

4

27

4

15

4

25

4

35

4

45

7936

4725 1.680

fx,y 1

x 1y 1

6. 20

20

fx,ydydx 4 2 8 6 20

8.

2

8

8x 1

2sin 2x

0

8

0

1 cos 2xdx

0

1

2

sin2x

2dx

2

2

31

3

1

x

y

0

20

sin2x cos2y

dy

dx

0

12 sin2xy 12 sin 2y

2

0dx

10.

1024

27

256

9

256

27

2y92

27

y6

1444

0

4

0 y

72

3 y

5

24dy

1

1

2

3

4

2 3 4

y

x

(2, 4)4

0

y12y

x2y2dxdy 40x

3y2

3 y

12ydy

12.

1

2e e1

ey 12 e2y11

0

10

e e2y1dy

x1 1

2

y x= + 1y x= + 1

y

10

0y1

exydxdy 10

1y0

exydxdy 10

exy0

y1dy 1

0

exy1y

0dy



8/41

14.

x

y

22

23

23

25

2

2

23

0

sinxdx

sinx cosy2

0dx

20

sinx sinydxdy

20

sinx sinydydx 16.

For the first integral, we obtain:

1

1

2

3

4

2 3 4

y

x

5 8 e4 e4 13.

e

4x

1

x

x2

2

4

0

40xey

4x

x0

dx 4x0

xe4 xdx

40

4x0

xeydydx 40

4y0

xeydxdy

18.

14 ln1 x24

0

1

4ln17

1

24

0

x

1 x2dx

1

24

0

y2

1 x2x

0 dx

x

2

3

4

1 2 3 4

1

y x=

y

2

0 4

y 2

y

1 x2dxdy 4

0 x

0

y

1 x2dydx

22.

x

2

3

4

1 2 3 4

1

y

40

8 dx 32

4

0 2

0 6

2ydydx

4

0 6y

y2

2

0dx 24.

1

1

2

2

y =x

y

x

2

0 x

0 4 dydx

2

0 4xdx

2x

2

2

0

8

20.

x4 4 x232 1

2 x4 x2 4 arcsinx

2 1

12x4 x232 6x4 x2 24 arctanx

22

2 4

22x24 x2 13 4 x232dx

22x2y 13y3

4x2

0dx

2

2

4

x2

0

x2 y 2dydx

x2 1 1

1

2

3

4

x y= 4 2x y= 4 2

y

20

4y2

4y 2x2 y 2dxdy



9/41

26.

1

1

2

2

y = 2 x

y

x

1

6x 23

2

0

4

3

20

1

22 x2 dx

20

2x0

2 x ydydx 202y xy y

2

2 2x

0

dx 28.

x

2

1 2

1 y x=

y

4

2y2 y4

4 2

0

20

y0

4 y 2dxdy 20

4y y3dy

30. 0

2ex2dx 4ex2

0 4

0

0

exy2dydx 0

2exy2

0dx

32. 1

0 x

0

1 x2dydx 13

34.

13x35

0

125

3

50

xyx

0dx 5

0

x2dx

V 50

x0

xdydx

x2 5

3

3

5

4

2

41

1

y x=

y

36.

4r3

3

2r2x 13x3r

0

42r

0

r2 x2dx

4r0

yr2 x2 y 2 r2 x2 arcsin yr2 x2r2x2

0dx

x

r

r

y r x= 2 2

yV 8r

0r2x2

0

r2 x2 y 2dydx

38.

32 64

3

32

5

256

15

16x 8x3

3

x5

5 2

0

20

16 8x2 x4dx

20

4 x24 x2dx

1 2

1

2

3

4

3 4

y

x

y = 4 x2

V 2

0

4x2

0

4 x2dydx 40.

x2 2

0

20

2dx

20

arctany

0dx

x

2

1 2

1

yV 2

0

0

1

1 y 2dydx



10/41

42.

5

2

2 y5

0

50

2dy

x2 5

3

3

5

4

2

41

1

y

V 50

0

sin2xdxdy 44. V 90

9y0

9 ydxdy 81

2

46. V 160

4y0

ln1 x ydxdy 38.25

48.

cab2 ab

3 ab

2

ab

6 abc

6

cab2 1 x

a2

x2b

2a

x3b

3a2

ab

6 1 x

a3

a

0

ca0

b1 xa xb

a 1 x

a b2

2b 1 x

a2

dx

ca0

y xya y 2

2bb1xa

0dx

V

Rfx,ydA

a

0 b1xa

0 c1

x

a

y

bdydx

z c1 xa y

b

x

ya

a

a

R

z

x

a

y

b

z

c 1

50.

x2 5

6

3

10

8

4

41

2

y e= x

y

101

dy y10

1 9

101

xlnyln y

0dy

ln 10

0 10

ex

1lny

dydx 10

1 ln y

0

1lny

dxdy 52.

2cos 2 2 sin 2 1

2cosy y siny2

0

220

y cosydy(2, 2)

y = x212

1

1

2

2

y

x

20

y cosy2ydy

2

0 2

12x2y cosydydx

2

0 2y

0y cosydxdy

54. Average 1

84

0

20

xydydx 1

84

0

2xdx x2

8 4

0 2

56.

e 12

e2 2e 1

2 ex1 12e2x1

0 2e2 12e2 e

1

2

Average1

121

0

1x

exydydx 210

ex1 e2xdx



11/41

58. The second is integrable. The first contains

which does not have an elementary antiderivation.

siny2dy 60. (a) The total snowfall in the countyR.(b) The average snowfall inR.

62. Average 1

15060

45

5040

192x 576y x2 5y 2 2xy 5000dxdy 13,246.67

64. for all and

P0 x 1, 1 y 2 10

21

1

4xydydx 1

0

3x

8dx

3

16.

fx,ydA 20

20

1

4xydydx 2

0

x

2dx 1

x,yfx,y 0

66. for all and

12 e2x ex11

0

1

2e2 e1

1

2

1

2e1 12 0.1998.

P0 x 1,x y 1 10

1x

exydydx 10

exy1

x dx 1

0

e2x ex1dx

0

limb

exyb

0dx

0

exdx limb

exb

0 1

fx,ydA

0

0

exydydx

x,yfx,y 0

68. Sample Program for TI-82:

Program: DOUBLE

: Input A

: Input B

: Input M

: Input C

: Input D

: Input N

:

:

:

: For

: For

:

:

:

: End

: End

: Disp V

V sin x y G H VC 0.5H2J 1 y

A 0.5G2I 1 xJ, 1, N, 1

I, 1, M, 1

D CN H

B AM G

0 V

70.

(a) 129.2018

(b) 129.2756

m 10, n 2020

40

20ex38dydx



12/41

72.

(a) 13.956

(b) 13.9022

m 6, n 441

21

x3 y3dxdy 74.

Matches a.

x

y

3

4

3 3

z

V 50

76. True

78.

Thus,

21

1

ydy lny

2

1 ln 2.

21

exy

y

0dy

21

0

exydxdy

0e

x

e

2x

x dx

0 2

1exydxdy

21

exydy 1x exy2

1

ex e2x

x

Section 13.3 Change of Variables: Polar Coordinates

2. Polar coordinates 4. Rectangular coordinates

6. R r, : 0 r 4 sin , 0 8. R r, : 0 r rcos 3, 0

10.

02 3 41

2

16

3

643 sin2

2 4

0

40

40

r2 sin cos drd 40

r3

3sin cos

4

0d 12.

02 31

2

4 1 1

e9

12 e9 12

0

20

30

rer2

drd 20

12er23

0d

Section 13.3 Change of Variables: Polar Coordinates 375


13/41

14.

16 1 cos 32

0

1

6

20

sin

21 cos 2d

01

( , ) = (0, 1)x y

2

20

1cos 0

sin rdrd 20

sin r2

2 1cos

0d

16. a0

a2x2

0

xdydx 20

a0

r2 cos drda3

32

0

cos d a3

3sin

2

0

a3

3

18.

40

22 3

3d 22

3

3

4

0

22 3

3

4

42

3

20

8y2

y

x2 y 2dxdy 40

22

0

r2drd

20.

6420

sin4 sin6d64

6sin5cos sin

3cos

4

3

8 sin cos

2

0 2

4

0 4yy2

0

x2dxdy 2

0 4 sin

0

r3 cos2drd 2

0

64 sin4cos2d

22.

625

16

625

8 sin2

4

0

40

625

4sin cos d

52 20

x0

xydydx 552 2

25x2

0

xydydx 40

50

r3 sin cos drd

24.

1 e2522

2 1 e252

22

1 e252d5 4 3 2

2

1

2

3

4

5

3

4

5

1 1 2 3 4

y

x

22

50

er22rdrd 2

2er22

5

0

d

26.

20

30

9r r3drd 20

92 r2 1

4r4

3

0d

81

42

0

d81

8

30

9x2

0

9 x2 y2dydx 20

30

9 r2rdrd



14/41

28.

74 7

4

440

7

4d

V 420

10

r2 3rdrd 420

r4

4

3r2

2 1

0

d 30.

4ln 4 34

220

ln 4 34d

2

2

0

r2

4

1 2 ln r

2

1

d

220

21

rln rdrd

R lnx2 y 2dA 2

0

21

ln r2rdrd

32. 20

515 d 1015V 2

0

41

16 r2rdrd 20

13 16 r23

4

1d

34.

(8 times the volume in the first octant)

820

a3

3d 8a

3

3

2

0

4a3

3

820

12 2

3a2 r232

a

0d

V 8

2

0 a

0

a2 r2rdrd

x2 y 2 z2 a2z a2 x2 y 2 a2 r2

36.

(a)

(b)

(c) V 220

121cos2

14

9

4r2 36rdrd 0.8000

Perimeter 2 0

14 1 cos22 cos2sin2d 5.21dr

d cos sin

xy

1

1

1

zr1

21 cos2

1

2

1

2cos2

Perimeter

r2 drd2

d.

9

4r2 36 z

9

4r2 36

0.7

1 1

0.7

1

4 r

1

21 cos2

9

4x2 y2 9 z

9

4x2 y2 9;

38. A 20

42

rdrd 20

6 d 12

40.

1

24 4 cos 1

2

1

4sin 2

2

0

1

28 4 4

9

2

20

2sin 0

rdrd1

22

0

2 sin 2

d1

22

0

4 4 sin sin2d1

22

04 4 sin 1 cos 22 d

Section 13.3 Change of Variables: Polar Coordinates 377


15/41

42. 840

3 cos 20

rdrd 4 40

9 cos2 2d 1840

1 cos 4d 18 14 sin 44

0

9

2

44. See Theorem 13.3. 46. (a) Horizontal or polar representative elements

(b) Polar representative element

(c) Vertical or polar

48. (a) The volume of the subregion determined by the point is base height Adding up the 20

volumes, ending with you obtain

(b)

(c) 7.4824103.5 179,621 gallons

5624013.5 1,344,759 pounds

5

4150 555 1250 2135 2025

5

46115 24013.5 ft3

3512 15 18 16 459 10 14 12

V 10

857 9 9 5 158 10 11 8 2510 14 15 11

45 10 812, 5 10 87.5, 16, 7

50. 4

0 4

0

5er rdrd 87.130 52.

Answer (a)

94 3 21

x

y

4

4

2

2

4

6

zVolume base height

54. True

56. (a) Let then

(b) Let then

e4x2

dx

eu2

1

2du

1

2.u 2x,

ex2

dx

eu2

2 12

du 12 2 .u 2x,

58.

For to be a probability density function,

k4

.

k

4 1

fx,y

0

0

kex2y2dydx 2

0

0

ker2

rdrd 20

k2 er2

0d 2

0

k

2d

k

4

60. (a)

(b)

(c) 220

4 cos 0

f rdrd

420

4x22

0

fdydx x

2

1

2

1

3

1

( 2) + 2 = 4x y 2y

420

24y2

2

fdxdy



16/41

Section 13.4 Center of Mass and Moments of Inertia

2.

1

120 93

243

4

149 x23

3 3

0

3

0

x9 x22

2dx

m 30

9x2

0

xydydx 30xy

2

2 9x2

0

dx

4.

1

281

2

81

4 54 2978

1

229 x232 9x2

2

x4

4 3

0

12

3

06x9 x2 9x x3dx

30

x

23 9 x2 9dx

m 30

39x2

3

xydydx 30xy

2

2 39x2

3

dx

6. (a)

y Mxm

ka2b36

ka2b24

2

3b

x My

m

ka3b26

ka2b24

2

3a

My a0

b0

kx2ydydx ka3b2

6

Mx a0

b0

kxy2dydx ka2b3

6

m a0

b0

kxydydx ka2b2

4(b)

y Mxm

kab2122a2 3b2

kab3a2 b2

b

4 2a2 3b2

a2 b2

x My

m

ka2b123a2 2b2

kab3a2 b2

a

4 3a2 2b2

a2 b2

My a0

b0

kx3 xy2dydx ka2b

123a2 2b2

Mx a0

b0

kx2y y3dydx kab2

122a2 3b2

m a0

b0

kx2 y 2dydx kab

3a2 b2

8. (a)

CONTINUED

x y Mxm

ka36

ka22

a

3

My Mx by symmetry

Mx

a

0

ax

0

kydydx ka3

6

xa

ay a x=

ym a2k

2

Section 13.4 Center of Mass and Moments of Inertia 379


17/41

10. Thex-coordinate changes by h units horizontally and kunits vertically. This is not true for variable densities.

12. (a)

y 4a

3by symmetry

x My

m

ka33

ka24

4a

3

k3 a2 x232a

0

ka3

3

ka0

xa2 x2dx

My a

0

a2x2

0

kxdydx

m a

0

a2x2

0

kdydx ka2

4(b)

x y M

y

m ka

5

5 8

ka4 8a

5

My Mx by symmetry

20

a0

kr4 sin drdka5

5

Mx a0

a2x2

0

kx2 y 2ydydx

20

a0

kr3drdka4

8

m a

0a2x2

0

kx2 y 2dydx

8. CONTINUED

(b)

y 2a

5by symmetry

x My

m

a515

a46

2a

5

a0

ax3 x 4 13 a3x a2x2 ax3 1

3x 4dx 13

a

0

a3x 3a2x2 6ax3 4x4dx a5

15

My a

0 ax

0

x3 xy2dydx

a0

x2y y3

3 ax

0dx a

0

ax2 x3 13 a x3dx a4

6

m a0

ax0

x2 y 2dydx

14.

y Mxm

16k

32k5

5

2

x My

m

32k

3

5

32k

5

3

My 20

x3

0

kx2dydx 32k

3

Mx 2

0

x3

0

kxydydx 16k

m 20

x3

0

kxdydx 20

kx4dx 32k

516.

x

2

3

4

1 2 3 4

1

y =4x

y

y My

m

24k

30k

4

5

x

My

m

84k

30k

14

5

My 41

4x0

kx3dydx 84k

Mx 41

4x0

kx2ydydx 24k

m 41

4x0

kx2dydx 30k



18/41

18.

x6 3 3 6

6

3

12

y x= 9 2

y

y Mxm

139,968k

35

35

23,328k 6

Mx 3

3

9x2

0

ky3dydx 139,968k

35

m 33

9x2

0

ky2dydx 23,328k

35

x 0 by symmetry 20.

x

1

L2

y = cosxL

t

y Mxm

kL

8

kL

8

x My

m

L2 2k

22

kL

L 2

2

My L2

0

cos xL0

kxdydx L 2 2k

22

Mx L20

cos xL0

kydydx kL

8

m L20

cos xL0

kdydx kL

22.

y Mxm

ka42 2 8

12

ka3

32 2a2

x My

m

ka42

8

12

ka3

32a

2

My R

kx2 y 2dA 40

a0

kr3 cos dka42

8

Mx R

kx2 y 2ydA 40

a0

kr3 sin dka42 2

8

0a

r a=y x=

2m

Rkx2 y2dA 4

0

a0

kr2drdka3

12

24.

y Mxm

k

6

2

k

1

3

x M

ym k1 2k

2

My e

1

ln x0

k

xxdydx k

Mx e1

ln x0

k

xydydx

k

6

m e1

ln x0

k

xdydx

k

2

2

2

3e1

3

1

x

y x= ln

y



19/41

26.

x My

m

5k

4

2

3k

5

6

5k

4

k

32

0

cos 32 1 cos2 3 cos 1 sin2 1

41 cos 22d

k3

2

0

cos 1 3 cos 3 cos2 cos3d

My RkxdA 2

0

1cos 0

kr2 cos drd

m RkdA 2

0

1cos 0

krdrd3k

2

01

r= 1 + cos

2y 0 by symmetry

28.

y Ixm bh312

bh2

h

66

6h

x Iym b3h12

bh2

b

66

6b

Iy b0

hhxb0

x2dydx b3h

12

Ix b0

hhxb0

y2dydx bh3

12

m

b

0

hhxb

0

dydx bh

230.

x y Ixm a4

8

2

a2

a

2

I0 Ix Ix a4

8

a4

8

a4

4

Iy Rx2dA

0

a0

r3 cos2drda4

8

Ix Ry 2dA

0

a0

r3 sin2drda4

8

m a2

2

32.

y Ixm ab3

4

1

ab

b

2

x Iym a3b

4

1

ab

a

2

I0 Iy Ix a3b

4

ab3

4

ab

4a2 b2

Iy 4 b0

abb2y 2

0

x2dxdy a3b

4

4b3

3a3a

2

2 xa2 x2 a2 arcsinx

a 1

8x2x2 a2a2 x2 a4 arcsinxa

a

0

ab3

4

4a0

b3

3a3a2 x232dx

4b3

3a3a

0

a2a2 x2 x2a2 x2dx

Ix 4a

0

baa2x2

0

y 2dydx

m

ab



20/41

34.

y Ixm 4ka515

2ka332a

2

5

2a

10

x Iym 2ka515

2ka33a

2

5

a

5

I0 Ix Iy 2ka5

5

Iy ka

aa

2x2

0

x2ydydx 2ka5

15

Ix k

a

aa2x2

0y3dydx

4ka5

15

ka0

a2 x2dx 2ka3

3

m 2ka0

a2x2

0

ydydx

ky 36.

y Ixmk60

k2425

x Iym k48

k24

1

2

I0 Ix Iy 9k

240

3k

80

Iy k

1

0 x

x2x3ydydx

k

21

0x5 x7dx

k

48

Ix k1

0

xx2

xy3dydx k

41

0

x5 x9dx k

60

m k10

xx2

xydydx k

21

0

x3 x5dx k

24

kxy

38.

y Ixm x 395

891

x

Iy

m

158

2079

35

6

395

891

I0 Ix Iy 316

2079

Iy 1

0

xx2

x2 y 2x2dydx 158

2079

Ix 1

0

xx 2

x2 y 2y 2dydx 158

2079

m 10

xx2

x2 y 2dydx 6

35

x2 y 2 40.

y Ixm 32,768k

65

21

512k

81365

65

x

Iy

m

2048k

45

21

512k

28

15

2105

15

I0 Ix Iy 321,536k

585

Iy 22

0

4xx3

kx2ydydx 2048k

45

Ix 22

0

4xx3

ky3dydx 32,768k

65

m 220

4xx3

kydydx 512k

21

ky

42. I 40

20

kx 62dydx 4

0

2kx 62dx 2k3 x 6

34

0

416k

3

44.

2k7a5

15

a5

8 ka556 15

60 2k1

4a5 2

3a5

1

5a5 2a3

a4

4

a4

16 a2

2 a3 a3

3

1

8 x2x2 a2a2 x2 a4 arcsinx

a a2

2 a2x x3

3 a

a

k14a4x 2a2x3

3

x5

5 2a

3 a2

2 xa2 x2 a2 arcsinx

a

aa

k14 a4 2a2x2 x 4 2a

3a2a2 x2 x2a2 x2

a2

2a2 x2dx

a

a

ky4

4

2ay3

3

a2y 2

2

a2x20

dx

I aaa

2x2

0

kyy a2dydx



21/41

46.

2k

3 32 32

192

5

128

7

1408k

105

k

322

16 12x2 6x4 x6dx k316x 4x3 6

5x

5

1

7x7

2

2

I 22

4x2

0

ky 22dydx 22

k3 y 13

4x2

0dx 2

2

k

32 x2 8dx

48.

will be the same.x,y

x,y k2 x. 50. Both and will decreaseyxx,y k4 x4 y.

52. Moment of inertia aboutx-axis.

Moment of inertia abouty-axis.Iy Rx2x,ydA

Ix R

y2x,ydA

54. Orient thexy-coordinate system so thatL is along they-axis andR is in the first quadrant. Then the volume of the solid is

By our positioning, Therefore, V 2rA.x r.

2xA.

2RxdARdA RdA

2RxdA

x

L

R( , )x y

y

V R 2xdA

56.

ya a

2

a3b12

L a2ab

a3L 2a

32L a

Iy b0

a0

y a22

dydx a3b

12

y a

2,A ab, h L

a

258.

ya a44

La2

a2

4L

a4

4

2

0

a4

4

sin2d

20

a0

r3 sin2drd

Iy aaa

2x2

a2x2y 2dydx

y 0,A a2, h L



22/41

Section 13.5 Surface Area

2.

2

2

31

3

1

x

R

y

S 3

0 3

0

14 dydx 3

0

314 dx 914

1 fx 2 fy

214

fx 2,fy 3

fx,y 15 2x 3y 4.

1

2

1

2

2 1 1

x

y x= 9 2

y x= 9 2

R

y

20

30

14 rdrd 914

S 339x29x2

14 dydx

1 fx 2 fy 2 14

fx 2, fy 3

R x,y:x2 y 2 9

fx,y 10 2x 3y

6.

square with vertices

34 2y1 4y2 ln2y 1 4y23

0

3

4637 ln6 37

S 30

30

1 4y 2dxdy 30

31 4y 2dy

1 fx 2 fy

21 4y2

fx 0, fy 2y

0, 0, 3, 0, 0, 3, 3, 3R

2

2

31

3

1

x

R

yfx,y y 2

8.

x

2

1 2

1 R

y = 2 x

y

12

53

8

5

2

332

2

5

352

2

2

5

21 y32 25 1 y522

0

S 202y

0

1 ydxdy 20

1 y 2 ydy

1 fx2fy

21 y

fx 0, fy y12

fx,y 2 2

3y32 10.

1

1

1 1

x

R

y

2

0

1

12

1732 1d

6

1717 1

20

112 1 4r2 322

0d

S 20

20

1 4r2rdrd

1 fx 2 fy

21 4x2 4y 2

fx 2x, fy 2y

fx,y 9 x2 y 2

Section 13.5 Surface Area 385


23/41

12.

204

0

1 r2rdrd2

31717 1

S

4

4

16x2

16x2

1 y2 x2dydx

1 fx 2 fy

21 y 2 x2

fx y, fy x

R x,y:x2 y2 162

2

2 2

x

x y2 2+ = 16

yfx,y xy

14. See Exercise 13.

20

a0

a

a2 r2rdrd 2a2S a

aa2x2a2x2

a

a2 x2 y 2dydx

16.

42

4

2

6

6

x

y x= 16 2

y

20

40

1 4r2rdrd

246565 1

S 4

0 16x

2

0

1 4x2 y 2dydx

1 fy2 fy

21 4x2 4y 2

z 16 x2 y 2 18.

1

1

1 1

x

y

x2+y2= 4

S 202

0

5rdrd 45

1 fx2fy

2

1

4x2

x2

y2

4y2

x2

y25

z 2x2 y2

20.

triangle with vertices

2

2

31

3

1

x

R

y x=

y

S 20

x0

5 4y2dydx 1

122121 55

1 fx 2 fy

25 4y2

0, 0, 2, 0, 2, 2R

fx,y 2x y2 22.

2

2

2 2

x

x y22+ = 16

y

2

0

4

0 1

4r2

drd

6565 16

S 4416x216x2

1 4x2 4y 2dydx

1 fx 2 fy

21 4x2 4y 2

fx 2x, fy 2y

0 x2 y 2 16

R x,y: 0 fx,y 16

fx,y x2 y 2



24/41

24.

S

1

0

1

0

1 x sinx2dydx 1.02185

1 fx 2 fy

21 x sinx2

fx x12 sinx, fy 0

R x,y: 0 x 1, 0 y 1

fx,y 23 x

32 cosx 26. Surface

Matches (c)

xy33

3

2

z

area 9

28.

1

0

1 y3dy 1.1114

S 10

10

1 y3dxdy

1 fx 2 fy

21 y3

fx 0, fy y32

R x,y: 0 x 1, 0 y 1

fx,y 25y

52 30.

S

4

0

x

0

1 13x2 y 2dydx

1 13x2 y 2

1 fx fy 21 2x 3y2 3x 2y2

fx 2x 3y, fy 3x 2y 3x 2y

R x,y: 0 x 4, 0 y x

fx,y x2 3xy y 2

32.

S 22

(2)x2(2)x2

1 4x2 y 2 sin2x2 y 2dydx

1 fx 2 fy

21 4x2 sin2x2 y 2 4y 2sin2x2 y 2 1 4sin2x2 y 2x2 y 2

fx 2x sinx2y 2, fy 2y sinx

2y 2

R x,y: x2 y 2 2fx,y cosx2 y 2

34.

S 40

x0

1 e2xdydx

1 e2x

1 fx 2 fy

21 e2x sin2y e2x cos2y

fx ex siny, fy e

x cosy

R x,y: 0 x 4, 0 y x

fx,y ex siny 36. (a) Yes. For example, letR be the square given by

and S the square parallel toR given by

(b) Yes. LetR be the region in part (a) and S the surface

given by

(c) No.

fx,y xy.

0 x 1, 0 y 1,z 1.

0 x 1, 0 y 1,

38.

S R1 fx 2 fy 2dA

Rk2 1 dA k2 1

RdA Ak2 1 r2k2 1

1 fx 2 fy

21 k2x2x2 y 2 k

2y 2

x2 y 2k2 1

fx,y kx2 y 2

Section 13.5 Surface Area 387


25/41

42. False. The surface area will remain the same for any vertical translation.

40. (a)

(c)

S 2500

150

1 fy2fx

2dydx 3087.58 sq ft

fx 0,fy 1

25

y2 8

25

y 16

15

fx,y 1

75y3

4

25y2

16

15y 25

z 1

75y3

4

25y2

16

15y 25 (b)

cubic feet

(d) Arc length

Surface area of roof 25030.8758 3087.58 sq ft

30.8758

100266.25 26,625

V 250150 175y3

4

25y2

16

15y 25dy

Section 13.6 Triple Integrals and Applications

2.

2

3

1

1

1

1

y 2z2dydz 2

9

1

1

y3z21

1dz

4

9

1

1

z2dz 427z31

1

8

27

11

11

11

x2y 2z2dxdydz 1

311

11

x3y 2z21

1dydz

4.

1

29

0

xy 2 3x3y3

0dy

2

189

0

y3dy 136y 49

0

729

4

90

y30

y29x2

0

zdzdxdy 1

29

0

y30

y 2 9x2dxdy

6.

41

1xlnz2

2 e2

1dx 4

1

2

xdx 2 ln x

4

1 2 ln 4

41

e2

1

1xz0

lnzdydzdx 41

e2

1

lnzy1xz

0dzdx 4

1

e2

1

lnz

xzdzdx

8. 20

y20

1y0

sinydzdxdy 20

y20

siny

ydxdy

1

22

0

sinydy 12 cosy2

0

1

2

10. 20

2x2

0

4y2

2x2y2ydzdydx 2

0

2x2

0

4y 2x2y 2y3dydx 162

15

12.

60

3(x2)0

1

2 6 x 2y

3 2

ex2y 2dydx 2.118

3

0 2(2y3)

0 62y3z

0

zex2y 2dxdzdy

6

0 (6x)2

0 (6x2y)3

0

zex2y 2dzdydx

14. 30

2x0

9x2

0

dzdydx



26/41

16.

44

16x2

16x280x

2y2

12x2y2dzdydx

z 8x2 y2 2z 16x2 y2 z2 2z z2 80z2 2z 80 0 z 8z 10 0

z 1

2x2 y2 2z x2 y2

18. 1

0 1

0 xy

0

dzdydx 1

0 1

0

xydydx 1

0

x

2dx x

2

4 1

0

1

4

20.

49x36 x2 324 arcsinx6 1

6x36 x232

6

0

4162 648

4603636 x2 x236 x2 1336 x232dx

460

36x2

0

36x2y

2

0

dzdydx 460

36x2

0

36 x2 y2dydx 46036y x2y y

3

3 36x2

0

dx

22.

20

18 9x 2x2 x3dx 18x 92x2 2

3x3

1

4x 4

2

0

50

3

2

0 2x

0 9x2

0dzdydx

2

0 2x

09 x2dydx

2

09 x22 xdx

24. Top plane:

Side cylinder:

30

9y2

0

6xy0

dzdxdy

x2 y2 9

x

y

6

3

6

63

zx y z 6 26. Elliptic cone:

40

4zy

2z22

0

dxdydz

x

y

5

5

4

3

3

2

2

1

1

z

4x2 z2 y 2

28.

40

2y0

y0

xyzdxdzdy 40

22y

2z0

dxdzdy 10421

20

(2z)2

0

y0

xyzdxdydz 20

4(2z)22z

0

xyzdxdydz

20

2z0

4x2

xyzdydxdz

2

0

2x

0

4

x2xyzdydzdx

40

y0

2x0

xyzdzdxdy

Q

xyzdV 20

4x22x

0

xyzdzdydx

x

y44

2

4

2

(2, 4)

zQ x,y,z: 0 x 2,x2 y 4, 0 z 2 x

Section 13.6 Triple Integrals and Applications 389


27/41

30.

60

10

1x2

0

xyzdydxdz

10

60

1x2

0

xyzdydzdx

60

10

1y2

0

xyzdxdydz

1

0 6

0 1y

2

0

xyzdxdzdy

10

1y2

0

60

xyzdzdxdy

Q

xyzdV 10

1x2

0

60

xyzdzdydx

x

y2

1

6

21

zQ x,y,z: 0 x 1,y 1 x2, 0 z 6

32.

y Mxz

m 2

Mxz k5

0

5x0

15153x3y0

y2dzdydx 125

4k

m k

5

0

5x

0

15153x3y

0

ydzdydz 125

8

k 34.

y Mxz

m

kab2c24

kabc6

b

4

Mxz kb

0

a[1(yb)]0

c[1(yb)(xa)]0

ydzdxdy kab2c

24

m k

b

0

a1(yb)

0

c1(yb)(xa)

0

dzdxdy kabc

6

36.

z Mxy

m

kabc33

kabc22

2c

3

y Mxz

m

kab2c24

kabc22

b

2

x Myz

m

ka2bc24

kabc22

a

2

Mxz ka0

b0

c0

yzdzdydx kab2c2

4

Myz ka

0 b

0 c

0

xzdzdydx ka2

bc2

4

Mxy ka0

b0

c0

z2dzdydx kabc3

3

m ka0

b0

c0

zdzdydx kabc2

2

38. will be greater than whereas and will be unchanged.yx85,z

40. and will all be greater than their original values.zx,y



28/41

42.

z Mxy

m

k

16k3

3

16

y Mxz

m

2k

16k3

3

8

x Myz

m

0

16k3 0

Mxy 2k2

0

4x2

0

y0

zdzdydx k

Mxz 2k2

0

4x2

0

y0

ydzdydx 2k

Myz k2

2

4x2

0

y0

xdzdydx 0

k20

4 x2dx 16k

3

m 2k20

4x2

0

y0

dzdydx

44.

z Mxy

m k12

4k2

4

y Mxzm kln 4

k ln 4

k20

10

1

y 2 12dydx k2

0

y2y 2 1 1

2arctany

1

0dx k14

82

0

dx k12

4

Mxy 2k2

0

10

1(y21)

0

zdzdydx

Mxz 2k2

0

10

1(y21)

0

ydzdydx 2k20

10

y

y 2 1dydx k2

0

ln 2dx kln 4

m 2k20

10

1(y21)

0

dzdydx 2k20

10

1

y 2 1dydx 2k4

2

0

dx k

x 0

46.

z Mxy

m

10k

10k 1

y Mxz

m

15k2

10k

3

4

x Myz

m

25k2

10k

5

4

Mxy k5

0

(35)x30

(115)(6012x20y)0

zdzdydx 10k

Mxz k5

0 (35)x3

0 (115)(6012x20y)

0

ydzdydx 15k2

Myz k50

(35)x30

(115)(6012x20y)0

xdzdydx 25k

2

m k50

(35)x30

(115)(6012x20y)0

dzdydx 10k

x2 5

3

3

5

4

2

41

1

y x= (5 - )35

yfx,y 1

1560 12x 20y

Section 13.6 Triple Integrals and Applications 391


29/41

48. (a)

(b)

Iz Iyz Ixz 7ka7

180

Iy Ixy Iyz a7k

30

Ix Ixy Ixz a7k

30

Iyz Ixz by symmetry

Ixz ka2

a2a2a2a2a2

y 2x2 y 2dzdydx kaa2a2a2a2

x2y 2 y 4dydx 7ka7

360

Ixy ka2

a2a2a2a2a2

z2x2 y 2dzdydx a3k

12a2a2a2a2

x2 y 2dydx a7k

72

Ix Iy Iz ka5

12

ka5

12

ka5

6

Ixz Iyz ka5

12by symmetry

Ixy ka2a2a2a2a2a2

z2dzdydx ka5

12

50. (a)

CONTINUED

Ix Ixz Ixy 2048k

9,Iy Iyz Ixy

8192k

21,Iz Iyz Ixz

63,488k

315

k40

20

1

2x216 8y 2 y4dydx

k

24

0

x216y 8y3

3

y5

5 2

0dx

k

24

0

256

15x2dx

8192k

45

Iyz k40

20

4y2

0

x2zdzdydx k40

20

1

2x24 y 22dydx

k40

20

1

216y 2 8y4 y6dydx

k

24

016y

3

3

8y5

5

y7

7 2

0dx

k

24

0

1024

105dx

2048k

105

Ixz k4

0 2

0 4y2

0

y 2zdzdydx k4

0 2

0

12y 24 y 22dydx

k

44

0

256y 256y3

3

96y5

5

16y7

7

y9

9 2

0dx k4

0

16,384

945dx

65,536k

315

k

44

0

20

256 256y 2 96y4 16y6 y8dydx

Ixy k40

20

4y2

0

z3dzdydx k40

20

1

44 y 24dydx



30/41

Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates

50. CONTINUED

(b)

Ix Ixz Ixy 48,128k

315, Iy Iyz Ixy

118,784k

315, Iz Ixz Iyz

11,264k

35

k40

20

4y2

0

4x2dzdydx k40

20

4y2

0

x2zdzdydx 4096k

9

8192k

45

4096k

15

Iyz k4

0

20

4y2

0

x24 zdzdydx

k40

20

4y2

0

4y 2dzdydx k40

20

4y2

0

y 2zdzdydx 1024k

15

2048k

105

1024k

21

Ixz 4

0 2

0 4y2

0

y 24 zdzdydx

k40

20

4y2

0

4z2dzdydx k40

20

4y2

0

z3dzdydx 32,768k

105

65,536k

315

32,768k

315

Ixy 40

20

4y2

0

z24 zdzdydx

52.

Iz Ixz Iyz 1

12ma2 c2

Iy Ixy Iyz 1

12mb2 c2

Ix Ixy Ixz 1

12ma2 b2

Iyz c2

c2a2a2b2b2

x2dzdydx abc2c2

x2dx abc3

12

1

12c2abc

1

12mc2

Ixz c2

c2a2a2b2b2

y 2dzdydx bc2c2a2a2

y 2dydx ba3

12c2c2

dx ba3c

12

1

12a2abc

1

12ma2

Ixy c2

c2a2a2b2b2

z2dzdydx b3

12c2c2a2a2

dydx 1

12b2abc

1

12mb2

54. 11

1x2

1x24x

2y2

0

kx2x2 y 2dzdydx 56. 6

58. Because the density increases as you move away from the axis of symmetry, the moment of intertia will increase.

2.

1

24

0

20

4r 4r2 r3drd1

24

0

2r2 4r3

3

r4

4 2

0

d2

34

0

d

6

4

0

2

0

2r

0

rzdzdrd 4

0

2

0rz

2

2 2r

0

drd

4. 20

0

20

e3

2ddd 20

0

13 e3

2

0

dd 20

0

1

31 e8

dd

2

61 e8

Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 393


31/41

6.

52

36

4

0

sin cos d 52

36sin2

2 4

0

52

144

1

34

0

sin cos sin sin3

3 4

0

d

1

34

0

40

sin cos cos 1 sin2dd

40

40

cos 0

2 sin cos ddd1

34

0

40

cos3sin cos dd

8. 20

0

sin 0

2 cos 2ddd8

9

10.

2

0

94

d 92

20

3r2

2

r4

4 3

0

d

y

x

2233

4

z20

30

3r2

0

rdzdrd 20

30

r3 r2drd

12.

468

3

117

32

0cos

0

d

yx 7

7

7 r= 5

r= 2

z20

0

52

2 sin ddd117

32

0

0

sin dd

14. (a)

(b) 20

60

40

3 sin2ddd 20

262 csc

4

3 sin2 ddd82

3 23

20

20

16r2

0

r2dzdrd82

3 23

16. (a) 20

10

1r2

0

rr2 z2dzdrd

8(b) 2

0

20

10

3 sin ddd

8

18.

(Volume of lower hemisphere) (Volume in the first octant)

128

3

642

3

64

32 2

128

3 4823

82

3

128

3 4823

2

013 16 r232

4

22d

V128

3 4

2

0

220

r2drd 20

422

r16 r2drd

4

yx7

7

7

zV2

343 4

2

0

220

r0

rdzdrd 20

42216r

2

0

rdzdrd



32/41

20.

8

32 2

2

0

1

3

4 r232 r3

3

2

0

d

20

20

r4 r2 r2drd

V 20

20

4r2

r

rdzdrd 22.

3k1 e4

20

6ke4 6kd

20

6ker2

2

0

20

20

12er

2

0

krdzdrd 20

20

12ker

2

rdrd

24.

z Mxy

m

kr02h2

12 3

r02hk

h

4

2kh2

r02

r04

12

2 kr0

2h2

12

2kh2

r02

20

r00

r02r 2r0r

2 r3drd

Mxy 4k2

0

r00

hr0rr00

zrdzdrd

m 1

3r0

2hk from Exercise 23

x y 0 by symmetry 26.

z Mxy

m

kr02h330

kr02h212

2h

5

1

30kr0

2h3

Mxy 4k2

0 r0

0 h(r

0r)r

0

0

z2rdzdrd

1

12kr0

2h2

m 4k20

r00

h(r0r)r00

zrdzdrd

x y 0 by symmetry

kz

28.

1

15r0

5kh

4kh1

30r0

5

2

4kh20

1

30r0

5d

4kh20

r05

5

r05

6 d

4kh20

r5

5

r6

6r0r0

0

d

4kh20

r00

r0 r

r0r4drd

4k

2

0

r0

0

hr0rr

0

0

r4dzdrd

Iz Q

x2 y2x,y,zdV 30.

3

2ma2

3

2ka4h

Iz 2k2

0

2a sin 0

h0

r3dzdrd

m ka2h

Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 395


33/41

34.

ka4

4

1

4k2a4

ka4121

4sin 2

2

0

ka4

2

0 sin2

d

2ka420

20

sin2dd

m 8k20

20

a0

3 sin2ddd 36.

z Mxy

m

kR4 r44

2kR3 r33

3R4 r4

8R3 r3

1

4kR4 r4 18 kR4 r4 cos 2

2

0

1

4kR4 r42

0

sin 2d

1

2kR4 r42

02

0

sin 2dd

Mxy 4k2

0

20

Rr

3 cos sin ddd

m k23R3 2

3r3 23 kR3 r3

x y 0 by symmetry

38.

4k

15R5 r5

2k5 R5 r5cos cos3

3 2

0

2k

5R5 r52

0

sin 1 cos2d

4k

5R5 r52

0

20

sin3dd

Iz 4k

2

0

2

0

R

r

4 sin3ddd 40.

z cos cos z

x2 y2 z2

y sin sin tan y

x

x sin cos 2 x2 y2 z2

42.

2

1

2

1

2

1

fsin cos , sin sin , cos 2

sin ddd

44. (a) You are integrating over a cylindrical wedge.

46. The volume of this spherical block can be determined as follows. One side is length

Another side is Finally, the third side is given by the length of an arc of angle in a

circle of radius Thus:

2 sin

V sin

sin .

.

x

y

i i isin

i i i

z.

32. (includes upper and lower cones)

1 22 43 b3 a3 23 2 2 b3 a3 43 b3 a3cos

4

0

4

3b3 a34

0

sin dy

x

aa

b

b

b

z

8

3b3 a34

0

20

sin dd

V 840

20

ba

2 sin ddd

(b) You are integrating over a spherical block.



34/41

Section 13.8 Change of Variables: Jacobians

2.

x

uy

v

y

ux

v

ad cb

y cu dv

x au bv 4.

x

uy

vy

ux

v v 2u vu 2u

y uv

x uv 2u

6.

x

uy

vy

ux

v 11 00 1

y v a

x u a 8.

x

uy

vy

ux

v 1v1 1

u

v2 1

v

u

v2

u v

v2

y u v

x u

v

3,66, 3

0,62, 2

3, 04, 1

0, 00, 0

u, vx,y10.

v x 4y

u x y

y

1

3 u v

x 1

34u v

(0, 0) (3, 0)

(3, 6)(0, 6)

11

1

2

3

4

5

6

1 2 4 5 6

v

u

12.

1523 26

3 120

152 23u3 26

3u

1

1

11

15

2 2u2 26

3 du

11152

v3

3 u2v

3

1

du

1

1

3

1

15

2v2 u2dvdu

1

1

3

1

6012 u v1

2u v12dvdu

R60xydA

x

uy

vy

ux

v

1

2 1

2 1

21

2 1

2

y 1

2u v, v x y

x 1

2u v, u x y

(1, 3)(1, 3)

(1, 1) (1, 1)

2 1

1

2

1 2

v

u

1, 11, 0

1, 31, 2

1, 32, 1

1, 10, 1

u, vx,y

Section 13.8 Change of Variables: Jacobians 397


35/41

18.

0

12 u3

3 1 cos 2v

2 2

dv 73

12 v 1

2sin 2v

0

74

12

R

x y2 sin2x ydA

0

2

u2 sin2v12dudv

x,y

u, v

1

2

x 1

2u v,

u x y 2,

2

x

x y = 0

x y =x y+ =

x y+ = 2

2

2

3

2

3

yu x y ,

y 1

2u v

v x y

v x y 0

20.

8

0

16v 32dv 2516v528

0

4096

52

R3x 2y2y x32dA

8

0 16

0

uv3218dudv

x

uy

vy

ux

v

1

43

8 1

81

4 1

8

x 1

4u v,

u 3x 2y 16,

x2

3

3

5

2

41

1

12

2 = 0y x

2 = 8y x

3 + 2 = 16x y

3 + 2 = 0x y

y

(0, 0)

(2, 3)

(2, 5)

(4, 2)

u 3x 2y 0,

y 1

8u 3v

v 2y x 8

v 2y x 0

16.

Ry sinxydA 4

1 4

1

vsin u1vdvdu

4

1

3 sin udu 3 cos u4

1 3cos 1 cos 4 3.5818

x

uy

vy

ux

v

1

v

y v

x u

v

14.

2

0

2u1 eu2du 2u2

2 ueu2 eu2

2

0 21 e2

R 4x yexydA

2

0

0

u2

4uev12dvdu

x

uy

vy

ux

v

1

2

y 1

2u v

x 1

2u v

u1 2

1

2

v u= 2

v



36/41

22.

4

112 ln1 v2

4

11

udu 12 ln 17 ln 2 ln u

4

1

1

2 ln17

2 ln 4

R xy

1 x2y2dA

4

1

4

1

v

1 v2 1

udvdu

x

uy

vy

ux

v

1

u

x u,

u x 4,

u x 1,

y v

u

v xy 4

v xy 1

x

2

3

4

1 2 3 4

1

x = 4

x = 1

xy = 4

xy = 1

y

24. (a)

Let and

Let

(b)

Let and

Let

2Aab2 0 0 4

2 4 2Aab

Aab2

0

1

0

cos2 rrdrdAab2r

sinr2

4

2cosr2

1

02

u rcos ,v rsin .

Rfx,ydA

1

1

1u2

1u2A cos2u2 v2abdvdu

y bv.x au

R:x2

a2

y2

b2 1

fx,y A cos 2x2

a2

y 2

b2

12398 7

16sin 2

2

0 12394 117

122

0

8 41 cos 22 9

4 1 cos 2

2 d 122

0

398 7

8cos 2d

122

0

8r2 4r4 cos2 94 r4 sin21

0d 12

2

0

8 4 cos2 94 sin2d

2

0

1

0

16 16r2 cos2 9r2 sin2 12rdrd

u rcos , v rsin .R16 x2 y 2dA

1

11u2

1u216 16u2 9v2 12dvdu

y 3v.x 4u

V

R

fx,ydA

R:x2

16

y 2

9 1

fx,y 16 x2 y 2

Section 13.8 Change of Variables: Jacobians 399


37/41

Review Exercises for Chapter 13

2. 2yy

x 2 y 2dx x3

3xy 2

2y

y

10y3

3

4.

2

0

4x 2 2x3 2x4dx

4

3

x3 1

2

x4 2

5

x5

2

0

88

15

2

0

2x

x2

x 2 2ydydx

2

0

x 2y y 2

2x

x2

dx

6. y4 y 2 4 arcsiny23

03

4

33

024y

2

24y2dxdy 23

0

4 y 2dy

8.

1

2

2

0

6 3ydy 126y 3

2y2

2

0 3

A 20

6y2y

dxdy

20

x0

dydx 32

62x0

dydx 20

6y2y

dxdy

10.

A 40

6xx2

x22x

dydx 40

8x 2x2dx 4x2 23x34

0

64

3

40

6xx2

x22x

dydx 01

11y11y

dydx 80

11y39y

dxdy 98

39y39y

dxdy

12. A 20

y21

0

dxdy 10

20

dydx 51

2x1

dydx 14

3

14. A

3

0

2yy2

y

dxdy

0

3

11x

x

dydx

1

0

11x

11x

dydx 9

2

16. Both integrations are over the common regionR shown in the figure. Analytically,

30

2x30

exydydx 53

5x0

exydydx 35 e5 e3 2

5 e5 e3 8

5e5

2

5

20

5y3y2

exydxdy 2

5

8

5e5

x

1

1

2

2

3

3

4

4

5

5

(3, 2)

y

26. See Theorem 13.5. 28.

x,y,z

u, v, w

4

0

1

1

4

0

0

1

1 17x 4u v, y 4v w, z u w

30.

1rcos2 rsin2 rx,y,z

r, ,z

cos

sin

0

rsin

rcos

0

0

0

1

x rcos , y rsin , z z



38/41

22.

Since we have

P 0.50

0.250

8xydydx 0.03125

k 8.k8 1,

kx4

8 1

0

k

8

10

kx3

2dx

10

x0

kxydydx 10kxy

2

2 x

0dx 24. False, 1

0

10

xdydx 21

21

xdydx

26. True, 10

10

1

1 x2 y2dxdy < 1

0

10

1

1 x2dxdy

4

28. 20

r4

4 4

0d 2

0

64 d 3240

16y20

x2 y 2dxdy 20

40

r3drd

30.

4

3R2 b232

8

3R2 b2322

0

d

8

32

0R2 r232

R

bd

V 820

Rb

R2 r2rdrd 32.

The polar region is given by and

Hence,

x

1

1

2

3

4

4, )( 12/ 138/ 13

8/ 13

x y2 2+ =16

23

xy =

y

arctan320

40

rcos rsin rdrd288

13

0 0.9828.

0 r 4

tan 1213

813

3

2 0.9828

18.

12x33

0

27

2

3

2

3

0

x2dx

30xy 12y2

x

0dx

V 30

x0

x ydydx 20. Matches (c)

x

y

2

2

1

2

3

z

Rev iew Exercises for Cha pte r 13 401


39/41

36.

y Ixm 16,384k315128k15 12821

x Iym 512k105128k15 47

m Rx,ydA 2

0

4x2

0

kydydx 128

15k

I0 Ix Iy 16,384k

315

512k

105

17,920

315k

512

9k

Iy Rx2x,ydA 2

0

4x2

0

kx2ydydx 512

105k

Ix Ry2x,ydA 2

0

4x2

0

ky3dydx 16,384

315k

38.

62 ln4 32 92

2 ln2

2

6

52

3 ln22 3

12418 2 ln4 18 1

121818 ln2 2212

122y2 4y2 2 ln2y 2 4y2 112 2 4y2322

0

S

2

0 2

y 2 4y2dxdy

2

0 22 4y2 y2 4y2dy

1 fx2 fy

2 2 4y2

fx 1,fy 2y

R x,y: 0 x 2, 0 y x

fx,y 16 x y2

34.

y Mxm 17kh

2

L80

127khL

51h140

x My

m

5khL2

24

12

7khL

5L

14

kh

2L

02x x

2

L

x3

L2dx kh

2 x2 x3

3L

x4

4L2L

0

kh

2

5L2

12

5khL2

24

My kL0

h22xLx2L2

0

xdydx

kh2

8

17L

10

17kh2L

80

kh2

8 4x 2x2

L

x3

L2

x 4

2L3

x5

5L4L

0

kh2

8

L

04 4xL

3x2

L2

2x3

L3

x4

L4dx

kh2

8L

02 xL

x2

L22

dx

Mx kL

0

h22xLx2L2

0

ydydx

x

h

L

y = 2 h2

xL

x2

L2( (

ym kL0

h22xLx2L2

0

dydx kh

2L

02 xL

x2

L2dx 7khL

12



40/41

44.

20

5 arctan 5cos 2

0d

25 arctan 5

20

20

arctan 5

0sin dd

50

25x2

0

25x2y2

0

1

1 x2 y2 z2dzdydx 2

0

20

50

2

1 2sin ddd

46. 20

4x2

0

4x2y2

0

xyzdzdydx 4

3

48.

84 2 sin 2 14sin3cos 3

41

2

1

4sin 2

2

0

29

2

220

32 sin2 4 sin4d 820

8 sin2 sin4d

220

2 sin 0

r16 r2drdV 220

2 sin 0

16r2

0

rdzdrd

50.

z Mxy

m

kc2a416

2kca33

3ca

32

y Mxz

m

kca48

2kca33

3a

16

x 0

Mxy 2k2

0

a0

crsin 0

rzdzdrd kc220

a0

r3 sin2drd1

4kc2a42

0

sin2d1

16kc2a4

Mxz 2k2

0

a

0

crsin

0

r2 sin dzdrd 2kc2

0

a

0

r3 sin2drd1

2kca4

2

0

sin2d1

8kca4

m 2k20

a0

crsin 0

rdzdrd 2kc20

a0

r2 sin drd2

3kca32

0

sin d2

3kca3

40. (a) Graph of

over regionR

x

y50

50

50

R

z

25 1 ex2y21000 cos2x2 y 2

1000

fx,y z

(b) Surface area

Using a symbolic computer program, you obtain surface

area sq. ft. 4,540

R

1 fxx,y2 fyx,y2dA

42. 1

22

0

20

r5drd16

32

0

d32

322

4x24x2

(x2y2)2

0

x2 y 2dzdydx 20

20

r22

0

r3dzdrd

Rev iew Exercises for Cha pte r 13 403


41/41

52.

z Mxy

m

81

4

1

162

1

8

20

30

12 r3 9

2rdrd

2

0

18 r4 9

4r2

3

0

d 818 2

0

81

4

Mxy 2

0 3

0 4

25r2zrdzdrd

2

0 5

3 25r2

25r2zrdzdrd

2

0 3

0 8 12 25 r2rdrd 0

x y 0 by symmetry

500

3 213 25 r232 2r2

3

0

500

3 2643 18

125

3 500

3

14

3 162

m 500

3 3

0

20

25r24

rdzddr500

3 3

0

20

r25 r2 4rddr

54.

4ka6

9

Iz k

0

20

a0

2 sin22 sin ddd

56.

8

15a

aa

1z2a2

1z2a21y

2z2a2

1y2z2a2

x2 y2dxdydz

Iz Qx2 y2dV

x2 y2 z2

a2 1 58.

Since represents a paraboloid with vertex

this integral represents the volume of the solid

below the paraboloid and above the semi-circle

in thexy-plane.y 4 x2

0, 0, 1,z 1 r2

0

20

1r2

0

rdzdrd

60.

2u2v 2u2v 8uv

x,yu, v

xu

yv yu

xv

62.

Boundary inxy-plane Boundary in uv-plane

4 arctan 5 4 arctan v5

51

4

1 v2dv 5

1

51

1

1 v2dudv

R x

1 x2y2dA 5

1

51

u

1 u2vu21ududv

v 5xy 5

v 1xy 1

u 5x 5

u 1x 1

x u,y v

u u x, v xy

y = 1x

y = 1x

x = 5

x = 1

1

1

2

3

4

5

4 5

y

x

x,y

u, vx

uy

vx

vy

u 11u 0

1

u


solucion larzon cap13

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