Solucionario Larson Cap 13

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C H A P T E R 13 Multiple IntegrationSection 13.1 Iterated Integrals and Area in the Plane . . . . . . . . . . . . . 133

Section 13.2 Double Integrals and Volume . . . . . . . . . . . . . . . . . . . 137 Section 13.3 Change of Variables: Polar Coordinates . . . . . . . . . . . . . 143 Section 13.4 Center of Mass and Moments of Inertia . . . . . . . . . . . . . 146 Section 13.5 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

Section 13.6 Triple Integrals and Applications . . . . . . . . . . . . . . . . . 157 Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates . . . . 162 Section 13.8 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . 166 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

C H A P T E R 13 Multiple IntegrationSection 13.1 Iterated Integrals and Area in the PlaneSolutions to Odd-Numbered Exercisesx

1.0

2x

y dy

2xy

1 2 y 2

x 0

3 2 x 2

2y

3.1

y dx x

2y

y ln x1

y ln 2y

0

y ln 2y

4

x2

5.0

x 2y dy

1 2 2 x y 2

4 0

x2

4x 2 2

x4

y

7.ey

y ln x dx x

1 y ln2 x 2

y ey

1 y ln2y 2x3 x3

ln2ey

y ln y 2

2

y2

x3

x3

9.0

ye u1

y x

dy dy, dv

xye e

y x 0

x0

e xe

y x

dy

x4 e

x2

x 2e

y x 0

x2 1

e

x2

x 2e

x2

y, du2

y x

dy, v 1 2 y 2

y x

1

2

1

1

11.0 0

x

y dy dx0

xy

dx0 0

2x

2 dx

x2

2x0

3

1

x

1

x

1

13.0 0

1

x2 dy dx0

y 1

x20

dx0

x 1

x2 dx

1 2 1 2 3

1

x2

3 2 0

1 3

2

4

2

15.1 0

x2

2y 2

1 dx dy1 2 1

1 3 x 3 64 3

4

2xy 2 8y 2

x0

dy 4 32

4 dy

191

6y 2 dy

4 19y 3

2y

3

2 1

20 3

1

1 0

y2

1

17.0

x

y dx dy0 1 0

1 2 x 2 1 1 2

xy0

1

y2

dy 1 y 2 1 3 y 6 1 2 1 2 31

y2

y 1

y 2 dy

y2

3 2 0

2 3

2

4 0

y2

19.0

2 4 y

2

dx dy 2

0

2x 4 y2sin

4 0

y2

2

2

dy0

2 dy

2y0

4

2

sin

2

21.0 0

r dr d0

r2 22

2

d0 0

1 sin2 2 1 42

d 1 cos 2 42 2

1 4

cos 20

d

2

2

sin 20

32

1 8

133

134

Chapter 131 x

Multiple Integration1 x

23.1 0

y dy dx1

y2 2

dx0

1 2

1

1 dx x2

1 2x

01

1 2

1 2

25.1 1

1 dx dy xy

1

1 ln x y

dy1 1

1 y

1 0 y

dy

Divergesy

8

3

8

3

8

8

27. A0 3 0 8

dy dx0 3

y0 8

dx0 3

3 dx

3x0 3

24

8 6

A0 0

dx dy0

x0

dy0

8 dy

8y0

24

4

2 x2 4 6 8

2

4 0

x2

2

4

x2

y

29. A0 2

dy dx0

y0

dx

4 3

y = 4 x2

40

x2 x3 34 y 2 0

dx 16 3 dx dy

2 1 x 1 2 3

4x4

1

A0 4 0 4 0

x0

y

4

4

dy0

4

y dy0

4

y

1 2

1 dy

2 4 34 2 0 x

y

3 2

4 0

2 8 34

16 32 x2

2

1

4

x2

31. A2 x 1 2 4

dy dxx2

y = 4 x2

y

33.03

dy dx0 4

y0

dx

(1, 3)

y2 1 x 2

dx

42

4 x 8 x x 3

x dx x2 24 0

y=x+21

0

42 1

x

2

x

2 dx

4x2 1 x 1 2

8 3

4

2 0

y

2

22

x 1 2 x 2

x 2 dx 1 3 x 3 dx dy1 2 4

dx dy0

8 3

2x3 y

9 24 0 4 y

Integration steps are similar to those above.y 4

2

A0 3 4 y y 2

23 4

dx dyy

32

y = (2

x )2

x0 3 4 y

dy

23

x0 4

dy

1 x1 2 3 4

y0

2 2y

4 2 4 3

y dy3

23

4 4 4 3

y dy4

1 2 y 2

y

3 2 0

y

3 2 3

9 2

Section 13.13 2x 3 5 5 0 5 x x

Iterated Integrals and Area in the Planea 0 0 a b a a2 x2 a b a a2 x2

135

35. A0 3 0

dy dx3 2x 3 5

dy dx y3 5 0

A 37. 4

dy dx0 2

y0

dx d

y0 3 0 0

dx 53

dx x

b a

a20

x 2 dx

ab0

cos2

2x dx 33

a sin , dx ab 22

a cos d cos 2 d ab 2 1 sin 2 22 0

x dx 1 2 x 25

10

1 2 x 32 5

5x0 y

53

ab 4 Therefore, A A 4b 0 0 a b

A0 2 3y 2 5

dx dyy

ab.b2 y2

x0 2 3y 2

dy 3y dy 2 5y 5 2 y 42

dx dy

ab 4

50 2

y

Therefore, A above.y

ab. Integration steps are similar to those

52y

5y dy 2

50b

y= b a

a2 x2

a

x

4 32

y= 2x 3

y=5x

1 x1 2 3 4 5

1

4

y

2

4

x2

39.0 0

f x, y dx dy, 0 x y, 0 y 44 0y

41.2 0 2

f x, y dy dx, 0 y 4 4y 3

4

x2,

2 x 2

4

y2

f x, y dy dxx 0 y2

dx dy

3 2 1 x 1 2 3 4 2 1 1 1 x 1 2

10

ln y

1

1

43.1 0

f x, y dx dy, 0 x ln y, 1 y 10ln 10 0y

45.1 x2

f x, y dy dx, x 2 y 1, 1 x 11 y

10

f x, y dy dxex

f x, y dx dy0y4

y

83

6 4 2 x 1 2 32 1 x 1 2 2

1361

Chapter 132 2 1

Multiple Integration1 1 1 y2 1 1 0 x2

47.0y

dy dx0 0 0

dx dy

2

49.0 y2y

dx dy1

dy dx

2

3

1

21

x

1x

1

1

2

3

2

x

4

4 0

x

2

4 y

y

2

1

1

2y

51.0y

dy dx0 2

dy dx0

dx dy

4

53.0y

dy dxx 2 0 0

dx dy

1

3

22

1x1 2 3 4

1

1

x

1

2

1

3

y

1

x

55.0 y2

dx dy0 x3

dy dx

5 122

x= 3 yy

x = y2

1

(1, 1)x

1

2

57. The first integral arises using vertical representative rectangles. The second two integrals arise using horizontal representative rectangles.5 0 x 50 x2 5

x 2y 2 dy dx0

1 2 x 50 3

x2

3 2

1 5 x dx 3

15625 245 0 y 5 2 0 50 y2 5

x 2y 2 dx dy0 5

x 2y 2 dx dy0

1 5 y dy 3

5 5

2

1 50 3

y2

3 2

y2 dy

15625 18

15625 18

15625 18

15625 24y

y=

50 x 2

(0, 5 2 )5

(5, 5) y=xx 5

Section 13.22 2 2 y 2

Double Integrals and Volume

137

59.0 x

x 1

y3 dy dx0 0 2

x 1 1 2

y3 dx dy0

1 1 2 1 3 2 1 3x

y3

x2 23 2

y

dy0 2

10

y3 y 2 dy

y3

0

1 27 9

1 1 9

26 9

1

1

1

x

1

61.0 y

sin x 2 dx dy0 1 0

sin x 2 dy dx0

y sin x 20 1 0

dx 1 cos 1 24

x sin x 2 dx0

1 cos x 2 2

1 1 2y 0

1 1 2

cos 1

0.2298

2

2x

63.0 x2

x3

3y 2 dy dx

1664 105

15.848

65.0

x

2 1 y

1

dx dy

ln 5

2

2.590

67. (a) x x8

y3 y

x1

3

y

4 2y x2x1 3

32y y

x2 32

4

x = y32

(8, 2)

(b)0 x2 32

x 2y

xy 2 dy dx2

x 2 4 6 8

x = 4 2y

(c) Both integrals equal 67520 693

97.43

2

4 0

x2

2

1 0

cos

69.0

exy dy dx

20.5648

71.0

6r 2 cos dr d

15 2

73. An iterated integral is a double integral of a function of two variables. First integrate with respect to one variable while holding the other variable constant. Then integrate with respect to the second variable. 75. The region is a rectangle. 77. True

Section 13.2For Exercise 13, 1 1 , , 2 2 3 1 , , 2 2 xi

Double Integrals and Volumeyi 1 and the midpoints of the squares are 7 1 , , 2 2 1 3 , , 2 2 3 3 , , 2 2 5 3 , , 2 2 7 3 , . 2 24 32

y

5 1 , , 2 2

1 x1 2 3 4

1. f x, y8

x

y xi yi x 1 24

f xi, y