# Solution manual calculus 8th edition varberg, purcell, rigdon ch13

13

CHAPTER

13.1 Concepts Review

Multiple Integrals
3.

= 2 A( R1 ) + 1A( R2 ) + 3 A( R3 )
= 2(2) + 1(2) + 3(2) = 12

n

1.

∫∫R f ( x, y)dA = ∫∫R1 2 dA + ∫∫R2 1 dA + ∫∫R3 3 dA

∑ f ( xk , yk )ΔAk

k =1

2. the volume of the solid under z = f(x, y) and
above R
3. continuous
4. 12

Problem Set 13.1
1.

∫∫R 2 dA + ∫∫R
1

2

4.

= 2 A( R1 ) + 3 A( R2 ) + 1A( R3 )
= 2(3) + 3(2) + 1(1) = 13

3 dA = 2 A( R1 ) + 3 A( R2 )

= 2(4) + 3(2) = 14

2.

∫∫R1 (–1)dA + ∫∫R2 2 dA = (–1) A( R1 ) + 2 A( R2 )

∫∫R1 2 dA + ∫∫R2 3 dA + ∫∫R3 1 dA

5. 3∫∫ f ( x, y )dA – ∫∫ g ( x, y )dA = 3(3) – (5) = 4
R

R

= (–1)(3) + 2(3) = 3

6. 2∫∫ f ( x, y )dA + 5∫∫ g ( x, y )dA
R

R

= 2(3) + 5(5) = 31

7.

∫∫R g ( x, y)dA – ∫∫R1 g ( x, y)dA = (5) – (2) = 3

8. 2∫∫ g ( x, y )dA + ∫∫ 3 dA = 2(2) + 3 A( R1 )
R1

R1

= 4 + 3(2) = 10

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9.

17.

[f(1, 1) + f(3, 1) + f(5, 1) + f(1, 3) + f(3, 3)
+ f(5, 3)](4) = [(10) + (8) + (6) + (8) + (6)
+ (4)](4) = 168
18.

10. 4(9 + 9 + 9 + 1 + 1 + 1) = 120
11. 4(3 + 11 + 27 + 19 + 27 + 43) = 520
⎡⎛ 41 ⎞ ⎛ 33 ⎞ ⎛ 25 ⎞ ⎛ 35 ⎞ ⎛ 27 ⎞ ⎛ 19 ⎞ ⎤
12. ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎥ (4)
⎣⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ ⎦
= 120

13. 4

(

)

2 + 4 + 6 + 4 + 6 + 8 ≈ 52.5665

14. 4(e + e3 + e5 + e3 + e9 + e15 ) ≈ 13109247
15.

19.

16.

20.

802

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Then c = 2 5 + 2 2 + 2(2) + 4(1) ≈ 15.3006

21.

z = 6 – y is a plane parallel to the x-axis. Let T be
the area of the front trapezoidal face; let D be the
distance between the front and back faces.
∫∫ (6 – y)dA = volume of solid = (T )( D)
R

⎡⎛ 1 ⎞

= ⎢⎜ ⎟ (6 + 5) ⎥ (1) = 5.5
⎣⎝ 2 ⎠

For C, take the sample point in each square to be
the point of the square that is farthest from the
origin. Then,
C = 2 13 + 2 10 + 2 8 + 4 5 + 2 2 ≈ 30.9652.
26. The integrand is symmetric with respect to the
y-axis (i.e. an odd function), so the value of the
integral is 0.
27. The values of a x ba y b and a x b + a y b are indicated

in the various square subregions of R. In each
case the value of the integral on R is the sum of
the values in the squares since the area of each
square is 1.

22.

a.

The integral equals –6.

z = 1 + x is a plane parallel to the y-axis.
∫∫ (1 + x)dA is the product of the area of a
R

trapezoidal side face and the distance between the
side faces.
⎡⎛ 1 ⎞

= ⎢⎜ ⎟ (1 + 3)(2) ⎥ (1) = 4
2
⎣⎝ ⎠

23.

b. The integral equals 6.

∫∫R 0 dA = 0 A( R) = 0
The conclusion follows.

24.

∫∫R m dA < ∫∫R f ( x, y)dA < ∫∫R M dA
(Comparison property)
Therefore, ma ( R ) < ∫∫ f ( x, y )dA < MA( R )
R

25.

28. Mass of the plate in grams
29. Total rainfall in Colorado in 2005; average
30. For each partition of R, each subrectangle
contains some points at which f(x, y) = 0 and
some points at which
f(x, y) = 1. Therefore, for each partition there are
sample points for which the Riemann sum is 0
and others for which the Riemann sum is
(1)[Area (R)] = 12.

For c, take the sample point in each square to be
the point of the square that is closest to the origin.

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31. To begin, we divide the region R (we will use the outline of the contour plot) into 16 equal squares. Then we can
approximate the volume by
16

V = ∫∫ f ( x, y )dA ≈ ∑ f ( xk , yk )Δ Ak .
k =1

R

Each square will have ΔA = (1 ⋅1) = 1 and we will use the height at the center of each square as f ( xk , yk ).
Therefore, we get
16

V ≈ ∑ f ( xk , yk ) = 20 + 21 + 24 + 29 + 22 + 23 + 26 + 32 + 26 + 27 + 30 + 35 + 32 + 33 + 36 + 42
k =1

= 458 cubic units

8.

∫0 ∫0

13.2 Concepts Review
1. iterated
2.

2

2

∫−1 ⎢⎣ ∫0

f ( x, y )dy ⎤ dx ;
⎥⎦

2⎡ 2

∫0 ⎢⎣ ∫−1 f ( x, y)dx ⎥⎦ dy

ln 3 ln 2 x y

ln 3

0

9.

π/ 2

∫0
=

Problem Set 13.2

∫0 [9 y − xy ]0 dx = ∫0 [ 27 − 3x ] dx
2

3

1

π⎛ 1 ⎞

y⎥
dy = ∫ ⎜ ⎟ sin y dy = 1
0 ⎝2⎠
⎦ x =0

e e dy dx = ∫

ln 3

0

[e x e y ]lny =20 dx
ln 3 x
e dx
0

[e x (2) – e x (1)]dx = ∫

3
= [e x ]ln
0 = 3 –1 = 2

4. is below the xy-plane

2

=∫

3. signed; plus; minus

1.

π ⎡⎛ 1 ⎞ 2
⎟ x sin
0 ⎢⎜
⎝2⎠

7.

10.

[– cos xy ]1y =0 dx = ∫

π/2

0

(1 – cos x)dx

π
– 1 ≈ 0.5708
2

1

∫0 [e

xy 1
] y =0 dx

1

= ∫ (e x – 1)dx = e – 2 ≈ 0.7183
0

2

3 ⎤

= ⎢ 27 x − x 2 ⎥ = 48
2 ⎦0

2.

1

2

11.

4.

5.

3

dx = ∫ 4 x 2 dx =
0

32
3

=

2

3 2[(1 +

dy
x =0

y )3 / 2 – y 3 / 2 ]
dy
3

(

)

12.

(

4 31 – 9 3
15

1

∫0 [–( xy + 1)

) ≈ 4.1097

–1 1
]x =0 dy

= 1 – ln 2 ≈ 0.3069

1⎛
1 ⎞
= ∫ ⎜1 –
⎟ dy
0
y +1⎠

3

2⎛ 9y

dx = ∫ ⎜
+ xy 2 ⎥
+ 3 y 2 ⎟ dy
1 ⎝ 2
2

⎦⎥ x =0
2

1

+ y )3 / 2 ⎤

3
⎦⎥

3
4 32 – 9 3 – 4
⎡ 4[(1 + y )5 / 2 – y 5 / 2 ] ⎤
=⎢
⎥ =
15
15
⎣⎢
⎦⎥ 0

4⎛

7⎞
115
⎛1⎞ 3⎤
∫–1 ⎢⎣ xy + ⎜⎝ 3 ⎟⎠ y ⎥⎦ dx = ∫–1 ⎜⎝ x + 3 ⎟⎠ dx = 6
y =1
2⎡x

2

0

2

⎦ y =1

4

∫1 ⎢⎢

=∫

1 ⎤
92

= ⎢9 x − x 3 ⎥ =
3 ⎦ −2 3

2 ⎡⎛ 1 ⎞ 2 2 ⎤
⎟x y ⎥
0 ⎢⎜
⎝2⎠

∫0 ⎢⎢

2

2
2
∫−2 ⎡⎣9 y − yx ⎤⎦ 0 dx = ∫−2 ⎡⎣9 − x ⎤⎦ dx
2

3.

3 ⎡ 2( x

y

13.

2

⎡ 9 y2

13 55
=⎢
+ y 3 ⎥ = 17 – =
= 13.75
4
4
⎣⎢ 4
⎦⎥1

∫0

ln 3 ⎛ 1 ⎞
2 ⎤
dx = ∫ ⎜ ⎟ (e x – 1)dx
⎢⎜ 2 ⎟ exp( xy ) ⎥
0
⎝2⎠
⎣⎝ ⎠
⎦ y =0

⎛1⎞
= 1 – ⎜ ⎟ ln 3 ≈ 0.4507
⎝2⎠

2

1 ⎛7
⎡⎛ 1 ⎞

16

6. ∫ ⎢⎜ ⎟ x3 + xy 2 ⎥ dy = ∫ ⎜ + y 2 ⎟ dy =
–1 ⎝ 3 ⎠
–1 ⎝ 3
3

⎦ x =1
1

1

ln 3 ⎡⎛ 1 ⎞

14.

1⎡

2

1 2
∫0 ⎢⎢ 2(1 + x 2 ) ⎥⎥ dx = ∫0 1 + x2 dx

⎦ y =0

y2

π
⎛π⎞
= [2 tan –1 x]10 = 2 ⎜ ⎟ – 0 =
2
⎝4⎠

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15.

π

∫0

3

π9
⎡1 2
2 ⎤
2
⎢ 2 y cos x ⎥ dx = ∫0 2 cos x dx

⎦0
π

9

⎡9

= ⎢ x + cos 2 x ⎥ =
8
4
⎣4
⎦0

22. V = ∫

2 3

0 0

( 25 − x2 − y 2 ) dy dx
3

2⎛
1 ⎞
= ∫ ⎜ 25 y − x 2 y − y 3 ⎟ dx
0⎝
3 ⎠0

=∫

2

0

( 75 − 3x2 − 9 ) dx = 124

1

16.

1 1 ⎡ x2 ⎤
1 1
e ⎥ dy = ∫ ( e − 1) dy

1

2
2 −1

⎦0
1
1
⎡ y ( e − 1) ⎤⎦ = e − 1
−1
2⎣

=

17.

1

∫0 0 dx = 0 (since

(

)

23. V = 3 4 1 + x 2 + y 2 dx dy
∫ ∫
0 0

3

3⎛
1

= ∫ ⎜ x + x3 + xy 2 ⎟ dy
0⎝
3
⎠0

xy 3 defines an odd

3⎛
64

= ∫ ⎜ 4 + + 4 y 2 ⎟ dy = 112
0⎝
3

function in y).
24. V = 3 2 5 xye − x dx dy
∫ ∫
2

18.

1

∫–1 ⎢⎣ x

2

1 ⎛
8⎞
⎛1⎞ ⎤
y + ⎜ ⎟ y3 ⎥
dx = ∫ ⎜ 2 x 2 + ⎟ dx
–1
3⎠
⎝ 3 ⎠ ⎦ y =0

2

1

⎡⎛ 2 ⎞
20
⎛8⎞ ⎤
= ⎢⎜ ⎟ x 3 + ⎜ ⎟ x ⎥ =
3
3

⎦ –1 3

19.

π/ 2 π/ 2

∫0 ∫0
π/ 2

∫0

=

=∫

[– cos( x + y )]πx =/ 02 dy

π/2 ⎡

⎛π

⎢ – cos ⎜ 2 + y ⎟ + cos y ⎥ dy

0

=∫

sin( x + y )dx dy

π/2

0

0 0

2

3
1
⎛ 1

= ∫ 5 y ⎜ − xe−2 x − e −2 x ⎟ dy
0
4
⎝ 2
⎠0

(

)

45 e4 − 1
3
1⎤
⎡ 5
= ∫ 5 y ⎢ − e −4 − ⎥ dy =
≈ 11.0439
0
4⎦
⎣ 4
4e 4

x
is a plane.
2
x – 2z = 0

25. z =

(sin y + cos y ) dy = [– cos y + sin y ]0π / 2

= (0 + 1) – (–1 + 0) = 2

20.

3

2 ⎡⎛ 1 ⎞

∫1

2⎛ 7 ⎞
2 3/ 2 ⎤
dy = ∫ ⎜ ⎟ y dy
⎢⎜ 3 ⎟ y (1 + x ) ⎥
1
⎝3⎠
⎣⎝ ⎠
⎦ x =0
2

⎡⎛ 7 ⎞ ⎤
= ⎢⎜ ⎟ y 2 ⎥ = 3.5
⎣⎝ 6 ⎠ ⎦1

21. V = ∫

26. z = 2 – x – y is a plane.
x+y+z=2

∫ ( 20 − x − y ) dy dx

2 3

0 0

3

2⎛
1 ⎞
= ∫ ⎜ 20 y − xy − y 2 ⎟ dx
0⎝
2 ⎠0
2⎛
9⎞
= ∫ ⎜ 60 − 3 x − ⎟ dy = 105
0⎝
2⎠

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27. z = x 2 + y 2 is a paraboloid opening upward with
z-axis.

33.

b d

∫a ∫c

b

d

a

c

g ( x)h( y )dy dx = ∫ g ( x) ∫ h( y )dy dx

d

b

c

a

= ∫ h( y ) dy ∫ g ( x)dx

(First step used linearity of integration with
respect to y; second step used linearity of
integration with respect to x; now commute.)
34.

∫0

ln 2

1

2

xe x dx ∫ y (1 + y 2 ) –1 dy
0

⎡⎛ 1 ⎞ 2 ⎤
= ⎢⎜ ⎟ e x ⎥
⎣⎝ 2 ⎠
⎦0

ln 2

1

⎡⎛ 1 ⎞
2 ⎤
⎢⎜ 2 ⎟ ln(1 + y ) ⎥
⎣⎝ ⎠
⎦0

⎤ ⎛1⎞
⎡ 1 ⎤ ⎡⎛ 1 ⎞
= ⎢ ⎥ ⎢⎜ ⎟ ln 2 ⎥ = ⎜ ⎟ ln 2 ≈ 0.1733
⎣ 2 ⎦ ⎣⎝ 2 ⎠
⎦ ⎝4⎠

28. z = 4 – y 2 is a parabolic cylinder parallel to the
x-axis.
35.

2
2
1
1
e dy dx = ⎛⎜ ∫ xe x dx ⎞⎟ ⎛⎜ ∫ ye y dy ⎞⎟
0
0

⎠⎝

1 1

x2 y 2

∫0 ∫0 xye

2

2
1
= ⎛⎜ ∫ xe x dx ⎞⎟ (Changed the dummy variable y
⎝ 0

to the dummy variable x.)

29.

30.

1

3 ⎡⎛ 1 ⎞ 2

∫1 ∫0 ( x + y + 1)dx dy = ∫1 ⎢⎣⎜⎝ 2 ⎟⎠ x + yx + x ⎥⎦ dy
x =0
3⎛
3⎞
= ∫ ⎜ y + ⎟ dy = 7
1⎝
2⎠
3 1

∫1 ∫0 (2 x + 3 y)dy dx = ∫1

36. V = ∫
=∫

2

1

2

37.

+ y 2 + 2) –1]dy dx

38.

1

1 ⎡

⎛1⎞
= ∫ ⎢ x 2 y + ⎜ ⎟ y3 + y ⎥
dx
–1
3
⎝ ⎠

⎦ y =0
1 ⎛
4⎞
10
= ∫ ⎜ x 2 + ⎟ dx =
–1 ⎝
3⎠
3

32.

2 2

∫0 ∫0 (4 – x

2

39.

)dx dy = ∫

2⎡

0

2

x3 ⎤
⎢ 4 x – ⎥ dy
3 ⎥⎦
⎢⎣
0

2

2 ⎛ 16 ⎞
32
⎡16 y ⎤
= ∫ ⎜ ⎟ dy = ⎢
=

0⎝ 3 ⎠
⎣ 3 ⎦0 3

806

Section 13.2

π

cos x cos y dx dy
π
cos y dy = ⎛⎜ ∫ cos x dx ⎞⎟
–π
⎝ –π

cos x dx ∫
π/ 2

31. x 2 + y 2 + 2 > 1
1

= ⎛⎜ 4∫
⎝ 0

1

∫–1 ∫0 [( x

π

–π

⎛ 3⎞ 2⎤
⎢ 2 xy + ⎜ 2 ⎟ y ⎥ dx
⎝ ⎠ ⎦ y =0

= ∫ (8 x + 24)dx = 36

π

1
2
⎤ ⎞
e –1⎞
⎥ ⎟ = ⎛⎜
⎟ ≈ 0.7381
2 ⎠
⎥ ⎟⎟

⎦0 ⎠

–π –π

4

2⎡

2 4

2

⎛ ⎡ x2
⎜ e
= ⎜⎢
⎜ ⎣⎢ 2

π

2

(

cos x dx ⎞⎟ = 4[sin x]0π / 2

)

2

2

= 16

2
1
dy = ⎛⎜ 2∫ x 2 dx ⎞⎟ ⎛⎜ 2 ∫ y 3 dy ⎞⎟
⎝ 0
⎠⎝ 0

⎛8⎞ ⎛1⎞ 8
= 2⎜ ⎟ 2⎜ ⎟ =
⎝3⎠ ⎝ 4⎠ 3
2

1

∫−2 x dx ∫−1 y
2

2

c

2f

2

c

2f

∫−2 ed x

3

1

3
hg dx ∫−1 y dy = 0 (since the second integral
equals 0).

∫−2 de x

1
2c 2 f
1 3
3
gh dx ∫−1 y dy = 2∫0 de x gh dx 2 ∫0 y dy

2
⎤ ⎡ ⎛ 1 ⎞⎤
2 dx + ∫ 3 dx ⎥ ⎢ 2 ⎜ ⎟ ⎥
3
⎦ ⎣ ⎝ 4 ⎠⎦
⎡1⎤
= 2 ⎡0 + 2 – 1 + 2 3 – 2 + 3 2 – 3 ⎤ ⎢ ⎥

⎦ ⎣2⎦
= 5 – 3 – 2 ≈ 1.8537
2
⎡ 1
= 2 ⎢ ∫ 0 dx + ∫ 1 dx + ∫
0
1

(

) (

3

2

) (

)

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40.

1

3

∫0 ∫0

1 ⎡ –1
1
1 ⎤
+
8 x( x 2 + y 2 + 1) –2 dx dy = ∫ [–4( x 2 + y 2 + 1) –1 ]x =30 dy = 4∫ ⎢
⎥ dy
0 4 + y2 1 + y2
0
⎣⎢
⎦⎥
1

⎡⎛ 1
⎡ 1

⎛1⎞ π⎞ ⎤
⎛ y⎞
⎛1⎞
= 4 ⎢ – arctan ⎜ ⎟ + arctan( y ) ⎥ = 4 ⎢⎜ – arctan ⎜ ⎟ + ⎟ – 0 ⎥ = π – 2 arctan ⎜ ⎟ ≈ 2.2143
2
2
2
2
4
⎝ ⎠
⎝ ⎠
⎝2⎠

⎦0
⎠ ⎦
⎣⎝

41. 0 ≤ ∫

b b

[ f ( x) g ( y ) – f ( y ) g ( x)]
a ∫a

2

dx dy = ∫

b b

b

b

b

a a
b

a

a

a

a

[ f 2 ( x) g 2 ( y ) – 2 f ( x) g ( x) f ( y ) g ( y ) + f 2 ( y ) g 2 ( x)]dx dy
b

b

a

a

= ∫ f 2 ( x)dx ∫ g 2 ( y )dy – 2∫ f ( x) g ( x)dx ∫ f ( y ) g ( y )dy + ∫ f 2 ( y )dy ∫ g 2 ( x)dx
b
b
b
= 2 ∫ f 2 ( x)dx ∫ g 2 ( x)dx – 2 ⎡ ∫ f ( x) g ( x)dx ⎤
⎢⎣ a
⎥⎦
a
a

2

2

b
b
b
Therefore, ⎡ ∫ f ( x) g ( x)dx ⎤ ≤ ∫ f 2 ( x)dx ∫ g 2 ( x)dx.
⎢⎣ a
⎥⎦
a
a

42. Since f is increasing, [y – x][f(y) – f(x)] > 0. Therefore,
0<∫

b b

a a

[ y – x][ f ( y ) – f ( x)]dx dy = ∫

b b

a a

b

= (b – a) ∫ yf ( y )dy –
a

b2 – a 2
2

b

∫a

f ( x)dx –

yf ( y )dx dy – ∫

b2 – a 2
2

b b

a a

b

∫a

yf ( x)dx dy – ∫

b b

a a

xf ( y )dx dy + ∫

b b

a a

xf ( x) dx dy

b

f ( y )dy + (b – a ) ∫ xf ( x)dx
a

b
b
b
b
= 2(b – a) ∫ xf ( x)dx – (b 2 – a 2 ) ∫ f ( x)dx = (b – a) ⎡ 2 ∫ xf ( x)dx – (b + a ) ∫ f ( x)dx ⎤
a
a
a
⎣⎢ a
⎦⎥
b

b

b

a

a

a

Therefore, (b + a ) ∫ f ( x)dx < 2 ∫ xf ( x)dx. Now divide each side by the positive number 2∫ f ( x)dx to obtain the
desired result.
Interpretation:
If f is increasing on [a, b] and f ( x) ≥ 0 , then the x-coordinate of the centroid (of the region between the graph of f
and the x-axis for x in [a, b]) is to the right of the midpoint between a and b.
Another interpretation:
If f(x) is the density at x of a wire and the density is increasing as x increases for x in [a, b], then the center of mass
of the wire is to the right of the midpoint of [a, b].

13.3 Concepts Review

3y

⎡ x3

3
dy = ∫ (9 y 3 + 3 y 3 )dy
3. ∫ ⎢ + y 2 x ⎥
–1 3
–1
⎢⎣
⎥⎦ x =0
3

1. A rectangle containing S; 0

= [3 y 4 ]3–1 = 243 – 3 = 240

2. φ1 ( x) ≤ y ≤ φ2 ( x)
3.

b φ2 ( x )

∫a ∫φ ( x )

f ( x, y )dy dx

4.

1

4.

1 1– x

∫0 ∫0

2 x dy dx;

= –32.2

1
3

5.

Problem Set 13.3
1.

2.

1

∫0 [ x

2

1

y ]3yx=0 dx = ∫ 3x3 dx =
0

2 ⎡⎛ 1 ⎞ 2 ⎤
⎟y ⎥
1 ⎢⎜
⎝2⎠

x

1 ⎡ 3 ⎛1⎞ 4⎤
⎡ 2
⎛1⎞ 4⎤
∫–3 ⎢⎣ x y – ⎜⎝ 4 ⎟⎠ y ⎥⎦ dx = ∫–3 ⎢⎣ x – ⎜⎝ 4 ⎟⎠ x ⎥⎦ dx
y =0
1

3
4

3 ⎡⎛ 1 ⎞ 2
⎟x
1 ⎢⎜
⎝2⎠

2y

3⎛ 3 ⎞

dy = ∫ ⎜ ⎟ y 2 exp( y 3 )dy
exp( y ) ⎥
1 ⎝2⎠
⎦ x=– y
3

⎛1⎞
= ⎜ ⎟ (e 27 – e) ≈ 2.660 × 1011
⎝2⎠

x –1

2⎛ 1 ⎞
1
dx = ∫ ⎜ ⎟ ( x –1) 2 dx =
1 ⎝2⎠
6
⎦ y =0

Instructor’s Resource Manual

Section 13.3

807

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6.

5⎡3

x

53 π
y ⎞⎤
dx = ∫
dx
⎢ x tan ⎜ x ⎟ ⎥
1
x4
⎝ ⎠ ⎦ y =0

∫1

–1 ⎛

1

1

15.

∫–1 ∫x

16.

∫0 ∫4 x ( x + y)dy dx = 6

2

xy dy dx = 0

5

3π ln 5
⎡ 3π ln x ⎤
=⎢
=
≈ 3.7921

4
⎣ 4 ⎦1

7.

1

∫1/ 2 [ y cos(πx
=–

8.

2

)]2y x=0 dx = ∫

1

1/ 2

2 x cos(πx 2 )dx

2
≈ –0.2251

π / 4 ⎡⎛ 1 ⎞ 2 ⎤

∫0

2 cos θ

⎢⎜ 2 ⎟ r ⎥
⎣⎝ ⎠ ⎦ r =

dθ = ∫

π/ 4

0

2

(cos 2 θ – 1)dθ

1 4

(2 – π)
=
≈ –0.1427
8
9.

π/9

∫0

[tan θ ]θ3r=π / 4 dr = ∫

π/9

0

(tan 3r – 1)dr

π/9

⎡ ln cos 3r

= ⎢–
– r⎥
3

⎦0

17.

( )

⎛ − ln 1
π ⎞ ⎛ ln(1)

2
=⎜
– ⎟–⎜–
– 0⎟
⎜ 3
9⎟ ⎝
3

3ln 2 – π
=
≈ –0.1180
9

10.

x

2⎡

∫0

2

2
2
− x2 ⎤
− x2
dx = ⎡ −e − x ⎤
⎢⎣ ye ⎥⎦ − x dx = ∫0 2 xe
⎢⎣
⎥⎦ 0

= 1 − e −4

11.

π/ 2

∫0

1

y
[e x cos y ]sin
x = 0 dy = ∫

π/ 2

0

= e – 2 ≈ 0.7183

∫0 ∫x

(esin y cos y – cos y )dy

x
2

1

( x 2 + 2 y )dy dx = ∫ [ x 2 y + y 2 ]
0

x
dx
y = x2

1

= ∫ [( x5 / 2 + x) – ( x 4 + x 4 )]dx
0

1

12.

13.

2⎡ y

2
3 ⎤x

⎡ 2 x7 / 2 x 2 2 x5 ⎤
2 1 2
=⎢
+

⎥ = + –
7
2
5
7
2 5
⎥⎦ 0
⎣⎢
27
=
≈ 0.3857
70

2

5
2x
⎡1 ⎤
dx = ⎢ x6 ⎥ = 3.5
⎢ ⎥ dx = ∫1
3
x
3
⎣18 ⎦1
⎣⎢ ⎦⎥ 0

∫1

4– x 2

2⎡

∫0

⎛1⎞ 2⎤
⎢ xy + ⎜ 2 ⎟ y ⎥
⎝ ⎠ ⎦ y =0

dx

18.

2⎡
16
⎛1⎞ ⎤
= ∫ ⎢ x(4 – x 2 )1/ 2 + 2 – ⎜ ⎟ x 2 ⎥ dx =
0
3
2
⎝ ⎠ ⎦

14.

π/ 2

∫π / 6 [3r

2

θ
cosθ ]sin
r = 0 dθ = ∫

= [sin 3 θ ]ππ // 62 =

808

π/ 2

π/6

3sin 2 θ cosθ dθ

2 3 x − x2

∫0 ∫x
=∫

2 −x

0

19.

2 2

3

( x 2 − xy )dy dx

( x2 − 4 x + 4) dx = − 8

∫0 ∫x 2(1 + x

2

15

) dy dx = 4 tan –1 2 – ln 5 ≈ 2.8192

2 –1

7
= 0.875
8

Section 13.3

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

20.

24.

2 −2 x + 4 ⎡

20
⎛1⎞ ⎤
⎢ 2 x + ⎜ 4 ⎟ y ⎥ dy dx = 3
⎝ ⎠ ⎦

∫0 ∫0

Since S is symmetric with respect to the origin
and the integrand is an odd function in x, the
value of the integral is 0.
25.

21.

2 (1/ 2) 36–9 x 2

3 (–2 / 3) x + 2

∫0 ∫0

∫0 ∫0

(6 – 2 x – 3 y )dy dx = 6

⎛1⎞
⎜ ⎟ (9 x + 4 y )dy dx = 10
⎝6⎠

26.

22.

3

4 (–3 / 4) x + 3

∫0 ∫0

(12 – 3 x – 4 y )dy dx = 24

∫0 ∫0

9– x 2

(9 – x 2 – y 2 )dy dx

3⎡

y3 ⎤
= ∫ ⎢ (9 − x 2 ) y − ⎥
0
3 ⎥⎦
⎢⎣

23.

=∫

π/2

0

=∫

y =0

dx = ∫
π/2

0

⎜ + 27 cos 2t +
⎝ 4

3 2(9 − x 2 )3 / 2

0

18cos3 t 3cos t dt = ∫

π / 2 ⎛ 81

0

9− x2

3

dx

54 cos 4 t dt

27 cos 4t ⎞
⎟ dt
4

π/2

5 44–

∫0 ∫0

2

y

5
⎛ 44– y ⎞
dy dx = ⎛⎜ ∫ 1 dx ⎞⎟ ⎜ ∫
dy ⎟
⎝ 0
⎠⎝ 0 2

4

y2 ⎤
= 5 ⎢ 2 y – ⎥ = 5(8 – 4) = 20
4 ⎥⎦
⎣⎢
0

Instructor’s Resource Manual

⎡ 81t 27 sin 2t 27 sin 4t ⎤
=⎢
+
+
2
16 ⎥⎦ 0
⎣ 4
81π
=
≈ 31.8086
8
(At the third step, the substitution x = 3 sin t was
used. At the 5th step the identity
⎛1⎞
cos 2 A = ⎜ ⎟ (1 + cos 2 A) was used a few times.)
⎝2⎠

Section 13.3

809

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

27.

31.

1

∫0 ∫0
28.

y

(1 – y )dx dy =

9 2 y

∫0 ∫0

4
15

2 (3 / 2) 4– x 2

∫0 ∫0

≈ 9.4248

⎡ ⎛5⎞ ⎤
⎢5 – ⎜ 9 ⎟ y ⎥ dx dy = 72
⎣ ⎝ ⎠ ⎦

2
⎢4 – x

⎛4⎞ ⎤
– ⎜ ⎟ y 2 ⎥ dy dx = 3π
⎝9⎠ ⎦

32.

Making use of symmetry, the volume is
29.

2∫∫ (16 – x 2 )1/ 2 dA = 2∫
R1

4 x

0 0

(16 – x 2 )1/ 2 dy dx

4

= 2 ∫ [(16 – x 2 )1/ 2 y ]xy =0 dx
0

4

⎡ –2(16 – x 2 )3 / 2 ⎤
4
= 2 ∫ (16 – x 2 )1/ 2 x dx = ⎢

0
3
⎣⎢
⎦⎥ 0

= 0+
1 x

∫0 ∫0 tan x

2

1

dy dx = ∫ [ y tan x 2 ]xy =0 dx
0

⎡ ln cos x 2
= ∫ x tan x dx = ⎢ –
0

2
⎣⎢
1

2(64) 128
=
≈ 42.6667
3
3

2

1

⎥ = ⎛ – 1 ⎞ ln(cos1)

⎝ 2⎠
⎦⎥ 0

33.

1 1

∫0 ∫ y f ( x, y)dx dy

≈ 0.3078

30.

810

1 1– x x – y

∫0 ∫0

e

⎛1⎞
dy dx = ⎜ ⎟ (e + e –1 – 2) ≈ 0.5431
⎝2⎠

Section 13.3

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4

x

1

y

34.

∫0 ∫x / 2

35.

∫0 ∫ y4

36.

f ( x, y )dy dx

f ( x, y )dx dy

1/ 2 y1/ 3

∫1/ 8 ∫1/ 2

1

0

38.

∫–1 ∫x2 –1 f ( x, y)dy dx

39.

∫0 ∫0

2 2– y

xy 2 dx dy + ∫

4 y –2

2 0

xy 2 dx dy =

256
15

≈ 17.0667

f ( x, y )dx dy + ∫

1

y1/ 3

1/ 2 y

f ( x, y )dx dy

40.

1

– x+2

∫–2 ∫x2
=–

xy dy dx – ∫

1/ 2

1/ 2

–1/ 2 x 2

xy dy dx

45
= –5.625
8

41. The integral over S of x 4 y is 0 since this is an
odd function of y. Therefore,

∫∫S ( x
37.

0

1

1 1

∫–1 ∫– x f ( x, y)dy dx + ∫0 ∫x f ( x, y)dy dx

Instructor’s Resource Manual

2

+ x 4 y )dA = ∫∫ x 2 dA
S

= 4 ⎛⎜ ∫∫ x 2 dA + ∫∫ x 2 dA ⎞⎟
S2
⎝ S1

2
⎛ 1 4– x 2

2
4– x 2 2
= 4⎜ ∫ ∫
x dy dx + ∫ ∫
x dy dx ⎟
2
⎜ 0 1– x

1 0

2 2
1 2
2
2

= 4 ⎜ ∫ x 4 – x dx – ∫ x 1 – x dx ⎟
0
⎝ 0

π/2
π/2
2
2

= 4 ⎜ 16 ∫ sin θ cos θ dθ – ∫ sin 2 φ cos 2 φ dφ ⎞⎟
0
⎝ 0

(using x = 2 sin θ in 1st integral; x = sin φ in 2nd)
π/ 2
15π
= 60∫ sin 2 θ cos 2 θ dθ =
0
4
(See work in Problem 42.)
≈ 11.7810

Section 13.3

811

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Then dx = 2 cos θ dθ
π
x = 2 ⇒θ =
2
x=0 ⇒ θ=0

42.

=

(2 – 2 )
2

(

=8 2– 2

z = f ( x, y ) = sin( xy 2 ) is symmetric with respect
to the x-axis, as is the annulus. Therefore, the
integral equals 0.

43.

(

*=8 2–

Therefore,

π/2

∫0

(2sin θ ) 2 (2 cos θ )2 cosθ dθ

) ∫0π / 2 sin 2 θ cos2 θ dθ
⎛ π ⎞ π(2 – 2 )
2)⎜ ⎟ =
2
⎝ 16 ⎠

(

)

⎡π 2 – 2 ⎤
2

⎥ = 2π 2 – 2 ≈ 3.6806
x
dA
=
4
∫∫S

2

* =∫

π/2

0

=∫

(

)

sin 2 θ cos 2 θ dθ

π/ 2 ⎡1

0

⎤ ⎡1

⎢ 2 (1 – cos 2θ ) ⎥ ⎢ 2 (1 + cos 2θ ) ⎥ dθ

⎦⎣

1 π/2
(1 – cos 2 2θ )dθ
4 ∫0
1 π 1 π/2 1
=

(1 + cos 4θ )dθ
4 2 4 ∫0 2
=

y2

2

∫0 ∫0

2

2

sin( y 3 ) dx dy = ∫ [ x sin( y 3 )]xy=0 dy
0

2

2 2
y sin( y 3 )dy
0

=∫

⎡ cos( y 3 ) ⎤
= ⎢–

3 ⎦⎥
⎣⎢
0

1 – cos8
=
≈ 0.3818
3
44. Let S ' be the part of S in the first quadrant.

π/2

=

π 1 π 1 ⎡ sin 4θ ⎤
+

8 8 2 8 ⎢⎣ 4 ⎥⎦ 0

=

π π
π
– +0 =
8 16
16

45. We first slice the river into eleven 100’ sections
parallel to the bridge. We will assume that the
cross-section of the river is roughly the shape of
an isosceles triangle and that the cross-sectional
area is uniform across a slice. We can then
approximate the volume of the water by
11
11
1
V ≈ ∑ Ak ( yk )Δy = ∑ ( wk )(d k )100
k =1
k =1 2
11

= 50 ∑ ( wk )(d k )
k =1

∫∫S ′ x

2

dA = ∫

2

0

4– x 2
4– x 2 / 2

x 2 dy dx

2
4 – x2 ⎞
⎟ dx
= ∫ x2 ⎜ 4 – x2 –
0

2 ⎟⎠

2
1 ⎞

= ∫ x 2 4 – x 2 ⎜1 –
⎟ dx
0
2⎠

(2 – 2 )
=
2

2 2
x
0

2

where wk is the width across the river at the left
side of the kth slice, and d k is the center depth of
the river at the left side of the kth slice. This
gives
V ≈ 50[300 ⋅ 40 + 300 ⋅ 39 + 300 ⋅ 35 + 300 ⋅ 31
+290 ⋅ 28 + 275 ⋅ 26 + 250 ⋅ 25 + 225 ⋅ 24
+205 ⋅ 23 + 200 ⋅ 21 + 175 ⋅19]
= 4,133, 000 ft 3

4 – x dx

⎛ π π⎞
Let x = 2 sin θ, θ in ⎜ – , ⎟ .
⎝ 2 2⎠

812

Section 13.3

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

46. Since f is continuous on the closed and bounded
set R, it achieves a minimum m and a maximum
M on R. Suppose ( x1 , y1 ) and ( x2 , y2 ) are such

2.

that f ( x1 , y1 ) = m and f ( x2 , y2 ) = M . Then,

π / 2 ⎡⎛ 1 ⎞ 2 ⎤
⎢⎜ 2 ⎟ r ⎥
0
⎝ ⎠

R

R

R

3.

mA ( R ) ≤ ∫∫ f ( x, y ) dA ≤ MA ( R )
1
m≤
f ( x, y ) dA ≤ M
A ( R ) ∫∫

h ( t ) = f ( x ( t ) , y ( t ) ) . Since f is continuous, so is

4.

R

h ( t0 ) = f ( x ( t0 ) , y ( t0 ) ) = f ( a, b ) , where
a = x ( t0 ) and b = y ( t0 ) . Thus,

⎦ r =0

2

0

3

θ

3

θ ) sin θ

π (1 – cos

3

2

π
θ⎤
θ
⎡1 2
⎢ 2 r cos 4 ⎥ dθ = ∫0 2 cos 4 dθ

⎦0
π

θ⎤

= ⎢8sin ⎥ = 4 2
4 ⎦0

6.

13.4 Concepts Review

1. a ≤ r ≤ b;α ≤ θ ≤ β

∫0

2. r dr dθ

=

π 2 3

0

∫0

∫∫ f ( x, y ) dA = f ( a, b ) ⋅ A ( R ) .

∫0 ∫0 r

π sin

1–cos θ
π ⎡⎛ 1 ⎞ 2

⎟ r sin θ ⎥
0 ⎢⎜
⎣⎝ 2 ⎠
⎦ r =0

π

R

3.

dθ = ∫

5.

R

or,

∫0 ⎢⎢ 3 ⎥⎥

π⎛ 1 ⎞
= ∫ ⎜ ⎟ (1 – cos θ ) 2 sin θ dθ
0 ⎝2⎠
4
=
3

1
f ( x, y ) dA . But
A ( R ) ∫∫

1
f ( x, y ) dA
A ( R ) ∫∫

3 ⎤ sin θ

⎛1 1⎞ ⎛ 1 1⎞ 4
=⎜ – ⎟–⎜– + ⎟ =
⎝3 9⎠ ⎝ 3 9⎠ 9

x = x ( t ) , y = y ( t ) , c ≤ t ≤ d . Let

f ( a, b ) =

2
⎜ ⎟ sin θ dθ
⎝2⎠

π

Let C be a continuous curve in the plane from
( x1 , y1 ) to ( x2 , y2 ) that is parameterized by

h ( t0 ) =

⎦ r =0

π/ 2⎛ 1 ⎞

0

⎡ – cos θ cos3 θ ⎤
=⎢
+

9 ⎥⎦
⎢⎣ 3
0

R

h. By the Intermediate Value Theorem, there
exists a t0 in ( c, d ) such that

π⎡r

=∫

R

dθ = ∫

π
≈ 0.3927
8

=

m ≤ f ( x, y ) ≤ M

∫∫ m dA ≤ ∫∫ f ( x, y ) dA ≤ ∫∫ M dA

sin θ

dr dθ

θ

2π 1 2
⎡1 2⎤
⎡1 3⎤
⎢ 2 r ⎥ dθ = ∫0 2 θ dθ = ⎢ 6 θ ⎥

⎦0

⎦0

4π 3
3

7.

4. 4π

Problem Set 13.4
1.

π / 2 ⎡⎛ 1 ⎞ 3

⎢⎜ 3 ⎟ r sin θ ⎥
0
⎝ ⎠

=∫

cos θ

⎦ r =0

π/ 2⎛ 1 ⎞

0

3
⎜ ⎟ cos θ sin θ dθ
⎝3⎠

1
=
≈ 0.0833
12

Instructor’s Resource Manual

2∫

π / 3 4 cos θ

0

∫2

⎡ 2π

r dr dθ = 2 ⎢ + 3 ⎥ ≈ 7.6529
⎣ 3

Section 13.4

813

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8.
2∫

3π / 2 2–4 sin θ

5π / 6 0

= 2∫

3π / 2

5π / 6

r dr dθ = 2 ∫

3π / 2 ⎡ r

5π / 6

2–4sin θ

2⎤
⎢ ⎥
⎣⎢ 2 ⎥⎦ 0

(6 – 8sin θ – 4 cos 2θ )dθ

= 2[6θ + 8cos θ – 2sin 2θ ]35ππ // 26

(

)

= 2 4π + 3 3 ≈ 35.525
π / 6 4sin θ

∫0 ∫0
=∫

π/6

0

r dr dθ = ∫

π/6

0

8sin 2 θ dθ = ∫

π/6

0

= [4θ – 2sin 2θ ]0π / 6 =

4sin θ

⎡ r2 ⎤
⎢ ⎥
⎣⎢ 2 ⎦⎥ 0

12.

4(1 – cos 2θ )dθ

– 3 ≈ 0.3623
3

9.
4∫

(1/ 2) cos −1 (4 / 9) 3 cos 2θ

∫2

0

r dr dθ

⎛4⎞
= 65 − 4 cos −1 ⎜ ⎟
⎝9⎠
≈ 3.6213

π / 2 a sin 2θ

∫0 ∫0

π /4⎡1 2⎤

2

π /4

13.

∫0

14.

⎡1 2⎤
∫0 ⎢⎣ 2 r ⎥⎦1 dθ = ∫0 4 dθ = 8π

a2π
r dr dθ =
8

⎢ 2 r ⎥ dθ = ∫0

⎦0

2 dθ =

π
2

10.

2 π 6–6 sin θ

∫0 ∫0

3

r dr dθ = 54π ≈ 169.6460

11.

814

Section 13.4

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

15.

θ
π /2⎡1 2⎤

∫0

π /2 1 2

⎢ 2 r ⎥ dθ = ∫0

⎦0

2

θ dθ

π /2

⎡1 ⎤
= ⎢ θ3⎥
⎣ 6 ⎦0

=

18.

π

3

48

θ2

3π / 2 ⎡ 1 2 ⎤
⎢2 r ⎥
0

dθ = ∫

⎡1

= ⎢ θ5⎥
⎣ 10 ⎦ 0

=

3π / 2 1

0

0
3π / 2

2

θ 4 dθ

243π 5
320

19.

16.

π /2⎡1 2⎤

∫0

cos θ

⎢2 r ⎥

⎦0

=∫

π /2 1 ⎡1

0

2 ⎢⎣ 2

+

dθ = ∫

π /2 1

0

2

cos 2 θ dθ

1

cos 2θ ⎥ dθ
2

π /2

1
⎡1

= ⎢ θ + cos 2θ ⎥
8
⎣4
⎦0

=

π
8

2∫

π 2 r2

0 0

e rdr dθ = π(e 4 − 1) ≈ 168.3836

20.

17.

π ⎡1 2⎤

sin θ

∫0 ⎢⎣ 2 r ⎥⎦ 0
=∫

dθ = ∫

π 1

0

2

sin 2 θ dθ

π 1 ⎡1

0

1

− sin 2θ ⎥ dθ

2 ⎣2 2

π/ 4 2

∫0 ∫0 (4 – r

π

π
1
⎡1

= ⎢ θ + cos 2θ ⎥ =
8
⎣4
⎦0 4

=∫

Instructor’s Resource Manual

⎣⎢

r dr dθ

2 3 / 2 ⎤2

)

–3

⎥ dθ
⎦⎥ 0
π/4

⎡ 8θ ⎤
⎜ ⎟ dθ = ⎢ ⎥
3
⎝ ⎠
⎣ 3 ⎦0

π/ 4⎛ 8 ⎞

0

21.

)

π / 4 ⎡ (4 – r

0

=∫

2 1/ 2

π/ 4 2

∫0 ∫0 (4 + r

=

≈ 2.0944
3

⎛π⎞
) r dr dθ = ⎜ ⎟ ln 2 ≈ 0.2722
⎝8⎠

2 –1

Section 13.4

815

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

22.

π/ 2 2

26.

7

∫0 ∫1 r sin θ r dr dθ = 3

π / 4 2 cos θ –1

∫0 ∫secθ
=∫

23.

π/4

0

r r dr dθ = ∫

π/4

0

2 cos θ
[r ]sec
θ dθ

(2 cos θ – secθ )dθ
π/ 4

= ⎡⎣ 2sin θ – ln secθ + tan θ ⎤⎦
0
= ⎡ 2 – ln

= 2 – ln

(

(

)

2 + 1 ⎤ – [0 – ln(1 + 0)]

)

2 + 1 ≈ 0.5328

27.
π/ 2 1

∫0 ∫0 (4 – r
=∫

π/2

=∫

π/2

0

)

r dr dθ

[–(4 – r 2 )1/ 2 ]10 dθ

0

24.

2 –1/ 2

(–

⎛π⎞
3 + 2 dθ = – 3 + 2 ⎜ ⎟ ≈ 0.4209
⎝2⎠

)

π/ 2 1

∫0 ∫0 [sin(r
≈ 0.3610

2

(

)

⎛π⎞
)]r dr dθ = ⎜ ⎟ (1 – cos1)
⎝4⎠

∫∫R ( x
=

2

π/2 3 2
r r dr dθ
0
0

+ y 2 )dA = ∫

81π
≈ 31.8086
8

28.

25.

π / 2 csc θ 2

∫π / 4 ∫0

r cos 2 θ r dr dθ =

1
≈ 0.0833
12

4∫∫ (18 – 2 x 2 – 2 y 2 )1/ 2 dA
R

= 4∫

π/2 2

0

∫0 (18 – 2r

2 1/ 2

)

r dr dθ

⎛π⎞
= ⎜ ⎟ (183 / 2 – 103 / 2 ) ≈ 46.8566
⎝3⎠

816

Section 13.4

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

29.

W = ∫∫ sin x 2 + y 2 dA + ∫∫ – sin x 2 + y 2 dA
S1

=∫

S2

2π ⎡ π

0

⎢⎣ ∫0

(sin r )r dr − ∫

π

(sin r )r dr ⎤ dθ
⎥⎦

= 2π[(π) – (–3π)] = 8π2 ≈ 78.9568

0

∫−5 ∫

⎡ y3 ⎤
( y )dy dx = ∫ ⎢ ⎥
−5 3
3x
⎢⎣ ⎥⎦

−x

−x

0

2

(

dx
3x

)

0

⎡ –1 – 3 3 x 4 ⎤
–1 – 3 3 3

=∫
x dx = ⎢
–5

3
12

⎦ –5
0

31. This can be done by the methods of this section,
but an easier way to do it is to realize that the
intersection is the union of two congruent
segments (of one base) of the spheres, so (see
Problem 20, Section 5.2, with d = h and a = r) the
(3a – d )
⎡⎛ 1 ⎞

volume is 2 ⎢⎜ ⎟ πd 2 (3a – d ) ⎥ = 2πd 2
.
3
3

32. 100 = ∫

∫0

10

ke – r /10 r dr dθ = 2π ∫ ke – r /10 r dr
0

Let u = r and dv = e – r /10 dr .

(1 + 3 3 ) 625 ≈ 322.7163
=

Then du = dr and v = –10e – r /10 .
10
10

= 2πk ⎜ ⎡ –10re – r /10 ⎤ + ∫ 10e – r /10 dr ⎟

0
0

10

= 2πk ⎜ –100e –1 – ⎡100e – r /10 ⎤ ⎟

⎦0 ⎠

12

30. a.

2 π 10

0

The solid bounded by the xy-plane and
z = sin x 2 + y 2 for x 2 + y 2 ≤ 4π2 is the
solid of revolution obtained by revolving
about the z-axis the region in the xz-plane
that is bounded by the x-axis and the graph
of z = sin x for 0 ≤ x ≤ 2π .

= 2πk (–100e –1 –100e –1 + 100)
= 200πk (1 – 2e –1 ), so k =

e
≈ 0.6023.
2π(e – 2)

33.

Regions A and B are congruent but region B
is farther from the origin, so it generates a
larger solid than region A generates.
Therefore, the integral is negative.
b.

V =∫

2π 2π

0

∫0

(sin r )r dr dθ = 2π ∫

Now use integration by parts.
= 2π(–2π) = –4π2 ≈ –39.4784

0

Volume = 4 ∫

π / 2 a sin θ

0

(sin r )r dr

∫0

a 2 – r 2 r dr dθ

1⎞ 3
3
3 ⎤
⎢⎜ – 3 ⎟ (a cos θ – a ) ⎥ dθ

⎣⎝

⎛ 4 ⎞ ⎡2 π⎤ ⎛ 2 ⎞
= ⎜ – ⎟ a3 ⎢ – ⎥ = ⎜ ⎟ a3 (3π – 4)
⎝ 3 ⎠ ⎣3 2⎦ ⎝ 9 ⎠

=∫

π / 2 ⎡⎛

0

c.

Instructor’s Resource Manual

Section 13.4

817

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

34. Normal vector to plane is 0, -sin a, cos a . Therefore, an equation of the plane is

(–sin α)y + (cos α)z = 0, or z = (tan α)y, or z = (tan α)(r sin θ).

Volume = 2∫

π/2 a

0

∫0

(tan α )r sin θ r dr dθ = 2(tan α ) ∫

π/2

0

⎡ a3 ⎤ ⎛ 2 ⎞
a
sin θ dθ ∫ r 2 dr = 2(tan α )[1] ⎢ ⎥ = ⎜ ⎟ a3 tan α
0
⎣⎢ 3 ⎦⎥ ⎝ 3 ⎠

35. Choose a coordinate system so the center of the sphere is the origin and the axis of the part removed is the z-axis.
Volume (Ring) = Volume (Sphere of radius a) – Volume (Part removed)

a2 – b2
a 2 – b2
4
4
= πa3 – 2∫ ∫
a 2 – r 2 r dr dθ = πa3 – 2(2π) ∫
(a 2 – r 2 )1/ 2 r dr
0
0
0
3
3
4
⎡1

= πa3 + 4π ⎢ (a 2 − r 2 )3 / 2 ⎥
3
⎣3
⎦0

36.

EF

2

= a 2 – b2

a 2 −b 2

=

4 3
1
4
πa + 4π (b3 – a3 ) = πb3
3
3
3

CD = a 2 – (h – b) 2

2

AB = b 2 – (h – b) 2

Area of left cross-sectional region = π[a 2 – (h – b) 2 ] – π[a 2 – b 2 ]
= π[b 2 – (h – b) 2 ] = area of right cross-sectional region
⎛4⎞
⎛1⎞
⎛1⎞
Volume = ⎜ ⎟ πb3 – ⎜ ⎟ π(2b – h) 2 [3b – (2b – h)] = ⎜ ⎟ πh 2 (3b – h)
⎝3⎠
⎝ 3⎠
⎝3⎠
h
1
2
Alternative: V = ∫ π ⎡⎢b 2 − ( t − b ) ⎤⎥ dt = π h 2 ( 3b − h )
0 ⎣

3

37.

b
π/ 2⎛
π/ 2⎛ 1 ⎞
⎡⎛ 1 ⎞
π
⎛ 1 ⎞⎤ ⎞
lim (1 + r 2 ) –2 r dr ⎤ dθ = ∫ ⎜ lim ⎢⎜ – ⎟ (1 + b 2 ) –1 – ⎜ – ⎟ ⎥ ⎟ dθ = ∫ ⎜ ⎟ dθ = ≈ 0.7854
0
0
→∞
b
⎣⎢b →∞ ∫0
⎦⎥
2
4
2
2

⎠⎦ ⎠
⎝ ⎠
⎣⎝

π/ 2 ⎡

∫0

38. A =

1 2
1
r2 (θ 2 – θ1 ) – r12 (θ 2 – θ1 )
2
2

1
(θ1 – θ 2 )(r22 – r12 )
2
1
= (θ 2 – θ1 )(r2 – r1 )(r2 + r1 )
2
r1 + r2
(r2 – r1 )(θ 2 – θ1 )
=
2
=

818

Section 13.4

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

39. Using the substitution u =
dx

du =

=

1

e−( x − μ )

∞ −u

2

σ 2

2.

we get

. Our integral then becomes

σ 2

∫−∞ σ

x−μ

e
π ∫0

2

2

/ 2σ 2

1

−∞

π

dx = ∫

e−u du
2

du

Using the result from Example 4, we see that
∞ −u 2
e du
0

1

∫−∞ σ
=

=

2

π

π

e

2

π
2

m=∫

–2 ∫0

. Thus we have

− ( x − μ )2 / 2σ 2

4– x 2

2

y dy dx =

16
3

M y = 0 (symmetry)

dx

Mx = ∫

2

–2 ∫0

4– x 2

yy dx dy = 2π

⎛ 3π ⎞
( x , y ) = ⎜ 0,

⎝ 8 ⎠

= 1.

3.

13.5 Concepts Review
1.

∫∫ x

2 4

y dA

S

2.

x 2 y 5 dA
∫∫ m
S

3.

∫∫ x

4 4

y dA

m=∫

π sin x

0 0

S

4. greater

=∫

π sin 2

0

Problem Set 13.5
1.

x

π⎡ y

y dy dx = ∫ ⎢ ⎥
0
⎢⎣ 2 ⎥⎦ 0

dx = ∫

π 1 – cos 2 x

0

2

2 ⎤ sin x

4

dx

dx

π

π
⎡ x sin 2 x ⎤
=⎢ –
=

8 ⎦0 4
⎣4

Mx = ∫

π sin x

0 0

=∫

π sin

0

3

x

3

π⎡ y

3 ⎤ sin x

yy dy dx = ∫ ⎢ ⎥
0
⎢⎣ 3 ⎥⎦ 0

dx =

dx

1 π
(1 – cos 2 x) sin x dx
3 ∫0
π

1⎡
cos3 x ⎤
4
= ⎢ – cos x +
⎥ =
3 ⎣⎢
3 ⎦⎥
9
0

m=∫

3 4

( y + 1)dx dy = 30
0 ∫0

My = ∫

3 4

x( y + 1)dx dy = 60

Mx = ∫

y ( y + 1)dx dy = 54

0 0
3 4

0 0

( x , y ) = (2, 1.8)

Instructor’s Resource Manual

4
9
π
4

y=

Mx
=
m

x=

π
(by symmetry)
2

=

16
≈ 0.5659;

Thus , M y = x ⋅ m =

π π

4 2

=

π2
8

Section 13.5

819

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4.

m=∫

1 x

0 0

My = ∫

x dy dx + ∫

2 1/ x

1

∫0

1 x 2

0 0

x dy dx + ∫

x dy dx =

4
3

2 1/ x 2
x dy dx
0

1

=

7
4

⎛1⎞
xy dy dx = ⎜ ⎟ (1 + 4 ln 2)
⎝8⎠
⎛ 21 ⎛ 3 ⎞

( x , y ) = ⎜ , ⎜ ⎟ (1 + 4 ln 2) ⎟ ≈ (1.3125, 0.3537)
16
32

Mx = ∫

1 x

0 0

xy dy dx + ∫

2 1/ x

1

∫0

5.

⎛1⎞
y 2 dy dx = ⎜ ⎟ (1 – e –3 )
⎝9⎠
−x
1 e
⎛1⎞
M x = ∫ ∫ y 3 dy dx = ⎜ ⎟ (1 − e −4 ) ≈ 0.0614
0 0
⎝ 16 ⎠

m=∫

1 e– x

0 0

⎛ 1 ⎞
xy 2 dy dx = ⎜ ⎟ (1 – 4e –3 ) ≈ 0.0297
⎝ 27 ⎠
⎛⎛ 1 ⎞

⎛9⎞
( x , y ) = ⎜ ⎜ ⎟ (e3 – 4)(e3 – 1) –1 , ⎜ ⎟ e –1 (e4 – 1)(e3 – 1) –1 ⎟ ≈ (0.2809, 0.5811)
3
16
⎝ ⎠
⎝⎝ ⎠

My = ∫

1 e– x

0 0

6.

820

Section 13.5

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

m=∫

1 ex

0 0

1⎡

y2 ⎤
(2 – x + y )dy dx = ∫ ⎢ (2 – x) y + ⎥
0
2 ⎥⎦
⎢⎣

ex

1⎛
e2 x ⎞
dx = ∫ ⎜ 2e x – xe x +
⎟ dx
0⎜
2 ⎟⎠

y =0

1

e2 x ⎤
e 2 + 8e – 13
= ⎢ 2e x – ( xe x – e x ) +
⎥ =
4 ⎥⎦
4
⎢⎣
0
Mx = ∫

1 ex

0 0

1⎡ 2
⎢y
0

(2 – x + y ) y dy dx = ∫

⎢⎣

ex

1⎡
xy 2 y 3 ⎤
xe 2 x e3 x ⎤
+ ⎥
+

dx = ∫ ⎢ e 2 x –
⎥ dx
0
2
3 ⎥⎦
2
3 ⎥⎦
⎢⎣
y =0

1

⎡ e2 x ⎛ xe2 x e2 x ⎞ e3 x ⎤
⎛ e 2 e 2 e 2 e3 ⎞ ⎛ 1
1 1 ⎞ 8e3 + 27e 2 – 53
= ⎜ – + + ⎟–⎜ –0+ + ⎟ =
=⎢
+
–⎜

⎜ 2
72
4
8 9 ⎟⎠ ⎝ 2
8 9⎠
8 ⎟⎠ 9 ⎥⎦
⎢⎣ 2 ⎜⎝ 4
0 ⎝
My = ∫

1 ex

0 ∫0

1⎡

xy 2 ⎤
(2 – x + y ) x dy dx = ∫ ⎢ 2 xy – x y +

0
2 ⎦⎥
⎣⎢
2

ex

1⎛
xe 2 x ⎞
dx = ∫ ⎜ 2 xe x – x 2 e x +
⎟ dx
0⎜
2 ⎟⎠

y =0

1

⎛ xe2 x e 2 x ⎞ ⎤
e2 – 8e + 33

=
= ⎢(2 xe x – 2e x ) – ( x 2 e x – 2 xe x + 2e x ) + ⎜

⎜ 4
8 ⎟⎠ ⎦⎥
8

⎣⎢
0
x=

My
m

=

e2 – 8e + 33
2(e2 + 8e –13)

≈ 0.5777; y =

7.

M x 8e3 + 27e2 – 53
=
≈ 1.0577
m 18(e2 + 8e –13)

8.

m=∫

π 2sin θ

0 0

Mx = ∫

r r dr dθ =

π 2 sin θ

0 0

m = 2∫

32
9

(r sin θ )r r dr dθ =

M y = 0 (symmetry)
( x , y ) = (0, 1.2)

Instructor’s Resource Manual

π 1+ cos θ

0 0

64
15

M y = 2∫

r r dr dθ =

π 1+ cos θ

0 0

3

(r cos θ )r r dr dθ =

M x = 0 (symmetry)
( x , y ) = (1.05, 0)

4

Section 13.5

821

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

9. m = ∫

π 21

0 1

r

r drdθ = ∫

π 2

0 1

drdθ = π

11.

π 21

π 3
cos θ dθ = 3
r cos θ r drdθ = ∫
0 2
r
M y = 0 by symmetry

Mx = ∫

0 1

( x , y ) = ⎛⎜ 0,

3⎞
π ⎟⎠
3 9

Ix = ∫

0 y2

y 2 ( x + y )dx dy

3 ⎛ 81 y 2
y6
7533
=∫ ⎜
+ 9 y3 –
− y 5 ⎟ dy =
≈ 269

0⎜ 2
2
28

Iy = ∫

9

9⎛
x3 ⎞
x ( x + y )dy dx = ∫ ⎜ x 7 / 2 + ⎟ dx
0⎜
2 ⎟⎠

x 2

0 0

10. m = ∫

2+ 2 cos θ

∫0

0

r r dr dθ

2 + 2 cos θ

⎡1 3 ⎤
r

0 ⎢⎣ 3 ⎥⎦
0
2π 1
⎡8 + 24 cos θ + 24 cos 2 θ + 8cos3 θ ⎤ dθ
=∫

0 3⎣
1
40π
= [ 0 + 24π + 16π ] =
3
3
M x = 0 by symmetry
=∫

My = ∫

0

2 + 2 cos θ

∫0

12.

r ( r cos θ ) r dr dθ
2+ 2 cos θ

⎡1 4

r cos θ ⎥
0 ⎢⎣ 4
⎦0
M y = 28π
=∫

41553
≈ 5194
8
305937
Iz = Ix + I y =
≈ 5463
56

=

2048
≈ 227.56
9
4
y 2
512
Iy = ∫ ∫
x y dx dy =
≈ 24.38
0 – y
21
15872
Iz = Ix + I y =
≈ 251.94
63
Ix = ∫

4

0 ∫–

28π
⎞ ⎛ 21 ⎞
,0⎟ = ⎜ ,0⎟
⎝ 40π / 3 ⎠ ⎝ 10 ⎠

( x , y ) = ⎛⎜

y

y

y 3 dx dy =

13.

⎛ 5⎞
dx dy = ⎜ ⎟ a5
⎝ 12 ⎠
⎛ 5⎞
⎛5⎞
I y = ⎜ ⎟ a5 ; I z = ⎜ ⎟ a5
12
⎝ ⎠
⎝6⎠
Ix = ∫

a a

( x + y) y
0 ∫0

822

Section 13.5

2

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

16. The density is proportional to the distance from
the x-axis, δ ( x, y ) = ky .

14.

(

1

)

1⎡ k
1k
k

1 − x2 =
m = ∫ ⎢ y 2 ⎥ dx = ∫
0⎣2
0
2
3
⎦x

Mx = ∫

1 x

0 0

=

Ix = ∫

a a– y

0 0

=

( x 2 + y 2 ) y 2 dx dy

(

)

1

7a6
1 a 3 2
2
2
5
(

3
6
4
)
a
y
a
y
+
ay

y
dy
=
180
3 ∫0

Iy =

k 1
3 ∫0

1

1⎡ k

ky 2 dy dx = ∫ ⎢ y 3 ⎥ dx
0⎣3
⎦x
k
1 − x3 dx =
4

7a 6
7a6
(Same result for a < 0)
; Iz =
180
90

15. The density is constant , δ ( x, y ) = k .

1 ⎡ kxy 2 ⎤
kxy dy dx = ∫ ⎢
⎥ dx

0 x
0⎢ 2 ⎥

⎦x
k 1
k
= ∫ x − x3 dx =
2 0
8
k /8 k / 4 ⎞ ⎛ 3 3 ⎞
( x , y ) = ⎛⎜ ,
⎟=⎜ , ⎟
⎝ k /3 k /3 ⎠ ⎝ 8 4 ⎠

My = ∫

1 1

(

)

2

2
⎡k ⎤
m = ∫ kx dx = ⎢ x 2 ⎥ = 2k
0
⎣ 2 ⎦0
2 x

Mx = ∫

0 0

=∫

x

2⎡k

ky dy dx = ∫ ⎢ y 2 ⎥ dx
0 ⎣2
⎦0
2

4k
⎡k ⎤
x 2 dx = ⎢ x3 ⎥ =
2
3
⎣ 6 ⎦0

2k

0

My = ∫

2 x

0 0

kx dy dx = ∫ [ kxy ]0 dx
2

x

0

2

8k
⎡k ⎤
= ∫ kx dx = ⎢ x3 ⎥ =
0
3
⎣ 3 ⎦0
2

17. The density is proportional to the squared

(

2

8k / 3 4 k / 3 ⎞ ⎛ 4 2 ⎞
,
,
=
( x , y ) = ⎛⎜
2k ⎟⎠ ⎜⎝ 3 3 ⎟⎠
⎝ 2k

)

distance from the origin, δ ( x, y ) = k x 2 + y 2 .
m=∫

9− x 2

3

−3 ∫0

(

)

k x 2 + y 2 dy dx
9− x 2

1 ⎤

= ∫ k ⎢ x2 y + y3 ⎥
−3 ⎣
3 ⎦0
3

dx

3 ⎡
1 ⎤
= ∫ k ⎢ 246 − 72 x 2 + 8 x 4 − x6 ⎥ dx
−3 ⎣
3 ⎦
3

8
1
25596k

= k ⎢ 246 x − 24 x3 + x5 − x7 ⎥ =
5
21 ⎦ −3
35

Mx = ∫

3

9− x 2

−3 0

(

)

ky x 2 + y 2 dy dx
9− x2

1 ⎤
⎡1
= ∫ k ⎢ x2 y2 + y 4 ⎥
−3 ⎣ 2
4 ⎦0
3

dx

3 ⎡ 6561 1377 x 2
225 4 17 x 6 x8 ⎤
x −
=∫ k⎢

+
+ ⎥ dx
−3
2
2
2
4 ⎥⎦
⎢⎣ 4
29160k
=
7

Instructor’s Resource Manual

Section 13.5

823

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

M y = 0 by symmetry

( x , y ) = ⎛⎜ 0,

29160k / 7 ⎞ ⎛ 450 ⎞
= 0,
25596k / 35 ⎟⎠ ⎜⎝ 79 ⎟⎠

20. The density is constant, δ ( r , θ ) = k .

18. The density is constant, δ ( x, y ) = k .
m=∫

π /2

π /2

k cos x dx = [ k sin x ]−π / 2 = 2k

−π / 2
π / 2 cos x

Mx = ∫

−π / 2 0

=∫

π /2 ⎡k

−π / 2 ⎢⎣ 2

ky dy dx

∫0

0

=∫

dx

=

kπ / 4 ⎞ ⎛ π ⎞
= 0,
2k ⎟⎠ ⎜⎝ 8 ⎟⎠

(

⎡k

π /2 θ

π /2

∫0 kr

2

=

kπ 3
48

sin θ dr dθ
θ

π /2 k 3

⎢ 3 r sin θ ⎥ dθ = ∫0 3 θ sin θ dθ

⎦0

k π 2 −8

)

4

My = ∫

π /2 θ

0

=∫

∫0 kr

2

cos θ dr dθ
θ

π /2⎡k 3

0

=

⎢ 2 r ⎥ dθ

⎦0

0

π /2⎡k 3

0

θ
π /2⎡k 2⎤

θ 2 dθ = ⎢ θ 3 ⎥
2
⎣ 6 ⎦0

0

=∫

k r dr dθ = ∫

π /2 k

0

k π /2

= ∫
cos 2 x dx =
2 −π / 2
4
M y = 0 by symmetry

( x , y ) = ⎛⎜ 0,

π /2 θ

Mx = ∫

cos x

y2 ⎥
⎦0

m=∫

(

π /2 k 3

⎢ 3 r cos θ ⎥ dθ = ∫0 3 θ cos θ dθ

⎦0

k π 3 − 24π + 48
24

(

)

) (

⎛ 2 π 3 − 24π + 48 12 π 2 − 8
=
x
,
y
,
( ) ⎜⎜
π3
π3

) ⎞⎟

19. The density is proportional to the distance from
the origin, δ ( r , θ ) = k ⋅ r .
3

π ⎡k

kr 2 dr dθ = ∫ ⎢ r 3 ⎥ dθ

0 1
0 ⎣3
⎦1
π 26
26kπ
k dθ =
=∫
0 3
3

m=∫

π 3

Mx = ∫

π 3

0 1

kr 2 r sin θ dr dθ
3

π ⎡k
π

= ∫ ⎢ r 4 sin θ ⎥ dθ = ∫ 20k sin θ dθ
0 ⎣4
0
⎦1
π

= [ −20k cos θ ]0 = 40k
M y = 0 by symmetry

( x , y ) = ⎛⎜ 0,

824

40k ⎞ ⎛ 60 ⎞
= 0,
26kπ / 3 ⎟⎠ ⎜⎝ 13π ⎟⎠

Section 13.5

Instructor’s Resource Manual

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21. m = ∫

a a

r =⎜ ⎟
⎝m⎠

22. m = ∫

24.

( x + y )dx dy = a3

0 0
1/ 2
⎛ Ix ⎞

a a– y

0 0

1/ 2

⎛ Iy ⎞
r =⎜ ⎟
⎝m⎠

1/ 2

⎛ 5⎞
=⎜ ⎟
⎝ 12 ⎠

a ≈ 0.6455a

⎛1⎞
( x 2 + y 2 ) dxdy = ⎜ ⎟ a 4
⎝6⎠
1/ 2

⎛ 7 ⎞
=⎜ ⎟
⎝ 30 ⎠

a ≈ 0.4830a

I = Iz = ∫

b/2

a/2

–b / 2 – a / 2

( x 2 + y 2 )k dx dy

⎛k ⎞
= ⎜ ⎟ (a3b + ab3 )
⎝ 12 ⎠

23.
25.

m = δπa 2
The moment of inertia about diameter AB is
I = Ix = ∫
=∫

2π δ a

0
4

δ a 4 ⎡ sin 2θ ⎤
⎢θ –

8 ⎣

2

1/ 2

⎛I ⎞
r =⎜ ⎟
⎝m⎠

26.

∫0

I x = ∫∫ δ y 2 dA

δ r 2 sin 2 θ r dr dθ

sin 2 θ
δ a4
dθ =
4
8

0

=

2π a

=

⎦0

⎛ δ a4π
=⎜ 4 2
⎜ δπa

∫0

= 2δ ∫

S
π / 2 2 a sin θ

(r sin θ ) 2 r dr dθ

0
0
π/2
4a 4 sin 6 θ
0

(1 – cos 2θ )dθ

= 2δ ∫

δ a4π

= 8a 4δ

4

(1)(3)(5) π 5a 4δπ
=
(2)(4)(6) 2
4

1/ 2

a
2

=

x = 0 (by symmetry)
M x = ∫∫ 1y dA = 2k ∫
S

π/2

–π / 2 0

=

2ka3
3

∫– π / 2 (1 + sin θ )

=

2ka3
3

∫– π / 2 (sin θ + 3sin

=

4ka3
3

∫0

π/2

3

a (1+ sin θ )

⎡ r3

(r sin θ )r dr dθ = 2k ∫
⎢ sin θ ⎥
–π / 2 3
⎢⎣
⎥⎦ r =0
π/ 2

sin θ dθ

π/2

π/2

a (1+ sin θ )

2

θ + 3sin 3 θ + sin 4 θ )dθ

(3sin 2 θ + sin 4 θ )dθ

4ka3 ⎛ 1 π 1 ⋅ 3 π ⎞ 5πka3
+
⎜3
⎟=
3 ⎝ 2 2 2⋅4 2 ⎠
4
M
5a
Therefore, y = x = .
m
6
=

Instructor’s Resource Manual

(using the symmetry property for odd and even functions.)
(using Formula 113)

Section 13.5

825

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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