Solution manual calculus 8th edition varberg, purcell, rigdon ch05

5

CHAPTER

Applications of the Integral
6. To find the intersection points, solve

5.1 Concepts Review
1.

b

∫a

f ( x)dx; − ∫

b

a

x + 4 = x2 − 2 .

f ( x)dx

x2 − x − 6 = 0
(x + 2)(x – 3) = 0
x = –2, 3
Slice vertically.
ΔA ≈ ⎡ ( x + 4) − ( x 2 − 2) ⎤ Δx = (− x 2 + x + 6)Δx
⎣
⎦

2. slice, approximate, integrate
3. g ( x) − f ( x); f ( x) = g ( x)
4.

∫c [ q( y) − p( y)] dy
d

3

3
1
⎡ 1
⎤
A = ∫ (− x 2 + x + 6) dx = ⎢ − x3 + x 2 + 6 x ⎥
−2
2
⎣ 3
⎦ −2

Problem Set 5.1
1. Slice vertically.

9
⎛
⎞ ⎛8
⎞ 125
⎜ −9 + + 18 ⎟ − ⎜ + 2 − 12 ⎟ =
2
6

⎝
⎠ ⎝3
⎠

ΔA ≈ ( x 2 + 1)Δx
2

2
⎡1
⎤
A = ∫ ( x 2 + 1)dx = ⎢ x3 + x ⎥ = 6
–1
3
⎣
⎦ −1

7. Solve x3 − x 2 − 6 x = 0 .

2. Slice vertically.
ΔA ≈ ( x3 − x + 2)Δx
2

33
2
1
⎡1
⎤
A = ∫ ( x3 − x + 2)dx = ⎢ x 4 − x 2 + 2 x ⎥ =
−1
2

⎣4
⎦ −1 4

A=∫

−2

A = A1 + A2
=∫

2

1
⎡1
⎤
( x + x + 2)dx = ⎢ x3 + x 2 + 2 x ⎥
2
⎣3
⎦ −2

0

0

ΔA ≈ −( x 2 + 2 x − 3)Δx = (− x 2 − 2 x + 3)Δx
1

32
⎡ 1
⎤

(− x 2 − 2 x + 3)dx = ⎢ − x3 − x 2 + 3x ⎥ =
−3
⎣ 3
⎦ −3 3

5. To find the intersection points, solve 2 – x 2 = x .
x + x−2 = 0
(x + 2)(x – 1) = 0
x = –2, 1
Slice vertically.
ΔA ≈ ⎡ (2 − x 2 ) − x ⎤ Δx = (− x 2 − x − 2)Δx
⎣
⎦
2

3

⎡ ⎛
8
⎞ ⎤ ⎡ 81
⎤
= ⎢ 0 − ⎜ 4 + − 12 ⎟ ⎥ + ⎢ − + 9 + 27 − 0 ⎥
3
⎠⎦ ⎣ 4
⎦
⎣ ⎝
16 63 253
= +
=
3 4

12

8. To find the intersection points, solve
− x + 2 = x2 .

x2 + x − 2 = 0
(x + 2)(x – 1) = 0
x = –2, 1
Slice vertically.
ΔA ≈ ⎡ (− x + 2) − x 2 ⎤ Δx = (− x 2 − x + 2)Δx
⎣
⎦
1

1
1
⎡ 1
⎤
A = ∫ (– x 2 – x + 2)dx = ⎢ – x3 – x 2 + 2 x ⎥
–2
3
2
⎣
⎦ −2
1
1
8
9
⎛
⎞ ⎛

⎞
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
⎝ 3 2
⎠ ⎝3
⎠ 2

Instructor’s Resource Manual

3

( x3 − x 2 − 6 x)dx + ∫ (− x3 + x 2 + 6 x)dx

1
1
⎡ 1
⎤
⎡1
⎤
+ ⎢ − x 4 + x3 + 3 x 2 ⎥
= ⎢ x 4 − x3 − 3 x 2 ⎥
4
3
4
3
⎣
⎦ −2 ⎣
⎦0

4. Slice vertically.

1

0

−2

2

⎛8
⎞ ⎛ 8
⎞ 40
= ⎜ + 2 + 4⎟ − ⎜ − + 2 − 4⎟ =
3
3
⎝
⎠ ⎝
⎠ 3

A=∫

ΔA1 ≈ ( x3 − x 2 − 6 x)Δx
ΔA2 ≈ −( x3 − x 2 − 6 x)Δx = (− x3 + x 2 + 6 x)Δx

3. Slice vertically.
ΔA ≈ ⎡ ( x 2 + 2) − (− x) ⎤ Δx = ( x 2 + x + 2)Δx
⎣
⎦
2

x( x 2 − x − 6) = 0

x(x + 2)(x – 3) = 0
x = –2, 0, 3
Slice vertically.

1

1
⎡ 1
⎤
(− x 2 − x + 2)dx = ⎢ − x3 − x 2 + 2 x ⎥
−2
3
2
⎣
⎦ −2
1
1
8
9
⎛
⎞ ⎛
⎞
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
⎝ 3 2
⎠ ⎝3
⎠ 2
A=∫

1

Section 5.1

295

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9. To find the intersection points, solve

13.

y +1 = 3 – y .
2

y2 + y − 2 = 0
(y + 2)(y – 1) = 0
y = –2, 1
Slice horizontally.
ΔA ≈ ⎡ (3 − y 2 ) − ( y + 1) ⎤ Δy = (− y 2 − y + 2)Δy
⎣
⎦
1

1
1
1
A = ∫ (– y 2 – y + 2)dy = ⎡⎢ – y 3 – y 2 + 2 y ⎤⎥
–2
2

⎣ 3
⎦ −2
⎛ 1 1
⎞ ⎛8
⎞ 9
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
3
2
3
⎝
⎠ ⎝
⎠ 2

ΔA ≈ −( x − 4)( x + 2)Δx = (− x 2 + 2 x + 8)Δx
3

3
⎡ 1
⎤
A = ∫ (− x 2 + 2 x + 8)dx = ⎢ − x3 + x 2 + 8 x ⎥
0
⎣ 3
⎦0
= –9 + 9 + 24 = 24
Estimate the area to be (3)(8) = 24.

10. To find the intersection point, solve y 2 = 6 − y .
y2 + y − 6 = 0
(y + 3)(y – 2) = 0
y = –3, 2

Slice horizontally.
ΔA ≈ ⎡ (6 − y ) − y 2 ⎤ Δy = (− y 2 − y + 6)Δy
⎣
⎦

14.

2
22
2
1
⎡ 1
⎤
A = ∫ ( − y 2 − y + 6)dy = ⎢ − y 3 − y 2 + 6 y ⎥ =
0
3
2
⎣ 3
⎦0

ΔA ≈ −( x 2 − 4 x − 5)Δx = (− x 2 + 4 x + 5)Δx

11.

4

⎡ 1
⎤
(− x 2 + 4 x + 5)dx = ⎢ − x3 + 2 x 2 + 5 x ⎥
−1

⎣ 3
⎦ −1
⎛ 64
⎞ ⎛1
⎞ 100
= ⎜ − + 32 + 20 ⎟ − ⎜ + 2 − 5 ⎟ =
≈ 33.33
3
3
3
⎝
⎠ ⎝
⎠
1
⎛ 1⎞
Estimate the area to be (5) ⎜ 6 ⎟ = 32 .
2
⎝ 2⎠

A=∫

1 ⎞
⎛
ΔA ≈ ⎜ 3 − x 2 ⎟ Δx
3 ⎠
⎝

4

15.

3

3⎛
1 ⎞
1 ⎤
⎡
A = ∫ ⎜ 3 − x 2 ⎟ dx = ⎢3 x − x3 ⎥ = 9 − 3 = 6
0 ⎝
3 ⎠
9 ⎦0
⎣

Estimate the area to be (3)(2) = 6.
12.

1
ΔA ≈ − ( x 2 − 7)Δx
4
2

2 1
1 ⎡1
⎤
A = ∫ − ( x 2 − 7)dx = − ⎢ x3 − 7 x ⎥
0
4
4 ⎣3
⎦0

1⎛8

⎞ 17
= − ⎜ − 14 ⎟ =
≈ 2.83
4⎝3
⎠ 6

ΔA ≈ (5 x − x 2 )Δx
3

3
1 ⎤
⎡5
A = ∫ (5 x − x 2 )dx = ⎢ x 2 − x3 ⎥ ≈ 11.33
1
3 ⎦1
⎣2
⎛ 1⎞
Estimate the area to be (2) ⎜ 5 ⎟ = 11 .
⎝ 2⎠

296

Section 5.1

⎛ 1⎞
Estimate the area to be (2) ⎜ 1 ⎟ = 3 .
⎝ 2⎠

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


16.

19.

ΔA ≈ [ x − ( x − 3)( x − 1) ] Δx

ΔA1 ≈ − x3 Δx

= ⎡ x − ( x 2 − 4 x + 3) ⎤ Δx = (− x 2 + 5 x − 3)Δx
⎣
⎦
To find the intersection points, solve
x = (x – 3)(x – 1).

ΔA2 ≈ x Δx
3

A = A1 + A2 = ∫

0

−3

0

3

− x3 dx + ∫ x3 dx
0

x2 − 5x + 3 = 0

3

⎡ 1 ⎤
⎡1 ⎤
⎛ 81 ⎞ ⎛ 81 ⎞ 81
= ⎢− x4 ⎥ + ⎢ x4 ⎥ = ⎜ ⎟ + ⎜ ⎟ =
⎣ 4 ⎦ −3 ⎣ 4 ⎦ 0 ⎝ 4 ⎠ ⎝ 4 ⎠ 2
= 40.5
Estimate the area to be (3)(7) + (3)(7) = 42.

5 ± 25 − 12
2
5 ± 13
x=
2

x=

17.
A=∫

5+ 13
2
5− 13

2

(− x 2 + 5 x − 3)dx

5+ 13

5
13 13
⎡ 1
⎤ 2
= ⎢ − x3 + x 2 − 3 x ⎥
=
≈ 7.81
−
5
13
2
6
⎣ 3
⎦
2

ΔA1 ≈ − x Δx
3

Estimate the area to be

ΔA2 ≈ 3 x Δx
A = A1 + A2 = ∫

0

−2

2
− 3 x dx + ∫ 3 x dx
0

1
(4)(4) = 8 .
2

20.

⎛ 3 3 2 ⎞ ⎛ 33 2 ⎞
⎡ 3
⎤
⎡3
⎤
= ⎢− x4 / 3 ⎥ + ⎢ x4 / 3 ⎥ = ⎜
⎟+⎜
⎟
⎣ 4
⎦ −2 ⎣ 4
⎦ 0 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
0

2

= 33 2 ≈ 3.78

Estimate the area to be (2)(1) + (2)(1) = 4.

18.
ΔA ≈ ⎣⎡ x − ( x − 4) ⎦⎤ Δx =

(

)

x − x + 4 Δx

To find the intersection point, solve
x = ( x − 4) .
x = ( x − 4)2
x 2 − 9 x + 16 = 0

ΔA ≈ −( x − 10)Δx = (10 − x )Δx
9

9
2
⎡
⎤
A = ∫ (10 − x ) dx = ⎢10 x − x3 2 ⎥
0
3
⎣
⎦0
= 90 – 18 = 72
Estimate the area to be 9 · 8 =72.

Instructor’s Resource Manual

9 ± 81 − 64
2
9 ± 17
x=
2
⎛
9 − 17
9 + 17 ⎞
is extraneous so x =
. ⎟⎟
⎜⎜ x =
2
2
⎝
⎠
x=

Section 5.1

297

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


A=∫

9+ 17
2

0

(

x2 + 5x + 6 = 0
(x + 3)(x + 2) = 0
x = –3, –2

)

x − x + 4 dx
9+ 17

1
⎡2
⎤ 2
= ⎢ x3 / 2 − x 2 + 4 x ⎥
2
⎣3
⎦0

2 ⎛ 9 + 17 ⎞
= ⎜⎜
⎟
3 ⎝ 2 ⎟⎠

3/ 2

2 ⎛ 9 + 17 ⎞
= ⎜⎜
⎟
3 ⎝ 2 ⎟⎠

3/ 2

A=∫

−2

23
17
−
≈ 15.92
4
4

Estimate the area to be

(− x 2 − 5 x − 6)dx

5
⎡ 1
⎤
= ⎢ − x3 − x 2 − 6 x ⎥
2
⎣ 3
⎦ −3

45
⎛8
⎞ ⎛
⎞ 1
= ⎜ − 10 + 12 ⎟ − ⎜ 9 − + 18 ⎟ = ≈ 0.17
3
2
⎝
⎠ ⎝
⎠ 6
1 ⎛
2⎞ 1
Estimate the area to be (1) ⎜ 5 − 4 ⎟ = .
2 ⎝
3⎠ 6

2

⎛ 9 + 17 ⎞
1 ⎛ 9 + 17 ⎞
− ⎜⎜
⎟⎟ + 4 ⎜⎜
⎟⎟
2⎝ 2 ⎠
⎝ 2 ⎠

+

−2

−3

1 ⎛ 1 ⎞⎛ 1 ⎞
1
⎜ 5 ⎟ ⎜ 5 ⎟ = 15 .
2 ⎝ 2 ⎠⎝ 2 ⎠
8

23.

21.

ΔA ≈ (8 y − y 2 )Δy
To find the intersection points, solve

ΔA ≈ ⎡ − x 2 − ( x 2 − 2 x) ⎤ Δx = (−2 x 2 + 2 x)Δx
⎣
⎦
To find the intersection points, solve

8 y − y2 = 0 .
y(8 – y) = 0
y = 0, 8

− x2 = x2 − 2 x .
2 x2 − 2 x = 0
2x(x – 1) = 0
x = 0, x = 1

8

8
1 ⎤
⎡
A = ∫ (8 y − y 2 ) dy = ⎢ 4 y 2 − y 3 ⎥
0
3 ⎦0
⎣
512 256
= 256 −
=
≈ 85.33
3
3
Estimate the area to be (16)(5) = 80.

1

1
⎡ 2
⎤
A = ∫ (−2 x 2 + 2 x )dx = ⎢ − x3 + x 2 ⎥
0
⎣ 3
⎦0
2
1
= − + 1 = ≈ 0.33
3
3

⎛ 1 ⎞⎛ 1 ⎞ 1
Estimate the area to be ⎜ ⎟⎜ ⎟ = .
⎝ 2 ⎠⎝ 2 ⎠ 4

24.

22.

ΔA ≈ (3 − y )( y + 1)Δy = (− y 2 + 2 y + 3)Δy

ΔA ≈ ⎡ ( x 2 − 9) − (2 x − 1)( x + 3) ⎤ Δx
⎣
⎦
= ⎡( x 2 − 9) − (2 x 2 + 5 x − 3) ⎤ Δx
⎣
⎦
= (− x 2 − 5 x − 6)Δx
To find the intersection points, solve

3

⎡ 1
⎤
(− y 2 + 2 y + 3)dy = ⎢ − y 3 + y 2 + 3 y ⎥
−1
⎣ 3
⎦ −1
⎛1
⎞ 32
= (−9 + 9 + 9) − ⎜ + 1 − 3 ⎟ =

≈ 10.67
⎝3
⎠ 3
⎛ 1⎞
Estimate the area to be (4) ⎜ 2 ⎟ = 10 .
⎝ 2⎠
A=∫

3

(2 x − 1)( x + 3) = x 2 − 9 .

298

Section 5.1

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25.

27.

ΔA ≈ ⎡(3 − y 2 ) − 2 y 2 ⎤ Δy = (−3 y 2 + 3)Δy
⎣
⎦
To find the intersection points, solve

ΔA ≈ ⎡ (−6 y 2 + 4 y ) − (2 − 3 y ) ⎤ Δy
⎣
⎦

2 y2 = 3 − y2 .

= (−6 y 2 + 7 y − 2)Δy
To find the intersection points, solve

3y2 − 3 = 0
3(y + 1)(y – 1) = 0
y = –1, 1

−6 y 2 + 4 y = 2 − 3 y.
6 y2 − 7 y + 2 = 0
(2 y − 1)(3 y − 2) = 0

−1

1 2
y= ,
2 3
2/3

7
⎡
⎤
(−6 y 2 + 7 y − 2)dy = ⎢ −2 y 3 + y 2 − 2 y ⎥
1/ 2

2
⎣
⎦1/ 2
1
16
14
4
1
7
⎛
⎞ ⎛
⎞
≈ 0.0046
= ⎜ − + − ⎟ − ⎜ − + − 1⎟ =
216
⎝ 27 9 3 ⎠ ⎝ 4 8 ⎠
Estimate the area to be
1 ⎛ 1 ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞⎛ 1 ⎞
1
.
⎜ ⎟⎜ ⎟ − ⎜ ⎟⎜ ⎟ =
2 ⎝ 2 ⎠⎝ 5 ⎠ 2 ⎝ 2 ⎠⎝ 6 ⎠ 120
A=∫

1

1

(−3 y 2 + 3)dy = ⎡ − y 3 + 3 y ⎤
⎣

⎦ −1
= (–1 + 3) – (1 – 3) = 4
Estimate the value to be (2)(2) = 4.
A=∫

2/3

26.

28.

ΔA ≈ ⎡(8 − 4 y 4 ) − (4 y 4 ) ⎤ Δy = (8 − 8 y 4 )Δy
⎣
⎦
To find the intersection points, solve
4 y4 = 8 − 4 y4 .
8 y4 = 8
y4 = 1
y = ±1
1

8 ⎤
⎡
(8 − 8 y 4 )dy = ⎢8 y − y 5 ⎥
−1
5 ⎦ −1
⎣
8⎞ ⎛
8 ⎞ 64
⎛

= ⎜ 8 − ⎟ − ⎜ −8 + ⎟ =
= 12.8
5⎠ ⎝
5⎠ 5
⎝
⎛ 1⎞
Estimate the area to be (8) ⎜ 1 ⎟ = 12 .
⎝ 2⎠

ΔA ≈ ⎡ ( y + 4) − ( y 2 − 2 y ) ⎤ Δy = (− y 2 + 3 y + 4)Δy
⎣
⎦
To find the intersection points, solve

A=∫

y2 − 2 y = y + 4 .
y2 − 3y − 4 = 0
(y + 1)(y – 4) = 0
y = –1, 4

1

4

3
⎡ 1
⎤
(− y 2 + 3 y + 4)dy = ⎢ − y 3 + y 2 + 4 y ⎥
−1

2
⎣ 3
⎦ −1
⎛ 64
⎞ ⎛1 3
⎞ 125
= ⎜ − + 24 + 16 ⎟ − ⎜ + − 4 ⎟ =
≈ 20.83
3
3
2
6
⎝
⎠ ⎝
⎠
Estimate the area to be (7)(3) = 21.
A=∫

4

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Section 5.1

299

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29.

30.

y = x3
y = x+6

2y + x = 0

An equation of the line through (–1, 4) and (5, 1)
1
7
is y = − x + . An equation of the line through
2
2
(–1, 4) and (2, –2) is y = –2x + 2. An equation of
the line through (2, –2) and (5, 1) is y = x – 4.
Two integrals must be used. The left-hand part of
the triangle has area
2 ⎡ 1
2 ⎛3
7
3⎞
⎤
∫−1 ⎢⎣− 2 x + 2 − (−2 x + 2) ⎥⎦ dx = ∫−1 ⎜⎝ 2 x + 2 ⎟⎠ dx .
The right-hand part of the triangle has area
5⎡ 1
5⎛ 3
7
15 ⎞

⎤
∫2 ⎣⎢− 2 x + 2 − ( x − 4)⎥⎦ dx = ∫2 ⎝⎜ − 2 x + 2 ⎠⎟ dx .
The triangle has area
2 ⎛3
5⎛ 3
3⎞
15 ⎞
∫−1⎜⎝ 2 x + 2 ⎟⎠ dx + ∫2 ⎜⎝ − 2 x + 2 ⎟⎠ dx

Let R1 be the region bounded by 2y + x = 0,
y = x + 6, and x = 0.
0 ⎡
⎛ 1 ⎞⎤
A( R1 ) = ∫ ⎢ ( x + 6) − ⎜ − x ⎟ ⎥ dx
−4 ⎣
⎝ 2 ⎠⎦
⎛3
⎞
⎜ x + 6 ⎟ dx
⎝2
⎠
Let R2 be the region bounded by y = x + 6,
=∫

0

−4

y = x3 , and x = 0.
2

2
A( R2 ) = ∫ ⎡ ( x + 6) − x3 ⎤ dx = ∫ (− x3 + x + 6)dx
⎦
0 ⎣
0
A( R ) = A( R1 ) + A( R2 )

=∫

0

−4

2
⎛3
⎞
3
⎜ x + 6 ⎟ dx + ∫0 (− x + x + 6)dx
⎝2
⎠
0

2

31.

9

∫−1 (3t

2

5

3 ⎤
15 ⎤
⎡3
⎡ 3
= ⎢ x2 + x ⎥ + ⎢− x2 + x ⎥
2 ⎦ −1 ⎣ 4
2 ⎦2
⎣4
27 27 27
=
+
=
= 13.5
4
4
2

2

1
⎡3
⎤
⎡ 1
⎤
= ⎢ x2 + 6 x ⎥ + ⎢− x4 + x2 + 6 x ⎥
2

⎣4
⎦ −4 ⎣ 4
⎦0
= 12 + 10 = 22

9

− 24t + 36)dt = ⎡t 3 − 12t 2 + 36t ⎤ = (729 – 972 + 324) – (–1 – 12 – 36) = 130
⎣
⎦ −1

The displacement is 130 ft. Solve 3t 2 − 24t + 36 = 0 .
3(t – 2)(t – 6) = 0
t = 2, 6
⎧⎪3t 2 − 24t + 36 t ≤ 2, t ≥ 6
V (t ) = ⎨
2
⎪⎩−3t + 24t − 36 2 < t < 6
9

∫−1 3t

2

− 24t + 36 dt = ∫

2

−1

2

6

9

2

6

(3t 2 − 24t + 36) dt + ∫ (−3t 2 + 24t − 36) dt + ∫ (3t 2 − 24t + 36) dt
6

9

= ⎡t 3 − 12t 2 + 36t ⎤ + ⎡ −t 3 + 12t 2 − 36t ⎤ + ⎡t 3 − 12t 2 + 36t ⎤ = 81 + 32 + 81 = 194
⎣
⎦ −1 ⎣
⎦2 ⎣
⎦6
The total distance traveled is 194 feet.

32.

300

3π / 2

1 ⎞ 3π
⎛1

⎞
⎡1 1
⎤
⎛ 3π 1 ⎞ ⎛
∫0 ⎜⎝ 2 + sin 2t ⎟⎠ dt = ⎣⎢ 2 t − 2 cos 2t ⎦⎥ 0 = ⎜⎝ 4 + 2 ⎟⎠ − ⎜⎝ 0 − 2 ⎟⎠ = 4 + 1
3π
1
3π
+ 1 ≈ 3.36 feet . Solve + sin 2t = 0 for 0 ≤ t ≤
.
The displacement is
4
2
2
1
7π 11π
7π 11π
,
sin 2t = − ⇒ 2t =
,
⇒t=
6 6
2
12 12
3π / 2

Section 5.1

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7 π 11π
3π
⎧1
0≤t≤
,
≤t≤
⎪⎪ 2 + sin 2t
1
12
12
2
+ sin 2t = ⎨
1
7
11
π
π
2
⎪− − sin 2t
⎪⎩ 2
12
12
3π / 2 1
7 π /12 ⎛ 1
11π /12 ⎛ 1

3π / 2 ⎛ 1
⎞
⎞
⎞
∫0 2 + sin 2t dt = ∫0 ⎜⎝ 2 + sin 2t ⎟⎠ dt + ∫7π /12 ⎜⎝ − 2 − sin 2t ⎟⎠ dt + ∫11π /12 ⎜⎝ 2 + sin 2t ⎟⎠ dt
7 π /12

⎡1 1
⎤
= ⎢ t − cos 2t ⎥
⎣2 2
⎦0

11π /12

3π / 2

⎡ 1 1
⎤
⎡1 1
⎤
+ ⎢ − t + cos 2t ⎥
+ ⎢ t − cos 2t ⎥
⎣ 2 2
⎦ 7 π /12 ⎣ 2 2
⎦11π /12

⎛ 7π
3 1⎞ ⎛ π
3 ⎞ ⎛ 7π

3 1 ⎞ 5π
= ⎜⎜
+
+ ⎟⎟ + ⎜⎜ − +
+
+ ⎟⎟ =
+ 3 +1
⎟⎟ + ⎜⎜
⎝ 24 4 2 ⎠ ⎝ 6 2 ⎠ ⎝ 24 4 2 ⎠ 12
5π
The total distance traveled is
+ 3 + 1 ≈ 4.04 feet.
12

33. s (t ) = ∫ v(t )dt = ∫ (2t − 4)dt = t 2 − 4t + C

Slicing the region horizontally, the area is
1
1
5
5
⎛ 1 ⎞
∫1/ 36 y dy + ⎜⎝ 36 ⎟⎠ (5) . Since 36 < 12 the

c.

Since s(0) = 0, C = 0 and s (t ) = t 2 − 4t. s = 12
when t = 6, so it takes the object 6 seconds to get
s = 12.
⎧4 − 2t 0 ≤ t < 2

2t − 4 = ⎨
⎩2t − 4 2 ≤ t

line that bisects the area is between y =
and y = 1, so we find d such that
1
1 1
5 1 1
∫d y dy = 12 ; ∫d y dy = ⎡⎣ 2 y ⎤⎦ d

2

2

2t − 4 dt = ⎡ −t 2 + 4t ⎤ = 4, so the object
⎣
⎦0
travels a distance of 4 cm in the first two seconds.

∫0

x

∫2

= 2−2 d ; 2−2 d =

x

2t − 4 dt = ⎡t 2 − 4t ⎤ = x 2 − 4 x + 4

⎣
⎦2

361
≈ 0.627 .
576
The line y = 0.627 approximately bisects the
area.

takes 2 + 2 2 ≈ 4.83 seconds to travel a total
distance of 12 centimeters.
6

6
1
5
⎡ 1⎤
A = ∫ x −2 dx = ⎢ − ⎥ = − + 1 =
1
6
6
⎣ x ⎦1

b. Find c so that

c −2

∫1 x

dx =

c

5
.
12

1
⎡ 1⎤
= ⎢− ⎥ = 1 −
∫
c
⎣ x ⎦1
1 5
12
1− = , c =
c 12
7
12
The line x =
bisects the area.
7
c −2
x dx
1

5
;
12

d=

x 2 − 4 x + 4 = 8 when x = 2 + 2 2, so the object

34. a.

1
36

35. Equation of line through (–2, 4) and (3, 9):
y=x+6
Equation of line through (2, 4) and (–3, 9):
y = –x + 6
0

3

–3

0

A( A) = ∫ [9 – (– x + 6)]dx + ∫ [9 – ( x + 6)]dx
0

3

= ∫ (3 + x)dx + ∫ (3 – x)dx
–3

0

0

3

1 ⎤
1 ⎤
9 9
⎡
⎡
= ⎢3 x + x 2 ⎥ + ⎢3 x – x 2 ⎥ = + = 9
2
2
⎣
⎦ −3 ⎣
⎦0 2 2

A( B ) = ∫

–2

–3

[(– x + 6) – x 2 ]dx
0

+ ∫ [(– x + 6) – ( x + 6)]dx
–2

=∫

–2

–3

0

(– x 2 – x + 6)dx + ∫ (–2 x)dx
–2

−2

0
1
37
⎡ 1
⎤
= ⎢ – x3 – x 2 + 6 x ⎥ + ⎡ – x 2 ⎤ =
⎣
⎦
−2
2
6
⎣ 3
⎦ −3

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A(C ) = A( B) =

π/6

37
(by symmetry)
6

0

2

–2

0

A( D) = ∫ [( x + 6) – x 2 ]dx + ∫ [(– x + 6) – x 2 ]dx
0

2

1
1
⎡ 1

⎤
⎡ 1
⎤
= ⎢ – x3 + x 2 + 6 x ⎥ + ⎢ – x3 – x 2 + 6 x ⎥
2
2
⎣ 3
⎦ −2 ⎣ 3
⎦0
44
=
3
A(A) + A(B) + A(C) + A(D) = 36

3

3
1 ⎤
⎡
A( A + B + C + D) = ∫ (9 – x 2 )dx = ⎢9 x – x3 ⎥
–3
3 ⎦ −3
⎣
= 36

⎡1
⎤
= ⎢ x + cos x ⎥
⎣2
⎦0

5π / 6

1 ⎤
⎡
+ ⎢ − cos x − x ⎥
2 ⎦π / 6
⎣

13π / 6

17 π / 6

1 ⎤
⎡1
⎤
⎡
+ ⎢ x + cos x ⎥
+ ⎢ − cos x − x ⎥
2 ⎦13π / 6
⎣2
⎦ 5π / 6 ⎣
⎛π
3 ⎞ ⎛
π⎞ ⎛
2π ⎞
= ⎜⎜ +
− 1⎟⎟ + ⎜ 3 − ⎟ + ⎜ 3 +
⎟
12

2
3
3 ⎠
⎝
⎠
⎝
⎝
⎠
π⎞ π 7 3
⎛
+⎜ 3 − ⎟ = +
− 1 ≈ 5.32
3 ⎠ 12
2
⎝

5.2 Concepts Review

36. Let f(x) be the width of region 1 at every x.
b

ΔA1 ≈ f ( x)Δx, so A1 = ∫ f ( x)dx .
a

Let g(x) be the width of region 2 at every x.
b

ΔA2 ≈ g ( x)Δx, so A2 = ∫ g ( x)dx .
a

Since f(x) = g(x) at every x in [a, b],
b

b

a

a

A1 = ∫ f ( x)dx = ∫ g ( x)dx = A2 .

1. πr 2 h
2. π( R 2 − r 2 )h
3. πx 4 Δx
4. π[( x 2 + 2)2 − 4]Δx

37. The height of the triangular region is given by
for 0 ≤ x ≤ 1 . We need only show that the
height of the second region is the same in order
to apply Cavalieri'’s Principle. The height of the
second region is
h2 = ( x 2 − 2 x + 1) − ( x 2 − 3 x + 1)
= x2 − 2 x + 1 − x2 + 3x − 1
= x for 0 ≤ x ≤ 1.
Since h1 = h2 over the same closed interval, we
can conclude that their areas are equal.

Problem Set 5.2
1. Slice vertically.
ΔV ≈ π( x 2 + 1) 2 Δx = π( x 4 + 2 x 2 + 1)Δx

2

V = π∫ ( x 4 + 2 x 2 + 1)dx
0

2

2
⎡1
⎤
⎛ 32 16
⎞ 206π
= π ⎢ x5 + x3 + x ⎥ = π ⎜ + + 2 ⎟ =
3
15
⎣5
⎦0
⎝ 5 3
⎠
≈ 43.14

38. Sketch the graph.
2. Slice vertically.
ΔV ≈ π(− x 2 + 4 x)2 Δx = π( x 4 − 8 x3 + 16 x 2 )Δx
3

V = π∫ ( x 4 − 8 x3 + 16 x 2 )dx
0

3

16 ⎤
⎡1
= π ⎢ x5 − 2 x 4 + x3 ⎥
3 ⎦0
⎣5
1
17π
.
for 0 ≤ x ≤
2
6
π 5π 13π 17π
x= , ,
,
6 6 6
6
The area of the trapped region is
π/6 ⎛ 1
5π / 6 ⎛
1⎞
⎞
∫0 ⎜⎝ 2 − sin x ⎟⎠ dx + ∫π / 6 ⎜⎝ sin x − 2 ⎟⎠ dx
13π / 6 ⎛ 1
17 π / 6 ⎛
1⎞
⎞
+∫
− sin x ⎟ dx + ∫
⎜ sin x − ⎟ dx

π
5π / 6 ⎜⎝ 2
13
/
6
2⎠
⎠
⎝

Solve sin x =

302

Section 5.2

⎛ 243
⎞
= π⎜
− 162 + 144 ⎟
5
⎝
⎠
153π
=
≈ 96.13
5

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3. a.

Slice vertically.

6.

ΔV ≈ π(4 − x ) Δx = π(16 − 8 x + x )Δx
2 2

2

4

2

V = π∫ (16 – 8 x 2 + x 4 )dx
0

=

256π
≈ 53.62
15

b. Slice horizontally.
x = 4– y
Note that when x = 0, y = 4.

ΔV ≈ π

(

4− y

)

2

ΔV ≈ π( x3 )2 Δx = πx6 Δx

Δy = π(4 − y )Δy

3

3
2187π
⎡1 ⎤
V = π∫ x 6 dx = π ⎢ x7 ⎥ =
≈ 981.52
0
7
7
⎣
⎦0

4

4

1 ⎤
⎡
V = π∫ (4 – y )dy = π ⎢ 4 y – y 2 ⎥
0
2 ⎦0
⎣
= π(16 − 8) = 8π ≈ 25.13

4. a.

7.

Slice vertically.
ΔV ≈ π(4 − 2 x)2 Δx
0≤ x≤2
2

2
⎡ 1
⎤
V = π∫ (4 − 2 x)2 dx = π ⎢ − (4 − 2 x)3 ⎥
0
⎣ 6
⎦0
32π
=
≈ 33.51
3

2

⎛ 1 ⎞
⎛1⎞
ΔV ≈ π ⎜ ⎟ Δx = π ⎜ ⎟ Δx
⎝x⎠
⎝ x2 ⎠

b. Slice vertically.
y
x = 2−
2

V = π∫

4

2

≈ 0.79

2

y⎞
⎛
ΔV ≈ π ⎜ 2 − ⎟ Δy
2
⎝
⎠
0 ≤ y ≤ 4

4

⎡ 1⎤
⎛ 1 1⎞ π
dx = π ⎢ − ⎥ = π ⎜ − + ⎟ =
2
⎣ x ⎦2
⎝ 4 2⎠ 4
x

1

8.
4

2
3
⎡ 2⎛
y⎞
y⎞ ⎤
V = π∫ ⎜ 2 − ⎟ dy = π ⎢− ⎜ 2 − ⎟ ⎥
0⎝
2⎠
3
2⎠ ⎥
⎣⎢ ⎝
⎦0
16π
=
≈ 16.76

3
4⎛

5.
ΔV ≈ π( x3 / 2 ) 2 Δx = πx3Δx
3

3
⎡1 ⎤
⎛ 81 16 ⎞
V = π∫ x3 dx = π ⎢ x 4 ⎥ = π ⎜ − ⎟
2
⎣ 4 ⎦2
⎝ 4 4⎠
65π
=
≈ 51.05
4

⎛ x2
ΔV ≈ π ⎜
⎜ π
⎝
V =∫

4

0

2

⎞
x4
Δx
⎟ Δx =
⎟
π
⎠
4

x4
1 ⎡1 ⎤
1024
dx = ⎢ x5 ⎥ =
≈ 65.19
π
π ⎣ 5 ⎦0
5π

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12.

9.

2

2
⎛ 1 ⎞
⎛2⎞
ΔV ≈ π ⎜ ⎟ Δy = 4π ⎜
Δy
⎜ y 2 ⎟⎟
⎝ y⎠
⎝
⎠

ΔV ≈ π ⎛⎜ 9 − x 2 ⎞⎟ Δx = π(9 − x 2 )Δx
⎝
⎠
V = π∫

3

−2

3

1 ⎤
⎡
(9 − x 2 )dx = π ⎢9 x − x3 ⎥
3 ⎦ −2

⎣

V = 4π ∫

2

⎡
8 ⎞ ⎤ 100π
⎛
= π ⎢(27 − 9) − ⎜ −18 + ⎟ ⎥ =
≈ 104.72
3 ⎠⎦
3
⎝
⎣

10.

6

=

6

⎡ 1⎤
⎛ 1 1⎞
dy = 4π ⎢ − ⎥ = 4π ⎜ − + ⎟
2
⎝ 6 2⎠
y

⎣ y ⎦2
1

4π
≈ 4.19
3

13.

ΔV ≈ π( x 2 / 3 ) 2 Δx = πx 4 / 3Δx
V = π∫

27

1

=

(

ΔV ≈ π 2 y

27

x

4/3

⎡3
⎤

⎛ 6561 3 ⎞
dx = π ⎢ x 7 / 3 ⎥ = π ⎜
− ⎟
7
7⎠
⎣
⎦1
⎝ 7

6558π
≈ 2943.22
7

)

2

Δy = 4πy Δy
4

4
⎡1 ⎤
V = 4π ∫ y dy = 4π ⎢ y 2 ⎥ = 32π ≈ 100.53
0
⎣ 2 ⎦0

14.

11.

ΔV ≈ π( y 2 ) 2 Δy = πy 4 Δy
3 4
y dy
0

V = π∫

304

3

243π
⎡1 ⎤
= π ⎢ y5 ⎥ =
≈ 152.68
5
⎣ 5 ⎦0

Section 5.2

ΔV ≈ π( y 2 / 3 )2 Δy = πy 4 / 3 Δy
27

6561π
⎡3
⎤
y 4 / 3 dy = π ⎢ y 7 / 3 ⎥ =
0
7
⎣7

⎦0
≈ 2944.57

V = π∫

27

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18. Sketch the region.

15.

y

y = 6 x2

8

y = 6x

−1

1

ΔV ≈ π( y 3 / 2 )2 Δy = πy 3Δy

V = π∫

9

0

9

6561π
⎡1 ⎤
y dy = π ⎢ y 4 ⎥ =
≈ 5153.00
4
4
⎣
⎦0
3

16.

x

−4

To find the intersection points, solve 6 x = 6 x 2 .
6( x 2 − x) = 0
6x(x – 1) = 0
x = 0, 1
ΔV ≈ π ⎡ (6 x) 2 − (6 x 2 ) 2 ⎤ Δx = 36π( x 2 − x 4 )Δx
⎣

⎦
1

1
1 ⎤
⎡1
V = 36π ∫ ( x 2 − x 4 )dx = 36π ⎢ x3 − x5 ⎥
0
3
5 ⎦0
⎣
⎛ 1 1 ⎞ 24π
= 36π ⎜ − ⎟ =
≈ 15.08
5
⎝3 5⎠

19. Sketch the region.

2

ΔV ≈ π ⎛⎜ 4 − y 2 ⎞⎟ Δy = π(4 − y 2 )Δy
⎝
⎠
2

1 ⎤
⎡
(4 − y 2 )dy = π ⎢ 4 y − y 3 ⎥
−2

3 ⎦ −2
⎣

V = π∫

2

8 ⎞ ⎤ 32π
⎡⎛ 8 ⎞ ⎛
= π ⎢⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎥ =
≈ 33.51
3
3 ⎠⎦
3
⎠ ⎝
⎣⎝
17. The equation of the upper half of the ellipse is
y = b 1−
V = π∫

a

x2
a2

b2

−a a2

or y =

b 2
a − x2 .
a

(a 2 − x 2 )dx
3 ⎤a

b2 π ⎡ 2
x
⎢a x − ⎥
2
3 ⎦⎥
a ⎣⎢
−a
2 ⎡⎛
3⎞ ⎛
b π
a
a3 ⎞ ⎤ 4
a3 − ⎟ − ⎜ − a3 + ⎟ ⎥ = ab 2 π
=
⎢
⎜
3 ⎟⎠ ⎜⎝
3 ⎟⎠ ⎥⎦ 3
a 2 ⎢⎣⎜⎝
=

To find the intersection points, solve

x
=2 x.
2

x2
= 4x
4

x 2 − 16 x = 0
x(x – 16) = 0
x = 0, 16
⎡
ΔV ≈ π ⎢ 2 x
⎢⎣

(

)

2

⎛ x⎞
−⎜ ⎟
⎝2⎠

2⎤

⎛
x2 ⎞

⎥ Δx = π ⎜ 4 x − ⎟ Δx
⎜
4 ⎟⎠
⎥⎦
⎝
16

⎛
⎡
x2 ⎞
x3 ⎤
V = π ∫ ⎜ 4 x − ⎟ dx = π ⎢ 2 x 2 − ⎥
0 ⎜
4 ⎟⎠
12 ⎦⎥
⎝
⎣⎢
0
1024 ⎞ 512π
⎛
= π ⎜ 512 −
≈ 536.17
⎟=
3 ⎠
3
⎝
16

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20. Sketch the region.

3 2
1
x + 3, y = x 2 + 5
16
16
Sketch the region.

22. y =

r

1 ⎤
⎡
(r 2 − x 2 )dx = π ⎢ r 2 x − x3 ⎥
r −h
3 ⎦ r −h
⎣

V = π∫

r

To find the intersection point, solve
3 2
1
x + 3 = x2 + 5 .
16
16
1 2
x −2 = 0
8

1
= πh 2 (3r − h)
3
21. Sketch the region.
y

x 2 − 16 = 0
(x + 4)(x – 4) = 0
x = –4, 4
2
2
4 ⎡⎛ 1
⎞ ⎛ 3
⎞ ⎤
V = π∫ ⎢⎜ x 2 + 5 − 2 ⎟ − ⎜ x 2 + 3 − 2 ⎟ ⎥ dx
0 ⎢⎝ 16
⎠ ⎝ 16
⎠ ⎦⎥

⎣

4

2

−1

1

2

4 ⎡⎛ 1
3
⎞
x4 − x2 + 9 ⎟
= π ∫ ⎢⎜
0 ⎣⎝ 256
8
⎠

x

⎛ 9 4 3 2 ⎞⎤
x − x + 1⎟⎥ dx
−⎜
8
⎝ 256
⎠⎦

y
y
.
To find the intersection points, solve =
4
2
y2 y
=
16 4
y2 − 4 y = 0
y(y – 4) = 0
y = 0, 4
2
⎡⎛
⎤
⎛ y y2 ⎞
y ⎞ ⎛ y ⎞2 ⎥
⎢
ΔV ≈ π ⎜
− ⎜ ⎟ Δy = π ⎜ −
⎟ Δy
⎟
⎜ 4 16 ⎟
⎢⎜ 2 ⎟ ⎝ 4 ⎠ ⎥
⎝
⎠
⎝
⎠
⎣
⎦

4

4⎛
1 5⎤
1
⎡
⎞
x
= π ∫ ⎜ 8 − x 4 ⎟ dx = π ⎢8 x −
0⎝
32 ⎠
160 ⎥⎦ 0
⎣
32 ⎞ 128π
⎛
= π ⎜ 32 − ⎟ =
≈ 80.42
5 ⎠
5
⎝

23.

4

⎡ y 2 y3 ⎤
4 ⎛ y y2 ⎞
V = π∫ ⎜ −
⎟ dy = π ⎢ − ⎥

0 ⎜ 4 16 ⎟
⎝
⎠
⎣⎢ 8 48 ⎥⎦ 0
2π
=
≈ 2.0944
3

The square at x has sides of length 2 4 − x 2 , as
shown.
2

⎛ 2 4 − x 2 ⎞ dx = 2 4(4 − x 2 )dx
⎜
⎟
∫−2
−2 ⎝
⎠

V =∫

2

2
⎡
x3 ⎤
⎡⎛ 8 ⎞ ⎛
8 ⎞ ⎤ 128
= 4 ⎢ 4 x − ⎥ = 4 ⎢ ⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎥ =

3
3
3 ⎠⎦
3
⎠ ⎝
⎣⎝
⎣⎢
⎦⎥ −2
≈ 42.67

306

Section 5.2

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24. The area of each cross section perpendicular to
1
the x-axis is (4) ⎛⎜ 2 4 − x 2 ⎞⎟ = 4 4 − x 2 .
⎠
2 ⎝
The area of a semicircle with radius 2 is
2

∫−2

Thus, the volume inside the “+” for two cylinders
of radius r and length L is
V = vol. of cylinders - vol. of common region
⎛2 ⎞
= 2(π r 2 L) − 8 ⎜ r 3 ⎟
⎝3 ⎠
16
= 2π r 2 L − r 3
3

4 − x 2 dx = 2π .
2

V = ∫ 4 4 − x 2 dx = 4(2π) = 8π ≈ 25.13
−2

25. The square at x has sides of length
V =∫

π/2

−π / 2

cos x .

cos xdx = [sin x]π−π/ 2/ 2 = 2

26. The area of each cross section perpendicular to

the x-axis is [(1 − x 2 ) − (1 − x 4 )]2 = x8 − 2 x 6 + x 4 .

1

V = ∫ ( x8 − 2 x6 + x 4 )dx
−1

1

16
2
1 ⎤
⎡1
= ⎢ x9 − x 7 + x5 ⎥ =
≈ 0.051
7
5 ⎦ −1 315
⎣9

30. From Problem 28, the volume of one octant of
2
the common region is r 3 . We can find the
3
volume of the “T” similarly. Since the “T” has
one-half the common region of the “+” in
Problem 28, the volume of the “T” is given by
V = vol. of cylinders - vol. of common region
⎛2 ⎞
=(π r 2 )( L1 + L2 ) − 4 ⎜ r 3 ⎟
⎝3 ⎠
With r = 2, L1 = 12, and L2 = 8 (inches), the
volume of the “T” is

V = vol. of cylinders - vol. of common region

27. The square at x has sides of length 1 − x 2 .

⎛2 ⎞
=(π r 2 )( L1 + L2 ) − 4 ⎜ r 3 ⎟
⎝3 ⎠
⎛2 ⎞
= (π 22 )(12 + 8) − 4 ⎜ 23 ⎟
⎝3 ⎠
64
= 80π −
in 3
3

3 ⎤1

⎡
1
x
2
V = ∫ (1 − x 2 )dx = ⎢ x − ⎥ =
≈ 0.67
0
3 ⎦⎥
3
⎣⎢
0

28. From Problem 27 we see that horizontal cross

sections of one octant of the common region are
squares. The length of a side at height y is
r − y where r is the common radius of the
cylinders. The volume of the “+” can be found
by adding the volumes of each cylinder and
subtracting off the volume of the common region
(which is counted twice). The volume of one
octant of the common region is
r 2
1 2 r
2
2
∫0 (r − y )dy = r y − 3 y |0
1
2
= r3 − r3 = r3
3
3
Thus, the volume of the “+” is
V = vol. of cylinders - vol. of common region
2

≈ 229.99 in 3

2

⎛2 ⎞
=2(π r 2 l ) − 8 ⎜ r 3 ⎟
⎝3 ⎠
128

⎛2
⎞
= 2π (22 )(12) − 8 ⎜ (2)3 ⎟ = 96π −
3
3
⎝
⎠
≈ 258.93 in 2

31. From Problem 30, the general form for the
volume of a “T” formed by two cylinders with
the same radius is
V = vol. of cylinders - vol. of common region
⎛2 ⎞
=(π r 2 )( L1 + L2 ) − 4 ⎜ r 3 ⎟
⎝3 ⎠
8
= π r 2 ( L1 + L2 ) − r 3
3

32. The area of each cross section perpendicular to

the x-axis is

1 ⎡1
π
2 ⎢⎣ 2

(

)

⎤
x − x2 ⎥
⎦

2

π 4
( x − 2 x5 / 2 + x).
8
π 1
V = ∫ ( x 4 − 2 x5 / 2 + x)dx
8 0
=

1

29. Using the result from Problem 28, the volume of
one octant of the common region in the “+” is
r 2
1 2 r
2
2
∫0 (r − y )dy = r y − 3 y |0
1
2
= r3 − r3 = r3
3
3

Instructor’s Resource Manual

=

π ⎡1 5 4 7 / 2 1 2 ⎤
9π
x − x
+ x ⎥ =
≈ 0.050
⎢
8 ⎣5
7
2 ⎦ 0 560

Section 5.2

307

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33. Sketch the region.

8

3
⎡ 24
⎤
= π ⎢ y5 / 3 − y 7 / 3 ⎥

7
⎣5
⎦0
⎛ 768 384 ⎞ 3456π
= π⎜
−
≈ 310.21
⎟=
7 ⎠
35
⎝ 5

b. Revolving about the line y = 8, the radius of

the disk at x is 8 − x3 = 8 − x3 / 2 .
4

a.

V = π∫ (8 − x3 / 2 )2 dx
0

Revolving about the line x = 4, the radius of

4

= π∫ (64 − 16 x3 / 2 + x3 )dx

the disk at y is 4 − 3 y 2 = 4 − y 2 / 3 .

0

4

32
1 ⎤
⎡
= π ⎢64 x − x5 / 2 + x 4 ⎥
5
4 ⎦0
⎣
1024
⎡
⎤ 576π
= π ⎢ 256 −
+ 64 ⎥ =
≈ 361.91
5
5
⎣
⎦

8

V = π∫ (4 − y 2 / 3 ) 2 dy
0

8

= π∫ (16 − 8 y 2 / 3 + y 4 / 3 )dy

0

8

24 5 / 3 3 7 / 3 ⎤
⎡
= π ⎢16 y −
y
+ y ⎥
5
7
⎣
⎦0
768 384 ⎞
⎛
= π ⎜ 128 −
+
⎟
5
7 ⎠
⎝
1024π
=
≈ 91.91
35

b. Revolving about the line y = 8, the inner

radius of the disk at x is 8 − x3 = 8 − x3 / 2 .
4

V = π∫ ⎡82 − (8 − x3 / 2 )2 ⎤ dx
⎦
0⎣
4

= π∫ (16 x3 / 2 − x3 )dx
0

4

1 ⎤
⎡ 32
⎛ 1024
⎞
= π ⎢ x5 / 2 − x 4 ⎥ = π ⎜
− 64 ⎟
5
4 ⎦0
⎣5
⎝
⎠
704π
=
≈ 442.34
5

34. Sketch the region.

35. The area of a quarter circle with radius 2 is
2

∫0

2

∫0

4 − y 2 dy = π .

⎡ 2 4 − y 2 + 4 − y 2 ⎤ dy
⎢⎣
⎥⎦

= 2∫

2

0

2

4 − y 2 dy + ∫ (4 − y 2 )dy
0

2

1 ⎤
⎡
⎛ 8⎞
= 2π + ⎢ 4 y − y 3 ⎥ = 2π + ⎜ 8 − ⎟

3 ⎦0
3⎠
⎣
⎝
16
= 2π + ≈ 11.62
3

36. Let the x-axis lie along the diameter at the base
perpendicular to the water level and slice
perpendicular to the x-axis. Let x = 0 be at the

center. The slice has base length 2 r 2 − x 2 and
hx
.
height
r
2h r
V=
x r 2 − x 2 dx
r ∫0
r
2h ⎡ 1 2
2h ⎛ 1 3 ⎞ 2 2
2 3/ 2 ⎤
=
− r −x
⎥ = r ⎜3r ⎟ = 3r h
r ⎢⎣ 3
⎝

⎠
⎦0

(

)

37. Let the x-axis lie on the base perpendicular to the
diameter through the center of the base. The slice
a.

Revolving about the line x = 4, the inner
radius of the disk at y is 4 − 3 y 2 = 4 − y 2 / 3 .
8⎡ 2
4
0 ⎢⎣

V = π∫
8

(

− 4− y

) ⎥⎦ dy

2/3 2 ⎤

= π∫ (8 y 2 / 3 − y 4 / 3 )dy
0

308

Section 5.2

at x is a rectangle with base of length 2 r 2 − x 2
and height x tan θ .
r

V = ∫ 2 x tan θ r 2 − x 2 dx
0

r

⎡ 2
⎤
= ⎢ − tan θ (r 2 − x 2 )3 / 2 ⎥
⎣ 3
⎦0
2
= r 3 tan θ
3

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38. a.

y
k
Slice horizontally.

1
3
3 2
r⋅
r=
r .
2
2
4
The center of an equilateral triangle is

is A =

x=4

2

2 3
1
⋅
r=
r from a vertex. Then the
3 2
3
height of a regular tetrahedron is

⎛ y⎞
⎛ y⎞
ΔV ≈ π ⎜⎜ 4 ⎟⎟ Δy = π ⎜⎜
⎟⎟ Δy
⎝ k⎠
⎝ k⎠
If the depth of the tank is h, then
V = π∫

h

0

=

2π

2

⎛ 1 ⎞
h = r2 − ⎜
r⎟ =
⎝ 3 ⎠

h

y
π ⎡ 2 3/ 2 ⎤
dy =

y ⎥
k
k ⎢⎣ 3
⎦0

V=

h3 / 2 .

3 k
The volume as a function of the depth of the
2π 3 / 2
y
tank is V ( y ) =
3 k

dy −m k
dy
=
= − m y and
dt
π
dt
k
which is constant.
π

y

39. Let A lie on the xy-plane. Suppose ΔA = f ( x)Δx

where f(x) is the length at x, so A = ∫ f ( x)dx .
Slice the general cone at height z parallel to A.
The slice of the resulting region is Az and ΔAz
is a region related to f(x) and Δ x by similar
triangles:
⎛ z⎞
⎛ z⎞
ΔAz = ⎜ 1 − ⎟ f ( x) ⋅ ⎜1 − ⎟ Δx
⎝ h⎠
⎝ h⎠
2

⎛ z⎞
= ⎜ 1 − ⎟ f ( x ) Δx
⎝ h⎠
⎛ z⎞
Therefore, Az = ⎜ 1 − ⎟
⎝ h⎠

2

∫

2

⎛ z⎞
f ( x)dx = ⎜ 1 − ⎟ A.
⎝ h⎠

2

2
3

r.

1
2 3
Ah =
r
3
12

40. If two solids have the same cross sectional area at
every x in [a, b], then they have the same volume.
41. First we examine the cross-sectional areas of
each shape.
Hemisphere: cross-sectional shape is a circle.

dV
= −m y .
b. It is given that
dt
dV
π 1/ 2 dy
From part a,
y
.
=

dt
dt
k

Thus,

2 2
r =
3

2

h⎛
z⎞
⎛ z⎞
ΔV ≈ Az Δz = A ⎜1 − ⎟ Δz V = A∫ ⎜ 1 − ⎟ dz
0⎝
h⎠
⎝ h⎠

The radius of the circle at height y is r 2 − y 2 .
Therefore, the cross-sectional area for the
hemisphere is
Ah = π ( r 2 − y 2 )2 = π (r 2 − y 2 )
Cylinder w/o cone: cross-sectional shape is a
washer. The outer radius is a constant , r. The
inner radius at height y is equal to y. Therefore,
the cross-sectional area is
A2 = π r 2 − π y 2 = π (r 2 − y 2 ) .
Since both cross-sectional areas are the same, we

can apply Cavaleri’s Principle. The volume of
the hemisphere of radius r is
V = vol. of cylinder - vol. of cone
1
= π r 2h − π r 2h
3
2 2
= πr h
3
With the height of the cylinder and cone equal to
r, the volume of the hemisphere is
2
2
V = π r 2 (r ) = π r 3
3
3

h

⎡ h ⎛ z ⎞3 ⎤
1
= A ⎢− ⎜ 1 − ⎟ ⎥ = Ah.
3
⎢⎣ 3 ⎝ h ⎠ ⎥⎦
0

a.

A = πr 2
1

1
V = Ah = πr 2 h
3
3

b. A face of a regular tetrahedron is an
equilateral triangle. If the side of an
equilateral triangle has length r, then the area

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Section 5.2

309

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


5.3 Concepts Review

3. a, b.

1. 2πx f ( x)Δx
2

2

0

0

2. 2π∫ x 2 dx; π∫ (4 − y 2 )dy
2

3. 2π∫ (1 + x) x dx
0

2

4. 2π∫ (1 + y )(2 − y )dy

c. ΔV ≈ 2πx x Δx = 2πx3 / 2 Δx

0

3

3
⎡2
⎤
d, e. V = 2π ∫ x3 / 2 dx = 2π ⎢ x5 / 2 ⎥
0
⎣5
⎦0

Problem Set 5.3

=

1. a, b.

36 3
π ≈ 39.18
5

4. a,b.

⎛1⎞
c. ΔV ≈ 2πx ⎜ ⎟ Δx = 2πΔx
⎝ x⎠

c. ΔV ≈ 2πx(9 − x 2 )Δx = 2π(9 x − x3 )Δx

d,e. V = 2π ∫ dx = 2π [ x ]1 = 6π ≈ 18.85
4

4

1

2. a, b.

3

3
1 ⎤
⎡9
d, e. V = 2π ∫ (9 x − x3 )dx = 2π ⎢ x 2 − x 4 ⎥
0

4 ⎦0
⎣2
81
81
81
π
⎛
⎞
= 2π ⎜ − ⎟ =
≈ 127.23
2
⎝ 2 4⎠

5. a, b.

c. ΔV ≈ 2πx ( x 2 )Δx = 2πx3Δx
1

1
π
⎡1 ⎤
d, e. V = 2π∫ x3 dx = 2π ⎢ x 4 ⎥ = ≈ 1.57
0
⎣ 4 ⎦0 2

c.

ΔV ≈ 2π(5 − x) x Δx

= 2π(5 x1/ 2 − x3 / 2 )Δx

5

d, e. V = 2π ∫ (5 x1/ 2 − x3 / 2 )dx
0

5

2
⎡10
⎤
= 2π ⎢ x3 / 2 − x5 / 2 ⎥
5
⎣3
⎦0
⎛ 50 5
⎞ 40 5
= 2π ⎜⎜
− 10 5 ⎟⎟ =
π ≈ 93.66
3
⎝ 3
⎠

310

Section 5.3

Instructor’s Resource Manual

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


6. a, b.

c. ΔV ≈ 2πx (3 x − x 2 )Δx = 2π(3 x 2 − x3 )Δx
3

3
1 ⎤
⎡
d, e. V = 2π ∫ (3 x 2 − x3 )dx = 2π ⎢ x3 − x 4 ⎥
0
4 ⎦0
⎣
81 ⎞ 27π
⎛
= 2π ⎜ 27 − ⎟ =
≈ 42.41
4⎠
2
⎝

9. a, b.
c. ΔV ≈ 2π(3 − x)(9 − x 2 )Δx
= 2π(27 − 9 x − 3 x 2 + x3 )Δx
3

d, e. V = 2π ∫ (27 − 9 x − 3 x 2 + x3 )dx
0

3

9
1 ⎤
⎡
= 2π ⎢ 27 x − x 2 − x3 + x 4 ⎥
2
4 ⎦0
⎣
81
81 ⎞ 135π
⎛
= 2π ⎜ 81 − − 27 + ⎟ =
≈ 212.06
2
4⎠
2
⎝

c. ΔV ≈ 2πy ( y 2 )Δy = 2πy 3Δy

7. a, b.

1

1
π
⎡1 ⎤
d, e. V = 2π ∫ y 3 dy = 2π ⎢ y 4 ⎥ = ≈ 1.57

0
⎣ 4 ⎦0 2

10. a, b.

⎡⎛ 1
⎤
⎞
c. ΔV ≈ 2πx ⎢⎜ x3 + 1⎟ − (1 − x) ⎥ Δx
4
⎠
⎣⎝
⎦
1
⎛
⎞
= 2 π ⎜ x 4 + x 2 ⎟ Δx
⎝4
⎠
1⎛1 4
x
0 ⎜⎝ 4

d, e. V = 2π ∫

c. ΔV ≈ 2πy

2⎞

+ x ⎟ dx

⎠
1

1 ⎤
⎡1
⎛ 1 1⎞
= 2π ⎢ x 5 + x 3 ⎥ = 2π ⎜ + ⎟
3 ⎦0
⎣ 20
⎝ 20 3 ⎠
23π
=
≈ 2.41
30

(

)

y + 1 Δ y = 2 π( y 3 / 2 + y ) Δ y

4

d, e. V = 2π ∫ ( y 3 / 2 + y ) dy
0

4

1 ⎤
⎡2

⎛ 64
⎞
= 2π ⎢ y 5 / 2 + y 2 ⎥ = 2π ⎜ + 8 ⎟
2 ⎦0
⎣5
⎝ 5
⎠
208π
=
≈ 130.69
5

8. a, b.
11. a, b.

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15.

c. ΔV ≈ 2π(2 − y ) y 2 Δy = 2π(2 y 2 − y 3 )Δy
2

2
1 ⎤
⎡2
d, e. V = 2π ∫ (2 y 2 − y 3 )dy = 2π ⎢ y 3 − y 4 ⎥
0
4 ⎦0
⎣3
⎛ 16
⎞ 8π
= 2π ⎜ − 4 ⎟ =
≈ 8.38
3
⎝
⎠ 3

12. a, b.

c. ΔV ≈ 2π(3 − y )

(

(

)

2 y + 1 Δy

V = 2π ∫

2

0

(3 + 3

3 ⎛ 1 ⎞
3 1
V = 2π∫ x ⎜ ⎟ dx = 2π ∫
dx
1 ⎝ x3 ⎠
1 x2

)

)

2 y1/ 2 − y − 2 y 3 / 2 dy
2

x3

b.

c.

d.

⎡
1
2 2 5/ 2 ⎤

= 2π ⎢3 y + 2 2 y 3 / 2 − y 2 −
y ⎥
2
5
⎣
⎦0

1

A=∫

= 2π 3 + 3 2 y1/ 2 − y − 2 y 3 / 2 Δy

d, e.

3

1

a.

16 ⎞ 88π
⎛
= 2π ⎜ 6 + 8 − 2 − ⎟ =
≈ 55.29
5⎠
5
⎝

dx

2
⎤
3 ⎡⎛ 1
⎞
V = π ∫ ⎢⎜ + 1⎟ − (−1)2 ⎥ dx
1 ⎢⎝ x 3
⎥⎦
⎠
⎣
3⎛ 1
2 ⎞
= π∫ ⎜
+ ⎟ dx
1 ⎝ x6
x3 ⎠
3
⎛ 1 ⎞
V = 2π ∫ (4 − x) ⎜ ⎟ dx
1
⎝ x3 ⎠
3⎛ 4
1 ⎞
= 2π ∫ ⎜ − ⎟ dx
1 ⎝ x3 x 2 ⎠

16.
13. a.

b

π ∫ ⎡ f ( x) − g ( x)
2

⎣

a

2⎤

⎦

dx

b.

2π∫ x [ f ( x) − g ( x) ] dx

c.

2π∫ ( x − a ) [ f ( x) − g ( x) ] dx

d.

2π∫ (b − x) [ f ( x) − g ( x) ] dx

a.

A = ∫ ( x3 + 1) dx

d
π∫ ⎡ f ( y ) 2 − g ( y ) 2 ⎤ dy
⎦
c ⎣

b.

V = 2π ∫ x( x3 + 1)dx = 2π ∫ ( x 4 + x)dx

y [ f ( y ) − g ( y ) ] dy

c.

2
V = π ∫ ⎡ ( x3 + 2)2 − (−1)2 ⎤
⎦
0 ⎣

14. a.

b

a

b

a

b

a

d

b.

2π ∫

c.

2π∫ (3 − y ) [ f ( y ) − g ( y ) ] dy

c

2

0

2

2

0

0

2

= π ∫ ( x 6 + 4 x3 + 3)dx

d

0

c

d.

2

V = 2π ∫ (4 − x)( x3 + 1)dx
0

2

= 2π ∫ (− x 4 + 4 x3 − x + 4)dx
0

312

Section 5.3

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


17. To find the intersection point, solve
y=

y=

y3
.
32

21. To find the intersection point, solve
sin( x 2 ) = cos( x 2 ) .
tan( x 2 ) = 1

y6
1024

π
4
π
x=
2
x2 =

y − 1024 y = 0
6

y ( y 5 − 1024) = 0
y = 0, 4
4 ⎛
y3 ⎞
V = 2π∫ y ⎜ y − ⎟ dy
0 ⎜

32 ⎟⎠
⎝

0

4⎛
y ⎞
= 2π ∫ ⎜ y 3 / 2 −
⎟ dy
0 ⎜
32 ⎟⎠
⎝

0

4

22. V = 2π ∫

⎛
y3 ⎞
18. V = 2π∫ (4 − y ) ⎜ y −
⎟ dy
⎜
0
32 ⎟⎠
⎝
4⎛
y3 y 4 ⎞
= 2π∫ ⎜ 4 y1/ 2 − y 3 / 2 −

+
⎟ dy
0 ⎜
8 32 ⎟⎠
⎝

= 2π ∫
= 2π ∫

0

23. a.

(2 x + x sin x)dx
2 x dx + 2π ∫

2π

0

x sin x dx
2π

⎞ dx
⎟
⎠

The curves intersect when x = 0 and x = 1.
1

1

0

0

V = π ∫ [ x 2 − ( x 2 )2 ] dx = π ∫ ( x 2 − x 4 )dx

2

20. y = ± a 2 − x 2 , − a ≤ x ≤ a

1

1 ⎤
⎡1
⎛ 1 1 ⎞ 2π
= π ⎢ x3 − x5 ⎥ = π ⎜ − ⎟ =
≈ 0.42
5 ⎦0
⎣3
⎝ 3 5 ⎠ 15

b.

1

1

0

0

V = 2π ∫ x( x − x 2 )dx = 2π ∫ ( x 2 − x3 )dx
1

1 ⎤
⎡1
⎛1 1⎞ π
= 2π ⎢ x3 − x 4 ⎥ = 2π ⎜ − ⎟ =
4 ⎦0
⎣3
⎝3 4⎠ 6
≈ 0.52

c.

Slice perpendicular to the line y = x. At
(a, a), the perpendicular line has equation
y = −( x − a ) + a = − x + 2a . Substitute
y = –x + 2a into y = x 2 and solve for x ≥ 0 .

x a 2 − x 2 dx
a

⎛1
⎞
⎡ 1
⎤
= 4πb ⎜ πa 2 ⎟ − 4π ⎢ − (a 2 − x 2 )3 2 ⎥ = 2π2 a 2b

2
3
⎝
⎠
⎣
⎦ −a
(Note that the area of a semicircle with radius a is
a
1 2
2
2
∫−a a − x dx = 2 πa .)

Instructor’s Resource Manual

)

2 − 1 ≈ 1.30

= 2π(4π2 ) + 2π(−2π) = 4π 2 (2π − 1) ≈ 208.57

b

−a

(

x(2 + sin x)dx

2π

b
⎡ 1
⎤
= 4π∫ x b 2 − x 2 dx = 4π ⎢ − (b 2 − x 2 )3 / 2 ⎥
a
3
⎣
⎦a
4
π
⎡1 2
⎤
= 4π ⎢ (b − a 2 )3 / 2 ⎥ =
(b 2 − a 2 )3 / 2
⎣3
⎦ 3

−a

π/2

= 2π ⎡ x 2 ⎤ + 2π [sin x − x cos x ]0
⎣ ⎦0

y = − b 2 − x 2 , and x = a. When R is revolved
about the y-axis, it produces the desired solid.
b
V = 2π∫ x ⎛⎜ b 2 − x 2 + b 2 − x 2 ⎞⎟ dx
a ⎝

⎠

a

2π

0
2π

19. Let R be the region bounded by y = b − x ,

a 2 − x 2 dx − 4π ∫

2π

0

4
⎡ 8 3 / 2 2 5 / 2 y 4 y5 ⎤
= 2π ⎢ y
− y
−
+
⎥
5
32 160 ⎥⎦
⎣⎢ 3
0
32 ⎞ 208π
⎛ 64 64

= 2π ⎜ − − 8 + ⎟ =
≈ 43.56
5
5 ⎠
15
⎝ 3

a
V = 2π∫ (b − x) ⎛⎜ 2 a 2 − x 2
−a
⎝

⎡ x cos( x 2 ) − x sin( x 2 ) ⎤ dx
⎣
⎦

⎡⎛ 1
1 ⎞ 1⎤
= 2π ⎢⎜
+
⎟− ⎥ = π
⎣⎝ 2 2 2 2 ⎠ 2 ⎦

4

2

x ⎡ cos( x 2 ) − sin( x 2 ) ⎤ dx
⎣
⎦

1
⎡1
⎤
= 2π ⎢ sin( x 2 ) + cos( x 2 ) ⎥
2
2
⎣
⎦0

⎡2
y5 ⎤
⎛ 64 32 ⎞ 64π
= 2π ⎢ y 5 / 2 −
⎥ = 2π ⎜ − ⎟ =
5
160
5 ⎠
5
⎝ 5
⎣⎢
⎦⎥ 0
≈ 40.21

a

π /2

= 2π ∫

4

= 4πb ∫

π/2

V = 2π ∫

x 2 + x − 2a = 0
−1 ± 1 + 8a
2
−1 + 1 + 8a
x=
2
x=

Section 5.3

313

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


Substitute into y = –x + 2a, so
1 + 4a − 1 + 8a
. Find an expression for
y=
2
r 2 , the square of the distance from (a, a) to

⎛ −1 + 1 + 8a 1 + 4a − 1 + 8a ⎞
,
⎜⎜
⎟⎟ .
2
2
⎝
⎠
⎡
−1 + 1 + 8a ⎤
r = ⎢a −
⎥
2
⎣
⎦

24. ΔV ≈ 4πx 2 Δx
r

r
4
⎡1 ⎤
V = 4π∫ x 2 dx = 4π ⎢ x3 ⎥ = πr 3
0
⎣ 3 ⎦0 3

2

V=

2

⎡ 1 + 4a − 1 + 8a ⎤
+ ⎢a −
⎥
2
⎣
⎦

⎡ 2a + 1 − 1 + 8a ⎤
=⎢
⎥
2
⎣
⎦

r2

S
r2

S Δx
r

∫0

r

x 2 dx =

S ⎡1 3 ⎤
1
x ⎥ = rS
2 ⎢⎣ 3
⎦0 3
r

2

5.4 Concepts Review

2

1. Circle

⎡ 2a + 1 − 1 + 8a ⎤
+ ⎢−
⎥
2
⎣
⎦
⎡ 2 a + 1 − 1 + 8a ⎤
= 2⎢
⎥
2
⎣
⎦

x2

25. ΔV ≈

x 2 + y 2 = 16 cos 2 t + 16sin 2 t = 16

2

2. x ; x 2 + 1

2

3.

= 2a 2 + 6a + 1 − 2a 1 + 8a − 1 + 8a

2
2
∫a [ f ′(t )] + [ g ′(t )] dt

b

4. Mean Value Theorem (for derivatives)

ΔV ≈ πr Δa
2

1

V = π ∫ (2a 2 + 6a + 1
0

Problem Set 5.4

− 2a 1 + 8a − 1 + 8a ) da
1

1
⎡2
⎤
= π ⎢ a3 + 3a 2 + a − (1 + 8a)3 / 2 ⎥
12
⎣3
⎦0

1.

f ( x) = 4 x3 / 2 , f ′( x ) = 6 x1/ 2
L=∫

5

1/ 3

1

−π ∫ 2a 1 + 8a da

1 + (6 x1/ 2 ) 2 dx = ∫

5

1/ 3

1 + 36 x dx

5

⎡1 2
⎤
= ⎢ ⋅ (1 + 36 x)3 / 2 ⎥
⎣ 36 3
⎦1/ 3
1
=
181 181 − 13 13 ≈ 44.23
54

0

⎡⎛ 2
9 ⎞ ⎛ 1 ⎞⎤
= π ⎢⎜ + 3 + 1 − ⎟ − ⎜ − ⎟ ⎥
4 ⎠ ⎝ 12 ⎠ ⎦
⎣⎝ 3

(

1

−π ∫ 2a 1 + 8a da

)

0

=

1
5π
− π ∫ 2a 1 + 8a da
0
2

To integrate

1

∫0 2a

1 + 8a da , use the

2.

f ( x) =
L=∫

2

1

substitution u = 1 + 8a.

1
9 1
1
∫0 2a 1 + 8a da = ∫1 4 (u − 1) u 8 du
1 9 3 / 2 1/ 2
=
(u
− u )du
32 ∫1

=∫

2

1

2 2
( x + 1)3 / 2 , f ′( x) = 2 x( x 2 + 1)1/ 2
3
2

1 + ⎡ 2 x ( x 2 + 1)1/ 2 ⎤ dx
⎣
⎦
2

4 x 4 + 4 x 2 + 1 dx = ∫ (2 x 2 + 1)dx
1

2

⎡2
⎤
⎛ 16
⎞ ⎛ 2 ⎞ 17
= ⎢ x3 + x ⎥ = ⎜ + 2 ⎟ − ⎜ + 1⎟ =
≈ 5.67
⎣3
⎦1 ⎝ 3
⎠ ⎝3 ⎠ 3

9

=

1 ⎡ 2 5/ 2 2 3/ 2 ⎤
u
− u ⎥
32 ⎢⎣ 5
3
⎦1

=

1 ⎡⎛ 486
⎞ ⎛ 2 2 ⎞ ⎤ 149
− 18 ⎟ − ⎜ − ⎟ ⎥ =
⎜
⎢
32 ⎣⎝ 5

⎠ ⎝ 5 3 ⎠ ⎦ 60

V=

314

5π 149π π
−
=
≈ 0.052
2
60
60

Section 5.4

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material
may be reproduced, in any form or by any means, without permission in writing from the publisher.


3.

f ( x) = (4 − x 2 / 3 )3 / 2 ,
f ′( x) =

3
⎛ 2
⎞

(4 − x 2 / 3 )1/ 2 ⎜ − x −1/ 3 ⎟
2
⎝ 3
⎠

g ( y) =

= − x −1/ 3 (4 − x 2 / 3 )1/ 2
L=∫

2

1 + ⎡ − x −1/ 3 (4 − x 2 / 3 )1/ 2 ⎤ dx
⎣
⎦

8

1

=∫

8

4x

1

y5
1

+
30 2 y 3

6. x =

−2 / 3

8

dx = ∫ 2 x

−1/ 3

1

L=∫

3

1

dx

8

⎡3
⎤
= 2 ⎢ x 2 / 3 ⎥ = 3(4 − 1) = 9
2
⎣

⎦1

=∫

y5
1
y4
3
+
, g ′( y ) =
−
3
30 2 y
6 2 y4
⎛ y4
3
1+ ⎜
−
⎜ 6 2 y4
⎝

2

⎞
⎟ dy
⎟
⎠
2

⎛ y4

3 ⎞
+
⎜
⎟ dy
⎜ 6 2 y4 ⎟
⎝
⎠

3
y8 1
9
+ +
dy = ∫
1
36 2 4 y8

3

1

3

4.

⎡ y5
3 ⎛ y4
3 ⎞
1 ⎤
=∫ ⎜
+

dy = ⎢ −
⎟
⎥
4
3
1 ⎜ 6
2 y ⎟⎠
⎝
⎣⎢ 30 2 y ⎦⎥1
⎛ 81 1 ⎞ ⎛ 1 1 ⎞ 1154
= ⎜ − ⎟−⎜ − ⎟ =
≈ 8.55
⎝ 10 54 ⎠ ⎝ 30 2 ⎠ 135

x 4 + 3 x3 1
f ( x) =
=
+
6x
6 2x
f ′( x) =
L=∫

⎛ x2
1
1+ ⎜
−
⎜ 2 2 x2
⎝

3

1

=∫

x2
1
−
2 2 x2
2

⎞
⎟ dx
⎟
⎠

3
x4 1
1
+ +
dx = ∫
1
4 2 4 x4

3

1

7.

⎛ x2
1
+
⎜
⎜ 2 2 x2
⎝

2

⎞
⎟ dx
⎟
⎠

3

⎡ x3 1 ⎤
3 ⎛ x2
1 ⎞
dx = ⎢ − ⎥
=∫ ⎜
+
⎟
1 ⎜ 2
2 x 2 ⎟⎠
⎝
⎣⎢ 6 2 x ⎦⎥1
⎛ 9 1 ⎞ ⎛ 1 1 ⎞ 14
= ⎜ − ⎟−⎜ − ⎟ =

≈ 4.67
⎝2 6⎠ ⎝6 2⎠ 3

dx 2 dy
=t ,
=t
dt
dt
L=∫

y4
1
y3 1
5. g ( y ) =
+
, g ′( y ) =
−
16 2 y 2
4 y3
L=∫

−2

−3

=∫

−2

−3

(t 2 )2 + (t )2 dt = ∫

1

t 4 + t 2 dt

0

1

(

)

1
1
⎡1
⎤
= ∫ t t 2 + 1 dt = ⎢ (t 2 + 1)3 / 2 ⎥ = 2 2 − 1
0
3
⎣
⎦0 3
≈ 0.61

2

⎛ y3 1 ⎞
1+ ⎜

−
⎟ dy
⎜ 4 y3 ⎟
⎝
⎠
−2
y6 1 1
+ +
dy = ∫
6
−3
16 2 y

1

0

2

⎛ y3 1 ⎞
+
⎜
⎟ dy
⎜ 4 y3 ⎟
⎝
⎠

8.

−2

⎛ y3 1 ⎞
⎡ y4
1 ⎤
dy = − ⎢ −
= ∫ −⎜
+
⎟
⎥
3⎟
2
−3 ⎜ 4
y ⎠
⎝
⎣⎢ 16 2 y ⎦⎥ −3
−2

⎡⎛ 1 ⎞ ⎛ 81 1 ⎞ ⎤ 595
= − ⎢⎜ 1 − ⎟ − ⎜ − ⎟ ⎥ =
≈ 4.13
⎣⎝ 8 ⎠ ⎝ 16 18 ⎠ ⎦ 144

dx
dy
= 6t ,
= 6t 2
dt
dt
L=∫

4

1

(6t )2 + (6t 2 )2 dt = ∫

4

1

36t 2 + 36t 4 dt
4

4

= ∫ 6t 1 + t 2 dt = ⎡ 2(1 + t 2 )3 / 2 ⎤
⎣
⎦1
1

(

)

= 2 17 17 − 2 2 ≈ 134.53

Instructor’s Resource Manual

Section 5.4

315

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may be reproduced, in any form or by any means, without permission in writing from the publisher.


9.

12. x = y +

3
2

3
g ( y ) = y + , g ′( y ) = 1
2
L=∫

3

1

3

1 + (1)2 = 2 ∫ dy = 2 2
1

3 5
= .
2 2

3 9
At y = 3, x = 3 + = .
2 2

At y = 1, x = 1 +

dx
dy
= 4 cos t ,
= −4sin t
dt
dt
L=∫

π

0

=∫

π

2

⎛9 5⎞
d = ⎜ − ⎟ + (3 − 1) 2 = 8 = 2 2
⎝2 2⎠

(4 cos t ) 2 + (−4sin t )2 dt
π

16 cos 2 t + 16sin 2 t dt = ∫ 4dt

0

0

= 4π ≈ 12.57

13.

10.

dx
dy
= 1,
= 2t
dt
dt
L=∫

2

0

−3

−2

with n = 8,

−3

2−0 ⎡
⎛1⎞
⎛1⎞
⎛3⎞
f (0) + 4 f ⎜ ⎟ + 2 f ⎜ ⎟ + 4 f ⎜ ⎟
3 × 8 ⎢⎣
⎝4⎠
⎝2⎠
⎝4⎠
⎛5⎞
⎛3⎞
⎛7⎞
+2 f (1) + 4 f ⎜ ⎟ + 2 f ⎜ ⎟ + 4 f ⎜ ⎟ + f (2)]
⎝4⎠
⎝2⎠
⎝4⎠
1
≈ [1 + 4 × 1.118 + 2 × 1.4142
12
+ 4 × 1.8028 + 2 × 2.2361
L≈

dx
dy
= 2 5 cos 2t ,
= −2 5 sin 2t
dt

dt

=∫

π/4

0

(2

5 cos 2t

) + ( −2
2

5 sin 2t

20 cos 2 2t + 20sin 2 2t dt = ∫

3

1

3

1 + (2)2 dx = 5 ∫ dx = 2 5
1

At x = 1, y = 2(1) + 3 = 5.
At x = 3, y = 2(3) + 3 = 9.

d = (3 − 1) + (9 − 5) = 20 = 2 5
2

2

dt

+ 4 × 2.6926 + 2 × 3.1623

2 5 dt

+ 4 × 3.6401 + 4.1231] ≈ 4.6468

14.

f ( x) = 2 x + 3, f ′( x) = 2
L=∫

)

π/ 4

0

5π
=
≈ 3.51
2

11.

1 + 4t 2 dt

Let f (t ) = 1 + 4t 2 . Using the Parabolic Rule

−2

0
π/ 4

2

0

−1
−1

L=∫

12 + (2t )2 dt = ∫

dx
dy
1
= 2t ,
=
dt
dt 2 t
L≈∫

4

1

2

4
1
⎛ 1 ⎞
2
(2t ) 2 + ⎜
⎟ dt = ∫1 4t + dt
4t
⎝2 t ⎠

Let f (t ) = 4t 2 +

2

1
. Using the Parabolic Rule
4t

with n = 8,

4 −1 ⎡
⎛ 11 ⎞
⎛ 14 ⎞
f (1) + 4 f ⎜ ⎟ + 2 f ⎜ ⎟
⎢

3× 8 ⎣
⎝8⎠
⎝8⎠
⎛ 17 ⎞
⎛ 20 ⎞
⎛ 23 ⎞
⎛ 26 ⎞
+4 f ⎜ ⎟ + 2 f ⎜ ⎟ + 4 f ⎜ ⎟ + 2 f ⎜ ⎟
8
8
8
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ 8 ⎠
⎤ 1
⎛ 29 ⎞
+4 f ⎜ ⎟ + f (4) ⎥ ≈ ( 2.0616 + 4 × 2.8118
⎝ 8 ⎠
⎦ 8
+2 × 3.562 + 4 × 4.312 + 2 × 5.0621 + 4 × 5.8122
2 × 6.5622 + 4 × 7.3122 + 8.0623) ≈ 15.0467
L≈

316

Section 5.4

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material
may be reproduced, in any form or by any means, without permission in writing from the publisher.


15.

dx
dy
= cos t ,
= −2sin 2t
dt
dt
L=∫

π 2

0

=∫

π 2

0

( cos t )2 + ( −2sin 2t )2 dt

dx
dy
= 1,
= sec 2 t

dt
dt
π 4

0

π 4

0

9a 2 cos 2 t sin 2 t (sin 2 t + cos 2 t ) dt
π/ 2

⎡ 1
⎤
3a cos t sin tdt = 3a ⎢ − cos 2 t ⎥
0
2
⎣
⎦0
(The integral can also be evaluated as

=

3a
2

π/ 2

⎡1

⎤
3a ⎢ sin 2 t ⎥
with the same result.)
⎣2
⎦0
The total length is 6a.

18. a.

( )

p = aθ
OT = length PT

b. From Figure 18 of the text,
PQ PQ
QC QC
sin θ =
=
and cos θ =
=
.
a
a
PC
PC
Therefore PQ = a sin θ and QC = a cos θ .
x = OT − PQ = aθ − a sin θ = a (θ − sin θ )
y = CT − CQ = a − a cos θ = a (1 − cos θ )

1 + sec 4 t dt

Let f (t ) = 1 + sec4 t . Using the Parabolic
⎛π ⎞
f (0) + 4 f ⎜ ⎟
⎢
3× 8 ⎣
⎝ 32 ⎠
⎛ 2π ⎞
⎛ 3π ⎞
⎛ 4π ⎞
⎛ 5π ⎞
+2 f ⎜
⎟+4f ⎜ ⎟+2f ⎜
⎟+4f ⎜
⎟
⎝ 32 ⎠
⎝ 32 ⎠
⎝ 32 ⎠
⎝ 32 ⎠
⎛ 6π ⎞
⎛ 7π ⎞
⎛ π ⎞⎤
+2 f ⎜
⎟+4f ⎜
⎟ + f ⎜ ⎟⎥
⎝ 32 ⎠
⎝ 32 ⎠
⎝ 4 ⎠⎦
≈

9a 2 cos 2 t sin 4 t + 9a 2 sin 2 t cos 4 t dt

π/ 2

c.

12 + (sec 2 t )2 dt = ∫

Rule with n = 8, L ≈

π/ 2

=∫

cos 2 t + 4sin 2 2t dt

π 2−0 ⎡

L=∫

=∫

0

⎛π ⎞
⎛ 2π ⎞
f (0) + 4 f ⎜ ⎟ + 2 f ⎜
⎟
3 × 8 ⎢⎣

16
⎝ ⎠
⎝ 16 ⎠
⎛ 3π ⎞
⎛ 4π ⎞
⎛ 5π ⎞
⎛ 6π ⎞
+4 f ⎜ ⎟ + 2 f ⎜
⎟+4f ⎜
⎟+2f ⎜
⎟
⎝ 16 ⎠
⎝ 16 ⎠
⎝ 16 ⎠
⎝ 16 ⎠
⎛ 7π ⎞
⎛ π ⎞⎤ π
+4 f ⎜
⎟ + f ⎜ ⎟⎥ ≈ [1 + 4 × 1.2441
16
⎝
⎠
⎝ 2 ⎠ ⎦ 48
+ 2 × 1.6892 + 4 × 2.0262 + 2 × 2.1213 + 4 × 1.9295
+2 × 1.4651 + 4 × 0.7898 + 0) ≈ 2.3241

16.

π/ 2

0

Let f (t ) = cos 2 t + 4sin 2 2t . Using the
Parabolic Rule with n = 8,
L≈

=∫

π 4−0 ⎡

π

[1.4142 + 4 × 1.4211 + 2 × 1.4425 + 4 × 1.4807
96
+2 × 1.5403 + 4 × 1.6288 + 2 × 1.7585
+4 × 1.9495 + 2.2361] ≈ 1.278

17.

19. From Problem 18,
x = a(θ − sin θ ), y = a (1 − cos θ )
dx
dy
= a(1 − cos θ ),
= a sin θ so
dθ
dθ
2

2

⎛ dx ⎞ ⎛ dy ⎞
2
2
⎜
⎟ +⎜
⎟ = [ a (1 − cos θ ) ] + [ a sin θ ]
⎝ dθ ⎠ ⎝ dθ ⎠
= a 2 − 2a 2 cos θ + a 2 cos 2 θ + a 2 sin 2 θ
= 2a 2 − 2a 2 cos θ = 2a 2 (1 − cos θ )
1 − cos θ
⎛θ ⎞
= 4a 2 sin 2 ⎜ ⎟ .
2
⎝2⎠
The length of one arch of the cycloid is
= 4a 2

2π

∫0

2π
⎛θ ⎞
⎛θ ⎞
4a 2 sin 2 ⎜ ⎟ dθ = ∫ 2a sin ⎜ ⎟ dθ
0
2
⎝ ⎠
⎝2⎠

2π

θ⎤
⎡
= 2a ⎢ −2 cos ⎥ = 2a(2 + 2) = 8a
2 ⎦0
⎣

dx
dy
= 3a cos t sin 2 t ,
= −3a sin t cos 2 t
dt
dt
The first quadrant length is L
=∫

π/ 2

0

(3a cos t sin 2 t ) 2 + (−3a sin t cos 2 t )2 dt

Instructor’s Resource Manual

Section 5.4

317

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material

may be reproduced, in any form or by any means, without permission in writing from the publisher.


20. a.

Using θ = ω t , the point P is at x = aω t − a sin(ω t ), y = a − a cos(ω t ) at time t.
dx
= aω − aω cos(ω t ) = aω (1 − cos(ω t ))
dt
dy
= aω sin(ω t )
dt
2

ds
⎡ dy ⎤ ⎡ dx ⎤
= ⎢ ⎥ +⎢ ⎥
dt
⎣ dt ⎦ ⎣ dt ⎦

2

= a 2ω 2 sin 2 (ω t ) + a 2ω 2 − 2a 2ω 2 cos(ω t ) + a 2ω 2 cos 2 (ω t ) = 2a 2ω 2 − 2a 2ω 2 cos(ω t )
1
ωt
ωt
(1 − cos(ω t )) = 2aω sin 2
= 2aω sin
2
2

2

= 2aω

ωt

π
= 1, which occurs when t = (2k + 1). The speed is a minimum when
2
ω
2k π
ωt
.
sin
= 0, which occurs when t =
2
ω

b. The speed is a maximum when sin

c.

21. a.

From Problem 18a, the distance traveled by the wheel is aθ, so at time t, the wheel has gone aθ = aω t miles.
Since the car is going 60 miles per hour, the wheel has gone 60t miles at time t. Thus, aω = 60 and the
maximum speed of the bug on the wheel is 2aω = 2(60) = 120 miles per hour.

dy
= x3 − 1

dx
L=∫

2

b.
2

1 + x3 − 1 dx = ∫ x3 / 2 dx

1

1

2

(

)

2
⎡2
⎤
= ⎢ x5 / 2 ⎥ = 4 2 − 1 ≈ 1.86
⎣5
⎦1 5

b.

L=∫

4π

0

2 − 2 cos t dt = ∫

4π

0

=∫

23.

π/6

318

1

1

0

0

1

4

Section 5.4

3

−2

π/3

x

f ( x) = 25 − x 2 , f ′( x) = −
A = 2π ∫

1 + 64sin cos x − 1 dx

π/3

0

⎡1 ⎤
= 12 37 π ⎢ x 2 ⎥ = 6 37π ≈ 114.66
⎣ 2 ⎦0

⎡ 8
⎤
8sin x cos 2 xdx = ⎢ − cos3 x ⎥
π/6
⎣ 3
⎦π / 6

1
= − + 3 ≈ 1.40
3

=∫

−1

0

f ( x) = 6 x, f ′( x) = 6

dy
= 64sin 2 x cos 4 x − 1
dx
L=∫

0

A = 2π∫ 6 x 1 + 36 dx = 12 37 π ∫ x dx

2π

2

1

at dt = ∫ at dt − ∫ at dt

a a

⎡a ⎤ ⎡a ⎤
= ⎢ t2 ⎥ − ⎢ t2 ⎥ = + = a
⎣ 2 ⎦ 0 ⎣ 2 ⎦ −1 2 2

⎛t⎞
2 sin ⎜ ⎟ dt
⎝2⎠

24.

π/3

a 2 t 2 cos 2 t + a 2t 2 sin 2 tdt

1

t⎤
⎛t⎞
⎡
L = 4 ∫ sin ⎜ ⎟ dt = ⎢ −8cos ⎥
0
2 ⎦0
⎝2⎠
⎣
= 8 + 8 = 16

22. a.

1

−1

⎛t⎞
sin ⎜ ⎟ is positive for 0 < t < 2 π , and
⎝2⎠
by symmetry, we can double the integral
from 0 to 2 π .
2π

1

−1

f ′(t ) = 1 − cos t , g ′(t ) = sin t

L=∫

dx
= − a sin t + a sin t + at cos t = at cos t
dt
dy
= a cos t − a cos t + at sin t = at sin t
dt

= 2π ∫

3

−2
3

= 2π ∫

−2

25 − x 2 1 +

25 − x 2
x2

25 − x 2

dx

25 − x 2 + x 2 dx
5dx = 10π[ x]3−2 = 50π ≈ 157.08

Instructor’s Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material
may be reproduced, in any form or by any means, without permission in writing from the publisher.


25.

f ( x) =

x3
, f ′( x) = x 2
3

7

A = 2π ∫

1

3

x
3

26.

f '( x ) = − x(r 2 − x 2 )−1/ 2

1 + x 4 dx
7

π
= 250 2 − 2 2
9

(

= 2π ∫

r

)

−r
r

= 2π ∫

f ( x) =

x6 + 2
2

=

−r

= 2π ∫

2

3 ⎛ x4
1 ⎞ x6 1
1
= 2π ∫ ⎜
+
+ +
dx
⎟
2⎟ 4
1 ⎜ 8
2
4x ⎠

4 x6
⎝
3 ⎛ x4
1 ⎞ ⎛ x3
1 ⎞
= 2π ∫ ⎜
+
⎟ ⎜ + 3 ⎟ dx
2
1 ⎜ 8
4 x ⎟⎠ ⎜⎝ 2 2 x ⎟⎠
⎝
3 ⎛ x7 3x
1 ⎞
= 2π ∫ ⎜
+ +
⎟ dx
1 ⎜ 16 16 8 x5 ⎟
⎝
⎠

r

−r
r

x4
1
x3
1

+
, f ′( x ) =
−
8 4 x2
2 2 x3

⎛ x3
3 ⎛ x4
1 ⎞
1 ⎞
+
1+ ⎜ −
A = 2π ∫ ⎜
⎟
⎟
2⎟
3⎟
⎜
1 ⎜ 8
4x ⎠
⎝
⎝ 2 2x ⎠

= 2π ∫

−r

2

r 2 − x 2 1 + ⎡ − x(r 2 − x 2 )−1/ 2 ⎤ dx

⎣
⎦
r 2 − x 2 1 + x 2 (r 2 − x 2 )−1 dx

( r 2 − x2 )(1 + x2 (r 2 − x2 )−1 )dx
r 2 − x 2 + x 2 dx
r

r 2 dx = 2π ∫ rdx = 2π rx |−r r = 4π r 2
−r

30. x = f (t ) = r cos t
y = g (t ) = r sin t
f '(t ) = −r sin t
g '(t ) = r cos t
π

A = 2π ∫ r sin t (−r sin t ) 2 + (r cos t )2 dt
0

π

= 2π ∫ r sin t r 2 sin 2 t + r 2 cos 2 tdt
0

π

= 2π ∫ r sin t r 2 dt
0

π

= 2π ∫ r 2 sin tdt = −2π r 2 cos t |π0

3

⎡ x8 3 x 2
1 ⎤
= 2π ⎢
+
−
⎥
4
⎣⎢128 32 32 x ⎦⎥1

0
2

= −2r (−1 − 1) = 4π r 2

⎡⎛ 6561 27
1 ⎞ ⎛ 1
3
1 ⎞⎤
= 2π ⎢⎜
+
−
+ − ⎟⎥
⎟−⎜
⎣⎝ 128 32 2592 ⎠ ⎝ 128 32 32 ⎠ ⎦

8429π
=
≈ 326.92
81

31. a.

The base circumference is equal to the arc
length of the sector, so 2πr = θ l. Therefore,
2πr
θ=
.
l

b. The area of the sector is equal to the lateral
surface area. Therefore, the lateral surface
1
1 ⎛ 2πr ⎞
area is l 2θ = l 2 ⎜
⎟ = πrl .
2
2 ⎝ l ⎠

dx
dy
= 1,
= 3t 2
dt
dt
1

A = 2π∫ t 3 1 + 9t 4 dt
0

1

π
⎡1
⎤
= 2π ⎢ (1 + 9t 4 )3 / 2 ⎥ =
10 10 − 1
⎣ 54
⎦ 0 27
≈ 3.56

28.

r

248π 2
≈ 122.43
9

8x

27.

A = 2π ∫

−r

⎡1
⎤
= 2π ⎢ (1 + x 4 )3 / 2 ⎥
18
⎣
⎦1
=

29. y = f ( x) = r 2 − x 2

(

)

dx
dy
= −2t ,
=2
dt
dt

c.

Assume r2 > r1 . Let l1 and l2 be the slant
heights for r1 and r2 , respectively. Then
A = πr2l2 − πr1l1 = πr2 (l1 + l ) − πr1l1 .
From part a, θ =

1

1

0

0

A = 2π∫ 2t 4t 2 + 4 dt = 8π∫ t t 2 + 1 dt
1

8π
⎡1
⎤
= 8π ⎢ (t 2 + 1)3 2 ⎥ =
(2 2 − 1) ≈ 15.32
⎣3
⎦0 3

2πr2 2πr2 2πr1
=
=
.
l2
l1 + l
l1

Solve for l1 : l1r2 = l1r1 + lr1
l1 (r2 − r1 ) = lr1
l1 =

lr1
r2 − r1

⎛ lr
⎞
⎛ lr ⎞
A = πr2 ⎜ 1 + l ⎟ − πr1 ⎜ 1 ⎟
r
−
r
⎝ 2 1 ⎠
⎝ r2 − r1 ⎠
⎡r + r ⎤
= π(lr1 + lr2 ) = 2π ⎢ 1 2 ⎥ l
⎣ 2 ⎦

Instructor’s Resource Manual

Section 5.4

319

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material
may be reproduced, in any form or by any means, without permission in writing from the publisher.