SECTION 7.3
7.3
■ Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the
y-axis.
1. y x 6x 10,
y x 6x 6
2
x 2y
3. y x ,
y 4,
2
4. y x x ,
2
2
5. y s4 x ,
■
■
x 0,
x4
■
■
■
■
■
■
■
9. x y 2,
x 0,
10. y x,
x 0,
11. y x 2,
y9
y 2,
■
■
xy2
y 0,
■
■
■
■
■
■
■
■
■ Set up, but do not evaluate, an integral for the volume
of the solid obtained by rotating the region bounded by the given
curves about the specified axis.
14 –22
14. y e x, y e x, x 1;
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about y 5
■
25. y x 4,
13. y sx,
■
22. x 4 y 2, x 8 2y 2;
24. y 0,
x0
about the y-axis
about the x-axis
■
■
■
■
■
■
■
■
■ Use a graph to estimate the x-coordinates of the points of
intersection of the given curves. Then use this information to estimate the volume of the solid obtained by rotating about the y-axis
the region enclosed by these curves.
y5
12. y 6y x 0,
■
about the x-axis
; 24 –25
xy2
2
■
20. x cos y, x 0, y 0, y 4;
solid. Describe the solid.
y 16
x 0,
about the y-axis
23. The integral x0 2 4 x sin4 x dx represents the volume of a
■
4
8. x s
y,
19. y sin x, y 0, x 2, x 3 ;
■
Use the method of cylindrical shells to find the volume
of the solid obtained by rotating the region bounded by the given
curves about the x-axis.
8 –13
about y 3
21. y x 7x 10, y x 2;
y0
about x 1
about y 1
2
y sx 2
■
16. y e x, y 0, x 1, x 0;
17. y e , x 0, y 2;
x0
6. y x 2 4x 3,
7. y x 2,
about the x-axis
18. y ln x, y 0, x e;
y 0,
2
15. y e x, x 0, y ;
x
y0
3
1
S Click here for solutions.
1–7
2. y x,
■
VOLUMES BY CYLINDRICAL SHELLS
A Click here for answers.
2
VOLUMES BY CYLINDRICAL SHELLS
■
■
y x x2 x4
y 3x x 3
■
■
■
■
■
■
■
■
■
26 –27 ■ The region bounded by the given curves is rotated about
the specified axis. Find the volume of the resulting solid by any
method.
26. y x 2 x 2, y 0;
27. y x 2 3x 2, y 0;
■
■
■
■
■
about the x-axis
about the y-axis
■
■
■
■
■
■
2
■
SECTION 7.3
VOLUMES BY CYLINDRICAL SHELLS
7.3
ANSWERS
E Click here for exercises.
S Click here for solutions.
2
1. 16π
17. V = 2π 1 (y ln y − ln y) dy
2. 64
π
15
18. V = 2π 0 (3e − ey − 3e + ye ) dy
1
3. 8π
3π
π/4
√
5. 16
π 5 5−1
3
20. V = 0
2πy cos y dy
4 4
3
2
21. V = π 2 x − 14x + 68x − 136x + 96 dx
2
2
22. V = −2 2π (5 − y) 4 − y dy
6. 16π
3
7. 45 π
23. Solid obtained by rotating the region under the curve
8. 4096
π
9
y = sin4 x, above y = 0, from x = 0 to x = π, about the
line x = 4
9. 609
π
2
10. 23 π
24. 4.05
11. 1944
π
5
25. 4.62
12. 216π
26. 81
π
10
13.
1
x
−x
14. V = 0 2πx e − e
π
27. 12 π
dx
15. V = 1 2πy · ln y dy
0
−x
16. V = −1 2π (1 − x) e
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y
19. V = 2π 2πx sin x dx
1
4. 10
π
5
π
6
y
dx
SECTION 7.3
7.3
■
VOLUMES BY CYLINDRICAL SHELLS
3
SOLUTIONS
E Click here for exercises.
4
−x2 + 6x − 6 − x2 − 6x + 10 dx
4 = 2π 2 x −2x2 + 12x − 16 dx
4
= 4π 2 −x3 + 6x2 − 8x dx
4
= 4π − 14 x4 + 2x3 − 4x2 2
1. V = 2 2πx
= 4π [(−64 + 128 − 64) − (−4 + 16 − 16)] = 16π
√
x − 2 − (x − 2) dx
1
√
= 0 2π (u + 2) ( u − u) du [u = x − 2]
1
= 2π 0 u3/2 − u2 + 2u1/2 − 2u du
1
= 2π 25 u5/2 − 13 u3 + 43 u3/2 − u2
0
= 2π 25 − 13 + 43 − 1 = 45 π
3
7. V = 2 2πx
16
16 5/4
√
4 y dy = 2π
y dy
0
16
= 89 π (512 − 0) =
= 2π 49 y 9/4
8. V = 0 2πy
0
5
2
9. V = 2 2πy · y dy = 2π 14 y
√
4
4
x − 12 x dx = 2π 0 x3/2 dx − π 0 x2 dx
4
4
π = 64
π
= 2π 25 x5/2 − π 13 x3 0 = 45 π (32) − 64
3
15
4
2. V = 0 2πx
0
4 5
2
1
=
π
2
4096
π
9
(625 − 16) =
1
10. V = 0 2πy [(2 − y) − y] dy = 4π 0 y (1 − y) dy
= 4π
1
2
y 2 − 13 y 3
1
0
= 4π
1
6
= 23 π
9
9
√
y dy = 4π 0 y 3/2 dy = 4π 25 y 5/2
9
11. V = 0 2πy · 2
= 85 π (243 − 0) =
2
2
3. V = 0 2πx 4 − x
dx = 2π
2
0
609
π
2
0
1944
π
5
4x − x3 dx
2
= 2π 2x2 − 14 x4 0 = 2π (8 − 4) = 8π
Note: If we integrated from −2 to 2, we would be generating
the volume twice.
1
dx = 2π 0 x3 − x4 dx
1
1
= 2π 14 x4 − 15 x5 0 = 2π 14 − 15 = 10
π
1
2
3
4. V = 0 2πx x − x
12. The two curves intersect at (0, 0) and (0, 6), so
V =
√
4√
5. V = 0 2πx 4 + x2 dx = π 0
x2 + 4 2x dx
3/2 4
√
= 2π
20 20 − 8
= 23 π x2 + 4
3
0
√
= 16
π
5
5
−
1
3
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4
3
2
4
3
−
3
2
=
0
6
2πy −y 2 + 6y dy = 2π − 14 y 4 + 2y 3 0
= 2π (−324 + 432) = 216π
6. V = 1 2πx −x + 4x − 3 dx
3
= 2π − 14 x4 + 43 x3 − 32 x2 1
1
= 2π − 81
+ 36 − 27
− −4 +
4
2
6
16π
3
4
■
SECTION 7.3
VOLUMES BY CYLINDRICAL SHELLS
1
13. V = 0 2πy (2 − y) − y
= 2π 1 −
1
3
−
1
2
1
dy = 2π y 2 − 13 y 3 − 14 y 4 0
25.
= 56 π
4
From the graph, it appears that the curves intersect at x = 0
1
x
−x
14. V = 0 2πx e − e
and at x ≈ 1.17, with 3x − x3 > x4 on (0, 1.17). So the
volume of the solid obtained by rotation about the y-axis is
1.17 V ≈ 2π 0 x 3x − x3 − x4 dx
1.17
= 2π x3 − 15 x5 − 16 x6 0 ≈ 4.62
dx
π
15. V = 1 2πy · ln y dy
0
26. Use disks:
−x
16. V = −1 2π (1 − x) e
dx
2
2
17. V = 1 2π (y − 1) ln y dy = 2π 1 (y ln y − ln y) dy
1
y
18. V = 0 2π (3 − y) (e − e ) dy
= 2π
1
0
(3e − ey − 3ey + yey ) dy
3π
19. V = 2π 2πx sin x dx
π/4
20. V = 0
2πy cos y dy
2
⇔ x2 − 6x + 8 = 0 ⇔
x = 2 or 4 ⇔ (x, y) = (2, 0) or (4, 2). Use washers:
2
4 V = π 2 −x2 + 7x − 10 − (x − 2)2 dx
4
= π 2 x4 − 14x3 + 68x2 − 136x + 96 dx
21. −x + 7x − 10 = x − 2
8 − 2y 2 − 4 − y 2 dy
2
= −2 2π (5 − y) 4 − y 2 dy
2
22. V = −2 2π (5 − y)
23. The solid is obtained by rotating the region under the curve
y = sin4 x, above y = 0, from x = 0 to x = π, about the
line x = 4.
24.
Copyright © 2013, Cengage Learning. All rights reserved.
2
π x2 + x − 2 dx
1 = π −2 x4 + 2x3 − 3x2 − 4x + 4 dx
1
= π 15 x5 + 12 x4 − x3 − 2x2 + 4x −2
= π 15 + 12 − 1 − 2 + 4 − − 32
+8+8−8−8
5
= π 33
+ 32 = 81
π
5
10
V =
From the graph, it appears that the curves intersect at x = 0
and at x ≈ 1.32, with x + x2 − x4 > 0 on (0, 1.32). So the
volume of the solid obtained by rotating the region about the
y-axis is
1.32 V ≈ 2π 0 x x + x2 − x4 dx
1.32
= 2π 13 x3 + 14 x4 − 16 x6 0 ≈ 4.05
1
−2
27. Use shells:
2
2πx −x2 + 3x − 2 dx
2
= 2π 1 −x3 + 3x2 − 2x dx
2
= 2π − 14 x4 + x3 − x2 1
= 2π (−4 + 8 − 4) − − 14 + 1 − 1 = 12 π
V =
1