Section 6.3 Volumes by Cylindrical Shells

Section 6.3 Volumes by Cylindrical Shells

47. A solid is obtained by rotating the shaded region about the specified line.

(a) Set up an integral using any method to find the volume of the solid.

(b) Evaluate the integral to find the volume of the solid.

466 CHAPTER 6 Applications of Integration

37. Use the Midpoint Rule with n = 5 to estimate the volume obtained by rotating about the y-axis the region under the curve y = JTτ言， 0 ~ x ~ 1

38. If the region shown in the figure is rotated about the y-axis to form a solid, use the Midpoint Rule with n = 5 to esti- mate the volume of the solid.

y

4

2

。

^{2 4 6 8 10 x}

39-42 Each integral represents the volume of a solid. Describe the solid.

39.

f : ^27TX

⁵

^你

40.

f戶的叫y

(4 Y + ²

41. 27T 1 一 ?-dy

Jl y~

42. fOl 27T(2 - x)(Y - Y) dx

圍的-44 U se a graph to 叫 mate the x-coo仙削es of the points of intersection of the given curves. Then use this information and a calculator or computer to estimate the volume of the solid obtained by rotating about the y-axis the region enclosed by these curves.

內 X

43. Y = x^{L -} 2x, y = --;;-一一­

x^L十 I

44. Y = e^SlIlX, y = χ2 - 4x 十 5

囝 45-46 Use a computer algebra 圳em t。而 nd the exact volume of the solid obtained by rotating the region bounded by the given curves about the speci自ed line.

45. Y = sin²x, y = sin⁴x， 。 還 x~ 甘 aboutx = π/2 46. Y = x³sin x^, y = 0, 0 ~ x '"三 7T; aboutx =一l

47-52 A solid is obtained by rotating the shaded region about the speci自 edline.

(a) Set up an integral using any method to find the volume of the solid.

(b) Evaluate the integral to 白 ndthe volume of the so1id.

47. About the y-axis 48. About the x-axis

y y

y= 立 x2

x

。

^x

49. About the x-axis 50. About the y-axis

y _句，

ιX

一一 A『 X

VP , '

J V

y=拉示 x

。

^{7T X x}

51. About the line x = - 2 52. About the line y = 3 y

。

2 ^{I X}

。

^{2 x}

53一59 The region bounded by the given curves is rotated about the speci自edaxis. Find the volume of the resulting solid by any method.

53. y = - x²

+

6x - 8, y = 0; about the y-axis 54. Y = - x²

+

6x - 8, y ⁼ 0; about the x-axis 55. y2 - x2 ₌ 1, y = 2; about the x-axis

56. y2 - x2 = 1, y = 2; about the y-axis 57. x²

+

(y - 1)2 = 1; about the y-axis

58. x = (y - 3)2, X = 4; about y = 1 59. x = (y 一 1)2， X - Y = 1; aboutx =一l

60. Let T be the triangular region with vertices (0,0), (1,0), and (1, 2), and let V be the volume of the solid generated when T is rotated about the line x = α ， where a > 1 Express a in terms of V.

61-63 Use cylindrical shells to find the volume of the solid.

61. A sphere of radius r

Solution: SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS ¤ 639

47. (a) Use shells. Each shell has radius , circumference 2, and height 1 1 + ² −

2.

 =

 1 0

2

 1

1 + ² − 2



.

(b)  = 1 0

2

 1

1 + ² − 2



 =

 1 0

2

 

1 + ² − 1 2²



 = 2

1

2ln1 + ²

 − 1 6³

1 0

= 2

1

2ln 2 − 1 6



− 0



= 



ln 2 −1 3



48. (a) Use washers. 2 − ²= ² ⇒ 2²= 2 ⇒  = 1 [  0]

 =1

0 [(2 − ²)²− (²)²] 

(b)  =1

0 [(2 − ²)²− (²)²]  =1

0 (4 − 4²)  = 

4 − 4 ·¹3³1

0= 4

1 − ¹3

− 0

= ⁸₃ 49. (a) Use disks.  =

0 √

sin 2



(b)  = 0 √

sin 2

 =

0  sin   = 

−cos 

0 = [−cos  − (−cos 0)] = (1 + 1) = 2

50. (a) Use shells. Each shell has radius  circumference 2, and height [(4 − ²) − ].

4 − ²=  ⇒ ²− 3 = 0 ⇒ ( − 3) = 0 ⇒  = 0 or  = 3

 =3

0 2[(4 − ²) − ] 

(b)  =3

0 2[(4 − ²) − ]  =3

0 2(3²− ³)  = 2

³− ¹4⁴3

0= 2

27 −⁸¹4

− 0

= ²⁷₂ 51. (a) Use shells. Each shell has radius  − (−2) =  + 2, circumference 2( + 2), and height ²− ³.

 =12

0 2( + 2)(²− ³) 

(b)  =12

0 2( + 2)(²− ³)  =12

0 2(−³− ⁴+ 2²)  = 2

−¹4⁴−¹5⁵+²₃³^12

0

= 2

−64¹ −160¹ +₁₂¹

− 0

= ₄₈₀⁵⁹

52. (a) Use shells. Each shell has radius 3 − , circumference 2(3 − ), and height [(3 − ²) − 2].

3 − ²= 2 ⇒ ²− 3 + 2 = 0 ⇒ ( − 1)( − 2) = 0 ⇒  = 1 or  = 2

 =2

1 2(3 − )[(3 − ²) − 2] 

(b)  =2

1 2(3 − )[(3 − ²) − 2]  =2

1 2(³− 6²+ 11 − 6) 

= 21

4⁴− 2³+¹¹₂²− 62 1= 2

(4 − 16 + 22 − 12) −1

4− 2 + ¹¹2 − 6

= ¹₂ 53. Use shells:

 =4

2 2(−²+ 6 − 8)  = 24

2(−³+ 6²− 8) 

= 2

−¹4⁴+ 2³− 4²4 2

= 2[(−64 + 128 − 64) − (−4 + 16 − 16)]

= 2(4) = 8

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55. The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. y²− x²= 1, y = 2; about x-axis.

Solution:

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS ¤ 591

38. Use disks:

 =4

2 (−²+ 6 − 8)²

= 4

2(⁴− 12³+ 52²− 96 + 64) 

= 1

5⁵− 3⁴+⁵²₃³− 48²+ 644 2

= ₅₁₂

15 −⁴⁹⁶15

= ¹⁶₁₅

39. Use washers: ²− ²= 1 ⇒  = ±√

²± 1

 =

 ^√3

−√ 3





(2 − 0)²−

²+ 1 − 02



= 2

 ^√3

0 [4 − (²+ 1)]  [by symmetry]

= 2

 ^√3

0 (3 − ²)  = 2

3 −¹3³^√3

0

= 2 3√

3 −√ 3

= 4√ 3 

40. Use disks: ²− ²= 1 ⇒  = ±

²− 1

 = 

 2 1

²− 12

 = 

 2 1

(²− 1) 

= ₁

3³− 2

1= ₈

3− 2

−₁

3− 1

= ⁴₃

41. Use disks: ²+ ( − 1)²= 1 ⇔  = ±

1 − ( − 1)²

 = 

 2 0

1 − ( − 1)²2

 = 

 2 0

(2 − ²) 

= 

²−¹3³2 0= 

4 −⁸3

= ⁴₃

42. Use shells:

 =5

1 2( − 1)[4 − ( − 3)²] 

= 25

1( − 1)(−²+ 6 − 5) 

= 25

1(−³+ 7²− 11 + 5) 

= 2

−¹4⁴+⁷₃³−¹¹2²+ 55 1

= 2275 12 −¹⁹12

= ¹²⁸₃ 

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

56. The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. y²− x²= 1, y = 2; about y-axis.

Solution:

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS ¤ 591

38. Use disks:

 =4

2 (−²+ 6 − 8)²

= 4

2(⁴− 12³+ 52²− 96 + 64) 

= ₁

5⁵− 3⁴+⁵²₃³− 48²+ 644 2

= 512 15 −⁴⁹⁶15

= ¹⁶₁₅

39. Use washers: ²− ²= 1 ⇒  = ±√

²± 1

 =

 ^√3

−√ 3





(2 − 0)²−

²+ 1 − 02



= 2

 ^√3

0 [4 − (²+ 1)]  [by symmetry]

= 2

 ^√3

0 (3 − ²)  = 2

3 −¹3³^√3

0

= 2 3√

3 −√ 3

= 4√ 3 

40. Use disks: ²− ²= 1 ⇒  = ±

²− 1

 = 

 2 1

²− 12

 = 

 2 1

(²− 1) 

= 1

3³− 2

1= 8 3 − 2

−1 3 − 1

= ⁴₃

41. Use disks: ²+ ( − 1)²= 1 ⇔  = ±

1 − ( − 1)²

 = 

 2 0

1 − ( − 1)²2

 = 

 2

0 (2 − ²) 

= 

²−¹3³2 0= 

4 −⁸3

= ⁴₃

42. Use shells:

 =5

1 2( − 1)[4 − ( − 3)²] 

= 25

1( − 1)(−²+ 6 − 5) 

= 25

1(−³+ 7²− 11 + 5) 

= 2

−¹4⁴+⁷₃³−¹¹2²+ 55 1

= 2275 12 −¹⁹12

= ¹²⁸₃ 

° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c

1

62. Use cylindrical shells to find the volume of the solid. The solid torus of Exercise 6.2.75 448 Chapter 6 Applications of Integration

50. A frustum of a pyramid with square base of side b, square top of side a, and height h

a

b

What happens if a − b? What happens if a − 0?

51. A pyramid with height h and rectangular base with dimen- sions b and 2b

52. A pyramid with height h and base an equilateral triangle with side a (a tetrahedron)

a

a a

53. A tetrahedron with three mutually perpendicular faces and three mutually perpendicular edges with lengths 3 cm, 4 cm, and 5 cm

54. The base of S is a circular disk with radius r. Parallel cross- sections perpendicular to the base are squares.

55. The base of S is an elliptical region with boundary curve 9x²14y²− 36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

56. The base of S is the triangular region with vertices s0, 0d, s1, 0d, and s0, 1d. Cross-sections perpendicular to the y-axis are equilateral triangles.

57. The base of S is the same base as in Exercise 56, but cross- sections perpendicular to the x-axis are squares.

58. The base of S is the region enclosed by the parabola y − 1 2 x² and the x-axis. Cross-sections perpendicular to the y-axis are squares.

59. The base of S is the same base as in Exercise 58, but cross- sections perpendicular to the x-axis are isosceles triangles with height equal to the base.

60. The base of S is the region enclosed by y − 2 2 x² and the x-axis. Cross-sections perpendicular to the y-axis are quarter-circles.

y

x y=2-≈

61. The solid S is bounded by circles that are perpendicular to the x-axis, intersect the x-axis, and have centers on the parabola y −¹₂s1 2 x²d, 21 < x < 1.

x

y

x y

62. The base of S is a circular disk with radius r. Parallel cross- sections perpendicular to the base are isosceles triangles with height h and unequal side in the base.

(a) Set up an integral for the volume of S.

(b) By interpreting the integral as an area, find the volume of S.

63. (a) Set up an integral for the volume of a solid torus (the donut-shaped solid shown in the figure) with radii r and R.

(b) By interpreting the integral as an area, find the volume of the torus.

R r

64. Solve Example 9 taking cross-sections to be parallel to the line of intersection of the two planes.

65. (a) Cavalieri’s Principle states that if a family of parallel planes gives equal cross-sectional areas for two solids

Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).

Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Solution:

592 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION

43.  + 1 = ( − 1)² ⇔  + 1 = ²− 2 + 1 ⇔ 0 = ²− 3 ⇔ 0 = ( − 3) ⇔  = 0 or 3.

Use disks:

 = 

 3 0

[( + 1) − (−1)]²− [( − 1)²− (−1)]²



= 

 3 0

[( + 2)²− (²− 2 + 2)²] 

= 

 3 0

[(²+ 4 + 4) − (⁴− 4³+ 8²− 8 + 4)]  = 

 3 0

(−⁴+ 4³− 7²+ 12) 

= 

−¹5⁵+ ⁴−⁷3³+ 6²3 0= 

−²⁴³5 + 81 − 63 + 54

= ¹¹⁷₅ 

44. Use cylindrical shells to find the volume  .

 =1

0 2( − )(2)  = 41

0( − ²) 

= 4₁

2²−¹3³1

0= 41 2 − ¹3

 Now solve for  in terms of  :

 = 4₁

2 −¹3

 ⇔ 

4 = 1 2 −1

3 ⇔ 1

2 =  4 +1

3 ⇔

 =  2 +2

3 45. Use shells:

 = 2 0 2√

²− ² = −2

0(²− ²)^12(−2) 

=

−2 · ²3(²− ²)^32

0= −⁴3(0 − ³) = ⁴₃³

46.  =+

− 2 · 2

²− ( − )²

=

−4( + )√

²− ² [let  =  − ]

= 4

−

√²− ² + 4

−√

²− ²

The first integral is the area of a semicircle of radius , that is,¹₂², and the second is zero since the integrand is an odd function. Thus,

 = 41 2²

+ 4 · 0 = 2²².

47.  = 2

  0





−

 + 



 = 2

  0



−²

 + 





= 2



−³ 3 +²

2

 0

= 2²

6 = ² 3

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2