Section 6.2 Volume

Section 6.2 Volume

44. Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.

y = x², x²+ y²= 1, y ≥ 0

(a) About the x-axis (b) About the y-axis.

Solution:

574 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION

(b) About  = 2:

²+ 4²= 4 ⇒ ²= 4 − 4² ⇒  = ± 4 − 4²

 =

 1

−1



2 −

−

4 − 4²2

− 2 −

4 − 4²2



= 2

 1 0

8

4 − 4² ≈ 7895684

[Notice that this is the same approximation as in part (a). This can be explained by Pappus’s Theorem in Section 8.3.]

34. (a) About the -axis:

 = ²and ²+ ²= 1 ⇒ ²+ ⁴= 1 ⇒ ⁴+ ²− 1 = 0 ⇒

²= −1 +√ 5

2 ≈ 0618 ⇒  = ± = ±

−1 +√ 5

2 ≈ ±0786.

 =

 

−



1 − ²2

− (²)²



 = 2

 

0 (1 − ²− ⁴) 

≈ 354459 (b) About the -axis:

 =

 ² 0

 (√ )²  +

 1

²



1 − ²2



= 

 ² 0

  + 

 1

²(1 − ²)  ≈ 099998

35.  = ln(⁶+ 2)and  =√

3 − ³intersect at  =  ≈ −4091,

 =  ≈ −1467, and  =  ≈ 1091.

 = 

 





ln(⁶+ 2)²

−

3 − ³2

 + 

 



3 − ³2

−

ln(⁶+ 2)²

 ≈ 89023

36.  = 1 + ^−3and  = arctan ²intersect at  =  ≈ −0570 and  =  ≈ 1391.

 = 

 





1 + ^−32

− (arctan ²)²



 ≈ 6923

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62. A frustum of a pyramid with square base of side b, square top of side a, and height h. What happens if a = b?

What happens if a = 0?

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

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Solution:

SECTION 6.2 VOLUMES ¤ 577

 = 

 − . Now

 =¹₃² −¹3²( − ) [by Exercise 47]

=¹₃² 

 − −¹3² 

 − 



 −  = 

 = 

( − )



=1

3³− ³

 −  =¹₃

²+  + ²

=¹₃

²+ ²+

(²) (²)

 =¹₃

1+ 2+√

12



where 1and 2are the areas of the bases of the frustum. (See Exercise 50 for a related result.)

49.²+ ²= ² ⇔ ²= ²− ²

 = 

 

−

²− ²

 = 



² −³ 3

^

−

= 



³−³ 3



−



²( − ) −( − )³ 3



= 2

3³−¹3( − )

3²− ( − )²

=¹₃

2³− ( − ) 3²−

²− 2 + ²

=¹₃

2³− ( − )

2²+ 2 − ²

=¹₃

2³− 2³− 2² + ²+ 2² + 2²− ³

=¹₃

3²− ³

= ¹₃²(3 − ), or, equivalently, ²



 −  3



50.An equation of the line is  = ∆

∆  + (-intercept) =2 − 2

 − 0  +

2 = −  2  + 

2.

 =

 0

()  =

 0

(2)²

=

 0

 2

 −  2  + 

2

2

 =

 0

 − 

  + 

2



=



0

( − )²

² ²+2( − )

  + ²





=

( − )²

3² ³+( − )

 ²+ ²

^

0

=¹₃( − )² + ( − ) + ² = ¹₃

²− 2 + ²+ 3



=¹₃

²+  + ²



[Note that this can be written as¹₃

1+ 2+√

12, as in Exercise 48.]

If  = , we get a rectangular solid with volume ². If  = 0, we get a square pyramid with volume¹₃².

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1

74. Cross-sections of the solid S in planes perpendicular to the x-axis are circles with diameters extending from the curve y =¹₂√

x to the curve y =√

x for 0 ≤ x ≤ 4.

458 CHAPTER 6 Applications of Integration

62. A frusturn of a pyrarnid with square base of side b, square top of side a^, and height h

y )_

/ /

b

α

What happens if a = b? What happens if a = O?

63. A pyrarnid with height h and rectangular base with dirnen- sions b and 2b

64. A pyrarnid with height h and base an equilateral triangle with side a (a tetrahedron)

G

α

a

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

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

67. The base of S is an elliptical region with boundary curve 9x2 + 4y2 = 36. Cross-sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

68. The base of S is the triangular region with vertices (0, 0), (l, 0), and (0, 1). Cross-sections perpendicular to the y-axis are equilateral triangles.

69. The base of S is the same base as in Exercise 68, but cross- sections perpendicular to the x-axis are squares.

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

71. The base of S is the same base as in Exercise 70, but cross- sectÍons perpendicular to the x-axis are isosceles triangles with height equal to the base.

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

y

73. The solid S is bounded by circles that are perpendicular

x

to the x-axis, intersect the x-axis, and have centers on the parabola y = ~(l - x²) ， 一 l 逗 X逗 l

74. Cross-sectÍons of the solid S in planes perpendicular to the x-axis are circles with diameters extending from the curve y = ~

J-;

to the curve y =

fx

^{for 0} ~ x

三

4

y 2

。

JL J

y=~vx

4 ^X

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

y

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

'y

Solution:

624 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION

72. The cross­section of the base corresponding to the coordinate  has length 2 = 2√2 − . [ = 2 − ² ⇔

 = ±√

2 −  ] The corresponding cross­section of the solid  is a quarter­circle with radius 2 √2 −  and area

() = ¹₄(2√

2 −  )²= (2 − ). Therefore,

 =

 2 0

()  =

 2

0 (2 − ) 

= 

2 − ¹2²2

0= (4 − 2) = 2

73. The cross­section of  at coordinate , −1 ≤  ≤ 1, is a circle centered at the point

¹₂(1 − ²)with radius ¹₂(1 − ²).

The area of the cross­section is

() = 1

2(1 − ²)²

= ^₄(1 − 2²+ ⁴) The volume of  is

 =

 1

−1

()  = 2

 1 0



4(1 − 2²+ ⁴)  = ^₂

 − ²3³+¹₅⁵¹

0= ^₂

1 −²3 +¹₅

= ^₂8 15

= ^4₁₅

74. The cross­section of  at coordinate , 0 ≤  ≤ 4, is a circle centered at the point

¹₂

1 2

√ +√



with radius

1 2

√ −¹2

√

. The area of the cross­section is () = 

1 2

√ −¹2

√2

=  · ¹4·

1 2

√2

= 

16. The volume of  is  =

 4 0

()  =

 4 0



16  =  32

²4

0= 

32(16 − 0) =  2. 75. (a) The torus is obtained by rotating the circle ( − )²+ ²= ²about

the ­axis. Solving for , we see that the right half of the circle is given by

 =  +

²− ²=  ()and the left half by  =  −

²− ²= ().

So

 = 

−

[ ()]²− [()]²



= 2 0

²+ 2

²− ²+ ²− ²

−

²− 2

²− ²+ ²− ²



= 2

0 4

²− ² = 8 0

²− ²

(b) Observe that the integral represents a quarter of the area of a circle with radius , so 8

0

²− ² = 8 ·¹4²= 2²².

76. (a)  =

−()  = 2

0 ()  = 2 0

1 2

2√

²− ²

 = 2 0

√²− ²

(b) Observe that the integral represents one quarter of the area of a circle of radius , so  = 2 ·¹4²= ¹₂².

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86. Suppose that a region R has area A and lies above the x-axis. When R is rotated about the x-axis, it sweeps out a solid with volume V1. When R is rotated about the line y = −k (where k is a positive number), it sweeps out a solid with volume V₂. Express V₂ in terms of V₁, k, and A.

Solution:

SECTION 6.3 VOLUMES BY CYLINDRICAL SHELLS ¤ 583 Trying to make this look more like the expression we want, we rewrite it as  = ¹₃

2²+

²−¹2²+₈₀³²⁴. But ²−¹2²+ ₈₀³²⁴=

 −¹4²²

− 40¹²⁴= ( − )²−²5

1 4²²

= ²−²5². Substituting this back into  , we see that  = ¹₃

2²+ ²− ²5², as required.

72. It suffices to consider the case where R is bounded by the curves  = () and  = () for  ≤  ≤ , where () ≤ () for all  in [ ], since other regions can be decomposed into subregions of this type. We are concerned with the volume obtained when R is rotated about the line  = −, which is equal to

2= 



[ () + ]²− [() + ]²



= 



[ ()]²− [()]²

 + 2

[ () − ()]  = ¹+ 2

6.3 Volumes by Cylindrical Shells

1. If we were to use the “washer” method, we would first have to locate the local maximum point ( ) of  = ( − 1)²using the methods of Chapter 4. Then we would have to solve the equation  = ( − 1)² for  in terms of  to obtain the functions  = 1()and  = 2() shown in the first figure. This step would be difficult because it involves the cubic formula. Finally we would find the volume using

 =  0

[1()]²− [²()]².

Using shells, we find that a typical approximating shell has radius , so its circumference is 2. Its height is , that is,

( − 1)². So the total volume is

 =

 1 0

2

( − 1)²

 = 2

 1 0

⁴− 2³+ ²

 = 2

⁵ 5 − 2⁴

4 +³ 3

¹

0

=  15

2. A typical cylindrical shell has circumference 2 and height sin(²).

 =^√

0 2 sin(²) . Let  = ². Then  = 2 , so

 = 

0 sin   = [− cos ]^0 = [1 − (−1)] = 2. For slicing, we would first have to locate the local maximum point ( ) of  = sin(²) using the methods of Chapter 4. Then we would have to solve the equation

 = sin

²for  in terms of  to obtain the functions  = 1()and

 = 2()shown in the second figure. Finally we would find the volume using  = 

0

[1()]²− [²()]². Using shells is definitely preferable to slicing.

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2