Math 241 Homework 12 Solutions Section 6.1 Problem 1. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections perpendicular to the axis on the interval 0 ≤ x ≤ 4 are squares whose diagonals run from the parabola y =- √ x to the parabola y = √ x Solution From the picture we have s 2 + s 2 =(2 √ x) 2 ⇔ 2s 2 = 4x ⇔ s 2 = 2x Since the area of the square is given by A = s 2 = 2x we have Volume = 4 0 2x dx = x 2 4 0 = 16 1
Transcript
Math 241 Homework 12 Solutions
Section 6.1
Problem 1. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. Thecross-sections perpendicular to the axis on the interval 0 ≤ x ≤ 4 are squares whose diagonals runfrom the parabola y = −√x to the parabola y = √
x
Solution
From the picture we have
s2 + s2 = (2√x)2⇔ 2s2 = 4x⇔ s2 = 2x
Since the area of the square is given by A = s2 = 2x we have
Volume = ∫4
02x dx = x2∣
4
0= 16
1
Problem 2. The solid lies between planes perpendicular to the x-axis at x = −1 and x = 1. Thecross-sections perpendicular to the x-axis are circular disks whose diameters run from the parabolay = x2 to the parabola y = 2 − x2.
Solution
We haveDiameter = 2 − 2x2 ⇒ Radius = 1 − x2
Since the area of a circle is given by πr2 we have
Volume = ∫1
−1(π − 2πx2 + πx4) dx
= (πx − 2πx3
3+ πx
5
5)RRRRRRRRRRR
1
−1
= (π − 2π
3+ π
5) − (−π + 2π
3− π
5)
= 2π − 4π
3+ 2π
5
= 30π − 20π + 6π
15
= 16π
15
2
Problem 5. The base of a solid is the region between the curve y = 2√
sinx and the interval [0.π]on the axis. The cross-sections perpendicular to the x-axis are
(a) equilateral triangles with bases running from the x-axis to the curve as shown in the accom-panying figure.
Volumes by SlicingFind the volumes of the solids in Exercises 1–10.
1. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections perpendicular to the axis on the interval 0 … x … 4 are squares whose diagonals run from the parabola y = - 2x to the parabola y = 2x.
2. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross-sections perpendicular to the x-axis are circular disks whose diameters run from the parabola y = x2 to the parabola y = 2 - x2.
y = x2
y = 2 − x2
2
0
x
y
3. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross-sections perpendicular to the x-axis between these planes are squares whose bases run from the semicircle y = - 21 - x2 to the semicircle y = 21 - x2.
4. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross-sections perpendicular to the x-axis between these planes are squares whose diagonals run from the semicircle y = - 21 - x2 to the semicircle y = 21 - x2.
5. The base of a solid is the region between the curve y = 22sin x and the interval 30, p4 on the x-axis. The cross-sections perpen-dicular to the x-axis are
a. equilateral triangles with bases running from the x-axis to the curve as shown in the accompanying figure.
0
p
y = 2"
sin x
x
y
b. squares with bases running from the x-axis to the curve.
6. The solid lies between planes perpendicular to the x-axis at x = -p>3 and x = p>3. The cross-sections perpendicular to the x-axis are
a. circular disks with diameters running from the curve y = tan x to the curve y = sec x.
b. squares whose bases run from the curve y = tan x to the curve y = sec x.
7. The base of a solid is the region bounded by the graphs of y = 3x, y = 6, and x = 0. The cross-sections perpendicular to the x-axis are
a. rectangles of height 10.
b. rectangles of perimeter 20.
8. The base of a solid is the region bounded by the graphs of y = 2x and y = x>2. The cross-sections perpendicular to the x-axis are
a. isosceles triangles of height 6.
b. semicircles with diameters running across the base of the solid.
9. The solid lies between planes perpendicular to the y-axis at y = 0 and y = 2. The cross-sections perpendicular to the y-axis are cir-cular disks with diameters running from the y-axis to the parabola x = 25y2.
10. The base of the solid is the disk x2 + y2 … 1. The cross-sections by planes perpendicular to the y-axis between y = -1 and y = 1 are isosceles right triangles with one leg in the disk.
1x2 + y2 = 1
0
y
x
11. Find the volume of the given right tetrahedron. (Hint: Consider slices perpendicular to one of the labeled edges.)
3
4
5
y
x
12. Find the volume of the given pyramid, which has a square base of area 9 and height 5.
3
5
3
y
x
13. A twisted solid A square of side length s lies in a plane perpen-dicular to a line L. One vertex of the square lies on L. As this square moves a distance h along L, the square turns one revolution about L to generate a corkscrew-like column with square cross-sections.
a. Find the volume of the column.
b. What will the volume be if the square turns twice instead of once? Give reasons for your answer.
Exercises 6.1
6.1 Volumes Using Cross-Sections 355
(b) squares with bases running from the x-axis to the curve.
Solution
(a)
From the pythagorean theorem we have
y2 + (√
sinx)2 = (2√
sinx)2⇔ y2 + sinx = 4 sinx
⇔ y2 = 3 sinx
⇒ y =√
3 sinx
Since the area of the triangle is given by
1
2(2
√x√
3 sinx) =√
3 sinx
We have
Volume = ∫π
0
√3 sinx dx
= −√
3 cosx∣π
0
= −√
3 cosπ +√
3 cos 0
=√
3 +√
3
= 2√
3
3
(b)
The area of the square is given by
(2√
sinx)2 = 4 sinx
We have
Volume = ∫π
04 sinx dx
= −4 cosx∣π0= −4 cosπ + 4 cos 0
= 4 + 4
= 8
4
Problem 8. The base of the solid is the disk x2+y2 ≤ 1. The cross-sections by planes perpendicularto the y-axis between y = −1 and y = 1 are isosceles right triangles with one leg in the disk.
Volumes by SlicingFind the volumes of the solids in Exercises 1–10.
1. The solid lies between planes perpendicular to the x-axis at x = 0 and x = 4. The cross-sections perpendicular to the axis on the interval 0 … x … 4 are squares whose diagonals run from the parabola y = - 2x to the parabola y = 2x.
2. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross-sections perpendicular to the x-axis are circular disks whose diameters run from the parabola y = x2 to the parabola y = 2 - x2.
y = x2
y = 2 − x2
2
0
x
y
3. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross-sections perpendicular to the x-axis between these planes are squares whose bases run from the semicircle y = - 21 - x2 to the semicircle y = 21 - x2.
4. The solid lies between planes perpendicular to the x-axis at x = -1 and x = 1. The cross-sections perpendicular to the x-axis between these planes are squares whose diagonals run from the semicircle y = - 21 - x2 to the semicircle y = 21 - x2.
5. The base of a solid is the region between the curve y = 22sin x and the interval 30, p4 on the x-axis. The cross-sections perpen-dicular to the x-axis are
a. equilateral triangles with bases running from the x-axis to the curve as shown in the accompanying figure.
0
p
y = 2"
sin x
x
y
b. squares with bases running from the x-axis to the curve.
6. The solid lies between planes perpendicular to the x-axis at x = -p>3 and x = p>3. The cross-sections perpendicular to the x-axis are
a. circular disks with diameters running from the curve y = tan x to the curve y = sec x.
b. squares whose bases run from the curve y = tan x to the curve y = sec x.
7. The base of a solid is the region bounded by the graphs of y = 3x, y = 6, and x = 0. The cross-sections perpendicular to the x-axis are
a. rectangles of height 10.
b. rectangles of perimeter 20.
8. The base of a solid is the region bounded by the graphs of y = 2x and y = x>2. The cross-sections perpendicular to the x-axis are
a. isosceles triangles of height 6.
b. semicircles with diameters running across the base of the solid.
9. The solid lies between planes perpendicular to the y-axis at y = 0 and y = 2. The cross-sections perpendicular to the y-axis are cir-cular disks with diameters running from the y-axis to the parabola x = 25y2.
10. The base of the solid is the disk x2 + y2 … 1. The cross-sections by planes perpendicular to the y-axis between y = -1 and y = 1 are isosceles right triangles with one leg in the disk.
1x2 + y2 = 1
0
y
x
11. Find the volume of the given right tetrahedron. (Hint: Consider slices perpendicular to one of the labeled edges.)
3
4
5
y
x
12. Find the volume of the given pyramid, which has a square base of area 9 and height 5.
3
5
3
y
x
13. A twisted solid A square of side length s lies in a plane perpen-dicular to a line L. One vertex of the square lies on L. As this square moves a distance h along L, the square turns one revolution about L to generate a corkscrew-like column with square cross-sections.
a. Find the volume of the column.
b. What will the volume be if the square turns twice instead of once? Give reasons for your answer.
Exercises 6.1
6.1 Volumes Using Cross-Sections 355
Solution Since the area of the triangle is given by
1
2(2
√1 − y2)2 = 2(1 − y2)
we have
Volume = ∫1
−12(1 − y2) dy
= ∫1
−1(2 − 2y2) dy
= (2y − 2y3
3)RRRRRRRRRRR
1
−1
= (2 − 2
3) − (−2 + 2
3)
= 4 − 4
3
= 8
3
5
Problem 11. Find the volume of the solid generated by revolving the shaded region about the x-axis.
14. Cavalieri’s principle A solid lies between planes perpendicular to the x-axis at x = 0 and x = 12. The cross-sections by planes perpendicular to the x-axis are circular disks whose diameters run from the line y = x>2 to the line y = x as shown in the accom-panying figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height 12.
x12
y
0
y = x
y =2x
Volumes by the Disk MethodIn Exercises 15–18, find the volume of the solid generated by revolv-ing the shaded region about the given axis.
15. About the x-axis 16. About the y-axis
x
y
0 2
1
x + 2y = 2
x
y
0
2
3
x =3y2
17. About the y-axis 18. About the x-axis
Q R
x
y
0
1
4x = tan yp
x
y
0
y = sin x cos x
2
21
p
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 19–28 about the x-axis.
19. y = x2, y = 0, x = 2 20. y = x3, y = 0, x = 2
21. y = 29 - x2, y = 0 22. y = x - x2, y = 0
23. y = 2cos x, 0 … x … p>2, y = 0, x = 0
24. y = sec x, y = 0, x = -p>4, x = p>4 25. y = e-x, y = 0, x = 0, x = 1
26. The region between the curve y = 2cot x and the x-axis from x = p>6 to x = p>2
27. The region between the curve y = 1>122x2 and the x-axis from x = 1>4 to x = 4
28. y = ex - 1, y = 0, x = 1, x = 3
In Exercises 29 and 30, find the volume of the solid generated by revolving the region about the given line.
29. The region in the first quadrant bounded above by the line y = 22, below by the curve y = sec x tan x, and on the left by the y-axis, about the line y = 22
30. The region in the first quadrant bounded above by the line y = 2, below by the curve y = 2 sin x, 0 … x … p>2, and on the left by the y-axis, about the line y = 2
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 31–36 about the y-axis.
31. The region enclosed by x = 25y2, x = 0, y = -1, y = 1
32. The region enclosed by x = y3>2, x = 0, y = 2
33. The region enclosed by x = 22 sin 2y, 0 … y … p>2, x = 0
34. The region enclosed by x = 2cos (py>4), -2 … y … 0, x = 0
35. x = 2>2y + 1, x = 0, y = 0, y = 3
36. x = 22y>( y2 + 1), x = 0, y = 1
Volumes by the Washer MethodFind the volumes of the solids generated by revolving the shaded regions in Exercises 37 and 38 about the indicated axes.
37. The x-axis 38. The y-axis
x
y
0−
y = 1y =
"
cos x
2p
2p
x
y
0 1
x = tan y4p
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 39–44 about the x-axis.
39. y = x, y = 1, x = 0
40. y = 22x, y = 2, x = 0
41. y = x2 + 1, y = x + 3
42. y = 4 - x2, y = 2 - x
43. y = sec x, y = 22, -p>4 … x … p>4 44. y = sec x, y = tan x, x = 0, x = 1
In Exercises 45–48, find the volume of the solid generated by revolv-ing each region about the y-axis.
45. The region enclosed by the triangle with vertices (1, 0), (2, 1), and (1, 1)
46. The region enclosed by the triangle with vertices (0, 1), (1, 0), and (1, 1)
47. The region in the first quadrant bounded above by the parabola y = x2, below by the x-axis, and on the right by the line x = 2
48. The region in the first quadrant bounded on the left by the circle x2 + y2 = 3, on the right by the line x = 23, and above by the line y = 23
In Exercises 49 and 50, find the volume of the solid generated by revolving each region about the given axis.
49. The region in the first quadrant bounded above by the curve y = x2, below by the x-axis, and on the right by the line x = 1, about the line x = -1
356 Chapter 6: Applications of Definite Integrals
Solution Since the radius is given by
r = 2 − x2
= 1 − x2
We have
Volume = ∫2
0π (1 − x
2)2
dx
= ∫2
0π (1 − x + x
2
4) dx
= ∫2
0(π − xπ + πx
2
4) dx
= (πx − πx2
2+ π
4⋅ x
3
3)RRRRRRRRRRR
2
0
= (2π − 2π + 2π
3)
= 2π
3
6
Problem 12. Find the volume of the solid generated by revolving the shaded region about the y-axis.
14. Cavalieri’s principle A solid lies between planes perpendicular to the x-axis at x = 0 and x = 12. The cross-sections by planes perpendicular to the x-axis are circular disks whose diameters run from the line y = x>2 to the line y = x as shown in the accom-panying figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height 12.
x12
y
0
y = x
y =2x
Volumes by the Disk MethodIn Exercises 15–18, find the volume of the solid generated by revolv-ing the shaded region about the given axis.
15. About the x-axis 16. About the y-axis
x
y
0 2
1
x + 2y = 2
x
y
0
2
3
x =3y2
17. About the y-axis 18. About the x-axis
Q R
x
y
0
1
4x = tan yp
x
y
0
y = sin x cos x
2
21
p
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 19–28 about the x-axis.
19. y = x2, y = 0, x = 2 20. y = x3, y = 0, x = 2
21. y = 29 - x2, y = 0 22. y = x - x2, y = 0
23. y = 2cos x, 0 … x … p>2, y = 0, x = 0
24. y = sec x, y = 0, x = -p>4, x = p>4 25. y = e-x, y = 0, x = 0, x = 1
26. The region between the curve y = 2cot x and the x-axis from x = p>6 to x = p>2
27. The region between the curve y = 1>122x2 and the x-axis from x = 1>4 to x = 4
28. y = ex - 1, y = 0, x = 1, x = 3
In Exercises 29 and 30, find the volume of the solid generated by revolving the region about the given line.
29. The region in the first quadrant bounded above by the line y = 22, below by the curve y = sec x tan x, and on the left by the y-axis, about the line y = 22
30. The region in the first quadrant bounded above by the line y = 2, below by the curve y = 2 sin x, 0 … x … p>2, and on the left by the y-axis, about the line y = 2
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 31–36 about the y-axis.
31. The region enclosed by x = 25y2, x = 0, y = -1, y = 1
32. The region enclosed by x = y3>2, x = 0, y = 2
33. The region enclosed by x = 22 sin 2y, 0 … y … p>2, x = 0
34. The region enclosed by x = 2cos (py>4), -2 … y … 0, x = 0
35. x = 2>2y + 1, x = 0, y = 0, y = 3
36. x = 22y>( y2 + 1), x = 0, y = 1
Volumes by the Washer MethodFind the volumes of the solids generated by revolving the shaded regions in Exercises 37 and 38 about the indicated axes.
37. The x-axis 38. The y-axis
x
y
0−
y = 1y =
"
cos x
2p
2p
x
y
0 1
x = tan y4p
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 39–44 about the x-axis.
39. y = x, y = 1, x = 0
40. y = 22x, y = 2, x = 0
41. y = x2 + 1, y = x + 3
42. y = 4 - x2, y = 2 - x
43. y = sec x, y = 22, -p>4 … x … p>4 44. y = sec x, y = tan x, x = 0, x = 1
In Exercises 45–48, find the volume of the solid generated by revolv-ing each region about the y-axis.
45. The region enclosed by the triangle with vertices (1, 0), (2, 1), and (1, 1)
46. The region enclosed by the triangle with vertices (0, 1), (1, 0), and (1, 1)
47. The region in the first quadrant bounded above by the parabola y = x2, below by the x-axis, and on the right by the line x = 2
48. The region in the first quadrant bounded on the left by the circle x2 + y2 = 3, on the right by the line x = 23, and above by the line y = 23
In Exercises 49 and 50, find the volume of the solid generated by revolving each region about the given axis.
49. The region in the first quadrant bounded above by the curve y = x2, below by the x-axis, and on the right by the line x = 1, about the line x = -1
356 Chapter 6: Applications of Definite Integrals
Solution
Volume = ∫2
0π (3y
2)2
dy
= ∫2
0
9πy2
4dy
= 9πy3
12∣2
0
= 3πy3
4∣2
0
= 24π
4
= 8π
7
Problem 13. Find the volume of the solid generated by revolving the shaded region about the y-axis.
14. Cavalieri’s principle A solid lies between planes perpendicular to the x-axis at x = 0 and x = 12. The cross-sections by planes perpendicular to the x-axis are circular disks whose diameters run from the line y = x>2 to the line y = x as shown in the accom-panying figure. Explain why the solid has the same volume as a right circular cone with base radius 3 and height 12.
x12
y
0
y = x
y =2x
Volumes by the Disk MethodIn Exercises 15–18, find the volume of the solid generated by revolv-ing the shaded region about the given axis.
15. About the x-axis 16. About the y-axis
x
y
0 2
1
x + 2y = 2
x
y
0
2
3
x =3y2
17. About the y-axis 18. About the x-axis
Q R
x
y
0
1
4x = tan yp
x
y
0
y = sin x cos x
2
21
p
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 19–28 about the x-axis.
19. y = x2, y = 0, x = 2 20. y = x3, y = 0, x = 2
21. y = 29 - x2, y = 0 22. y = x - x2, y = 0
23. y = 2cos x, 0 … x … p>2, y = 0, x = 0
24. y = sec x, y = 0, x = -p>4, x = p>4 25. y = e-x, y = 0, x = 0, x = 1
26. The region between the curve y = 2cot x and the x-axis from x = p>6 to x = p>2
27. The region between the curve y = 1>122x2 and the x-axis from x = 1>4 to x = 4
28. y = ex - 1, y = 0, x = 1, x = 3
In Exercises 29 and 30, find the volume of the solid generated by revolving the region about the given line.
29. The region in the first quadrant bounded above by the line y = 22, below by the curve y = sec x tan x, and on the left by the y-axis, about the line y = 22
30. The region in the first quadrant bounded above by the line y = 2, below by the curve y = 2 sin x, 0 … x … p>2, and on the left by the y-axis, about the line y = 2
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 31–36 about the y-axis.
31. The region enclosed by x = 25y2, x = 0, y = -1, y = 1
32. The region enclosed by x = y3>2, x = 0, y = 2
33. The region enclosed by x = 22 sin 2y, 0 … y … p>2, x = 0
34. The region enclosed by x = 2cos (py>4), -2 … y … 0, x = 0
35. x = 2>2y + 1, x = 0, y = 0, y = 3
36. x = 22y>( y2 + 1), x = 0, y = 1
Volumes by the Washer MethodFind the volumes of the solids generated by revolving the shaded regions in Exercises 37 and 38 about the indicated axes.
37. The x-axis 38. The y-axis
x
y
0−
y = 1y =
"
cos x
2p
2p
x
y
0 1
x = tan y4p
Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises 39–44 about the x-axis.
39. y = x, y = 1, x = 0
40. y = 22x, y = 2, x = 0
41. y = x2 + 1, y = x + 3
42. y = 4 - x2, y = 2 - x
43. y = sec x, y = 22, -p>4 … x … p>4 44. y = sec x, y = tan x, x = 0, x = 1
In Exercises 45–48, find the volume of the solid generated by revolv-ing each region about the y-axis.
45. The region enclosed by the triangle with vertices (1, 0), (2, 1), and (1, 1)
46. The region enclosed by the triangle with vertices (0, 1), (1, 0), and (1, 1)
47. The region in the first quadrant bounded above by the parabola y = x2, below by the x-axis, and on the right by the line x = 2
48. The region in the first quadrant bounded on the left by the circle x2 + y2 = 3, on the right by the line x = 23, and above by the line y = 23
In Exercises 49 and 50, find the volume of the solid generated by revolving each region about the given axis.
49. The region in the first quadrant bounded above by the curve y = x2, below by the x-axis, and on the right by the line x = 1, about the line x = -1
356 Chapter 6: Applications of Definite Integrals
Solution
Volume = ∫1
0π (tan
πy
4)2
dy
= ∫1
0π (sec2
πy
4− 1) dy
=⎛⎝π ( 4
π) tan(πy
4) − πy
⎞⎠
RRRRRRRRRRRR
1
0
= 4 tan(π4) − π
= 4 − π
Problem 15. Find the volumes of the solids generated by revolving the regions bounded by the linesand curves about the x-axis: y = x2, y = 0, x = 2
Solution
Volume = ∫2
0π(x2)2 dx
= ∫2
0πx4 dx
= πx5
5∣2
0
= 32π
5
8
Problem 21. The region in the first quadrant bounded above by the line y =√
2, below by the curvey = secx tanx, and on the left by the y-axis, about the line y =
√2
Solution To find the x-value of the intersection point we solve
secx tanx =√
2⇔ sinx
cos2 x=√
2
⇔ sinx =√
2(1 − sin2 x)⇔
√2 −
√2 sin2 x = sinx
⇔√
2 sin2 + sinx −√
2 = 0
⇔ 2 sin2 x +√
2 sinx − 2 = 0
⇔ (2 sinx −√
2)(sinx +√
2) = 0
⇒ sinx =√
2
2
⇒ x = π4
The radius is√
2 − secx tanx so the area is given by
∫π/4
0π(
√2 − secx tanx)2 dx = π∫
π/4
0(2 − 2
√2 secx tanx + sec2 x tan2 x) dx
= π (2x − 2√
2 secx + tan3 x
3)RRRRRRRRRRR
π/4
0
= π (π2− 2
√2 sec
π
4+ tan3(π/4)
3) − π (0 − 2
√2 sec 0 + tan3 0
3)
= π (π2− 2��
√2 ⋅ 2
��
√2+ 1
3) + π(2
√2)
= π (π2− 12
3+ 1
3+ 2
√2)
= π (π2− 11
3+ 2
√2)
9
Problem 41. Find the volume of the solid generated by revolving the region in the first quadrantbounded above by the curve y = x2, below by the x-axis, and on the right by the line x = 1, about theline x = −1
Solution We have that R = 2 and r = 1 +√y so using the washer method we have
Volume = ∫1
0π(4 − (1 +√
y)2) dy
= π∫1
0(4 − 1 − 2
√y − y) dy
= (3y − 4
3y3/2 − y
2
2)RRRRRRRRRRR
1
0
= π (3 − 4
3− 1
2)
= π (18
6− 8
6− 3
6)
= 7π
6
10
Problem 43. Find the volume of the solid generated by revolving the region the region bounded byy = √
x and the lines y = 2 and x = 0 about
(a) the x-axis
(b) the y-axis
(c) the line y = 2
(d) the line x = 4
Solution
(a) We have R = 2 and r = √x so using the washer method we have
Volume = ∫4
0π(4 − x) dx
= π (4x − x2
2)RRRRRRRRRRR
4
0
= π (4(4) − 16
2)
= 8π
(b)
Volume = ∫2
0πy4 dy
= πy5
5∣2
0
= 32π
5
11
(c) R is given by 2 −√x so by the disk method we have
Volume = ∫4
0π(2 −
√x)2 dx
= ∫4
0π(4 − 4
√x + x) dx
= π (4x − 8
3x3/2 + x
2
2)RRRRRRRRRRR
4
0
= π (16 − 64
3+ 8)
= π (72 − 64
3)
= 8π
3
(d) We have R = 4 and r = 4 − y2 so the volume is
Volume = ∫2
0π(42 − (4 − y2)2) dy
= π∫2
0(16 − (16 − 8y2 + y4)) dy
= π∫2
0(8y2 − y4) dy
= π (8y3
3− y
5
5)RRRRRRRRRRR
2
0
= π (64
3− 32
5)
= 224π
15
12
Section 6.2
Problem 1. Use the shell method to find the volume of the solid generated by revolving the shadedregion about the indicated axis
The shell method gives the same answer as the washer method when both are used to calculate the volume of a region. We do not prove that result here, but it is illustrated in Exercises 37 and 38. (Exercise 45 outlines a proof.) Both volume formulas are actually special cases of a general volume formula we will look at when studying double and triple integrals in Chapter 15. That general formula also allows for computing volumes of solids other than those swept out by regions of revolution.
Summary of the Shell MethodRegardless of the position of the axis of revolution (horizontal or vertical), the steps for implementing the shell method are these.
1. Draw the region and sketch a line segment across it parallel to the axis of revolution. Label the segment’s height or length (shell height) and distance from the axis of revolution (shell radius).
2. Find the limits of integration for the thickness variable.
3. Integrate the product 2p (shell radius) (shell height) with respect to the thickness variable (x or y) to find the volume.
Revolution About the AxesIn Exercises 1–6, use the shell method to find the volumes of the solids generated by revolving the shaded region about the indicated axis.
1. 2.
x
y
0 2
1
y = 1 + x2
4
x
y
0 2
2y = 2 − x2
4
3. 4.
x
y
0 2
x = y2
y = "
2"
2
x
y
0 3
x = 3 − y2
y = "
3"
3
5. The y-axis 6. The y-axis
x
y
0
1
2
x = "
3
"
3
y = "
x2 + 1
x
y
0
3
5"
x3 + 9
9xy =
Revolution About the y-AxisUse the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises 7–12 about the y-axis.
7. y = x, y = -x>2, x = 2
8. y = 2x, y = x>2, x = 1
9. y = x2, y = 2 - x, x = 0, for x Ú 0
10. y = 2 - x2, y = x2, x = 0
11. y = 2x - 1, y = 2x, x = 0
12. y = 3>122x2, y = 0, x = 1, x = 4
Exercises 6.2
6.2 Volumes Using Cylindrical Shells 363
Solution We have that r(x) = x and h(x) = 1 + x2
4
Volume = ∫2
02πx(1 + x
2
4) dx
= ∫2
0(2πx + πx
3
2) dx
= (πx2 + πx4
8)RRRRRRRRRRR
2
0
= 4π + 16π
8
= 4π + 2π
= 6π
13
Problem 3. Use the shell method to find the volume of the solid generated by revolving the shadedregion about the indicated axis
The shell method gives the same answer as the washer method when both are used to calculate the volume of a region. We do not prove that result here, but it is illustrated in Exercises 37 and 38. (Exercise 45 outlines a proof.) Both volume formulas are actually special cases of a general volume formula we will look at when studying double and triple integrals in Chapter 15. That general formula also allows for computing volumes of solids other than those swept out by regions of revolution.
Summary of the Shell MethodRegardless of the position of the axis of revolution (horizontal or vertical), the steps for implementing the shell method are these.
1. Draw the region and sketch a line segment across it parallel to the axis of revolution. Label the segment’s height or length (shell height) and distance from the axis of revolution (shell radius).
2. Find the limits of integration for the thickness variable.
3. Integrate the product 2p (shell radius) (shell height) with respect to the thickness variable (x or y) to find the volume.
Revolution About the AxesIn Exercises 1–6, use the shell method to find the volumes of the solids generated by revolving the shaded region about the indicated axis.
1. 2.
x
y
0 2
1
y = 1 + x2
4
x
y
0 2
2y = 2 − x2
4
3. 4.
x
y
0 2
x = y2
y = "
2"
2
x
y
0 3
x = 3 − y2
y = "
3"
3
5. The y-axis 6. The y-axis
x
y
0
1
2
x = "
3
"
3
y = "
x2 + 1
x
y
0
3
5"
x3 + 9
9xy =
Revolution About the y-AxisUse the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises 7–12 about the y-axis.
7. y = x, y = -x>2, x = 2
8. y = 2x, y = x>2, x = 1
9. y = x2, y = 2 - x, x = 0, for x Ú 0
10. y = 2 - x2, y = x2, x = 0
11. y = 2x - 1, y = 2x, x = 0
12. y = 3>122x2, y = 0, x = 1, x = 4
Exercises 6.2
6.2 Volumes Using Cylindrical Shells 363
Solution We have r(y) = y and h(y) = y2
Volume = ∫√2
02πy3 dy
= πy4
2∣
√2
0
= 4π
2
= 2π
14
Problem 5. Use the shell method to find the volume of the solid generated by revolving the shadedregion about the indicated axis
The shell method gives the same answer as the washer method when both are used to calculate the volume of a region. We do not prove that result here, but it is illustrated in Exercises 37 and 38. (Exercise 45 outlines a proof.) Both volume formulas are actually special cases of a general volume formula we will look at when studying double and triple integrals in Chapter 15. That general formula also allows for computing volumes of solids other than those swept out by regions of revolution.
Summary of the Shell MethodRegardless of the position of the axis of revolution (horizontal or vertical), the steps for implementing the shell method are these.
1. Draw the region and sketch a line segment across it parallel to the axis of revolution. Label the segment’s height or length (shell height) and distance from the axis of revolution (shell radius).
2. Find the limits of integration for the thickness variable.
3. Integrate the product 2p (shell radius) (shell height) with respect to the thickness variable (x or y) to find the volume.
Revolution About the AxesIn Exercises 1–6, use the shell method to find the volumes of the solids generated by revolving the shaded region about the indicated axis.
1. 2.
x
y
0 2
1
y = 1 + x2
4
x
y
0 2
2y = 2 − x2
4
3. 4.
x
y
0 2
x = y2
y = "
2"
2
x
y
0 3
x = 3 − y2
y = "
3"
3
5. The y-axis 6. The y-axis
x
y
0
1
2
x = "
3
"
3
y = "
x2 + 1
x
y
0
3
5"
x3 + 9
9xy =
Revolution About the y-AxisUse the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises 7–12 about the y-axis.
7. y = x, y = -x>2, x = 2
8. y = 2x, y = x>2, x = 1
9. y = x2, y = 2 - x, x = 0, for x Ú 0
10. y = 2 - x2, y = x2, x = 0
11. y = 2x - 1, y = 2x, x = 0
12. y = 3>122x2, y = 0, x = 1, x = 4
Exercises 6.2
6.2 Volumes Using Cylindrical Shells 363
Solution We have r(x) = x and h(x) =√x2 + 1
Volume = ∫√3
02πx
√x2 + 1 dx
Let u = x2 + 1⇒ du = 2x dx
= π∫4
1
√u du
= π⎛⎝
2u3/2
3
⎞⎠
RRRRRRRRRRRR
4
1
= 2π
3(43/2 − 1)
= 14π
3
Problem 7. Use the shell method to find the volumes of the solid generated by revolving the regionbounded by the curves and lines about the y-axis: y = x, y = −x/2, x = 2
Solution We have r(x) = x and h(x) = x + x2= 3x
2
Volume = ∫2
02πx(3x
2) dx
= π∫2
03x2 dx
= πx3∣2
0
= 8π
15
Problem 12. Use the shell method to find the volumes of the solid generated by revolving the regionbounded by the curves and lines about the y-axis: y = 3/(2√x), y = 0, x = 1, x = 4
Solution We have h(x) = 3
2√x
and r(x) = x which gives
Volume = ∫4
1�2πx( 3
�2√x)
= ∫4
13π
√x dx
= 2πx3/2∣4
1
= 16π − 2π
= 14π
16
Problem 13. Let f(x) =⎧⎪⎪⎨⎪⎪⎩
(sinx)/x, 0 < x ≤ π1, x = 0
(a) Show that xf(x) = sinx, 0 ≤ x ≤ π
(b) Find the volume of the solid generated by revolving the shaded region about the y-axis in theaccompanying figure.
13. Let ƒ(x) = e (sin x)>x, 0 6 x … p
1, x = 0
a. Show that x ƒ(x) = sin x, 0 … x … p.
b. Find the volume of the solid generated by revolving the shaded region about the y-axis in the accompanying figure.
x
y
0
1
y =1, x = 0
, 0 < x ≤ psin xx
p
14. Let g(x) = e (tan x)2>x, 0 6 x … p>40, x = 0
a. Show that x g(x) = (tan x)2, 0 … x … p>4.
b. Find the volume of the solid generated by revolving the shaded region about the y-axis in the accompanying figure.
x
y
0
y =0, x = 0
, 0 < x ≤ tan2 x
x 4
4
4p
p
p
Revolution About the x-AxisUse the shell method to find the volumes of the solids generated by revolving the regions bounded by the curves and lines in Exercises 15–22 about the x-axis.
15. x = 2y, x = -y, y = 2
16. x = y2, x = -y, y = 2, y Ú 0
17. x = 2y - y2, x = 0 18. x = 2y - y2, x = y
19. y = 0 x 0 , y = 1 20. y = x, y = 2x, y = 2
21. y = 2x, y = 0, y = x - 2
22. y = 2x, y = 0, y = 2 - x
Revolution About Horizontal and Vertical LinesIn Exercises 23–26, use the shell method to find the volumes of the solids generated by revolving the regions bounded by the given curves about the given lines.
23. y = 3x, y = 0, x = 2
a. The y-axis b. The line x = 4
c. The line x = -1 d. The x@axis
e. The line y = 7 f. The line y = -2
24. y = x3, y = 8, x = 0
a. The y-axis b. The line x = 3
c. The line x = -2 d. The x@axis
e. The line y = 8 f. The line y = -1
25. y = x + 2, y = x2
a. The line x = 2 b. The line x = -1
c. The x@axis d. The line y = 4
26. y = x4, y = 4 - 3x2
a. The line x = 1 b. The x@axis
In Exercises 27 and 28, use the shell method to find the volumes of the solids generated by revolving the shaded regions about the indi-cated axes.
27. a. The x-axis b. The line y = 1
c. The line y = 8>5 d. The line y = -2>5
x
y
0
1
1
x = 12(y2 − y3)
28. a. The x-axis b. The line y = 2
c. The line y = 5 d. The line y = -5>8
x
y
2
(2, 2)
10
2
x =y2
2
x = −y4
4y2
2
Choosing the Washer Method or Shell MethodFor some regions, both the washer and shell methods work well for the solid generated by revolving the region about the coordinate axes, but this is not always the case. When a region is revolved about the y-axis, for example, and washers are used, we must integrate with respect to y. It may not be possible, however, to express the integrand in terms of y. In such a case, the shell method allows us to integrate with respect to x instead. Exercises 29 and 30 provide some insight.
29. Compute the volume of the solid generated by revolving the region bounded by y = x and y = x2 about each coordinate axis using
a. the shell method. b. the washer method.
30. Compute the volume of the solid generated by revolving the tri-angular region bounded by the lines 2y = x + 4, y = x, and x = 0 about
a. the x-axis using the washer method.
b. the y-axis using the shell method.
c. the line x = 4 using the shell method.
d. the line y = 8 using the washer method.
In Exercises 31–36, find the volumes of the solids generated by revolving the regions about the given axes. If you think it would be better to use washers in any given instance, feel free to do so.
364 Chapter 6: Applications of Definite Integrals
Solution
(a) Multiplying each piece by x gives
xf(x) =⎧⎪⎪⎨⎪⎪⎩
sinx, 0 < x ≤ πx, x = 0
But when x = 0 we have sin 0 = 0 thus xf(x) = sinx for all values of x
(b) We have r(x) = x and h(x) = f(x) so
Volume = ∫π
02πxf(x) dx
= ∫π
02π sinx dx
= −2π cosx∣π0= 2π + 2π
= 4π
17
Problem 15. Use the shell method to find the volume of the solid generated by revolving the regionbounded by the curves and lines about the x-axis: x = √
y, x = −y, y = 2
Solution We have h(y) = √y − (−y) and r(y) = y so the volume is given by
Volume = ∫2
02πy(√y + y) dy
= 2π∫2
0(y3/2 + y2) dy
= 2π (2
5y5/2 + y
3
3)RRRRRRRRRRR
2
0
= 2π (2
5⋅ 4
√2 + 8
3)
= 2π⎛⎝
8√
2
5+ 8
3
⎞⎠
= 2π⎛⎝
24√
2
15+ 40
15
⎞⎠
= 16π
15(3
√2 + 5)
18
Problem 23. Use the shell method to find the volume of the solid generated by revolving the shadedregion about the indicated axis:
(a) The x-axis
(b) The line y = 1
(c) The line y = 8/5
(d) The line y = −2/5
Solution
(a) We have r(y) = y and h(y) = 12y2 − 12y3 so
Volume = ∫1
02πy(12y2 − 12y3) dy
= 2π∫1
0(12y3 − 12y4) dy
= 2π (3y4 − 12y5
5)RRRRRRRRRRR
1
0
= 2π (3 − 12
5)
= 2π (3
5)
= 6π
5
(b) We have r(y) = 1 − y and h(y) = 12y2 − 12y3 so
Volume = ∫1
02π(1 − y)(12y2 − 12y3) dy
= 2π∫1
0(12y2 − 24y3 + 12y4) dy
= 2π (4y3 − 6y4 + 12y5
5)RRRRRRRRRRR
1
0
= 2π (4 − 6 + 12
5)
= 4π
5
19
(c) We have r(y) = 8
5− y and h(y) = 12y2 − 12y3
Volume = ∫1
02π (8
5− y)(12y2 − 12y3) dy
= 2π∫1
0(96y2
5− 156y3
5+ 12y4) dy
(d) r(y) = y + 2
5and h(y) = 12y2 − 12y3
Volume = ∫1
02π (y + 2
5)(12y2 − 12y3) dy
20
Problem 24.
21
Problem 25. Compute the volume of the solid generated by revolving the region bounded by y = xand y = x2 about each coordinate axis using
