6 Volumes of Solids Of Revolution II-x

1. Volume and Solids of Revolution II

2. Volume and Solids of Revolution II
Following are more examples of solids whose
cross–sectional areas are tractable so that we may
use the Cavalieri’s principle to find their volumes.

3. Volume and Solids of Revolution II
One group of such solids has cross–sectional areas
that are formed from the linear cross–sections of the
base regions of the solids.
Following are more examples of solids whose
cross–sectional areas are tractable so that we may
use the Cavalieri’s principle to find their volumes.

4. Volume and Solids of Revolution II
A solid has a semi–circular
disc of radius r as base.
Its cross-sections,
perpendicular to the
diameter of the base, are
isosceles right triangles as
shown. Find its volume.
Example A.
One group of such solids has cross–sectional areas
that are formed from the linear cross–sections of the
base regions of the solids.
Following are more examples of solids whose
cross–sectional areas are tractable so that we may
use the Cavalieri’s principle to find their volumes.

5. Volume and Solids of Revolution II
Example A.
A solid has a semi–circular
disc of radius r as base.
Its cross-sections,
perpendicular to the
diameter of the base, are
isosceles right triangles as
shown. Find its volume.
r
One group of such solids has cross–sectional areas
that are formed from the linear cross–sections of the
base regions of the solids.
Following are more examples of solids whose
cross–sectional areas are tractable so that we may
use the Cavalieri’s principle to find their volumes.

6. Volume and Solids of Revolution II
Example A.
A solid has a semi–circular
disc of radius r as base.
Its cross-sections,
perpendicular to the
diameter of the base, are
isosceles right triangles as
shown. Find its volume.
r
One group of such solids has cross–sectional areas
that are formed from the linear cross–sections of the
base regions of the solids.
Following are more examples of solids whose
cross–sectional areas are tractable so that we may
use the Cavalieri’s principle to find their volumes.

7. Volume and Solids of Revolution II
We calculate the volume of half of the solid.
r

8. Volume and Solids of Revolution II
r
isosceles right triangles
We calculate the volume of half of the solid.

9. Volume and Solids of Revolution II
rr
isosceles right triangles
We calculate the volume of half of the solid.

10. Volume and Solids of Revolution II
r
We calculate the volume of half of the solid.
Set the x measurement as shown from x = 0 to x = r.
rr
isosceles right triangles
x=0
x=r
x

11. Volume and Solids of Revolution II
√r2 – x2
r
We calculate the volume of half of the solid.
Set the x measurement as shown from x = 0 to x = r.
Given an x, the cross-sectional length is √r2 – x2.
rr
isosceles right triangles
x=0
x=r
x

12. Volume and Solids of Revolution II
So the cross-section of the solid is a right isosceles
triangle with area ½ (√r2 – x2)2 = ½ (r2 – x2).
√r2 – x2
r
We calculate the volume of half of the solid.
Set the x measurement as shown from x = 0 to x = r.
Given an x, the cross-sectional length is √r2 – x2.
rr
isosceles right triangles
x=0
x=r
x

13. Volume and Solids of Revolution II
So the cross-section of the solid is a right isosceles
triangle with area ½ (√r2 – x2)2 = ½ (r2 – x2).
So the volume of the entire solid (two halves) is
∫x=0
r
dxr2 – x2
√r2 – x2
r
We calculate the volume of half of the solid.
Set the x measurement as shown from x = 0 to x = r.
Given an x, the cross-sectional length is √r2 – x2.
rr
isosceles right triangles
x=0
x=r
x

14. Volume and Solids of Revolution II
So the cross-section of the solid is a right isosceles
triangle with area ½ (√r2 – x2)2 = ½ (r2 – x2).
So the volume of the entire solid (two halves) is
∫x=0
r
dxr2 – x2
= r2x – x3
3 x=0
r
√r2 – x2
r
We calculate the volume of half of the solid.
Set the x measurement as shown from x = 0 to x = r.
Given an x, the cross-sectional length is √r2 – x2.
rr
isosceles right triangles
x=0
x=r
x

15. Volume and Solids of Revolution II
So the cross-section of the solid is a right isosceles
triangle with area ½ (√r2 – x2)2 = ½ (r2 – x2).
So the volume of the entire solid (two halves) is
∫x=0
r
dxr2 – x2
= r2x – x3
3 x=0
r
= 2r3
3
x=0
x=r
x√r2 – x2
r
We calculate the volume of half of the solid.
Set the x measurement as shown from x = 0 to x = r.
Given an x, the cross-sectional length is √r2 – x2.
rr
isosceles right triangles

16. Volume and Solids of Revolution II
Note that if the vertical leg of the triangle is above
the circular arc instead of above the diameter,

17. Volume and Solids of Revolution II
Note that if the vertical leg of the triangle is above
the circular arc instead of above the diameter,
(These two solids have the same cross-sections hence the same volume.)
r

18. Volume and Solids of Revolution II
Note that if the vertical leg of the triangle is above
the circular arc instead of above the diameter, the
resulting solid looks different but has the same
volume as before because they have the same
cross-sections.
(These two solids have the same cross-sections hence the same volume.)
r
‘

19. There are other ways to find volumes by the
Fundamental Theorem of Calculus.
Volume and Solids of Revolution II

20. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces.
Volume and Solids of Revolution II

21. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.
Volume and Solids of Revolution II

22. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.
For example, a solid of revolution is generated by a
template,
Volume and Solids of Revolution II

23. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.
For example, a solid of revolution is generated by a
template,
Volume and Solids of Revolution II

24. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.
For example, a solid of revolution is generated by a
template, then a strip of cross–section that is parallel
to the axis of rotation will generate a shell that is
approximately a cylindrical shell.
Volume and Solids of Revolution II

25. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.
For example, a solid of revolution is generated by a
template, then a strip of cross–section that is parallel
to the axis of rotation will generate a shell that is
approximately a cylindrical shell.
Volume and Solids of Revolution II

26. There are other ways to find volumes by the
Fundamental Theorem of Calculus. Basically, we
subdivide the solid into small manageable pieces
and sum the volumes of all these pieces. Then we
cut the solid into smaller and smaller such pieces
and pass the sums into a definite integral.
For example, a solid of revolution is generated by a
template, then a strip of cross–section that is parallel
to the axis of rotation will generate a shell that is
approximately a cylindrical shell.
Volume and Solids of Revolution II

27. The volume of a cylindrical shell depends on the
radius r, the thickness Δx, and the height h.
Δx
r
h
Volume and Solids of Revolution II

28. The volume of a cylindrical shell depends on the
radius r, the thickness Δx, and the height h.
To approximate the volume of a cylindrical shell,
slice it open and unroll it into a flat slab.
Δx
r
h
Volume and Solids of Revolution II
Δx

29. The volume of a cylindrical shell depends on the
radius r, the thickness Δx, and the height h.
To approximate the volume of a cylindrical shell,
slice it open and unroll it into a flat slab.
Δx
r
h
Volume and Solids of Revolution II
Δx
It’s almost a rectangular box.

30. The volume of a cylindrical shell depends on the
radius r, the thickness Δx, and the height h.
To approximate the volume of a cylindrical shell,
slice it open and unroll it into a flat slab.
Δx
r
h
Volume and Solids of Revolution II
Δx
It’s almost a rectangular box.
2πr
Δx
h≈
Δx

31. The volume of a cylindrical shell depends on the
radius r, the thickness Δx, and the height h.
To approximate the volume of a cylindrical shell,
slice it open and unroll it into a flat slab.
It’s almost a rectangular box. We use the volume of
the rectangular box which is 2πr*Δx*h to approximate
the volume of this slab.
Δx
r
h
Volume and Solids of Revolution II
Δx
2πr
Δx
h≈
Δx

32. Δx
r
h
2πr
Δx
h≈
Volume and Solids of Revolution II
Therefore the volume of the cylindrical shell is
roughly 2πrhΔx.

33. Δx
r
h
2πr
Δx
h≈
Volume and Solids of Revolution II
Therefore the volume of the cylindrical shell is
roughly 2πrhΔx.
x=a x=b
y=f(x)
Suppose a solid of revolution is formed by rotating a
template that is bounded by y = f(x), from x = a to
x = b, around a vertical axis.

34. Δx
r
h
2πr
Δx
h≈
Volume and Solids of Revolution II
Therefore the volume of the cylindrical shell is
roughly 2πrhΔx.
x=a x=b
y=f(x)
Suppose a solid of revolution is formed by rotating a
template that is bounded by y = f(x), from x = a to
x = b, around a vertical axis.
x=a
x=b
y=f(x)

35. x=a x=b
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi].
*
xi
Volume and Solids of Revolution II
*

36. x=a x=b
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi]. Let Δx be the width of the subinterval.
*
xi
Volume and Solids of Revolution II
*

37. x=a x=b
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi]. Let Δx be the width of the subinterval.
Then the shell generated by the strip over [xi–1, xi]
is approximately the cylindrical shell with height f(xi),
width Δx, and radius D(xi).
*
xi
*
*
Volume and Solids of Revolution II
*

38. x=a x=b
y=f(xi)
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi]. Let Δx be the width of the subinterval.
Then the shell generated by the strip over [xi–1, xi]
is approximately the cylindrical shell with height f(xi),
width Δx, and radius D(xi).
*
xi
D(x )
*
*
Volume and Solids of Revolution II
*
*
*

39. x=a x=b
y=f(xi)
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi]. Let Δx be the width of the subinterval.
Then the shell generated by the strip over [xi–1, xi]
is approximately the cylindrical shell with height f(xi),
width Δx, and radius D(xi).
*
xi
D(x )
*
*
x=a x=b
D(xi )
Volume and Solids of Revolution II
*
*
**
y=f(xi)*
xi*

40. x=a x=b
y=f(xi)
Partition [a, b] into n equal size subintervals
{x0=a, x1, x2, ..xi, .. xn=b}, xi be an arbitrary point in
[xi–1, xi]. Let Δx be the width of the subinterval.
Then the shell generated by the strip over [xi–1, xi]
is approximately the cylindrical shell with height f(xi),
width Δx, and radius D(xi).
*
xi
D(x )
2πD(xi)f(xi)Δx.
*
*
x=a x=bxi
D(xi )
* *
Hence the volume of this cylindrical shell is
Volume and Solids of Revolution II
*
*
**
y=f(xi)*
*

41. ≈ 2πD(xi)f(xi)Δx.* *
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

42. ≈ 2πD(xi)f(xi)Δx.* *
The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

43. ≈ 2πD(xi)f(xi)Δx.* *
Volume of V ≈
i=1
n
∑
The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
2πD(xi)f(xi)Δx* *
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

44. ≈ 2πD(xi)f(xi)Δx.* *
Volume of V ≈
i=1
n
∑
In fact the volume is its limit as Δx0 or as n ∞.
The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
2πD(xi)f(xi)Δx* *
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*

45. ≈ 2πD(xi)f(xi)Δx.* *
Volume of V ≈
i=1
n
∑
Therefore the volume of V = lim
i=1
n
∑
n ∞
The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
2πD(xi)f(xi)Δx* *
2πD(xi)f(xi)Δx* *
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*
In fact the volume is its limit as Δx0 or as n ∞.

46. ≈ 2πD(xi)f(xi)Δx.* *
Volume of V ≈
i=1
n
∑
Therefore the volume of V = lim
i=1
n
∑
n ∞
By the FTC, the volume of V = ∫x=a
b
The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
2πD(xi)f(xi)Δx* *
2πD(xi)f(xi)Δx* *
2πD(x)f(x)dx
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*
In fact the volume is its limit as Δx0 or as n ∞.
radius* height

47. ≈ 2πD(xi)f(xi)Δx.* *
Volume of V ≈
i=1
n
∑
Therefore the volume of V = lim
i=1
n
∑
n ∞
By the FTC, the volume of V = ∫x=a
b
The sum of the volumes of all such cylindrical shells
approximates the volume of the solid of revolution V.
2πD(xi)f(xi)Δx* *
2πD(xi)f(xi)Δx* *
2πD(x)f(x)dx
This known as the shell method.
Volume and Solids of Revolution II
x=a x=b
y=f(xi)
xi
D(x )
x=a x=b
D(xi )
*
*
**
y=f(xi)*
xi*
In fact the volume is its limit as Δx0 or as n ∞.
radius* height

48. Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x

49. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2.
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x

50. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2.
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3

51. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
x

52. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2)
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x

53. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2) which is the height of the
cylindrical shell it forms.
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x

54. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2) which is the height of the
cylindrical shell it forms.
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x

55. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2) which is the height of the
cylindrical shell it forms. The radius is D(x) =
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x

56. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2) which is the height of the
cylindrical shell it forms. The radius is D(x) = (3 – x).
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x
3–x

57. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2) which is the height of the
cylindrical shell it forms. The radius is D(x) = (3 – x).
Hence the volume is
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x
3–x
∫x=0
2
2π(3 – x)(2x – x2)dx =

58. Set 2x = x2, we get x = 0, 2, so the area spans from
x = 0 to x = 2. Given a vertical cross–section at x,
its length is (2x – x2) which is the height of the
cylindrical shell it forms. The radius is D(x) = (3 – x).
Hence the volume is
Volume and Solids of Revolution II
Example B.
a. Revolve the area as shown around x = 3, find
the volume it formed by the shell method.
y = x2
y = 2x
x=3
2x–x2
x
y = x2
y = 2x
x=3
2x–x2
x
3–x
∫x=0
2
2π(3 – x)(2x – x2)dx =16π
3

59. Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1

60. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1

61. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1

62. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1. Hence we need to solve
each equation for x.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1

63. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1. Hence we need to solve
each equation for x.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.

64. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1. Hence we need to solve
each equation for x.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.
x = √yx = y/2

65. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1. Hence we need to solve
each equation for x.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.
x = √yx = y/2
The template spans from y = 0 to y = 4.

66. We are to track a horizontal
cross–section, parallel to the axis
of rotation, as it orbits around
y = –1. Hence we need to solve
each equation for x.
Volume and Solids of Revolution II
b. Revolve the area as shown around y = –1,
find the volume it formed by the shell method.
y = x2
y = 2x
y = –1
y = x2  x = f(y) = √y, and
y = 2x  x = g(y) = y/2.
x = √yx = y/2
The template spans from y = 0 to y = 4.
For a fixed y, the horizontal cross–section forms
a cylindrical shell as it revolves around y = –1.

67. Volume and Solids of Revolution II
y = –1
x = √y
x = y/2
y

68. Volume and Solids of Revolution II
y = –1
x = √y
x = y/2
y = –1
x = √y
x = y/2
y
y

69. Volume and Solids of Revolution II
At y, the horizontal cross–section
has length (√y – y/2).
y = –1
x = √y
x = y/2
y
y = –1
x = √y
x = y/2
y
√y – y/2

70. Volume and Solids of Revolution II
The radius of rotation is (y + 1).
At y, the horizontal cross–section
has length (√y – y/2).
y = –1
x = √y
x = y/2
y
y = –1
x = √y
x = y/2
y
√y – y/2
y + 1

71. Volume and Solids of Revolution II
The radius of rotation is (y + 1).
At y, the horizontal cross–section
has length (√y – y/2).
Therefore the volume of the solid is
∫y=0
4
2π(√y – y/2)(y + 1)dy = 104π
15
y = –1
x = √y
x = y/2
y
y = –1
x = √y
x = y/2
y
√y – y/2
y + 1