This document provides an overview of using the cylindrical shell method to calculate volumes of revolution. It defines cylindrical shells as formed by rotating rectangles parallel to the axis of rotation. The volume of a shell is given as 2πrhΔr, where r is the radius, h is the height, and Δr is the thickness. Several examples are worked out in detail, calculating the volumes of solids formed by rotating regions between curves about different axes using the cylindrical shells method. Guidelines are also given for determining whether to use shells or washers/disks based on the region geometry.

1. Chapter 6: Applications of Integration
Section 6.3: Volumes by Cylindrical Shells
Alea Wittig
SUNY Albany

2. Outline
Cylindrical Shells
Volume of a Cylindrical Shell
Examples
Cylindrical Shells Vs Washers
Example

3. Cylindrical Shells

4. ▶ Consider the problem of finding the volume of the solid
obtained by rotating about the y−axis the region bounded by
y = 2x2 − x3 and y = 0.
▶ To use the washer/disk
method with a vertical axis
of rotation, we would need
to find the inner and outer
radii by determining the left
and right curves.
▶ But how would we
determine xR and xL?
▶ xR and xL in this case correspond to the same function,
y = 2x2 − x3, which additionally is very difficult to write as a
function of y (g(y) = x).
▶ Too hard!

5. ▶ Idea: Form n rectangles "parallel" to the axis of rotation,
which in this example is a vertical line, specifically, the y-axis.
▶ We take "vertical" rectangles in the sense that the rectangle
has width △x and height y = f (x).
▶ Rotating about the y−axis we get a new type of solid called a
cylindrical shell.

6. ▶ The shape of a shell versus washer is different in the following
sense:
▶ A cylindrical shell looks like
a paper towel roll.
▶ The thickness of the wall of
the roll corresponds to the
width of the rectangle
being rotated (△x or △y).
▶ A washer looks more like,
well, a washer.
▶ The thickness of the washer
corresponds to the width of
the rectangle being rotated
(△x or △y).

7. Volume of a Cylindrical Shell

8. ▶ The volume of a cylindrical
shell is obtained by
subtracting the inner volume
V1 from the outer volume V2
V1 = πr2
1 h
V2 = πr2
2 h
Vshell = V2 − V1
= π(r2
2 − r2
1 )h
= π(r2 + r1)(r2 − r1)h
= 2π
(r2 + r1)
2
(r2 − r1)h
▶ Let △r = r2 − r1
▶ Let r denote the average value of r1 and r2, ie,
r =
(r2 + r1)
2

9. Vshell = 2π
(r2 + r1)
2
(r2 − r1)h
= 2πr · △r · h
▶ When we take n → ∞, the
thickness △r will become
arbitrarily close to 0.
▶ So the wall of the shell is
going to become arbitrarily
thin like a thin sheet of
paper rolled up.
▶ So we can think of r as just
the radius and 2πr as the
circumference of the shell.
Vshell = 2πr
|{z}
circumference
· h
|{z}
height
· △r
|{z}
thickness

10. ▶ Notice that if we unfurl the cylindrical shell, we have
something like a rectangular box with dimensions
length = 2πx
width = △x
height = f (x)
So the volume is
V = length · height · width
= 2πx
|{z}
circumference
· f (x)
|{z}
height
· △x
|{z}
thickness

11. ▶ Now we want to sum up the volumes of the shells to
approximate the volume of the solid so
V ≈
n
X
i=1
Vi =
n
X
i=1
2πri hi △r where n = number of shells
▶ We intuitively expect that as we take n → ∞ we have
V = lim
n→∞
n
X
i=1
2πri hi △r
If, as in the first example, we have a solid obtained by rotating
about the y-axis the region under the curve y = f (x) from a
to b, then ri = xi , hi = f (xi ), and △r = △x so that the
volume of the solid is
V =
Z b
a
2πxf (x)dx where 0 ≤ a < b

12. Examples

13. Example 1
Now let’s use cylindrical shells to solve that first problem:
Find the volume of the solid obtained by rotating the region
between y = 2x2 − x3 and y = 0 about the y−axis.
▶ From the sketch we see
that the height of a shell is
f (x) = 2x2 − x3, the radius
is x, and the curve
intersects y = 0 (the
x−axis) at x = 0 and
x = 2.
V =
Z 2
0
2π x
|{z}
radius
(2x2
− x3
)
| {z }
height
dx

14. Example 1
V =
Z 2
0
2πx(2x2
− x3
)dx
= 2π
Z 2
0
x(2x2
− x3
)dx
= 2π
Z 2
0
2x3
− x4
dx
= 2π
2x4
4
−
x5
5

17. 2
0
= 2π
x4
2
−
x5
5

20. 2
0
= 2π
24
2
−
25
5
−
*0
04
2
−
05
5
= 2π
8 −
32
5
= 2π
40
5
−
32
5
= 2π
8
5
=
16π
5

21. Remark
▶ In general, to use cylindrical shells we shouldn’t rely on the
formula
V =
Z b
a
2πxf (x)dx
because we won’t always have that type of geometry.
▶ Rather, we should remember the more general formula
V =
Z
2πr
|{z}
circumference
h
|{z}
height
dr
|{z}
thickness
where dr is
▶ dx if axis of rotation is a vertical line
▶ dy if axis of rotation is a horizontal line
▶ And we must find the radius r, the height h, and the limits of
integration by sketching the graph.

22. Example 2
Find the volume of the solid obtained by rotating about the y−axis
the region between y = x and y = x2.

23. Example 2
2/3
▶ Not shown in the image but the curves intersect when
x2 = x =⇒ x2 − x = 0 =⇒ x(x − 1) = 0 =⇒ x = 0, 1.
▶ To get cylindrical shells we form rectangles parallel to the
axis of rotation. In this case the axis of rotation is vertical (the
y-axis). So we form rectangles of width △x.
▶ A sample rectangle has width △x and height yt − yB = x − x2.
▶ Rotating the rectangle we get a cylindrical shell of
thickness=△x, height =x − x2 and radius= x.
▶ Note that △x will become dx in the integrand.

24. Example 2
3/3
V =
Z 1
0
2πx
|{z}
circumference
(x − x2
)
| {z }
height
dx
|{z}
thickness
=
Z 1
0
2πx(x − x2
)dx
= 2π
Z 1
0
x2
− x3
dx
= 2π
x3
3
−
x4
4

27. 1
0
= 2π
1
3
−
1
4
−
0 − 0
!
= 2π
4
12
−
3
12
= 2π
1
12
=
π
6

28. Example 3
1/3
Use cylindrical shells to find the volume of the solid obtained by
rotating about the x−axis the region under the curve y =
√
x from
0 to 1.

29. Example 3
2/3
▶ The axis of rotation is horizontal so we take horizontal
rectangles, ie, rectangles of width △y.
▶ Since △y → dy in the integrand, this tells us we will be
integrating with respect to y.
▶ A sample rectangle has width △y and height = 1 − y2.
▶ Rotating the rectangle we get a cylindrical shell of thickness =
△y, height = 1 − y2 and radius = y.
▶ at x = 0, y =
√
0 = 0 and at x = 1, y =
√
1 = 1 so we are
integrating from y = 0 to y = 1.

30. Example 3
3/3
V =
Z 1
0
2πy
|{z}
circumference
(1 − y2
)
| {z }
height
dy
|{z}
thickness
= 2π
Z 1
0
y(1 − y2
)dy
= 2π
Z 1
0
y − y3
dy
= 2π
y2
2
−
y4
4

35. 1
0
= 2π
1
2
−
1
4
−
0 − 0
!
= 2π
1
4
=
π
2

36. Example 4
1/4
Find the volume of the solid obtained by rotating the region
bounded by y = x − x2 and y = 0 about the line x = 2.

37. Example 4
2/4
▶ First we must graph the curve y = x − x2.
▶ It’s graph is a transformation of the parabola y = x2.
▶ To see the transformation(s) we first complete the square.
y = x − x2
= −(x2
− x)
= − x2
− x +
−1
2(−1)
!2!
+
−1
2(−1)
!2
= −(x2
− x +
1
4
) +
1
4
= −(x −
1
2
)2
+
1
4
▶ So the transformation is:
▶ Shift right 1
2 unit.
▶ Shift up 1
4 unit.
▶ Flip over the x axis.

38. Example 4
3/3

39. Example 4
4/4
V =
Z 1
0
2π(2 − x)
| {z }
circumference
(x − x2
)
| {z }
height
dx
|{z}
thickness
= 2π
Z 1
0
(2 − x)(x − x2
)dx
= 2π
Z 1
0
2x − 2x2
− x2
+ x3
dx
= 2π
Z 1
0
2x − 3x2
+ x3
dx
= 2π
2x2
2
−
3x3
3
+
x4
4
!

44. 1
0
= 2π
1 − 1 +
1
4
−
0 − 0 + 0
=
π
2

45. Cylindrical Shells Vs Washers

46. How do we know when to use the cylindrical shell method or
the washer/disk method?
▶ Is the region more easily described by top and bottom
boundary curves of the form y = f (x), or by left and right
boundaries curves of the form x = g(y)?
▶ top/bottom - then integration will be wrt x and rectangles will
be drawn vertically.
▶ left/right - then integration will be wrt y and rectangles drawn
horizontally and int wrt y.
▶ Draw a sample rectangle in the region. Imagine the rectangle
revolving; it becomes either a disk/washer or a shell.

47. Example

48. Example 5
1/5
Find the volume of the solid
obtained by rotating the region
bounded by y = x2 and y = 2x
about the line x = −1 using
a. x as the variable of
integration and
b. y as the variable of
integration

49. Example 5
2/5
a. Integrating with respect to x we have got to take rectangles of
width △x because this will become dx in the integrand..
▶ This is shown in image (b).
▶ Rotating the vertical rectangle about a vertical line we get a
cylindrical shell.
b. Integrating with respect to y we have got to take rectangles of
width △y because this will become dy in the integrand.
▶ This is shown in image (c).
▶ Rotating the horizontal rectangle about a vertical line we
get a washer.

50. Example 5
3/5
▶ Now set up and compute the integrals on your own as an
exercise. You should get the same volume of course whether
you are using washers or shells.

51. Example 5
4/5
a. Integrating wrt x → rectangles of width △x → cylindrical
shells (since axis of rotation is vertical)
V =
Z b
a
[circumference] · [height] · [thickness]
=
Z 2
0
2π(x + 1)(2x − x2
)dx
= 2π
Z 2
0
2x2
− x3
+ 2x − x2
dx
= 2π
Z 2
0
x2
+ 2x − x3
dx
= 2π
x3
3
+
2x2
2
−
x4
4

56. 2
0
= 2π
8
3
+ 4 −
16
4
= 2π
8
3
+ 4 − 4
=
16π
3

57. Example 5
5/5
b. Integrating wrt y → rectangles of width △y → washers (since
axis of rotation is vertical)
V =
Z d
c
π[r2
outer − r2
inner]dy
=
Z 4
0
π[(1 +
√
y)2
− (1 +
y
2
)2
]dy
= π
Z 4
0
1 + 2
√
y + y − (1 + y +
y2
4
)dy
= π
Z 4
0
2
√
y −
y2
4
dy
= π
2y
3
2
3
2
−
y3
12

60. 4
0
= π
4y
3
2
3
−
y3
12

63. 4
0
= π
4
√
64
3
−
64
12
=
32π
3
−
16π
3
=
16π
3