Calculus Of One Real Variable –
By Pheng Kim Ving 
12.4 
Return
To Contents
Go To Problems & Solutions
region R bounded
by f, y = 0, x = a,
and x = b is revolved about the yaxis, it generates a solid S, as shown in Fig. 1.2.
We wish to find the volume V
of S. If we
use the slice method as discussed in Section
12.3 Part 3, a typical slice will be
perpendicular to the yaxis,
the integration will be along the yaxis,
the increment will be dy,
and the integrand will be a
function of y. That
entails solving the equation y
= f(x) for x to get an equation of the form x = g( y).
That task may be easy,
or hard, or impossible.
So in this section we're going to investigate another
approach, one that doesn't require solving the equation y = f(x) for
x. Let x be an arbitrary point in [a, b],
as shown in Fig. 1.1. Erect a thin vertical rectangle at x of width dx
and height
Fig. 1.1
The colored region is revolved about the
yaxis.

Fig. 1.2
The Solid Of Revolution And A
Cylindrical Shell.

Fig. 1.3
The cylindrical shell is reproduced here
for clarity.

f(x). The rectangle is gray
colored. When the region R
is revolved about the yaxis,
the vertical line segment at x
sweeps
out a cylindrical cross section, and the rectangle sweeps out a cylindrical
shell, as shown gray colored in Fig. 1.2.
Note that the rectangular strip is parallel to the yaxis, which is the axis of revolution,
and the cylindrical shell has its
axis along the axis of revolution. Let dV
be the volume of the shell. The volume V
of the solid S can be
approximated by
the sum of the differential volume
elements of such shells. Each shell in Fig. 1.2 corresponds to a rectangle
in Fig. 1.1.
The rectangles run from a
to b. Thus so
do the shells. Consequently V
is the integral of dV
from x = a to x = b.
Hence
the integration will be along the xaxis
and the integrand will be a function of x
(an expression involving f(x), as is
the case for the slice method; see the volume formula in Section
12.3 Part 1). This shows that we won't have to solve the
equation y = f(x) for x.
Now let's find the volume V.
The cylindrical shell is reproduced in Fig. 1.3. Its volume dV is:
This approach of finding the volume of revolution by using
cylindrical shells is called, well, the method of cylindrical
shells. For the sake of simplicity, it's also called the shell method.
A solid generated by revolving a disk about an axis that is
on its plane and external to it is called a torus (a
doughnutshaped solid). In Section
12.3 Problem & Solution 7 we've found the volume of the torus using the
slice
method. The torus is shown in Fig. 1.4. Find its volume using the shell method.
Fig. 1.4 A Torus. 
Solution
Fig. 1.5 The torus can be generated by
revolving this disk about the 
EOS
Remarks 1.1
i. The shell method gives the same result as does the slice method.
ii. Instead of a rectangle, we simply draw a strip,
and call it a rectangle and treat it as such when finding the volume of
the generated cylindrical shell. This
is because drawing a strip is simpler than drawing a rectangle and there's no
harm
in doing so.
iii. The axis of revolution is the yaxis.
iv. The rectangle is parallel to the axis of revolution.
vi. The rectangular strip is perpendicular to the xaxis. As a consequence the
increment is dx. Hence
the integration is
along the xaxis and we must express both the radius and
the height of the shell in terms of x.
Go To Problems & Solutions Return To Top Of Page
2. Axis Of Revolution Is Horizontal And/Or Doesn't Coincide With A

Example 2.1
In Section
12.3 Part 4 we used the slice method to find the volume of the solid
generated by revolving the plane region
bounded by y = x^{2} and y = 3 about the line y = –1. Now use the shell method to find that
volume.
Solution
Fig. 2.1
Plane region bounded by y = x^{2}
and y = 3 is revolved about line y = –1.

EOS
Remarks 2.1
i. The shell method gives the same result as does the slice method.
ii. The shell method can also be employed when the axis of revolution doesn't coincide with a coordinate axis.
iii. The axis of revolution is parallel to the xaxis. The axis of revolution
is horizontal, while in the case for the torus in
Example 1.1
above it's vertical.
iv. The rectangle is parallel to the axis of revolution.
vi. The rectangular strip is perpendicular to the yaxis. Consequently the
increment is dy, the
integration is along the
yaxis,
and we must express both the radius and the height of the shell in terms of y.
Go To Problems & Solutions Return To Top Of Page
3. General Case 
We've seen that the shell method can be used whether the
axis of revolution is vertical or horizontal, and whether it
coincides with a coordinate axis or not. In the case where the curve is y = f(x),
a boundary of the region is the xaxis,
and the axis of revolution coincides with the yaxis (and thus is vertical ) as discussed in Part 1, the radius and height of
the shell are x
and f(x) respectively and the increment
is dx. One or
more of these aspects may no longer be true in
other cases, as shown for one case in Part 2. However
the basic formula for the volume element is the same:
Let r and h be the radius and height of the shell respectively. Thus in general:

Problems & Solutions 
1. Use the shell method to find the volume of the
solid generated by revolving the plane region bounded by y = x^{2}, y = 9,
and x = 0 about the yaxis.
Solution
Note
The volume of this solid was also found in Section
12.3 Part 3 using the slice method. For this solid, the slice and shell
methods
require roughly the same amount of work.
Solution
Note
The volume of this solid with a hole was also found in Section
12.3 Part 2 using the slice method. There, we had to take the
hole into consideration in calculating the volume of the washer slice. Here,
using the shell method, the hole has no role in our
calculation of the volume of the shell.
3. Utilize the shell method to find the volume of the
solid generated by revolving the triangular region bounded by y = x,
y = 0, and x = 2 about the line x = 3.
Solution
4. The plane region bounded by x = y ^{2} and y = – x + 2 is revolved about the line y = 1. Find the volume of the
generated solid by using the shell
method.
Solution
Let's find the points of intersection of x = y^{2} and y = – x + 2 by solving the system:
Note
The volume of this solid was also found in Section
12.3 Problem & Solution 5 using the slice method. For this
solid, the
shell method is simpler than the slice method.
5. Find the volume of the solid generated by
revolving the plane region bounded by y
= e ^{x}, y = 0, x = 0, and x
= 1 about
the yaxis.
Solution
Return To Top Of Page Return To Contents