Calculus Of One Real Variable – By Pheng Kim Ving Chapter 12: Applications Of The Integral – Section 12.4: Finding Volumes By Using Cylindrical Shells 12.4 Finding Volumes By Using Cylindrical Shells

 1. Cylindrical Shells region R bounded by f, y = 0, x = a, and x = b is revolved about the y-axis, 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 y-axis, the integration will be along the y-axis, 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 # The colored region is revolved about the y-axis. # The Solid Of Revolution And A Cylindrical Shell. # The cylindrical shell is reproduced here for clarity.

f(x). The rectangle is gray colored. When the region R is revolved about the y-axis, 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 y-axis, 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 x-axis 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.

Example 1.1 – The Torus

A solid generated by revolving a disk about an axis that is on its plane and external to it is called a torus (a
doughnut-shaped 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  y-axis. 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 y-axis.

iv. The rectangle is parallel to the axis of revolution. vi. The rectangular strip is perpendicular to the x-axis. As a consequence the increment is dx. Hence the integration is
along the x-axis and we must express both the radius and the height of the shell in terms of x.

 2. Axis Of Revolution Is Horizontal And/Or Doesn't Coincide With A     Coordinate Axis

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 = x2 and y = 3 about the line y = –1. Now use the shell method to find that volume.

Solution  # Plane region bounded by y = x2 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 x-axis. 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 y-axis. Consequently the increment is dy, the integration is along the
y-axis, and we must express both the radius and the height of the shell in terms of y.

 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 x-axis,
and the axis of revolution coincides with the y-axis (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 = x2, y = 9,
and x = 0 about the y-axis.

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 = y2 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 y-axis.

Solution  