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# Section7.3The Shell Method¶ permalink

Often a given problem can be solved in more than one way. A particular method may be chosen out of convenience, personal preference, or perhaps necessity. Ultimately, it is good to have options.

The previous section introduced the Disk and Washer Methods, which computed the volume of solids of revolution by integrating the cross–sectional area of the solid. This section develops another method of computing volume, the Shell Method. Instead of slicing the solid perpendicular to the axis of rotation creating cross-sections, we now slice it parallel to the axis of rotation, creating “shells.”

Consider Figure 7.3.1, where the region shown in (a) is rotated around the $y$-axis forming the solid shown in (b). A small slice of the region is drawn in (a), parallel to the axis of rotation. When the region is rotated, this thin slice forms a cylindrical shell, as pictured in part (c) of the figure. The previous section approximated a solid with lots of thin disks (or washers); we now approximate a solid with many thin cylindrical shells.

To compute the volume of one shell, first consider the paper label on a soup can with radius $r$ and height $h\text{.}$ What is the area of this label? A simple way of determining this is to cut the label and lay it out flat, forming a rectangle with height $h$ and length $2\pi r\text{.}$ Thus the area is $A = 2\pi rh\text{;}$ see <<Unresolved xref, reference "fig_soupcan"; check spelling or use "provisional" attribute>> (a).

Do a similar process with a cylindrical shell, with height $h\text{,}$ thickness $\Delta x\text{,}$ and approximate radius $r\text{.}$ Cutting the shell and laying it flat forms a rectangular solid with length $2\pi r\text{,}$ height $h$ and depth $\dx\text{.}$ Thus the volume is $V \approx 2\pi rh\dx\text{;}$ see <<Unresolved xref, reference "fig_soupcan"; check spelling or use "provisional" attribute>> (b). (We say “approximately” since our radius was an approximation.)

By breaking the solid into $n$ cylindrical shells, we can approximate the volume of the solid as \begin{equation*} V = \sum_{i=1}^n 2\pi r_ih_i\dx_i, \end{equation*} where $r_i\text{,}$ $h_i$ and $\dx_i$ are the radius, height and thickness of the $i^\text{ th }$ shell, respectively.

This is a Riemann Sum. Taking a limit as the thickness of the shells approaches 0 leads to a definite integral.

##### Key Idea7.3.6The Shell Method

Let a solid be formed by revolving a region $R\text{,}$ bounded by $x=a$ and $x=b\text{,}$ around a vertical axis. Let $r(x)$ represent the distance from the axis of rotation to $x$ (i.e., the radius of a sample shell) and let $h(x)$ represent the height of the solid at $x$ (i.e., the height of the shell). The volume of the solid is \begin{equation*} V = 2\pi\int_a^b r(x)h(x)\ dx. \end{equation*}

Special Cases:

1. When the region $R$ is bounded above by $y=f(x)$ and below by $y=g(x)\text{,}$ then $h(x) = f(x)-g(x)\text{.}$

2. When the axis of rotation is the $y$-axis (i.e., $x=0$) then $r(x) = x\text{.}$

Let's practice using the Shell Method.

##### Example7.3.7Finding volume using the Shell Method

Find the volume of the solid formed by rotating the region bounded by $y=0\text{,}$ $y=1/(1+x^2)\text{,}$ $x=0$ and $x=1$ about the $y$-axis.

Solution

With the Shell Method, nothing special needs to be accounted for to compute the volume of a solid that has a hole in the middle, as demonstrated next.

##### Example7.3.9Finding volume using the Shell Method

Find the volume of the solid formed by rotating the triangular region determined by the points $(0,1)\text{,}$ $(1,1)$ and $(1,3)$ about the line $x=3\text{.}$

Solution

When revolving a region around a horizontal axis, we must consider the radius and height functions in terms of $y\text{,}$ not $x\text{.}$

##### Example7.3.13Finding volume using the Shell Method

Find the volume of the solid formed by rotating the region given in Example 7.3.9 about the $x$-axis.

Solution

At the beginning of this section it was stated that “it is good to have options.” The next example finds the volume of a solid rather easily with the Shell Method, but using the Washer Method would be quite a chore.

##### Example7.3.18Finding volume using the Shell Method

Find the volume of the solid formed by revolving the region bounded by $y= \sin(x)$ and the $x$-axis from $x=0$ to $x=\pi$ about the $y$-axis.

Solution

We end this section with a table summarizing the usage of the Washer and Shell Methods.

##### Key Idea7.3.23Summary of the Washer and Shell Methods

Let a region $R$ be given with $x$-bounds $x=a$ and $x=b$ and $y$-bounds $y=c$ and $y=d\text{.}$

 Washer Method Shell Method \text{ Horizontal Axis } $\ds \pi\int_a^b \big(R(x)^2-r(x)^2\big)\ dx$ $\ds 2\pi\int_c^d r(y)h(y)\ dy$ \text{ Vertical Axis } $\ds\pi \int_c^d\big(R(y)^2-r(y)^2\big)\ dy$ $\ds 2\pi\int_a^b r(x)h(x)\ dx$

As in the previous section, the real goal of this section is not to be able to compute volumes of certain solids. Rather, it is to be able to solve a problem by first approximating, then using limits to refine the approximation to give the exact value. In this section, we approximate the volume of a solid by cutting it into thin cylindrical shells. By summing up the volumes of each shell, we get an approximation of the volume. By taking a limit as the number of equally spaced shells goes to infinity, our summation can be evaluated as a definite integral, giving the exact value.

We use this same principle again in the next section, where we find the length of curves in the plane.

# Subsection7.3.1Exercises

Terms and Concepts

In the following exercises, a region of the Cartesian plane is shaded. Use the Shell Method to find the volume of the solid of revolution formed by revolving the region about the $y$-axis.