Study Guide - Volumes of Revolution: Cylindrical Shells (2024)

Again, we are working with a solid of revolution. As before, we define a region [latex]R,[/latex] bounded above by the graph of a function [latex]y=f\left(x\right),[/latex] below by the [latex]x\text{-axis,}[/latex] and on the left and right by the lines [latex]x=a[/latex] and [latex]x=b,[/latex] respectively, as shown in [link](a). We then revolve this region around the y-axis, as shown in [link](b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of [latex]x[/latex] were revolved around the [latex]x\text{-axis}[/latex] or a line parallel to it.

As we have done many times before, partition the interval [latex]\left[a,b\right][/latex] using a regular partition, [latex]P=\left\{{x}_{0},{x}_{1}\text{,…},{x}_{n}\right\}[/latex] and, for [latex]i=1,2\text{,…},n,[/latex] choose a point [latex]{x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].[/latex] Then, construct a rectangle over the interval [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] of height [latex]f\left({x}_{i}^{*}\right)[/latex] and width [latex]\text{Δ}x.[/latex] A representative rectangle is shown in [link](a). When that rectangle is revolved around the y-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.

To calculate the volume of this shell, consider [link].

The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius [latex]{x}_{i}[/latex] and inner radius [latex]{x}_{i-1}.[/latex] Thus, the cross-sectional area is [latex]\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}.[/latex] The height of the cylinder is [latex]f\left({x}_{i}^{*}\right).[/latex] Then the volume of the shell is

[latex]\begin{array}{cc}\hfill {V}_{\text{shell}}& =f\left({x}_{i}^{*}\right)\left(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}\right)\hfill \\ & =\pi f\left({x}_{i}^{*}\right)\left({x}_{i}^{2}-{x}_{i-1}^{2}\right)\hfill \\ & =\pi f\left({x}_{i}^{*}\right)\left({x}_{i}+{x}_{i-1}\right)\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =2\pi f\left({x}_{i}^{*}\right)\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\left({x}_{i}-{x}_{i-1}\right).\hfill \end{array}[/latex]

Note that [latex]{x}_{i}-{x}_{i-1}=\text{Δ}x,[/latex] so we have

[latex]{V}_{\text{shell}}=2\pi f\left({x}_{i}^{*}\right)\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\text{Δ}x.[/latex]

Furthermore, [latex]\frac{{x}_{i}+{x}_{i-1}}{2}[/latex] is both the midpoint of the interval [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] and the average radius of the shell, and we can approximate this by [latex]{x}_{i}^{*}.[/latex] We then have

[latex]{V}_{\text{shell}}\approx 2\pi f\left({x}_{i}^{*}\right){x}_{i}^{*}\text{Δ}x.[/latex]

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ([link]).

In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height [latex]f\left({x}_{i}^{*}\right),[/latex] width [latex]2\pi {x}_{i}^{*},[/latex] and thickness [latex]\text{Δ}x[/latex] ([link]). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

[latex]{V}_{\text{shell}}\approx f\left({x}_{i}^{*}\right)\left(2\pi {x}_{i}^{*}\right)\text{Δ}x,[/latex]

which is the same formula we had before.

To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

[latex]V\approx \sum _{i=1}^{n}\left(2\pi {x}_{i}^{*}f\left({x}_{i}^{*}\right)\text{Δ}x\right).[/latex]

Here we have another Riemann sum, this time for the function [latex]2\pi xf\left(x\right).[/latex] Taking the limit as [latex]n\to \infty [/latex] gives us

[latex]V=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}\left(2\pi {x}_{i}^{*}f\left({x}_{i}^{*}\right)\text{Δ}x\right)={\int }_{a}^{b}\left(2\pi xf\left(x\right)\right)dx.[/latex]

This leads to the following rule for the method of cylindrical shells.

Now let’s consider an example.

As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the [latex]x\text{-axis},[/latex] when we want to integrate with respect to [latex]y.[/latex] The analogous rule for this type of solid is given here.

For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall that we found the volume of one of the shells to be given by

[latex]\begin{array}{cc}\hfill {V}_{\text{shell}}& =f\left({x}_{i}^{*}\right)\left(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}\right)\hfill \\ & =\pi f\left({x}_{i}^{*}\right)\left({x}_{i}^{2}-{x}_{i-1}^{2}\right)\hfill \\ & =\pi f\left({x}_{i}^{*}\right)\left({x}_{i}+{x}_{i-1}\right)\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =2\pi f\left({x}_{i}^{*}\right)\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\left({x}_{i}-{x}_{i-1}\right).\hfill \end{array}[/latex]

This was based on a shell with an outer radius of [latex]{x}_{i}[/latex] and an inner radius of [latex]{x}_{i-1}.[/latex] If, however, we rotate the region around a line other than the [latex]y\text{-axis},[/latex] we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line [latex]x=\text{−}k,[/latex] where [latex]k[/latex] is some positive constant. Then, the outer radius of the shell is [latex]{x}_{i}+k[/latex] and the inner radius of the shell is [latex]{x}_{i-1}+k.[/latex] Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line [latex]x=\text{−}k,[/latex] the volume of a shell is given by

[latex]\begin{array}{cc}\hfill {V}_{\text{shell}}& =2\pi f\left({x}_{i}^{*}\right)\left(\frac{\left({x}_{i}+k\right)+\left({x}_{i-1}+k\right)}{2}\right)\left(\left({x}_{i}+k\right)-\left({x}_{i-1}+k\right)\right)\hfill \\ & =2\pi f\left({x}_{i}^{*}\right)\left(\left(\frac{{x}_{i}+{x}_{i-2}}{2}\right)+k\right)\text{Δ}x.\hfill \end{array}[/latex]

As before, we notice that [latex]\frac{{x}_{i}+{x}_{i-1}}{2}[/latex] is the midpoint of the interval [latex]\left[{x}_{i-1},{x}_{i}\right][/latex] and can be approximated by [latex]{x}_{i}^{*}.[/latex] Then, the approximate volume of the shell is

[latex]{V}_{\text{shell}}\approx 2\pi \left({x}_{i}^{*}+k\right)f\left({x}_{i}^{*}\right)\text{Δ}x.[/latex]

The remainder of the development proceeds as before, and we see that

[latex]V={\int }_{a}^{b}\left(2\pi \left(x+k\right)f\left(x\right)\right)dx.[/latex]

We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the [latex]x\text{-term}[/latex] in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.

For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. [link] describes the different approaches for solids of revolution around the [latex]x\text{-axis}.[/latex] It’s up to you to develop the analogous table for solids of revolution around the [latex]y\text{-axis}.[/latex]

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

For the following exercise, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand.

For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the [latex]x\text{-axis}[/latex] and are rotated around the [latex]y\text{-axis}.[/latex]

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and [latex]y=0[/latex] around the [latex]x\text{-axis}.[/latex]

For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.

For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume.

For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.