第6章積分の応用

In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells, we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation.

Again, we are working with a solid of revolution. As before, we define a region $R,$ bounded above by the graph of a function $y=f(x),$ below by the $x\text{-axis,}$ and on the left and right by the lines $x=a$ and $x=b,$ respectively, as shown in Figure 6.25(a). We then revolve this region around the y-axis, as shown in Figure 6.25(b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of $x$ were revolved around the $x\text{-axis}$ or a line parallel to it.

As we have done many times before, partition the interval $\left[a,b\right]$ using a regular partition, $P=\left\{x_0,x_1\text{,…},x_n\right\}$ and, for $i=1,2\text{,…},n,$ choose a point $x_i^{*}\in \left[x_{i-1},x_i\right].$ Then, construct a rectangle over the interval $\left[x_{i-1},x_i\right]$ of height $f(x_i^{*})$ and width $\text{Δ}x.$ A representative rectangle is shown in Figure 6.26(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 Figure 6.27.

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 $x_i$ and inner radius $x_{i-1}.$ Thus, the cross-sectional area is $\pi x_i^2-\pi x_{i-1}^2.$ The height of the cylinder is $f(x_i^{*}).$ Then the volume of the shell is

Note that $x_i-x_{i-1}=\text{Δ}x,$ so we have

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

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 (Figure 6.28).

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 $f(x_i^{*}),$ width $2\pi x_i^{*},$ and thickness $\text{Δ}x$ (Figure 6.28). 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

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

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

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 $x\text{-axis},$ when we want to integrate with respect to $y.$ 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

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

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

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

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 $x\text{-term}$ 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.