Section 7.3 The Shell Method
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 βshells.βKey Idea 7.3.4. The Shell Method.
Let a solid be formed by revolving a region R, bounded by x=a and x=b, 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
When the region R is bounded above by y=f(x) and below by y=g(x), then h(x)=f(x)βg(x).
When the axis of rotation is the y-axis (i.e., x=0) then r(x)=x.
Example 7.3.5. Finding volume using the Shell Method.
Find the volume of the solid formed by rotating the region bounded by y=0, y=1/(1+x2), x=0 and x=1 about the y-axis.
This is the region used to introduce the Shell Method in Figure 7.3.2, but is sketched again in Figure 7.3.6 for closer reference. A line is drawn in the region parallel to the axis of rotation representing a shell that will be carved out as the region is rotated about the \(y\)-axis. (This is the differential element.)
The distance this line is from the axis of rotation determines \(r(x)\text{;}\) as the distance from \(x\) to the \(y\)-axis is \(x\text{,}\) we have \(r(x)=x\text{.}\) The height of this line determines \(h(x)\text{;}\) the top of the line is at \(y=1/(1+x^2)\text{,}\) whereas the bottom of the line is at \(y=0\text{.}\) Thus \(h(x) = 1/(1+x^2)-0 = 1/(1+x^2)\text{.}\) The region is bounded from \(x=0\) to \(x=1\text{,}\) so the volume is
This requires substitution. Let \(u=1+x^2\text{,}\) so \(du = 2x\, dx\text{.}\) We also change the bounds: \(u(0) = 1\) and \(u(1) = 2\text{.}\) Thus we have:
\begin{align*} \amp = \pi\int_1^2 \frac{1}{u}\, du\\ \amp = \pi\ln(u)\Big|_1^2\\ \amp = \pi\ln(2) \approx 2.178 \,\text{units}^3\text{.} \end{align*}Note: in order to find this volume using the Disk Method, two integrals would be needed to account for the regions above and below \(y=1/2\text{.}\)
Example 7.3.7. Finding volume using the Shell Method.
Find the volume of the solid formed by rotating the triangular region determined by the points (0,1), (1,1) and (1,3) about the line x=3.
The region is sketched in Figure 7.3.8.(a) along with the differential element, a line within the region parallel to the axis of rotation. In Figure 7.3.8.(b), we see the shell traced out by the differential element, and in Figure 7.3.8.(c) the whole solid is shown.
The height of the differential element is the distance from \(y=1\) to \(y=2x+1\text{,}\) the line that connects the points \((0,1)\) and \((1,3)\text{.}\) Thus \(h(x) = 2x+1-1 = 2x\text{.}\) The radius of the shell formed by the differential element is the distance from \(x\) to \(x=3\text{;}\) that is, it is \(r(x)=3-x\text{.}\) The \(x\)-bounds of the region are \(x=0\) to \(x=1\text{,}\) giving
Example 7.3.9. Finding volume using the Shell Method.
Find the volume of the solid formed by rotating the region given in Example 7.3.7 about the x-axis.
The region is sketched in Figure 7.3.10.(a) with a sample differential element. In Figure 7.3.10.(b) the shell formed by the differential element is drawn, and the solid is sketched in Figure 7.3.10.(c). (Note that the triangular region looks βshort and wideβ here, whereas in the previous example the same region looked βtall and narrow.β This is because the bounds on the graphs are different.)
The height of the differential element is an \(x\)-distance, between \(x=\frac12y-\frac12\) and \(x=1\text{.}\) Thus \(h(y) = 1-(\frac12y-\frac12) = -\frac12y+\frac32\text{.}\) The radius is the distance from \(y\) to the \(x\)-axis, so \(r(y) =y\text{.}\) The \(y\) bounds of the region are \(y=1\) and \(y=3\text{,}\) leading to the integral
Example 7.3.11. Finding 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=Ο about the y-axis.
The region and a differential element, the shell formed by this differential element, and the resulting solid are given in Figure 7.3.12.
The radius of a sample shell is \(r(x) = x\text{;}\) the height of a sample shell is \(h(x) = \sin(x)\text{,}\) each from \(x=0\) to \(x=\pi\text{.}\) Thus the volume of the solid is
This requires Integration By Parts. Set \(u=x\) and \(dv=\sin(x) \, dx\text{;}\) we leave it to the reader to fill in the rest. We have:
\begin{align*} \amp = 2\pi\Big[-x\cos(x) \Big|_0^{\pi} +\int_0^{\pi}\cos(x) \, dx \Big]\\ \amp = 2\pi\Big[\pi + \sin(x) \Big|_0^{\pi}\,\Big]\\ \amp = 2\pi\Big[\pi + 0 \Big]\\ \amp = 2\pi^2 \approx 19.74 \,\text{units}^3\text{.} \end{align*}Note that in order to use the Washer Method, we would need to solve \(y=\sin x\) for \(x\text{,}\) requiring the use of the arcsine function. We leave it to the reader to verify that the outside radius function is \(R(y) = \pi-\arcsin y\) and the inside radius function is \(r(y)=\arcsin y\text{.}\) Thus the volume can be computed as
This integral isn't terrible given that the \(\arcsin^2 y\) terms cancel, but it is more onerous than the integral created by the Shell Method.
Key Idea 7.3.13. Summary 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.
Washer Method | Shell Method | ||
Horizontal Axis | Οβ«ba(R(x)2βr(x)2)dx | 2Οβ«dcr(y)h(y)dy | |
Vertical Axis | Οβ«dc(R(y)2βr(y)2)dy | 2Οβ«bar(x)h(x)dx |
Exercises Exercises
Terms and Concepts
1.
T/F: A solid of revolution is formed by revolving a shape around an axis.
2.
T/F: The Shell Method can only be used when the Washer Method fails.
3.
T/F: The Shell Method works by integrating cross-sectional areas of a solid.
4.
T/F: When finding the volume of a solid of revolution that was revolved around a vertical axis, the Shell Method integrates with respect to x.
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.
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 x-axis.
In the following exercises, a region of the Cartesian plane is described. Use the Shell Method to find the volume of the solid of revolution formed by rotating the region about each of the given axes.
13.
Region bounded by: y=βx, y=0 and x=1.
Rotate about:
the y-axis
x=1
the x-axis
y=1
14.
Region bounded by: y=4βx2 and y=0.
Rotate about:
x=2
x=β2
the x-axis
y=4
15.
The triangle with vertices (1,1), (1,2) and (2,1).
Rotate about:
the y-axis
x=1
the x-axis
y=2
16.
Region bounded by y=x2β2x+2 and y=2xβ1.
Rotate about:
the y-axis
x=1
x=β1
17.
Region bounded by y=1/βx2+1, x=1 and the x and y-axes.
Rotate about:
the y-axis
x=1
18.
Region bounded by y=2x, y=x and x=2.
Rotate about:
the y-axis
x=2
the x-axis
y=4