In principle, any volume of revolution that can be calculated by using the shell method can be calculated using the disk or washer method, but once in a while, the shell method gives a clearer picture and may lead to an integral that's much easier to evaluate, as we'll see in the first example, below. It's worth learning.
The idea is that any solid of revolution, whether it has a hole or not, can be approximated as a set of empty cylinders (just the "wall") of some thickness Δr (r for radius), stacked within one-another (concentric—sharing the same center. Here's an example of what building a volume of revolution from shells might look like:
The shell method relies upon your knowledge of the surface area of the wall, or side, of a right-regular cylinder. This figure ↓ will help you
rationalize why that area is the circumference of the top/bottom times the height. Remember that a cylinder is just two discs with a rectangle wrapped around them.
Then the shell method is just multiplying that area by an infinitessimal thickness, dx or dy, depending on the axis of revolution of the figure, and integrating. The example below shows very clearly how the shell method works, and why it's better than washers in this case (and many others).
r is the radius of the shell, h is the height and dr is the thickness. Remember that you're not filling the solid with what's inide each cylinder (they're empty), but rather with the cylinder walls.
This example should show you how finding the volume of revolution of some functions is much easier with shells. Take the function f(x) = sin(x) between 0 and pi. Revolve it around the y-axis to make the shape below.
Our goal is to find the volume with washers – just disks with holes in them. That's a little tricky to set up, but do-able. If we integrate (stack up washers) along the y-axis, the inner radius is given by y = sin(x), but we want it as x = sin-1(y) in order to integrate along y.
Now the outer radius is a little trickier. Take a look at the red type in the figure below. The outer radius is just y = sin(π - x), and solving for y gives us x = π - sin-1(y).
So here is the expression for the volume of one of our washers:
Well, that is a terrible function to integrate. Give it a try if you'd like, but I'm going to skip below and try shells.
First, here's a typical shell for this problem:
The volume of a shell is V = 2πrh·dr, so the volume of the shells for this figure is
... which gives us this much simpler integral
This integral can be solved by parts, by making these substitutions:
The intermediate integral (I've left off the limits) is:
Integration is easy:
Evaluate the limits:
And the volume is just
So you see, sometimes the problem is much easier by taking a slightly different approach. There are quite a few examples where integration of volumes by concentric shells is much easier than by the method of washers. Indeed, sometimes shells just make the problem possible.
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