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).

Notice that when we're using the shell method, we're not filling the solid figure with little *cylinders*, but rather with the (infinitessimally thin) *walls* of cylinders (shells). It's an important distinction, and it might take you a while to wrap your head around it – but keep trying!

$$V = 2\pi \int_z^b (r \cdot h) \, dr$$

**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

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:

$$ \begin{align} V_{washer} &= \pi(R^2 - r^2) \ dy \\ &= \pi [(\pi - sin^{-1}(y))^2 - (sin^{-1}(y))^2] \, dy \end{align}$$

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 = 2πrhÂ·dr**, so the volume of the shells for this figure is

$$ \begin{align} V_{shell} &= 2 \pi r h \, dx \\[5pt] &= 2 \pi x \cdot sin(x) \, dx \end{align}$$

... which gives us this much simpler integral

$$V = 2\pi \int_0^{\pi} x\cdot sin(x) \, dx$$

This integral can be solved by parts, by making these substitutions:

$$ \begin{matrix} let \; u = x && dv = sin(x) \, dx \\ then \; du = dx && v = -cos(x) \end{matrix}$$

The intermediate integral (I've left off the limits) is:

$$\rightarrow \; 2\pi \left[ -x cos(x) + \int cos(x) \, dx \right]$$

Integration is easy:

$$\rightarrow \; 2\pi \left[ -x cos(x) + sin(x) \right]_0^{\pi}$$

Evaluate the limits:

$$= 2\pi \left[ -\pi cos(\pi) + sin(\pi) + cos(0) - sin(0) \right]$$

And the volume is just

$$ \begin{align} &= 2\pi \left[ \pi + 0 - 1 - 0 \right] \\ \\ &= 2\pi (\pi - 1) = 13.5 \; units^3 \end{align}$$

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*.

When you're integrating a volume of revolution with a hole in it, and the integral setup with washers produces a difficult integral, try shells. Often you'll find a much simpler integrand.

Consider the region between $f(x) = x^2$ and $g(x) = x,$ shown below. Revolve that region about the y-axis (resulting in the figure shown in the second graph), and calculate its volume (Note: It's not the volume of the empty area of the bowl, which would just be the volume of a cone, but rather the volume of the material that composes the bowl).

**Solution**

The radius of any cylinder is x. Its height is the difference between $y = x$ and $y = x^2$ at that point, so the volume of each cylinder is

$$ \begin{align} V_{cylinder} &= 2\pi r h \cdot dx \\[5pt] &= 2 \pi x (x - x^2) \cdot dx \end{align}$$

Now to find the volume, we just integrate this volume element between x = 0 and x = 1, where the functions intersect.

$$ \begin{align} V &= 2\pi \int_0^1 x (x - x^2) \, dx \\[5pt] &= 2\pi \int_0^1 x^2 - x^3 \, dx \\[5pt] &= 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 \\[5pt] &= 2\pi \left[\frac{1}{3} - \frac{1}{4} - 0 + 0\right] = \bf \frac{\pi}{6} \; units^3 \end{align}$$

Here is what the revolved figure would look like:

Calculate the volume of the solid obtained by revolving the region bounded by the x-axis, the line $x = 4$ and the function $f(x) = \sqrt{x - 2}$ about the y-axis.

**Solution**

The radius of any cylinder (shell) is x. Its height is the function $y = \sqrt{x}.$ The volume of each cylinder is

$$ \begin{align} V_{cylinder} &= 2\pi r h \cdot dx \\[5pt] &= 2 \pi x \sqrt{x} \cdot dx \end{align}$$

Now to find the volume, we just integrate this volume element between x = 2 and x = 4.

$$ \begin{align} V &= 2\pi \int_2^4 x \sqrt{x} \, dx \\[5pt] &= 2\pi \int_0^1 x^{3/2} \, dx \\[5pt] &= 2\pi \left[ \frac{2}{5}x^{5/2} \right]_2^4 \\[5pt] &= \frac{4\pi}{5} \left[ 4^{2/5} - 2^{2/5} \right] \\[5pt] &= \frac{4\pi}{5} [1.741 - 1.320] = \bf 1.06 \; units^3 \end{align}$$

Here is what the revolved figure would look like. It's rendered a little bit transparent so you can see the hole through the middle.

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