Volumes by Shells

There are some examples where it is impossible, or at least inconvenient, to find the volume of a revolved object using disks or washers. Take, for example, the object you get when you revolve the region bounded by $y=2x^2-x^3$ and $y=0$ around the $y$-axis.

The issue is this: if we tried to do this by slicing, we'd use washers. However, the inner radius and outer radius would both be defined by the same function, and we'd have to solve this function for $x$.

Instead, we can use a different approach to divide this volume up into simple figures. This approach uses nested shells to fill up the volume.

It can be hard to visualize this whole process, but the image above shows a representative shell. If we can find the volume of each shell, the approximate total volume will be the sum of those volumes. As before, when we make the shells infinitesimally thin, the sum will turn into an integral, and we'll get the exact volume.

1. Slice

Each slice is a thin cylindrical shell, like a paper towel roll. The issue is to find the volume of this shell -- to do this, picture cutting and unrolling the shell.

Once it is unrolled, it's easy to see that the volume is simply length times width times height. All we have to do is figure out what the length, width, and height are in terms of the original shell.

Length

The length is the circumference of the shell, which is $2\pi r$. This can be tricky to see at first, but since the radius is the distance from the central axis ($x=0$) and the edge of the shell (at an arbitrary $x$ value), the radius is simply $x$. Thus the length is $2\pi x.$

Width, or thickness

The thickness of the wall is simply $\Delta x$.

Height

The height of the shell is based on the original function -- it's the distance between $y=0$ and $y=2x^2-x^3$. Thus, the height is $2x^2-x^3$.

Volume

The volume of one slice, then, is $\begin{aligned} V_{slice} &= 2\pi r h\ \Delta x\ &= 2\pi x(2x^2-x^3)\ \Delta x\ \end{aligned}$

2. Approximate

The approximate volume will be the sum of the volumes of the slices: $\begin{aligned} V &\approx \sum_{i=1}^n 2\pi r h \ \Delta x\ &\approx \sum_{i=1}^n 2\pi x(2x^2-x^3) \ \Delta x \end{aligned}$

3. Integrate

As $\Delta x \to 0$, or $n \to \infty$, this sum turns into an integral: $\begin{aligned} V &= \int_a^b 2 \pi r h \ dx\ &= \int_0^2 2 \pi x(2x^2-x^3) \ dx \end{aligned}$

Note that the limits of integration are based on where we start and stop drawing shells.

We can evaluate this integral to find the volume: $\begin{aligned} V &= \int_0^2 2 \pi x(2x^2-x^3) \ dx\ &= \int_0^2 4\pi x^3 - 2\pi x^4\ dx\ &= \pi x^4 - \dfrac{2}{5}\pi x^5 \bigg|_0^2\ &= 16\pi - \dfrac{64}{5}\pi\ &= \ans{\dfrac{16\pi}{5}} \end{aligned}$

Conclusion

In general, when we use shells to find the volume of a revolved solid, the integral will look like $\ans{V = \int_a^b 2\pi r h \ dx}$

All we have to do, then, is to find the radius $r$ (often simply equal to $x$) and the height $h$ (generally defined by the original function), as well as the limits of integration (where we start and stop drawing shells).