Example of a solid whose volume can only be computed either the washer/disk or the shell method

Question

Does anyone have any suggestions of regions in $\mathbb{R}^2$ whose volume generated by rotating the region about a specific axis or vertical or horizontal line in $\mathbb{R^2}$ can only be computed either by the disk/washer method or the shell method but not by both?

I am looking for examples to my students so that they would recognize the advantages/disadvantages of each method.

• I am looking for regions whose resulting integral for the volume would be impossible (or close to impossible) to evaluate under 1 method but possible for the other method

Answer 1

I don’t think that you really need (near-)impossibility in order to make the point: just choose a region that makes one method straightforward and the other unpleasantly messy. Many simple non-convex regions work well, like the region $R$ bounded by the $x$-axis, the $y$-axis, the line $x=3$, and the curve $y=x^2-2x+2$. Revolving $R$ about a vertical axis $x=a$, where $a\le 0$ or $a\ge 3$, is straightforward with shells but rather messy with washers. On the other hand, revolving $R$ about $y=a$, where $a\le 0$ or $a\ge 5$, is straightforward with washers but rather messy with shells.

Answer 2

Here's a silly example: let $S$ be the solid obtained by rotating the region under the graph of $f(x) = x^{-3} \sin^2\left( \frac 1x \right)$, $\frac{1}{10000\pi} \le x \le \frac{1}{2\pi}$, around the $y$-axis. The volume (using shells) is $$2\pi \int_{1/10000\pi}^{1/2\pi} x^{-2} \sin^2\left(\frac{1}{x}\right) \, dx = 2\pi \int_{2\pi}^{10000\pi} \sin^2 u \, du = 9998\pi^2.$$

The computation using washers is left as an exercise for the interested reader.