## 175- Disk- vs. Shell method

Here is the computation I wanted to show at the end of today’s lecture: Consider a function $f(x)$ that is continuous and invertible on an interval $\null[a,b]$. Rotate about the $y$-axis the region bounded by $y=f(a)$, $x=b$, and $y=f(x)$. Then the values assigned to the volume of the resulting solid by applying either the disk method or the shell method coincide.

To see this, let’s call $D$ the expression obtained using the disk (or washer) method and $S$ the expression obtained using the shell method.

Then $D=\int_{f(a)}^{f(b)} \pi[b^2-(f^{-1}(y))^2]\,dy$ and $S=\int_a^b 2\pi x(f(x)-f(a))\,dx$.

As I mentioned in lecture, the idea to show the equality of both expressions is to, more generally, consider functions $D(t)$ and $S(t)$ that compute the values given by the disk and shell methods, respectively, for the volume of the solid obtained by rotating about the $y$-axis the region bounded by $y=f(a)$, $y=f(x)$, and $x=t$. Here, $t$ is a new variable that varies in $\null[a,b]$. Hence, $D(t)=\int_{f(a)}^{f(t)} \pi[t^2-(f^{-1}(y))^2]\,dy$ and $S(t)=\int_a^t 2\pi x(f(x)-f(a))\,dx$. We will show that $D(t)=S(t)$ for all $t\in[a,b]$.

The trick is to proceed indirectly and, rather than looking at $D(t)$ and $S(t)$, we examine their derivatives. Of course, if $D(t)=S(t)$ for all $t\in[a,b]$, then we must also have $D'(t)=S'(t)$. Let’s check that this is indeed the case:

We have $S'(t)=2\pi t (f(t)-f(a))$. To compute $D'(t)$, let’s simplify $D(t)$ a bit before differentiating: $D(t)=\int_{f(a)}^{f(t)} \pi t^2\,dy-\int_{f(a)}^{f(t)}\pi f^{-1}(y)^2\,dy$, so $D(t)=\pi t^2(f(t)-f(a))-\pi\int_{f(a)}^{f(t)} f^{-1}(y)^2\,dy$ (Notice that $t$ does not depend on $y$, so we can take $t^2$ out of the integral sign.)

Then $D'(t)=2\pi t(f(t)-f(a))+\pi t^2 f'(t)-(\pi\int_{f(a)}^{f(t)} f^{-1}(y)^2\,dy)'$ so $D'(t)=2\pi t(f(t)-f(a))+\pi t^2 f'(t)- \pi f^{-1}(f(t))^2f'(t)$, or $D'(t)=2\pi t(f(t)-f(a))$, since $f^{-1}(f(t))=t$.

We have found that $S'(t)=2\pi t (f(t)-f(a))=D'(t)$.

This does not automatically imply that $S(t)=D(t)$, but almost: Let $h(t)=S(t)-D(t)$. Then $h'(t)=0$ (for all $t\in(a,b)$), so $h(t)$ is constant.

Finally, notice that $h(a)=S(a)-D(a)=0-0=0$. Since $h$ is constant, it follows that $S(t)=D(t)$ for all $t\in[a,b]$. And we are done.

Remark 1. We will see a different argument later in the course, once we study integration by parts.

Remark 2. One can also check that the disk and shell methods provide the same result when we rotate about the $y$-axis the region bounded by $y=f(x)$, $y=f(b)$ and $x=a$. To check your understanding of the argument above, it may be useful to try to work this case out on your own; the algebra is somewhat simpler than the computations I just detailed.