I recently saw a 'derivation' of the shell method of integration for volumes in a book that went like this:
To find the element of volume contained in a shell of inner radius $r = x$ and out radius $R = x + \Delta x$, length $y$, we have:
$$\begin{align*}\Delta V &= \pi(R^2-r^2)y\\ &=\pi y(x^2+2x\Delta x + \Delta x^2 – x^2)\\&=2\pi xy\Delta x + \pi y\Delta x^2\end{align*}$$
As $\Delta x$ is very small, $(\Delta x)^2$ is negligible, hence $$\Delta V = 2\pi xy\Delta x\\\therefore V = 2\pi \int_a^bxy\,dx $$
I completely understand this, but I'm unsatisfied with the reasoning that $\Delta x^2$ is negligible. Formally, why are we allowed to ignore the $\Delta x^2$ term?
Best Answer
It looks like you're dealing with the case of a cylinder (i.e. the height is always $y$). In general, the height of a shell is a function of $x$, say $f(x)$, so the volume of a shell is $$\Delta V=2\pi xf(x)\Delta x + \pi f(x) (\Delta x)^2.$$ Now say the range of $x$ values is $[a,b]$. If we partition this into $n$ pieces, the left endpoint of the $k$th piece is $(k-1)\cdot \frac{(b-a)}{n}$. This is the inner radius of the $k$th shell. Now the total volume of the $k$ shells is $$\sum_{k=1}^n 2\pi\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot f\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot \frac{1}{n} + \pi f\left((k-1)\cdot\frac{(b-a)}{n}\right) \cdot \left(\frac{1}{n}\right)^2.$$ Here I've just substituted $(k-1)\cdot \frac{(b-a)}{n}$ for $x$ and $\frac{1}{n}$ for $\Delta x$ and then summed over the $n$ shells. Now as the number of pieces in the partition $n$ grows, this approaches the volume of the solid. The claim is that $$\small \lim_{n\to\infty}\sum_{k=1}^n 2\pi\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot f\left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot \frac{1}{n} + \pi f\left((k-1)\cdot\frac{(b-a)}{n}\right) \cdot \left(\frac{1}{n}\right)^2$$ $$= 2\pi\int_a^bxf(x) \ dx.$$ The limit can be broken into two parts. The first is $$\lim_{n\to\infty}\sum_{k=1}^n 2\pi \left((k-1)\cdot\frac{(b-a)}{n}\right)\cdot f\left((k-1)\cdot \frac{(b-a)}{n}\right)\cdot \frac{1}{n}.$$ Assuming $xf(x)$ is Riemann integrable on $[a,b]$, this converges to $2\pi \int_a^bxf(x) \ dx$. The second part of the limit is $$\lim_{n\to\infty} \sum_{k=1}^n\pi f\left((k-1)\cdot \frac{(b-a)}{n}\right)\cdot \left(\frac{1}{n}\right)^2.$$ Now we want to show this limit is zero (this is what is meant by "$(\Delta x)^2$ is negligible"). We can re-write it as $$\lim_{n\to\infty} \left(\frac{1}{n}\right)\left( \pi\sum_{k=1}^n f\left((k-1)\cdot \frac{(b-a)}{n}\right)\cdot\frac{1}{n}\right).$$ Now provided $f$ is Riemann integrable on $[a,b]$, we have $$\lim_{n\to\infty} \pi \sum_{k=1}^n f\left((k-1)\frac{(b-a)}{n}\right) \cdot \frac{1}{n}=\pi\int_a^b f(x) \ dx<\infty.$$ We also have $$\lim_{n\to\infty} \frac{1}{n}=0.$$ Since the previous two limits exist and are finite, the limit of the product splits into the product of the limits: $$\lim_{n\to\infty} \left(\frac{1}{n}\right)\left( \pi\sum_{k=1}^n f\left((k-1)\cdot \frac{(b-a)}{n}\right)\cdot\frac{1}{n}\right) = \left(\lim_{n\to\infty} \frac{1}{n}\right) \left(\lim_{n\to\infty} \pi \sum_{k=1}^n f\left((k-1)\frac{(b-a)}{n}\right) \cdot \frac{1}{n}\right)= 0 \cdot \pi \int_a^b f(x) \ dx = 0.$$ So this second part is indeed negligible.