Derivation of Shell Method

Question

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?

Answer 1

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.

Answer 2

This is not a formal method that has the absolute rigour of the answer first given, but this is how I learnt to deal with it, intuitively.

For a sufficiently small $\Delta x$, the outermost shell and the innermost shell are (practically) equal.

Suppose you had a circular sponge cake, with thickness of $\Delta x$. If you unravel the first layer of the sponge cake (i.e. the outer "shell" of cake) and flatten the sponge cake, what you have is (practically) a rectangular prism with thickness $\Delta x$ and length (approximately) equal to the circumference of the sponge cake. The height remains unchanged by this unraveling.

So the shell has a volume which approximates, tends to, and coincides with, a rectangular prism of thickness $\Delta x$, length $2\pi x$ and height $y$ in the limit.

Answer 3

A.

Without loss of generality, consider a surface confined to the I and IV quadrants of $x$-$z$ plane, where $h{(x)}$ is a real-valued, continuous function that gives the height of the surface at $x$. Next, consider a volume in three dimensions that is obtained by rotating said surface around the $z$-axis. Again, without loss of generality, given non-negative $a$ and $b$, the volume is given by $$V = 2\,\pi\,\int_a^b{x\,h{(x)}\, dx}.$$

B.

Divide the volume into $n$ shells of volume $V_i$. Given that the inner radius and outer radius of the $i^\text{th}$ shell are $x_{i}$ and $x_{i+1}$, respectively, and $h(x)$ is the height, the volume can be bounded by the volume of two cylindrical tubes each with the same inner and outer radius of the $i^\text{th}$ shell but with distinct heights as $$ 2\,\pi \, \overline{x}_{i} \,h{(x_l)}\, \Delta{x_{i}} \leq V_i \leq 2\,\pi \, \overline{x}_{i} \,h{(x_h)}\, \Delta{x_{i}}$$ where $$ \Delta{x_{i}} = \left(x_{i+1} - x_{i}\right);$$ $$\overline{x}_{i} = \dfrac{\left(x_{i+1} + x_{i}\right)}{2}; $$ $$h_l = \inf \left\{ h(x) \mid x\in [x_{i},x_{i+1}] \right\};$$ $$h_h = \sup \left\{ h(x) \mid x\in [x_{i},x_{i+1}] \right\}; \text{ and }$$ $ x_l, x_h \in [x_{i},x_{i+1}]$ are each a value of $x$ such that $$h(x_l) = h_l \text{ and }$$ $$ h(x_h) = h_h.$$

C.

Here we have a real-valued, continuous function $$f(x) = 2\,\pi \, \overline{x}_{i} \,h{(x)}$$ that is defined for all $x$ in the interval $[x_{i},x_{i+1}]$.

By the intermediate-value theorem [1], since ${\displaystyle V_i}$ is a number between ${\displaystyle f(x_l)}\, \Delta{x_{i}}$ and ${\displaystyle f(x_h)}\, \Delta{x_{i}}$, then there is a ${\displaystyle x_i^*\in (x_{l},x_{h})}$ such that ${\displaystyle f(x_i^*)\, \Delta{x_{i}}= V_i} $. Yet, since $x_l,x_h \in [x_{i},x_{i+1}]$, it is also true that ${\displaystyle x_i^*\in (x_{i},x_{i+1})}.$

D.

From [2], since ${\displaystyle f:[a,b]\rightarrow \mathbb {R} }$ be a function defined on a closed interval ${\displaystyle [a,b]}$ of the real numbers, ${\displaystyle \mathbb {R} }$, and given ${\displaystyle P=\left\{[x_{0},x_{1}],[x_{1},x_{2}],\dots ,[x_{n-1},x_{n}]\right\}}$, be a partition of in interval $I$, where ${\displaystyle a=x_{0}<x_{1}<x_{2}<\cdots <x_{n}=b}$. A Riemann sum of $f$ over $I$ with partition $P$ is defined as $${\displaystyle V =\sum _{i=1}^{n}V_i = \sum _{i=1}^{n}f(x_{i}^{*})\,\Delta x_{i}}.$$

Further, in the limit as the width of the subintervals go to zero $$\int_a^b \! f(x) \, dx = \lim_{\|\Delta x\|\rightarrow0} \sum_{i=1}^{n} f(x_i^*) \,\Delta x_i.$$

E.

Since $$\lim_{\|\Delta x\|\rightarrow 0}{ \bar{x}_i} = x,$$ we find that $$V = 2\,\pi \, \int_a^b { x \,h{(x)}}\, dx$$

BIBLIOGRAPHY:

[1] https://en.wikipedia.org/wiki/Intermediate_value_theorem

[2] https://en.wikipedia.org/wiki/Riemann_sum#Definition