Shell integration

Consider the function ${\displaystyle f(x)}$ which describes our cross-section of the solid, now the integral of the function can be described as a Riemann integral: ${\displaystyle \int \limits _{a}^{b}f(x)dx=\lim _{n\to \infty }\sum _{i=1}^{n}f(a+i\Delta x)\Delta x}$

Where ${\displaystyle \Delta x={\frac {b-a}{n}}}$ is a small difference in ${\displaystyle x}$

The Riemann sum can be thought up as a sum of a number n of rectangles with ever shrinking bases, we might focus on one of them:

${\displaystyle f(a+k\Delta x)\Delta x}$

Now, when we rotate the function around the axis of revolution, it is equivalent to rotating all of these rectangles around said axis, these rectangles end up becoming a hollow cylinder, composed by the difference of two normal cylinders. For our chosen rectangle, its made by obtaining a cylinder of radius ${\displaystyle a+(k+1)\Delta x}$ with height ${\displaystyle f(a+k\Delta x)}$ , and subtracting it another smaller cylinder of radius ${\displaystyle a+k\Delta x}$, with the same height of ${\displaystyle f(a+k\Delta x)}$ , this difference of cylinder volumes is:

${\displaystyle \pi (a+(k+1)\Delta x)^{2}f(a+k\Delta x)-\pi (a+k\Delta x)^{2}f(a+k\Delta x)}$

${\displaystyle =\pi f(a+k\Delta x)((a+(k+1)\Delta x)^{2}-(a+k\Delta x)^{2})}$

By difference of squares , the last factor can be reduced as:

${\displaystyle \pi f(a+k\Delta x)(2a+2k\Delta x+\Delta x)\Delta x}$

The third factor can be factored out by two, ending up as:

${\displaystyle 2\pi f(a+k\Delta x)(a+k\Delta x+{\frac {\Delta x}{2}})\Delta x}$

This same thing happens with all terms, so our total sum becomes:

${\displaystyle \lim _{n\to \infty }2\pi \sum _{i=1}^{n}f(a+i\Delta x)(a+i\Delta x+{\frac {\Delta x}{2}})\Delta x}$

In the limit of ${\displaystyle n\rightarrow \infty }$, we can clearly identify that:

• ${\displaystyle f(a+i\Delta x)}$ as ${\displaystyle \Delta x}$ tends to 0 ends up becoming ${\displaystyle f(x)}$
• ${\displaystyle (a+i\Delta x+{\frac {\Delta x}{2}})}$ becomes ${\displaystyle x}$ itself, going from a to b (ignoring the last term which vanishes)
• ${\displaystyle \Delta x}$ becomes the infinitesimal ${\displaystyle dx}$

Thus, at the limit of infinity, the sum becomes the integral:

${\displaystyle 2\pi \int \limits _{a}^{b}xf(x)dx}$

QED ${\displaystyle \square }$.