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: ${\displaystyle \int \limits _{a}^{b}f(x)dx=\lim _{n\to \infty }\sum _{i=1}^{n}f(a+i{\frac {b-a}{n}})({\frac {b-a}{n}})}$

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

${\displaystyle f(a+k{\frac {b-a}{n}})({\frac {b-a}{n}})}$

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 ${\displaystyle a+(k+1){\frac {b-a}{n}}}$ with height ${\displaystyle f(a+k{\frac {b-a}{n}})}$ , and substracting it another smaller cylinder of radius ${\displaystyle a+k{\frac {b-a}{n}}}$, with the same height of ${\displaystyle f(a+k{\frac {b-a}{n}})}$ , this difference of cylinder volumes is:

${\displaystyle \pi f(a+k{\frac {b-a}{n}})((a+(k+1){\frac {b-a}{n}})^{2}-(a+k{\frac {b-a}{n}})^{2})}$

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

${\displaystyle \pi f(a+k{\frac {b-a}{n}})(2a+2k{\frac {b-a}{n}}+{\frac {b-a}{n}}){\frac {b-a}{n}}}$

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

${\displaystyle 2\pi f(a+k{\frac {b-a}{n}})(a+k{\frac {b-a}{n}}+{\frac {b-a}{2n}}){\frac {b-a}{n}}}$

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

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

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

• ${\displaystyle f(a+i{\frac {b-a}{n}})}$ as ${\displaystyle {\frac {b-a}{n}}}$ goes to 0 ends up becoming ${\displaystyle f(x)}$
• ${\displaystyle (a+i{\frac {b-a}{n}}+{\frac {b-a}{2n}})}$ becomes ${\displaystyle x}$ itself, going from a to b (ignoring the last term which is now infinitesimal)
• ${\displaystyle {\frac {b-a}{n}}}$ becomes an infinitesimal

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

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

QED ${\displaystyle \square }$.