# How do you determine the volume of a solid created by revolving a function around an axis?

Jan 7, 2017

$\text{Volume} = \pi {\int}_{a}^{b} f {\left(x\right)}^{2} \mathrm{dx}$

#### Explanation:

Given a function $f \left(x\right)$ and an interval $\left[a , b\right]$ we can think of the solid formed by revolving the graph of $f \left(x\right)$ around the $x$ axis as a horizontal stack of an infinite number of infinitesimally thin disks, each of radius $f \left(x\right)$.

The area of a circle is $\pi {r}^{2}$, so the area of the circle at a point $x$ will be $\pi f {\left(x\right)}^{2}$.

The volume of the solid is then the infinite sum of the infinitesimally thin disks over the interval $\left[a , b\right]$

So:

$\text{Volume} = {\int}_{a}^{b} \pi f {\left(x\right)}^{2} \mathrm{dx} = \pi {\int}_{a}^{b} f {\left(x\right)}^{2} \mathrm{dx}$