# How do you find the surface area of a solid of revolution?

If the solid is obtained by rotating the graph of $y = f \left(x\right)$ from $x = a$ to $x = b$, then the surface area $S$ can be found by the integral
$S = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$