5.2 Surface areas of solids of revolution

In the previous section we discussed how to compute the volume of a solid of revolution. We now show how to compute the surface area of such a solid.

Suppose the solid is generated by rotating the graph of the function $y=f(x)$ around the $x$-axis and then taking the portion between $x=a$ and $x=b$. See the figure below.

Select an infinitesimally small disc of the solid, of thickness $\delta x$. This cuts out a ribbon on the surface of the solid. Its radius is $f(x)$, and hence its circumference is $2\pi f(x)$. By Pythagoras, the width of the ribbon is approximately equal to $\sqrt{{(\delta x)}^2+{(\delta y)}^2}=\sqrt{1+{\left(\frac{\delta y}{\delta x}\right)}^2}\delta x$. (Recall also the formula for the length of a curve.) Therefore, the area of the ribbon is approximately $2\pi f(x)\sqrt{1+{\left(\frac{\delta y}{\delta x}\right)}^2}\delta x$.

It follows that the surface area of the solid is given by

Example. Let $y=f(x)=\sqrt{r^2-x^2}$, $-r\le x\le r$, be rotated about the $x$-axis. The surface obtained is a sphere of radius $r$. Find its surface area (using $y^{\prime }=\frac{-x}{\sqrt{r^2-x^2}}$).

The area equals

Similarly, if we are given a parametrized curve $(x(t),y(t))$ instead, then the surface area obtained by rotating the curve about the $x$-axis for $t\in [a,b]$ (with $x(t)>0$ in this interval) is given by