# Math Insight

### Volume of surfaces of revolution

Another way of computing volumes of some special types of solid figures applies to solids obtained by rotating plane regions about some axis.

If we rotate the plane region described by $f(x)\le y\le g(x)$ and $a\le x\le b$ around the $x$-axis, the volume of the resulting solid is \begin{align*} \hbox{ volume }&=\int_a^b \pi (g(x)^2-f(x)^2)\;dx\\ &=\int_{\hbox{left limit}}^{\hbox{right limit}} \pi (\hbox{upper curve}^2-\hbox{lower curve}^2)\;dx \end{align*} It is necessary to suppose that $f(x)\ge 0$ for this to be right.

This formula comes from viewing the whole thing as sliced up into slices of thickness $dx$, so that each slice is a disk of radius $g(x)$ with a smaller disk of radius $f(x)$ removed from it. Then we use the formula $$\hbox{ area of disk }= \pi \hbox{ radius}^2$$ and ‘add them all up’. The hypothesis that $f(x)\ge 0$ is necessary to avoid different pieces of the solid ‘overlap’ each other by accident, thus counting the same chunk of volume twice.

If we rotate the plane region described by $f(x)\le y\le g(x)$ and $a\le x\le b$ around the $y$-axis (instead of the $x$-axis), the volume of the resulting solid is \begin{align*} \hbox{ volume }&=\int_a^b 2\pi x(g(x)-f(x))\;dx\\ &=\int_{\hbox{left}}^{\hbox{ right}} 2\pi x(\hbox{ upper - lower})\;dx \end{align*}

This second formula comes from viewing the whole thing as sliced up into thin cylindrical shells of thickness $dx$ encircling the $y$-axis, of radius $x$ and of height $g(x)-f(x)$. The volume of each one is $$\hbox{ (area of cylinder of height g(x)-f(x) and radius x) }\cdot dx= 2\pi x(g(x)-f(x))\;dx$$ and ‘add them all up’ in the integral.

As an example, let's consider the region $0\le x \le 1$ and $x^2\le y \le x$. Note that for $0\le x \le 1$ it really is the case that $x^2\le y \le x$, so $y=x$ is the upper curve of the two, and $y=x^2$ is the lower curve of the two. Invoking the formula above, the volume of the solid obtained by rotating this plane region around the x-axis is \begin{align*} \hbox{ volume }&=\int_{\hbox{left}}^{\hbox{ right}} \pi (\hbox{upper}^2-\hbox{lower}^2)\;dx\\ &=\int_0^1 \pi ((x)^2-(x^2)^2)\; dx = \pi [ x^3/3 - x^5/5]_0^1 = \pi(1/3 - 1/5) \end{align*}

On the other hand, if we rotate this around the $y$-axis instead, then \begin{align*} \hbox{ volume }&=\int_{\hbox{left}}^{\hbox{ right}} 2\pi x (\hbox{upper}-\hbox{lower})\;dx\\ &= \int_0^1 \, 2\pi x (x-x^2)\, dx = \pi\int_0^1 \, {2x^3 \over 3} - {2x^4 \over 4} \, dx = [{2x^3 \over 3} - {2x^4 \over 4}]_0^1 = {2\over 3} - {1\over 2} = {1\over 6} \end{align*}

#### Exercises

1. Find the volume of the solid obtained by rotating the region $0\leq x\leq 1, 0\leq y\leq x$ around the $y$-axis.
2. Find the volume of the solid obtained by rotating the region $0\leq x\leq 1, 0\leq y\leq x$ around the $x$-axis.
3. Set up the integral which expresses the volume of the doughnut obtained by rotating the region $(x-2)^2+y^2\leq 1$ around the $y$-axis.