### 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

- 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.
- 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.
- 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.

#### Thread navigation

##### Calculus Refresher

- Previous: Volumes by cross sections
- Next: Integration by parts