# Calculus/Volume of solids of revolution

In this section we cover solids of revolution and how to calculate their volume. A solid of revolution is a solid formed by revolving a 2-dimensional region around an axis. For example, revolving the semi-circular region bounded by the curve $y={\sqrt {1-x^{2}}}$ and the line $y=0$ around the $x$-axis produces a sphere. There are two main methods of calculating the volume of a solid of revolution using calculus: the disk method and the shell method.

## Disk Method

Consider the solid formed by revolving the region bounded by the curve $y=f(x)$ , which is continuous on $[a,b]$ , and the lines $x=a$ , $x=b$ and $y=0$ around the $x$-axis. We could imagine approximating the volume by approximating $f(x)$ with the stepwise function $g(x)$ shown in figure 2, which uses a right-handed approximation to the function. Now when the region is revolved, the region under each step sweeps out a cylinder, whose volume we know how to calculate, i.e.
    $V_{\rm {cylinder}}=\pi r^{2}h$
where $r$ is the radius of the cylinder and $h$ is the cylinder's height. This process is reminiscent of the Riemann process we used to calculate areas earlier. Let's try to write the volume as a Riemann sum and from that equate the volume to an integral by taking the limit as the subdivisions get infinitely small.
Consider the volume of one of the cylinders in the approximation, say the $k$-th one from the left. The cylinder's radius is the height of the step function, and the thickness is the length of the subdivision. With $n$ subdivisions and a length of $b-a$ for the total length of the region, each subdivision has width
    $\Delta x={\frac {b-a}{n}}$
Since we are using a right-handed approximation, the $k$-th sample point will be
    $x_{k}=k\Delta x$
So the volume of the $k$-th cylinder is
    $V_{k}=\pi f(x_{k})^{2}\Delta x$
Summing all of the cylinders in the region from $a$ to $b$ , we have
    $V_{\rm {approx}}=\sum _{k=1}^{n}\pi f(x_{k})^{2}\Delta x$
Taking the limit as $n$ approaches infinity gives us the exact volume
    $V=\lim _{n\to \infty }\sum _{k=1}^{n}\pi f(x_{k})^{2}\Delta x$
which is equivalent to the integral
    $V=\int \limits _{a}^{b}\pi f(x)^{2}dx$
* Example: Volume of a Sphere
* Let's calculate the volume of a sphere using the disk method. Our generating region will be the region bounded by the curve $f(x)={\sqrt {r^{2}-x^{2}}}$ and the line $y=0$ . Our limits of integration will be the $x$-values where the curve intersects the line $y=0$ , namely, $x=\pm r$ . We have
* ${\begin{aligned}V_{\rm {sphere}}&=\int \limits _{-r}^{r}\pi (r^{2}-x^{2})dx\\&=\pi \left(\int \limits _{-r}^{r}r^{2}dx-\int \limits _{-r}^{r}x^{2}dx\right)\\&=\pi \left(r^{2}x{\bigg |}_{-r}^{r}-{\frac {x^{3}}{3}}{\bigg |}_{-r}^{r}\right)\\&=\pi {\Bigr^{2}{\biglr-(-r){\bigr )}-{\tfrac {1}{3}}{\biglr^{3}-(-r)^{3}{\bigr )}{\Big )}\\&=\pi \left(2r^{3}-{\frac {2r^{3}}{3}}\right)\\&=\pi {\frac {6r^{3}-2r^{3}}{3}}\\&={\frac {4\pi }{3}}r^{3}\end{aligned}}$

### Exercises

$r$
$h$
$y=r-{\frac {r}{h}}x$
$y=0$
$x=0$
$x$
${\frac {\pi r^{2}h}{3}}$
${\frac {\pi r^{2}h}{3}}$
$y=x^{2}$
$x=1$
$y=0$
$x$
${\frac {\pi }{5}}$
${\frac {\pi }{5}}$
Solutions

## Washer Method

The washer method is an extension of the disk method to solids of revolution formed by revolving an area bounded between two curves around the $x$-axis. Consider the solid of revolution formed by revolving the region in figure 3 around the $x$-axis. The curve $f(x)$ is the same as that in figure 1, but now our solid has an irregularly shaped hole through its center whose volume is that of the solid formed by revolving the curve $g(x)$ around the $x$-axis. Our approximating region has the same upper boundary, $f_{\rm {step}}(x)$ as in figure 2, but now we extend only down to $g_{\rm {step}}(x)$ rather than all the way down to the $x$-axis. Revolving each block around the $x$-axis forms a washer-shaped solid with outer radius $f_{\rm {step}}(x)$ and inner radius $g_{\rm {step}}(x)$ . The volume of the $k$-th hollow cylinder is
    ${\begin{aligned}V_{k}&=\pi \cdot f(x_{k})^{2}\Delta x-\pi \cdot g(x_{k})^{2}\Delta x\\&=\pi {\biglf(x_{k})^{2}-g(x_{k})^{2}{\bigr )}\Delta x\end{aligned}}$
where $\Delta x={\frac {b-a}{n}}$ and $x_{k}=k\Delta x$ . The volume of the entire approximating solid is
    $V_{\rm {approx}}=\sum _{k=1}^{n}\pi {\biglf(x_{k})^{2}-g(x_{k})^{2}{\bigr )}\Delta x$
Taking the limit as $n$ approaches infinity gives the volume
    ${\begin{aligned}V&=\lim _{n\to \infty }\sum _{k=1}^{n}\pi {\biglf(x_{k})^{2}-g(x_{k})^{2}{\bigr )}\Delta x\\&=\int \limits _{a}^{b}\pi {\biglf(x)^{2}-g(x)^{2}{\bigr )}dx\end{aligned}}$

### Exercises

$y=R-{\frac {R}{h}}x$
$y=r$
$x=0$
$x$
$\pi h\left({\frac {R^{2}}{3}}-r^{2}\right)$
$\pi h\left({\frac {R^{2}}{3}}-r^{2}\right)$
$y=x^{2}$
$y=x^{3}$
$x=1$
$y=0$
$x$
${\frac {2\pi }{35}}$
${\frac {2\pi }{35}}$
Solutions

## Shell Method

The shell method is another technique for finding the volume of a solid of revolution. Using this method sometimes makes it easier to set up and evaluate the integral. Consider the solid of revolution formed by revolving the region in figure 5 around the $y$-axis. While the generating region is the same as in figure 1, the axis of revolution has changed, making the disk method impractical for this problem. However, dividing the region up as we did previously suggests a similar method of finding the volume, only this time instead of adding up the volume of many approximating disks, we will add up the volume of many cylindrical shells. Consider the solid formed by revolving the region in figure 6 around the $y$-axis. The $k$-th rectangle sweeps out a hollow cylinder with height ${\Big |}f(x_{k}){\Big |}$ and with inner radius $x_{k}$ and outer radius $x_{k}+\Delta x$ , where $\Delta x={\frac {b-a}{n}}$ and $x_{k}=k\Delta x$ , the volume of which is
    $V_{k}$ $=\pi {\bigl(x_{k}+\Delta x)^{2}-x_{k}^{2}{\bigr )}{\Big |}f(x_{k}){\Big |}$ $=\pi {\bigl(x_{k}^{2}+2x_{k}\Delta x+\Delta x^{2})-x_{k}^{2}{\bigr )}{\Big |}f(x_{k}){\Big |}$ $=\pi (2x_{k}\Delta x+\Delta x^{2}){\Big |}f(x_{k}){\Big |}$
The volume of the entire approximating solid is
    $V_{\rm {approx}}=\sum _{k=1}^{n}\pi (2x_{k}\Delta x+\Delta x^{2}){\Big |}f(x_{k}){\Big |}$
Taking the limit as $n$ approaches infinity gives us the exact volume
    $V$ $=\lim _{n\to \infty }\sum _{k=1}^{n}\pi (2x_{k}\Delta x+\Delta x^{2}){\Big |}f(x_{k}){\Big |}$ $=\pi \cdot \lim _{n\to \infty }\left(\sum _{k=1}^{n}2x_{k}\Delta x{\Big |}f(x_{k}){\Big |}+\sum _{k=1}^{n}\Delta x^{2}{\Big |}f(x_{k}){\Big |}\right)$
Since $|f|$ is continuous on $[a,b]$ , the Extreme Value Theorem implies that $|f|$ has some maximum, $M$ , on $[a,b]$ . Using this and the fact that $\Delta x^{2}{\Big |}f(x_{k}){\Big |}>0$ , we have
    ${\sum _{k=1}^{n}2x_{k}\Delta x{\Big |}f(x_{k}){\Big |}\leq \sum _{k=1}^{n}2x_{k}\Delta x{\Big |}f(x_{k}){\Big |}+\sum _{k=1}^{n}\Delta x^{2}{\Big |}f(x_{k}){\Big |}\leq \sum _{k=1}^{n}2x_{k}\Delta x{\Big |}f(x_{k}){\Big |}+\sum _{k=1}^{n}\Delta x^{2}M}$
But
    $\lim _{n\to \infty }\sum _{k=1}^{n}\Delta x^{2}M$ $=\lim _{n\to \infty }\sum _{k=1}^{n}\left({\frac {b-a}{n}}\right)^{2}M$ $=\lim _{n\to \infty }{\frac {(b-a)^{2}}{n}}M$ $=0$
So by the Squeeze Theorem
    $\pi \cdot \lim _{n\to \infty }\left(\sum _{k=1}^{n}2x_{k}\Delta x{\Big |}f(x_{k}){\Big |}+\sum _{k=1}^{n}\Delta x^{2}{\Big |}f(x_{k}){\Big |}\right)=\pi \cdot \lim _{n\to \infty }\sum _{k=1}^{n}2x_{k}\Delta x{\Big |}f(x_{k}){\Big |}$
which is just the integral
    $\int \limits _{a}^{b}2\pi x{\Big |}f(x){\Big |}dx$

### Exercises

$r$
$h$
$y$
${\frac {\pi r^{2}h}{3}}$
${\frac {\pi r^{2}h}{3}}$
$y=x^{2}$
$x=1$
$y=0$
$y$
${\frac {\pi }{2}}$
${\frac {\pi }{2}}$
Solutions
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