Solid of revolution

We discuss how the idea of the area as a sum of bar areas can be extended to calculate volumes of solids. The kind of solids we will discuss are called solids of revolution because they are created by rotating (or revolving) a curve $f$ around the $x$-axis:

The rotating curve forms the surface of the solid. The formula for calculating its volume from $a$ to $b$ is

Note that you have to square the function $f$, which results in a new function ($f(x)\cdot f(x)$), and it is this new function which you have to integrate! A typical mistake is to first finding the integral of $\int_a^b f(x)\, dx$, and then square the result - but this is wrong!

Example 1

The graph of the function $f(x)=\sqrt{x}$ (for $0\leq x\leq 4$) is rotated about the $x$-axis to form a solid of revolution. Draw the solid and determine its volume.

Solution

$V=\pi \int_0^4 (\sqrt{x})^2\, dx = \pi \int_0^4 x\, dx = \pi \cdot (\frac{1}{2}4^2-\frac{1}{2}0^2)=\underline{8\pi}$

To understand why the volume formula is correct, cover the area from $a$ to $b$ under the graph of $f$ with a huge number of very thin bars of width $\Delta x$. A bar at location $x$ will have the height $f(x)$. Now, imagine that when we rotate the graph of $f$ about the $x$-axis, we also rotate the bars as well. Each bar turns then into a {\it disk} whose width is simply the width of the bar, and whose radius is the height of the bar. So the bar at position $x$ has the volume

where

is the radius of the disk.

The volume of the solid is approximately the sum of the disk volumes

The approximation turns into equality for thinner and thinner bars, that is, for $\Delta x\rightarrow 0$. So the formula is indeed correct.

Here are two additional exercises.

Exercise 1

Q1

The graph of the function $f(x)=\frac{1}{\sqrt{x}}$ (for $1\leq x\leq u$) is rotated about the $x$-axis to form a solid of revolution. Find a value of $u$ such that the solid has the volume $V=20$.

Q2

Use calculus to prove that the volume of a sphere of radius $r$ is given by the formula

$$V=\frac{4}{3}\pi r^3$$

Hint: Find the function equation of a half-circle of radius $r$.

Solution

A1

Applying the volume formula, we have

$$\begin{array}{lll} V &=&\pi \int_1^u \left(\frac{1}{\sqrt{x}}\right)^2\, dx \\ &=& \pi \int_1^u \frac{1}{x}\, dx \\ &=& \pi (\ln(u)-\ln(1))=20\end{array}$$

As $\ln(1)=0$, we have to solve the equation

$$\ln(u)=\frac{20}{\pi} \rightarrow u=e^{20/\pi}=\underline{581.84...}$$

A2

The sphere of radius $r$ is a solid of revolution, as it is obtained by rotating the half-circle of radius $r$ about the $x$-axis. To find the function equation $f$ of the half-circle, let $f(x)$ be the height of the half-circle at $x$ (see figure).

From the theorem of Pythagoras we know that

$$f(x)^2+x^2=r^2 \rightarrow f(x)=\sqrt{r^2-x^2}$$

The square of this function is

$$f(x)^2=(\sqrt{r^2-x^2})^2=r^2-x^2$$

and the antiderivative of this squared function is

$$F(x)=r^2 x-\frac{1}{3}x^3$$

Half the volume of the sphere is therefore

$$\begin{array}{lll} V_{1/2}&=&\pi \int_0^r (\sqrt{r^2-x^2})^2\, dx \\ &=& \pi (F(r)-F(0)) \\ &= &\pi(r^2 r -\frac{1}{3} r^3)\\ &=& \pi \frac{2}{3}r^3\end{array}$$

and the Volume of the sphere is therefore $\frac{4}{3}\pi r^3$.