Solid of revolution

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

The rotating curve forms the surface of the solid.

Theorem 1

The graph of a function is rotated about the $x$ -axis to form a solid of revolution. The volume $V$ from $a$ to $b$ of this solid is given by

Equation 1

V=\pi \int_a^b \left(f(x)\right)^2 \, dx

Volume of a solid of revolution

Proof

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

r^2 \pi \cdot \Delta x

where

r=f(x)

is the radius of the disk.

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

\begin{array}{lll} V & \approx & f(x_1)^2 \pi\, \Delta x + ... + f(x_n)^2 \pi\, \Delta x \\ &= &\pi (f(x_1)^2 \Delta x + ... + f(x_n)^2 \Delta x)\\ &\approx &\pi (\int_a^b f(x)^2 \, dx) \end{array}

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

Warning

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

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

Here are two additional exercises.

Exercise 1

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 $V=20$ .
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

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
$\ln(u)=\frac{20}{\pi} \rightarrow u=e^{20/\pi}=\underline{581.84...}$
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 $\frac{4}{3}\pi r^3$ .