# Determining the Surface Area of a Solid of Revolution

## Key Questions

• First of all, you are missing a bound. We will assume that the other bound is $y = 0$ or the $x$-axis. The answer is $\frac{15 \pi}{2}$.

The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter ($x$ in this case). And we are not, so this integration should be done with cylindrical shells.

Always draw a diagram to verify what is the parameter and what is the function.

You should note that $y$ is not always a parameter of $x$. For instance, $x = {y}^{2}$, $x$ is now a parameter of $y$.

The formula for cylindrical shells is:

$V = {\int}_{a}^{b} 2 \pi r h \mathrm{dr}$
$h$ is represented by $y$, we have $y = {x}^{2}$ and $y = 0$
$r$ is represented by $x$
$V = {\int}_{1}^{2} 2 \pi x \left({x}^{2} - 0\right) \mathrm{dx}$

Now that the substitutions are done, we can solve:

$V = 2 \pi {\int}_{1}^{2} {x}^{3} \mathrm{dx}$
$= 2 \pi \frac{{x}^{4}}{4} {|}_{1}^{2}$
$= 2 \pi \frac{\left[{2}^{4} - {1}^{4}\right]}{4}$
$= \frac{15 \pi}{2}$

• The answer is $\frac{\pi}{2} \left[{e}^{2} - 1\right]$.

Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method. The disk method is:

$V = {\int}_{a}^{b} A \mathrm{dx}$
$= {\int}_{a}^{b} \pi {r}^{2} \mathrm{dx}$
$= {\int}_{a}^{b} \pi {\left[f \left(x\right)\right]}^{2} \mathrm{dx}$

We have the known values:

$f \left(x\right) = {e}^{x}$
$a = 0$
$b = 1$

And now we can substitute:

$V = {\int}_{0}^{1} \pi {\left({e}^{x}\right)}^{2} \mathrm{dx}$
$= \pi {\int}_{0}^{1} {e}^{2 x} \mathrm{dx}$
$= \pi \frac{{e}^{2 x}}{2} {|}_{0}^{1}$
$= \frac{\pi}{2} \left[{e}^{2} - {e}^{0}\right]$
$= \frac{\pi}{2} \left[{e}^{2} - 1\right]$

• If the solid is obtained by rotating the graph of $y = f \left(x\right)$ from $x = a$ to $x = b$, then the surface area $S$ can be found by the integral

$S = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$