# How do you find the surface area of the solid obtained by rotating about the x-axis the region bounded by y=e^x on the interval 0<=x<=1 ?

Sep 16, 2014

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:

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