# Let R be the region enclosed by y= e^(2x), y=0, and y=2. What is the volume of the solid produced by revolving R around the x-axis?

##### 1 Answer
Jan 18, 2017

$\frac{3}{4} \pi = 2.3562$ cubic units. nearly.

#### Explanation:

y = 2 meets $y = {e}^{2 x}$, at $\left(\frac{1}{2} \ln 2 , 2\right) .$

The x-axis y = 0 is the asymptote to $y = {e}^{2 x} > 0$.

Now,

volume $= \pi \int {y}^{2} \mathrm{dx}$, for x from $- \infty \to \frac{1}{2} \ln 2$

$- \pi \int {e}^{4 x} \mathrm{dx}$, for x from $- \infty \to \frac{1}{2} \ln 2$

$= \frac{\pi}{4} \left[{e}^{4 x}\right]$, between the limits

$= \frac{\pi}{4} \left[{e}^{2 \ln 2} - {e}^{0}\right]$

$= \frac{\pi}{4} \left[4 - 1\right)$, using ${e}^{\ln} a = a$.

$= \frac{3}{4} \pi = 2.3562$ cubic units. nearly.

The graph shows the area that is revolved.

graph{(y-.1)(e^(2x)-y)(x-.5ln2+.0001y)=0 [-1.5, 1.5, 0, 2,1]}