# How do you find the volume bounded by y = e^x , y=0 , x = 0, x = ln2 revolved about the x=axis?

##### 1 Answer
Jun 19, 2018

$\frac{3 \pi}{2} \approx 4.71$

#### Explanation:

$V = \pi {\int}_{a}^{b} {y}^{2} \mathrm{dx}$

${y}^{2} = {\left({e}^{x}\right)}^{2} = {e}^{2 x}$

${\int}_{0}^{\ln} 2 {e}^{2 x} \mathrm{dx} = {\left[{e}^{2 x} / 2\right]}_{0}^{\ln} 2$

$\pi {\left[{e}^{2 x} / 2\right]}_{0}^{\ln} 2 = \pi \left({e}^{2 \ln 2} / 2 - {e}^{2 \left(0\right)} / 2\right) = \pi \left(\frac{4}{2} - \frac{1}{2}\right) = \frac{3 \pi}{2} \approx 4.71$