# What is the surface area of the solid created by revolving f(x) = xe^(2x) , x in [2,3] around the x axis?

Jan 1, 2017

Volume : $\frac{61 {e}^{12} - - 23 {e}^{8}}{32} \pi$ = 30974.5 cubic units, nearly. The graph is not to scale. I would try to get surface area, soon.

#### Explanation:

graph{x^2e^(2x) [-5, 5, -10000, 10000]}

The volume = $\pi \int {f}^{2} \mathrm{dx}$, with x from 2 to 3

$= \pi \int {x}^{2} {e}^{4 x} \mathrm{dx}$, with x from 2 to 3

Now,

$\int {x}^{2} {e}^{4 x} \mathrm{dx}$

$= \frac{1}{4} \int {x}^{2} d \left({e}^{4 x}\right)$

=1/4(x^2e^(4x)--int e^(4x)d(x^2))

$= \frac{1}{4} {x}^{2} {e}^{4} x - \frac{1}{8} \int x d \left({e}^{4 x}\right)$

=1/4x^2e^4x-1/8(xe^(4x)-inte^(4x) dx))#

$= \frac{1}{4} {x}^{2} {e}^{4} x - \frac{1}{8} x {e}^{4 x} + \frac{1}{32} {e}^{4 x} + C$

Substituting the limits for the difference, the value here is

$\frac{61 {e}^{12} - - 23 {e}^{8}}{32}$.

So, the volume is

$\frac{61 {e}^{12} - - 23 {e}^{8}}{32} \pi$ = 30974.5 cubic units, nearly.