# What is the surface area of the solid created by revolving f(x) = x^2+e^x , x in [2,4] around the x axis?

Nov 13, 2017

$V = \pi 1044 \approx 3279.82$

#### Explanation:

First, take a look at the graph $f \left(x\right) = {x}^{2} + {e}^{x} , x \in \left[2 , 4\right]$

I've drawn vertical lines at the endpoints to show where the solid would revolve around the $x$-axis. It looks like the object would ultimately resemble a circular bell.

The formula for find the volume of a shape like this is

$V = \pi {\int}_{\text{lower")^("upper}} {\left[f \left(x\right)\right]}^{2} \mathrm{dx}$

In this case, you can plug in all the values and find the integral in a straight forward fashion.

$V = \pi {\int}_{2}^{4} {\left[{x}^{2} + {e}^{x}\right]}^{2} \mathrm{dx}$

$V = \pi \left({\int}_{2}^{4} \left[{x}^{4} + 2 {x}^{2} {e}^{x} + {e}^{2 x}\right] \mathrm{dx}\right)$

$V = \pi {\left[\frac{1}{5} {x}^{5} + 2 {x}^{2} {e}^{x} + 4 x {e}^{x} + 2 {e}^{2 x}\right]}_{2}^{4}$

$V = \pi \left[\frac{1}{5} {\left(4\right)}^{5} + 2 {\left(4\right)}^{2} {e}^{4} + 4 \left(4\right) {e}^{x} + 2 {e}^{2 \cdot 4}\right] - \pi \left[\frac{1}{5} {\left(2\right)}^{5} + 2 {\left(2\right)}^{2} {e}^{2} + 4 \left(2\right) {e}^{2} + 2 {e}^{2 \cdot 2}\right]$

$V = \pi 1044 \approx 3279.82$