# What is the surface area of the solid created by revolving f(x) =2x^3-2x , x in [2,4] around the x axis?

##### 1 Answer
Jul 22, 2017

$A = 216 \pi$

#### Explanation:

The area is given by the following integral :

$A = {\int}_{2}^{4} 2 \pi f \left(x\right) \mathrm{dx} = 2 \pi {\int}_{2}^{4} \left(2 {x}^{3} - 2 x\right) \mathrm{dx} =$

$2 \pi {\left[{x}^{4} / 2 - {x}^{2}\right]}_{2}^{4} = 2 \pi \left({4}^{4} / 2 - {4}^{2} - {2}^{4} / 2 + {2}^{2}\right) =$

$2 \pi \cdot 108 = 216 \pi$