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

Feb 26, 2017

$2 \pi {\int}_{2}^{4} \left(2 {x}^{3} - 2\right) \sqrt{1 + 36 {x}^{4}} \mathrm{dx} \approx 49266.1936$

#### Explanation:

The surface area of a solid created by rotating a function $f$ around the $x$ axis on $x \in \left[a , b\right]$is given by:

$S = 2 \pi {\int}_{a}^{b} f \left(x\right) \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

Here, $f \left(x\right) = 2 {x}^{3} - 2$ so $f ' \left(x\right) = 6 {x}^{2}$. Then:

$S = 2 \pi {\int}_{2}^{4} \left(2 {x}^{3} - 2\right) \sqrt{1 + {\left(6 {x}^{2}\right)}^{2}} \mathrm{dx}$

$S = 2 \pi {\int}_{2}^{4} \left(2 {x}^{3} - 2\right) \sqrt{1 + 36 {x}^{4}} \mathrm{dx}$

This cannot be easily integrated, so plug this into a calculator.

$S = 49266.1936 \ldots$