# What is the surface area of the solid created by revolving f(x) = (3x-1)^2 , x in [1,3] around the x axis?

May 28, 2018

The question seems rather odd.

#### Explanation:

The formula for computing the surface area is

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

In your case, $f \left(x\right) = {\left(3 x - 1\right)}^{2}$, which implies

$f ' \left(x\right) = 2 \left(3 x - 1\right) \setminus \cdot 3 = 6 \left(3 x - 1\right) = 18 x - 6$

Plug $f$ and $f '$ into to the formula:

$S = 2 \pi \setminus \int {\left(3 x - 1\right)}^{2} \setminus \sqrt{1 + {\left(18 x - 6\right)}^{2}} \mathrm{dx}$

This integral yields a very complicate solution, which I don't believe your homework could ever ask for. For completeness sake, you can check it from WolframAlpha