Question

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

Answer 1

The question seems rather odd.

Explanation:

The formula for computing the surface area is

`S = 2pi\int f(x)\sqrt(1+ (f'(x))^2) dx`

In your case, `f(x) = (3x-1)^2`, which implies

`f'(x) = 2(3x-1)\cdot 3 = 6(3x-1)=18x-6`

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

`S = 2pi\int (3x-1)^2\sqrt(1+ (18x-6)^2) 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