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

1 Answer
KillerBunny
May 28, 2018

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%5E2sqrt(1%2B(18x-6)%5E2)

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