Question

What is the surface area of the solid created by revolving `f(x) = 2x^2-x+5 , x in [1,2]` around the x axis?

Answer 1

`SA \approx 181.265`

Explanation:

The surface area of the solid created by revolving a positive function, `f(x)`, for `x in [a,b]` around the x-axis is always

`\int_a^b 2pif(x)sqrt(1+f'(x)^2)dx`

The derivation of this is fairly simple. It involves the surface area of a frustrum which is `2pirl`. I would highly recommend looking up the derivation, as it make this formula make a lot more sense. This link is pretty good, http://tutorial.math.lamar.edu/Classes/CalcII/SurfaceArea.aspx

Anyways, in the case of this problem, the first thing to notice is that `f(x)` is negative over the interval indicated.

`SA = \int_1^2 2pif(x)sqrt(1+f'(x)^2)dx`

`= \int_1^2 2pi(2x^2-2x+5)sqrt(1+(4x-2)^2)dx`

at this point I would use a calculator to find the value of this integral which would end up being the answer.

`SA \approx 181.265`