Question

What is the surface area of the solid created by revolving `f(x) = 2x^2-4x+8 , x in [1,2]` around the x axis?

Answer 1

approximately `100.896`

Explanation:

The surface area `S` created by revolving the function `f(x)` around the `x`-axis on the interval `x in[a,b]` can be found through:

`S=2piint_a^bf(x)sqrt(1+[f'(x)]^2)dx`

Using `f(x)=2x^2-4x+8` and `f'(x)=4x-4` on the interval `x in[1,2]` gives:

`S=2piint_1^2(2x^2-4x+8)sqrt(1+(4x-4)^2)dx`

Actually integrating this is very complex and far beyond the scope of this problem, so put this into a calculator--be mindful of parentheses.

The answer you receive (don't forget to multiply by `2pi`) should be

`Sapprox100.896`