Question

What is the surface area of the solid created by revolving `f(x) =2x^3-2x , x in [2,4]` around the x axis?

Answer 1

`A=216pi`

Explanation:

The area is given by the following integral :

`A=int_2^4 2pif(x)dx=2piint_2^4(2x^3-2x)dx=`

`2pi[x^4/2-x^2]_2^4=2pi(4^4/2-4^2-2^4/2+2^2)=`

`2pi*108=216pi`