# I need help with solving area of a surface of revolution. Calculus 2

*label*Calculus

*timer*Asked: Feb 11th, 2015

**Question description**

The question is x=y^4/2+1/(16y^2) from 1<y<3. I took the derivative and put it in the formula but I got stuck. I go to integral of (y^4/2+1/(16y^2)sqrt(4y^6+1/2+1/(64y^6)) I think that is a perfect square but idk what it is.