HallsofIvy said:The surface is z= \sqrt{x^2+ y^2}, the upper nappe of a cone. In cylindrical coordinates, that is z= r. So good parametic equations would be x= r cos(\theta), y= r sin(\theta), z= r which means that the vector equation would be
\vec{r}(r, \theta)= r cos(\theta)\vec{i}+ r sin(\theta)\vec{j}+ r \vec{k}. Since z goes from 0 to 1, r goes from 0 to 1 and theta, of course, from 0 to 2\pi.
Does that help?
HallsofIvy said:Once you have done the cross product, and the integral, yes, that should be correct. I would recommend you do the integrations with respect to \theta first. Most of those trig functions, integrated from 0 to 2\pi will give 0.
asi123 said:10x a lot.