Odd, if someone expects you to be able to do a problem like this then surely they expect you to be able to integrate over a surface area! Perhaps you need to review your text.
Since we are given that r=1, we have x= cos(\theta), y= sin(\theta), and z= z. The "position vector" of any point on the surface is cos(\theta)\vec{i}+ sin(\theta)\vec{j}+ z\vec{j}.
The derivative with respcect to \theta is -sin(\theta)\vec{i}+ cos(\theta)\vec{j} and the derivative with respect to z is \vec{k}. The "fundamental vector product" is the cross product of those two vectors:cos(\theta)\vec{i}+ sin(\theta)\vec{j} and the length of that gives the "differential of surface area". \sqrt{cos^2(\theta)+ sin^2(\theta)}= 1so d\sigma = d\theta dy.
