How to use a surface integral to derive the surface area of a cylinder? Use a surface integral to derive the expression for the surface area of a cylinder of radius a and height h.

What I got so far was 
$$ r(u,v) = vcosu  \hat i + vsin \hat j + v \hat k $$
I am setting up a double integral :
$$  \int_0^a\int_0^h \vec r_u \times \vec r_v dudv $$
My cross product is only turrning out to be $v^2$
I am not looking for the answer, just some guidence as to where I should go from here. 

Your parametrization should be
$$
\vec r(u,v)=a\cos u\hat i+a\sin u\hat j+v\hat k\\
u\in[0,2\pi), \quad v\in[0,h]
$$
So $|\vec r_u\times \vec r_v|=a$ and your integral should be
$$
\iint_{\text{surface}}|\vec r_u\times \vec r_v|\,du\,dv
$$
