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 $$ \iint_{\text{surface}}|\vec r_u\times \vec r_v|\,du\,dv $$