Find the surface area of that part of the cylinder given by $\mathbf{r}(u,v) = 3\cos u \mathbf{i} + 3\sin u\mathbf{j} + v\mathbf{k}$ over the region where $0\leq u \leq2\pi$ and $0\leq v \leq2$.
The answer is $12\pi$.
I am trying to solve this but I am not understanding how to do this. I am thinking about either Stoke's theorem or the Divergence theorem. In both cases, we have to take the derivative with respect to x and y and z, but it is in terms with u and v.
Also, how would you find the normal vector for Stoke's theorem? Or is it better to just use the Divergence Theorem?
It's not clear why you would use Stoke's theorem or divergence, although sometimes people set relatively trivial exercises like this in order to give you a way to check your application of the theorem. (But in that case I'd expect there to have been more guidance on how to set up the problem; for example, "Use this vector field ... .")
The quantities $u$ and $v$ are just parameters that you can vary within the given intervals; for each possible pair of values $(u,v)$ you can plug those values into the formula $3\cos u \mathbf{i} + 3\sin u\mathbf{j} + v\mathbf{k}$ and the result will be a point on the desired surface.
Try drawing a picture of the surface. You can start by drawing just the points for which $v = 0$ (these all have the form $3\cos u \mathbf{i} + 3\sin u\mathbf{j}$, so they lie in the $x,y$ plane). Every other point is obtained by adding $v\mathbf{k}$ for some $v$ in $(0,2]$ to one of the $x,y$ points, so those points are all directly above the first set of points, that is, take your graph in the $x,y$ plane and build a wall of height $2$ on it.
It's a very simple surface (hint: you already know it's part of a cylinder) and you should be able to find its area easily using high-school geometry.