Surface area of S, parameterized as $r(u,v)=( 2\cos(u),2\sin(u),v )\;\;,0 ≤ u ≤ 2π, 0 ≤ v ≤2.$ I first found the tangent vectors, $r_{u}$ and $r_{v}$. Then compute $$\lvert r_{u} \ \times \ r_{v}\lvert$$ and I got $$\sqrt{4(\cos^{2}u+\sin^{2}u)}.$$ I simplified it. $$\sqrt{4}=2,$$ because v is the axis of the cylinder, so I figured do I need a double integral to deal with this? $$\int_{0}^{2\pi}\int_{0}^{2}2dvdu?$$ Is this correct?
 A: Actually there is a better (more general) way of solving this using the First Fundamental Form of differential geometry:
$$(I_{kl}(u,v)) = \begin{pmatrix} E(u,v) & F(u,v) \\ F(u,v) & G(u,v) \end{pmatrix} =
\begin{pmatrix} g_{11}(u,v) & g_{12}(u,v) \\ g_{21}(u,v) & g_{22}(u,v) \end{pmatrix}$$
and then apply the following: 
$$A(\Omega) = \iint_\Omega \sqrt{EG-F^2} dudv$$
Using this approach you can solve the areas of more complex surfaces!
EDIT: 
Following what you were doing we get
$$A = \int_0^{2\pi}du\int_0^2dv \lVert r_u \times r_v\rVert = \int_0^{2\pi}du\int_0^2dv \sqrt{4(cos^2u + sin^2u)} = \int_0^{2\pi}du\int_0^2dv \ 2 $$
where we used the trigonometric equality $cos^2u + sin^2u  = 1$.
To me this would be an easier way than trying to change to polars as the integrals in the cartesian form can be solved without any issues.
Hope it helps! 
A: Your double integral looks OK.
Since you are extruding a circle radius $2$  to form a cylinder of length/height $=2$.
A short way to find area by Pappu's theorem is 
$$ 2 \cdot \pi\cdot 2 \cdot 2 = 8 \pi $$
