How do I calculate surface integral? For the vector field $a = [-z^2–2z,-2xz+2y^2,-2xz-2z^2]^T$ and the area $F$ on the cylinder $x^2 + z^2 = 4$ , which is above the ground plane $z = 0$ , in front of the plane $x = 0$ and between the cross plane $y = 0$ and lies to the their parallel plane $y = 2$ , calculate the following integral:
$\int_{F}^{} \! a\cdot dn \, = ?$
So I use that:
$x=2cos(u)$
$y=v$
$z=2sin(u)$
and than I calculate normal vector. 
I get integral 
$\int_{0}^{2}\int_{0}^{\Pi/2}\begin{pmatrix}-z^2–2z\\-2xz+2y^2\\-2xz-2z^2\end{pmatrix}\cdot \begin{pmatrix}2sin(u)\\ 0\\ 2sin(u)\end{pmatrix}dudv $
$\int_{0}^{2}\int_{0}^{\Pi/2}\begin{pmatrix}-(2sin(u))^2–4sin(u)\\-2(2cos(u))(2sin(u))+2v^2\\-2(2cos(u))(2sin(u))-2(2sin(u))^2\end{pmatrix}\cdot \begin{pmatrix}2sin(u)\\ 0\\ 2sin(u)\end{pmatrix}dudv $
$\int_{0}^{2}\int_{0}^{\Pi/2}\begin{pmatrix}-8sin^3(u)–8sin^2(u)\\0\\-16cos(u)sin^2(u))-16sin^2(u)\end{pmatrix} $
I was to lazy to replace x,y and z, but at the end I get wrong solution and I need a lot of time to calculate, is there any better method?

Use the Divergence theorem, and for your triple integral, work with cylindrical coordinates. From there, it's just a matter of evaluating the iterated integral. Just set up the integral bounds and you are good to go. 
