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?
 A: 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. 
A: If you evaluate the dot product correctly, you get
\begin{align*}
    \int_0^2\int_0^{\pi/2} &\left(-8\sin^3 u -8 \sin^2 u - 16 \cos u \sin^2 u - 16 \sin^2 u\right) \,du\,dv\\
    &=(-8)\int_0^2 1 \,dv \cdot \int_0^{\pi/2} \left(\sin^3 u + \sin^2u + 2 \cos u \sin^2 u + 2 \sin^2 u\right)\,du \\
    &=(-16)\int_0^{\pi/2} \left((1-\cos^2 u)\sin u + 2 \cos u \sin^2 u + 3 \sin^2 u\right)\,du \\
    &=(-16)\int_0^{\pi/2} \left(\sin u  -\cos^2 u \sin u + 2 \cos u \sin^2 u + 3 \sin^2 u\right)\,du \\
\end{align*}
Now
\begin{align*}
    \int_0^{\pi/2} \sin u \,du &= \left.-\cos u\right|^{\pi/2}_0 = 1 \\
    \int_0^{\pi/2} \cos^2 u \,\sin u \,du 
        &= \int_0^1 y^2\,dy = \frac{1}{3}\quad \text{(substitute $y=\cos u$)}\\
    \int_0^{\pi/2} \sin^2u \cos u\,du &= \frac{1}{3}\quad \text{(substitute $y=\sin u$)}\\
    \int_0^{\pi/2} \sin^2u\,du
        &=\int_0^{\pi/2} \left(\frac{1-\cos 2u}{2}\right)\,du
         =\left[\frac{u}{2} - \frac{1}{4}\sin 2u\right]^{\pi/2}_0 = \frac{\pi}{4}
\end{align*}
So the integral is
$$
(-16)\left[1 -\frac{1}{3} + \frac{2}{3} + 3 \cdot \frac{\pi}{4}\right]
= -\frac{64}{3} - 12\pi
$$
(verified with Wolfram Alpha.)
