Flux of a vector field I've been trying to solve a flux integral with Gauss' theorem so a little input would be appreciated.
Problem statement: Find the flux of ${\bf{F}}(x,y,z) = (x,y,z^2)$ upwards through the surface ${\bf r}(u,v) = (u \cos v, u \sin v, u), \hspace{1em} (0 \leq u \leq 2; 0 \leq v \leq \pi)$
OK. I notice that $z = u$ so $0 \leq z \leq 2$. Furthermore I notice that $x^2 + y^2 = z^2$ so $x^2 + y^2 \leq 4$. It makes sense to use cylindrical coordinates so $(0 \leq r \leq 2)$ and $(0 \leq \theta \leq 2 \pi)$. Finally $div {\bf F} = 2(z+1)$.With this in mind I set up my integral
\begin{align*}
2\int ^{2 \pi} _0 \int ^2 _0 \int _0 ^2 (z+1)rdrdzd\theta &= \int ^{2 \pi} _0 \int ^2 _0[(z+1)r^2]_0 ^2 dzd\theta \\
&= 4\int ^{2 \pi} _0 \int ^2 _0 z + 1 dzd\theta\\
&= 4\int ^{2 \pi} _0  [1/2 z^2 + z]_0 ^2 d\theta \\
&= 16 \int _0 ^{2 \pi}d\theta \\
&= 32 \pi
\end{align*}
And I'm not sure how to continue from this point so if anyone can offer help it would be appreciated.
Thanks!

I am not convinced that your integration limits are in order. Domain of integration is the volume below a half cone. So I would proceed as follows
$$2\int_{0}^{\pi}\int_{0}^{2}\int_{0}^{r}\left(z+1\right)rdzdrd\theta=2\int_{0}^{\pi}\int_{0}^{2}\left(\frac{r^{3}}{2}+r^{2}\right)drd\theta=2\int_{0}^{\pi}\left[\left.\left(\frac{r^{4}}{8}+\frac{r^{3}}{3}\right)\right|_{0}^{2}\right]d\theta=2\pi\left(2+\frac{8}{3}\right)=\frac{28\pi}{3}$$
Then by Gauss' theorem you will have calculated the flux
EDIT: arithmetical error in the second transition

