# 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!

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You seem to have integrated over a cylinder instead of a cone... – anon May 30 '12 at 20:52
Before I have checked the solution, your last line compels me to answer you with a quote: "Begin at the beginning,' the King said gravely, 'and go on till you come to the end: then stop.' " (Because it seemed you had reached a solution, and I had no idea what you would want to do after that.) – rschwieb May 30 '12 at 20:57

$$2\int_{0}^{\pi}\int_{0}^{2}\int_{0}^{r}\left(z+1\right)rdzdrd\theta=2\int_{0}^{\pi}\int_{0}^{2}\left(\frac{r^{3}}{2}+r^{2}\right)drd\theta=2\int_{0}^{\pi}\left[\left.\left(\frac{r^{4}}{8}+\frac{r^{3}}{3}\right)\right|_{0}^{2}\right]d\theta=2\pi\left(2+\frac{8}{3}\right)=\frac{28\pi}{3}$$