Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
add comment

1 Answer

up vote 1 down vote accepted

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

enter image description here

share|improve this answer
    
Thank you for the help. –  docjay May 31 '12 at 11:37
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.