Yo guys, I'm stumped on how to parameterize this surface and then compute an integral over it(adsbygoogle = window.adsbygoogle || []).push({});

I'm supposed to compute [tex]\int\int_S \vec{r}[/tex] over the surface formed by the x-y plane and [tex]z=4-(x^2+y^2)[/tex], but I don't know to put it together and do it

(No matter how you work with x and y, z will always be zero for the flat part, and whether you use the normal or that [tex]\bigl(\frac{\delta S}{\delta u}\times\frac{\delta S}{\delta v}\bigr)[/tex], the integrand will end up being zero for the bottom)

For the curved part, I tried using [tex]S: \cases{ x=r \cos(\theta), y=r \sin(\theta), z=4-r^2}[/tex], which gave me [tex]\bigl(\frac{\delta S}{\delta \theta}\times\frac{\delta S}{\delta r}\bigr) = 2 r \sin\theta\vec{i} - 2 r \cos\theta\vec{j} - r\vec{k}[/tex], and then I computed [tex]\int_{\theta=0}^{2\pi}\int_{r=0}^{2}\vec{r}(r\cos\theta,r\sin\theta,4-r^2)\cdot\bigl(2 r \sin\theta\vec{i} - 2 r \cos\theta\vec{j} - r\vec{k}\bigr)[/tex]

But I came up with [tex]-8\pi[/tex] rather than [tex]24\pi[/tex].. I think the negative can come from how I arbitrarily decided to look at d theta cross dr instead of the other way around, sso I was looking at the flux in instead of the flux out, but I don't know how to explain the fact that I ended up with 4 - 8 integrated from zero to 2pi rather than 4 + 12 from zero to 2pi

**Physics Forums - The Fusion of Science and Community**

# Need help on easy surface integral

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Need help on easy surface integral

Loading...

**Physics Forums - The Fusion of Science and Community**