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 | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Need help on easy surface integral

**Physics Forums | Science Articles, Homework Help, Discussion**