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Need help on easy surface integral

by KillaMarcilla
Tags: integral, surface
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KillaMarcilla
#1
Dec6-03, 02:41 PM
P: 56
Yo guys, I'm stumped on how to parameterize this surface and then compute an integral over it

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
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KillaMarcilla
#2
Dec6-03, 08:39 PM
P: 56
So I should have taken ds/dr ds/dtheta instead of the other way around to get the right sign, and I should have taken r dr dtheta rather than just dr dtheta, but even when I do that I still get the wrong answer

hmm.. I can't figure out how to get this one right without just using the divergence theorem
Hurkyl
#3
Dec6-03, 09:12 PM
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Well, your surface is a sphere, so it's normal has to be proportional to [itex]\vec{r}[/itex]... does that help you spot your mistake?

gnome
#4
Dec6-03, 10:00 PM
P: 1,046
Need help on easy surface integral

orig. posted by Hurkyl
Well, your surface is a sphere...
Isn't it a paraboloid?
Hurkyl
#5
Dec6-03, 10:12 PM
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Yes it is, ignore me!
HallsofIvy
#6
Dec7-03, 07:12 AM
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Thanks
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Don't worry, Hurkyl, we do!
KillaMarcilla
#7
Dec7-03, 10:43 PM
P: 56
Don't worry; I figured it out on my own

Man, this site is so crappy for getting help on anything specific

Well, it's no big deal

If someone comes across this in the archives or something and wants me to tell them how to solve the problem, they can drop a line to killamarcilla@bigpen.us
meister
#8
Dec8-03, 10:01 PM
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Originally posted by KillaMarcilla
Man, this site is so crappy for getting help on anything specific
Because we don't just tell you the answer...?
Hurkyl
#9
Dec8-03, 10:10 PM
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I was busy at the time I answered; I thought I saw a quick hint, so I posted it.


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