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April 28th, 2017, 01:52 PM   #1
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Divergence Theorem Conflict

Problem: Use the Divergence Theorem to compute the net outward flux of the field F=<4x, y, z> across the surface​ S, where S is the surface of the paraboloid z=4−x^2−y^2​, for z≥​0, plus its base in the​ xy-plane.

This was solved easily using the volume method

$\displaystyle \int\int\int\nabla\cdot F dV$

$\displaystyle \nabla\cdot F = 6$

Volume of the paraboloid is calculated using shells

$\displaystyle 2\pi\int_{0}^{2}y(4-y^2)dy=8\pi$

$\displaystyle 6*8\pi=48\pi$

I wanted to see it via surface integral and that is where I ran into problems, here is the work I did:

$\displaystyle \int\int (F\cdot n) dS$

$\displaystyle r(u,v)=<v\cos(u),v\sin(u),v^2>$

Where $\displaystyle 0\leq u \leq 2\pi$ and $\displaystyle 0\leq v \leq \sqrt h$ where $\displaystyle h=z$

$\displaystyle t_{u}= <-v\sin(u),v\cos(u),0>$
$\displaystyle t_{v}= <\cos(u),\sin(u),2v>$

$\displaystyle t_{u}\times t_{v}=<2v^2\cos(u),2v^2\sin(u),-v>$

$\displaystyle F=<4v\cos(u),v\sin(u),v^2>$

$\displaystyle F\cdot (t_{u}\times t_{v})=8v^3\cos^2(u)+2v^3\sin^2(u)-v^3$

$\displaystyle \int_{0}^{2\pi}\int_{0}^{2}F\cdot (t_{u}\times t_{v})dvdu$

$\displaystyle \int_{0}^{2\pi}2v^4\cos^2(u)+\frac{v^4}{2}\sin^2(u )-\frac{v^4}{4}\bigg\rvert_{v=0}^{v=2}du$

$\displaystyle \int_{0}^{2\pi}32\cos^2(u)+8\sin^2(u)-4 du$

$\displaystyle 32\bigg(\frac{u}{2}+\frac{ \sin(2u)}{4}\bigg)+8\bigg(\frac{u}{2}-\frac{ \sin(2u)}{4}\bigg)-4u\bigg\rvert_{u=0}^{u=2\pi}=32\pi$

This is obviously incorrect, I realize that the problem is me and my formulation, but I can't determine where the mistake is. I did gather that if the sign on $\displaystyle 4u$ was different it would have added to $\displaystyle 48\pi$, but I can't find a way to make that happen. Please help, thank you.

Last edited by skipjack; April 28th, 2017 at 06:29 PM.
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April 28th, 2017, 05:24 PM   #2
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I've been spending a lot more time on this one, and I think it has to do with the fact that my surface method didn't account for the base, which is specified in the problem.

I tried a few different calculations to add to the 32pi, but I get 24pi going through the base alone. Not sure what I'm missing here.

Last edited by skipjack; April 28th, 2017 at 06:18 PM.
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April 29th, 2017, 04:17 AM   #3
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The base is a disk in the xy-plane so a normal vector, and then , is in the z direction. But in the xy-plane, the z component of is 0. The flux through that disk is 0.
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April 29th, 2017, 04:24 AM   #4
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That makes plenty of sense, but I still can't find where the missing 16pi is.
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