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 April 28th, 2017, 01:52 PM #1 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 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: $\displaystyle \int\int (F\cdot n) dS$ $\displaystyle r(u,v)=$ 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.
 April 28th, 2017, 05:24 PM #2 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 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.
 April 29th, 2017, 04:17 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The base is a disk in the xy-plane so a normal vector, and then $d\vec{S}$, is in the z direction. But in the xy-plane, the z component of $\vec{F}$ is 0. The flux through that disk is 0.
 April 29th, 2017, 04:24 AM #4 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 That makes plenty of sense, but I still can't find where the missing 16pi is.

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