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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 , is in the z direction. But in the xy-plane, the z component of 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. Tags conflict, divergence, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post xRock3r Calculus 2 December 30th, 2016 04:22 AM vysero Calculus 0 December 5th, 2014 09:21 PM alan0354 Applied Math 0 August 10th, 2013 09:14 AM Tutu Algebra 15 May 19th, 2012 06:39 AM aaron-math Calculus 5 December 17th, 2011 06:27 PM

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