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March 26th, 2019, 07:15 AM   #1
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Divergence theorem

Attached Images 20190326_192514.jpg (87.7 KB, 16 views) March 26th, 2019, 10:59 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,636 Thanks: 1472 Gauss's Theorem The flux integral of a field through a closed surface is equal to the integral of the divergence of the field over the volume enclosed by the surface. I think this problem is omitting that the surface in question is also bounded by the first (or any) quadrant. Otherwise I can't make the answer match any of the options. Thanks from topsquark March 26th, 2019, 11:47 AM #3 Member   Joined: Apr 2017 From: India Posts: 76 Thanks: 0 Suppose the surface in question is bounded by the first quadrant, then I will be more than happy to see the procedure you applied to reach any of the conclusive answers. I can dispute that assignment question in my university, but for that I would require your help. "You are an awesome mathematician."  March 26th, 2019, 12:32 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,636 Thanks: 1472 It's pretty straightforward $\nabla \cdot F = x^2 + y^2$ We again change to cylindrical coordinates $\displaystyle \int_0^{\frac \pi 2}\int_0^2 \int_{r^2}^{6-r}r^2 \cdot r~dz~dr~d\theta =\dfrac{52\pi}{15}$ Thanks from topsquark and shashank dwivedi March 26th, 2019, 08:39 PM #5 Member   Joined: Apr 2017 From: India Posts: 76 Thanks: 0 Thank you for the brilliant answer. However, I will be more than interested in knowing why the limits of r in this case has been taken from 0 to 2. When I tried, I took it from 0 to 6 as z was smaller than {6 - sqrt(x^2 + y^2)}. Can you please help me with the reason for why limits of r taken from 0 to 2? Last edited by shashank dwivedi; March 26th, 2019 at 08:52 PM. March 26th, 2019, 09:06 PM   #6
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Quote:
 Originally Posted by shashank dwivedi Thank you for the brilliant answer. However, I will be more than interested in knowing why the limits of r in this case has been taken from 0 to 2. When I tried, I took it from 0 to 6 as z was smaller than {6 - sqrt(x^2 + y^2)}. Can you please help me with the reason for why limits of r taken from 0 to 2?
$r^2 \leq z \leq 6 - r$

$r^2 = 6 - r$

$r^2 +r - 6 = 0$

$(r+3)(r-2) = 0$

$r = 2,~-3$

$r > 0 \Rightarrow r \neq -3$

$r = 2$ Tags 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 sarajoveska Calculus 0 August 27th, 2017 06:12 AM phrack999 Calculus 3 April 29th, 2017 05:24 AM xRock3r Calculus 2 December 30th, 2016 05:22 AM vysero Calculus 0 December 5th, 2014 10:21 PM aaron-math Calculus 5 December 17th, 2011 07:27 PM

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