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

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 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?
$r^2 \leq z \leq 6 - r$

$r^2 = 6 - r$

$r^2 +r - 6 = 0$

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

$r = 2,~-3$

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

$r = 2$

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