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March 26th, 2019, 06:15 AM   #1
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Divergence theorem

How the divergence theorem is used in this question?. Please help.
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March 26th, 2019, 09:59 AM   #2
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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.
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March 26th, 2019, 10:47 AM   #3
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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."
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March 26th, 2019, 11:32 AM   #4
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It's pretty straightforward

$\nabla \cdot F = x^2 + y^2$

We again change to cylindrical coordinates

$\displaystyle \int_0^{\frac \pi 2}\int_0^2 \int_{r^2}^{6-r}r^2 \cdot r~dz~dr~d\theta =\dfrac{52\pi}{15}$
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March 26th, 2019, 07:39 PM   #5
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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 07:52 PM.
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March 26th, 2019, 08:06 PM   #6
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Quote:
Originally Posted by shashank dwivedi View Post
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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