March 26th, 2019, 06:15 AM  #1 
Member Joined: Apr 2017 From: India Posts: 56 Thanks: 0  Divergence theorem
How the divergence theorem is used in this question?. Please help.

March 26th, 2019, 09:59 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1310 
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. 
March 26th, 2019, 10:47 AM  #3 
Member Joined: Apr 2017 From: India Posts: 56 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, 11:32 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1310 
It's pretty straightforward $\nabla \cdot F = x^2 + y^2$ We again change to cylindrical coordinates $\displaystyle \int_0^{\frac \pi 2}\int_0^2 \int_{r^2}^{6r}r^2 \cdot r~dz~dr~d\theta =\dfrac{52\pi}{15}$ 
March 26th, 2019, 07:39 PM  #5 
Member Joined: Apr 2017 From: India Posts: 56 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 07:52 PM. 
March 26th, 2019, 08:06 PM  #6  
Senior Member Joined: Sep 2015 From: USA Posts: 2,408 Thanks: 1310  Quote:
$r^2 = 6  r$ $r^2 +r  6 = 0$ $(r+3)(r2) = 0$ $r = 2,~3$ $r > 0 \Rightarrow r \neq 3$ $r = 2$  

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