Gauss Divergence Theorem - Silly doubt - Almost solved


by come2ershad
Tags: divergence, doubt, gauss, silly, solved, theorem
come2ershad
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#1
Oct3-10, 04:25 AM
P: 16
1. The problem statement, all variables and given/known data

The problem statement has been attached with this post.

2. Relevant equations

I considered u = ux i + uy j and unit normal n = nx i + ny j.


3. The attempt at a solution

I used gauss' divergence theorem. Then it came as integral [(dux/dx) d(omega)] + integral [(duy/dy) d(omega)] = integral [(ux nx d(gamma)] + integral [(uy ny d(gamma)]

My question is can I separate the x and y components and write as separate equations as given in the problem? Is that right?
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come2ershad
come2ershad is offline
#2
Oct3-10, 06:43 AM
P: 16
anybody?
HallsofIvy
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#3
Oct3-10, 08:27 AM
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PF Gold
P: 38,879
You can, although it is not trivial to prove it. Break up the boundary so that the components of the vector can be written as functions of x on each piece and use x itself as parameter. That will give the first equation.

Then break up the boundary so the components of the vector can be written as functions of y on each piece and use y itself as parameter. That will give the second equation.

That partitioning is used in the proof of the theorem. You might want to look at the proof in any Calculus text.

come2ershad
come2ershad is offline
#4
Oct3-10, 08:33 AM
P: 16

Gauss Divergence Theorem - Silly doubt - Almost solved


Thanks for the reply.


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