Gauss Divergence Theorem


by HeheZz
Tags: divergence, gauss, theorem
HeheZz
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#1
May11-10, 02:46 AM
P: 13
1. The problem statement, all variables and given/known data

Verify Gauss Divergence Theorem ∭∇.F dxdydz=∬F. (N)dA
Where the closed surface S is the sphere x^2+y^2+z^2=9 and the vector field F = xz^2i+x^2yj+y^2zk


3. The attempt at a solution

I have tried to solve the left hand side which appear to be (972*pi)/5
However, I cant seems to solve the right hand side to get the same answer.
I substitute x = 3sin(theta)cos(phi), y=3sin(theta)sin(phi), z=3cos(theta)
Therefore N=9sin^2(theta)cos(phi)i+9sin^2(theta)cos(phi)j+9cos(theta)
and F = F=27sin(θ)cos^2(θ)cos(φ)+27sin^3(θ)cos^2(φ)sin(φ)+27sin^2(θ)cos(θ)sin^2 (φ)
then I used ∫(0-2pi)∫(0-pi) F. (N) dθdφ
I got the final answer as (324*pi)/5 which does not match with left hand side.
Hope anyone can help here plz. Thanks!
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vela
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#2
May11-10, 03:12 AM
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Your N is wrong. Describe to me what you think N is.
HeheZz
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#3
May11-10, 03:21 AM
P: 13
N is the normal vector?

where r(θ,φ) = (3sinθcosφ, 3sinθsinφ, 3cosθ)

and N = rθ X rφ

Thus, N=9sin^2(θ)cos(φ)i+9sin^2(θ)cos(φ)j+9cos(θ)sin(θ)k

vela
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#4
May11-10, 03:37 AM
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Gauss Divergence Theorem


Yes, it's a normal vector. More important, it's the unit normal vector. Since you're using a sphere, it will just be the radial unit vector.

Edit: Oh, I see what you're doing. That's not just the normal vector but the normal vector scaled by the part of the area element. I think you're just cranking out the integral wrong, but let me try calculate it here to make sure.
HeheZz
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#5
May11-10, 04:30 AM
P: 13
Do you mean the range values when i am integrating? I dont understand what to use. Isnt it 0-pi for the inner integral and 0-2pi for the outer integral?
vela
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#6
May11-10, 04:33 AM
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Quote Quote by HeheZz View Post
N is the normal vector?

where r(θ,φ) = (3sinθcosφ, 3sinθsinφ, 3cosθ)

and N = rθ X rφ

Thus, N=9sin^2(θ)cos(φ)i+9sin^2(θ)cos(φ)j+9cos(θ)sin(θ)k
The y-component of your N should have sin(φ), not cos(φ).
vela
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#7
May11-10, 04:36 AM
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Quote Quote by HeheZz View Post
Do you mean the range values when i am integrating? I dont understand what to use. Isnt it 0-pi for the inner integral and 0-2pi for the outer integral?
Yeah, you did everything right except you made a mistake on the y-component of N. If you fix that, you should get the right answer. It worked out for me.
HeheZz
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#8
May11-10, 04:50 AM
P: 13
OK! I got it! Tnx for the help very much!! I didnt realise this mistake and was really stress over it.. Thanks again for the help and I got the answer :D


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