
#1
May1110, 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 ∫(02pi)∫(0pi) 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! 



#2
May1110, 03:12 AM

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PF Gold
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Your N is wrong. Describe to me what you think N is.




#3
May1110, 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 



#4
May1110, 03:37 AM

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PF Gold
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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. 



#5
May1110, 04:30 AM

P: 13

Do you mean the range values when i am integrating? I dont understand what to use. Isnt it 0pi for the inner integral and 02pi for the outer integral?




#6
May1110, 04:33 AM

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#7
May1110, 04:36 AM

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#8
May1110, 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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