Solving Gauss Divergence Theorem on a Closed Surface

Click For Summary

Homework Help Overview

The discussion revolves around verifying the Gauss Divergence Theorem for a closed surface, specifically a sphere defined by the equation x² + y² + z² = 9, with a given vector field F. Participants are attempting to equate the volume integral of the divergence of F to the surface integral over the sphere.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the normal vector N and its components in relation to the parameterization of the sphere. There are attempts to compute both sides of the divergence theorem and questions about the correctness of the integration limits and the components of N.

Discussion Status

The discussion includes clarifications about the nature of the normal vector and its proper formulation. Some participants have identified potential errors in the calculations, particularly regarding the components of N, and there is a sense of progress as one participant reports resolving their confusion after receiving feedback.

Contextual Notes

There are indications of confusion regarding the integration limits and the correct formulation of the normal vector components, which are critical for the evaluation of the surface integral. Participants are working within the constraints of the problem as stated, without additional context or information.

HeheZz
Messages
11
Reaction score
0

Homework Statement



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


The Attempt at a Solution



I have tried to solve the left hand side which appear to be (972*pi)/5
However, I can't 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 please. Thanks!
 
Physics news on Phys.org
Your N is wrong. Describe to me what you think N is.
 
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
 
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.
 
Last edited:
Do you mean the range values when i am integrating? I don't understand what to use. Isnt it 0-pi for the inner integral and 0-2pi for the outer integral?
 
HeheZz said:
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).
 
HeheZz said:
Do you mean the range values when i am integrating? I don't 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.
 
OK! I got it! Tnx for the help very much! I didnt realize this mistake and was really stress over it.. Thanks again for the help and I got the answer :D
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
8
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K