Using the Divergence Theorem on the surface of a sphere

  • #1
TheGreatDeadOne
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Homework Statement:
Let a sphere of radius r_0 be centered at the origin, and r′ the position vector of a point p′ within the sphere or under its surface S. Let the position vector r be an arbitrary fixed point P.
Relevant Equations:
Divergence theorem (it's not really necessary, but I'm trying to solve it that way)
The integral that I have to solve is as follows:

[tex]\oint_{s} \frac{1}{|r-r'|}da', \quad\text{ integrating with respect to r '}[/tex], integrating with respect to r'

Then I apply the divergence theorem, resulting in:

[tex]\iiint \limits _{v} \nabla \cdot \frac{1}{|r-r'|}dv' = \int_{0}^{r_0}\int_{0}^{\pi}\int_{0}^{2\pi} \frac{(r-r')}{|r-r'|^3}r'^2\sin{\theta}d\theta d\phi dr'[/tex]

Integrating θ and φ we have:

[tex]4\pi \int_{0}^{r_0} \frac{(r-r')}{|r-r'|^3}r'^2 dr'[/tex]

And in that part I got stuck. I didn't try to solve it by integrating directly, because I'm still a bit confused. I know the result is:
[tex]\frac{4\pi r^{2}_{0}}{r} \quad\text{for} \quad r\geq r_0 \quad and \quad 4\pi r^{2}_{0}\quad for \quad r\leq r_0[/tex]
 

Answers and Replies

  • #2
Delta2
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I am a bit confused on which variation of divergence theorem you are using. Can you write it explicitly?

The final result is a scalar or a vector? It seems it is a scalar by what you say at the last sentence, but the divergence theorem as you set it up in the above equation it seems to give a vector result (r and r' are vectors right)?
 
  • #3
ehild
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Homework Statement:: Let a sphere of radius r_0 be centered at the origin, and r′ the position vector of a point p′ within the sphere or under its surface S. Let the position vector r be an arbitrary fixed point P.
Relevant Equations:: Divergence theorem (it's not really necessary, but I'm trying to solve it that way)

The integral that I have to solve is as follows:

[tex]\oint_{s} \frac{1}{|r-r'|}da', \quad\text{ integrating with respect to r '}[/tex], integrating with respect to r'
What exactly is the vector-vector function you have to integrate over the surface of the sphere?
 
  • #4
TheGreatDeadOne
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I am a bit confused on which variation of divergence theorem you are using. Can you write it explicitly?

The final result is a scalar or a vector? It seems it is a scalar by what you say at the last sentence, but the divergence theorem as you set it up in the above equation it seems to give a vector result (r and r' are vectors right)?
it's a scalar (lol I saw the error here, i forgot to fix it here)
 
  • #5
TheGreatDeadOne
11
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What exactly is the vector-vector function you have to integrate over the surface of the sphere?
I am a bit confused on which variation of divergence theorem you are using. Can you write it explicitly?

The final result is a scalar or a vector? It seems it is a scalar by what you say at the last sentence, but the divergence theorem as you set it up in the above equation it seems to give a vector result (r and r' are vectors right)?
Sorry, I forgot to organize and correct something here. 1 / | r-r '| is a scalar, in my papers I was disregarding the module, forget that part of the divergence theorem.
 
  • #6
Delta2
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Better we take things from start, you say you want to calculate $$\oint \frac{1}{|r-r'|}da'$$. The integrand is a scalar as you said, but what about da' is it scalar or vector (with direction the normal to the surface element da').

I have in mind two variations of the divergence theorem:
  1. $$\iiint_V \nabla\cdot\vec{F} dV=\oint_{\partial V} \vec{F}\cdot d\vec{A}$$
  2. $$\iiint_V \nabla f dV=\oint_{\partial V} f d\vec{A}$$
In 1. ##\vec{F}## is a vector but because we take the divergence in the LHS (and the dot product in the RHS) the final result is scalar.
in 2.## f## is a scalar but because we take the gradient of ##f ## in the LHS (and the multiplication of## f## by the vector surface element ##d\vec{A}## in the RHS) the final result is a vector

So which one are you using. I think you are using 2. with ##f=\frac{1}{|\vec{r}-\vec{r'}|}## but then the final result is a vector not a scalar.

But then again if I strictly follow your notation, you seem to take the divergence of a scalar, you write $$\nabla\cdot\frac{1}{|r-r'|}$$ while you should probably mean the gradient there.

I have to say this, in this thread and in the other thread with the gradient, your notation is a bit ambiguous, you seem to use the symbol r (or r') to denote both the position vector and the magnitude of the position vector interchangeably.
 
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