Integration in polar coordinates

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Homework Statement:

In spherical poler coordinates the volume integral over a sphere of radius R of $$\int^R_0\frac{\hat r}{r^2}dv=\int_{surface}\frac{\hat r}{r^2}•\vec ds$$

$$=4\pi$$

Relevant Equations:

In spherical poler coordinates the volume integral over a sphere of radius R of $$\int^R_0\frac{\hat r}{r^2}dv=\int_{surface}\frac{\hat r}{r^2}•\vec ds$$

$$=4\pi$$
In spherical poler coordinates the volume integral over a sphere of radius R of $$\int^R_0\vec \nabla•\frac{\hat r}{r^2}dv=\int_{surface}\frac{\hat r}{r^2}•\vec ds$$
$$=4\pi=4\pi\int_{-\inf}^{inf}\delta(r)dr$$
How can it be extended to get $$\vec \nabla•\frac{\hat r}{r^2}=4\pi\delta^3(r)??$$
 

Answers and Replies

  • #2
haruspex
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I am unclear on what you are given and what you are trying to show.

$$\int^R_0\frac{\hat r}{r^2}dv$$
The bounds don't match the integration variable (dv). You mean
$$\int_V\frac{\hat r}{r^2}dv$$
$$\int^R_0\vec \nabla•\frac{\hat r}{r^2}dv=\int_{surface}\frac{\hat r}{r^2}•\vec ds$$
This seems to be identical to your earlier equation except for the Del on the left. They can't both be true.
 
  • #3
Orodruin
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This seems to be identical to your earlier equation except for the Del on the left. They can't both be true.
The only possibly relevant interpretation is
$$
\int_{r < R} \nabla \cdot \frac{\hat r}{r^2} dV
$$
Without the divergence operator, the expression is a vector that obviously cannot equal a scalar.

Of course, OP should be careful in formulating the original post such that this ambiguity does not arise.
 
  • #4
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The only possibly relevant interpretation is
$$
\int_{r < R} \nabla \cdot \frac{\hat r}{r^2} dV
$$
Without the divergence operator, the expression is a vector that obviously cannot equal a scalar.

Of course, OP should be careful in formulating the original post such that this ambiguity does not arise.
Ya in this case the Rhs equals $$4\pi$$.
What I want to clarify is $$\vec \nabla•\frac{\hat r}{r^2}=4\pi\delta^3(r)$$
 
  • #5
Orodruin
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Ya in this case the Rhs equals $$4\pi$$.
What I want to clarify is $$\vec \nabla•\frac{\hat r}{r^2}=4\pi\delta^3(r)$$
The divergence is zero everywhere except in r=0, as can be easily verified by using the expression for divergence in spherical coordinates. In order to integrate to ##4\pi##, the divergence therefore must be ##4\pi\delta^{(3)}(\vec r)##.
 

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