Integration in polar coordinates

[Apashanka]

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 $$\vec \nabla•\frac{\hat r}{r^2}=4\pi\delta^3(r)??$$

 

[haruspex]

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.

 

[Orodruin]

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.

 

[Apashanka]

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 $$\vec \nabla•\frac{\hat r}{r^2}=4\pi\delta^3(r)$$

 

[Orodruin]

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 $4\pi\delta^{(3)}(\vec r)$.