# Integration in polar coordinates

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

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haruspex
Homework Helper
Gold Member
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
Staff Emeritus
Homework Helper
Gold Member
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.

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
Staff Emeritus
Ya in this case the Rhs equals $$4\pi$$.
What I want to clarify is $$\vec \nabla•\frac{\hat r}{r^2}=4\pi\delta^3(r)$$