Dirac delta function in 2d polar coordinates

In 3 d spherical coordinates we know that
$$\triangledown \cdot \frac{\hat{\textbf{r}}}{r^2}=4π\delta^3(\textbf{r})$$
Integration over all## R^3## is 4π
So when we remove the third dimensions and enter 2d polar coordinates then
$$\triangledown \cdot \frac{\hat{\textbf{r}}}{r}=2π\delta^2(\textbf{r})$$
So the integral over ##R^2## is 2π?

First of all, note that your expressions are not written in any particular coordinates (apart from the interpretation of ##r## as a coordinate function and ##\hat{\boldsymbol r}## as the corresponding unit vector. In particular, the delta functions are not the product of the coordinate delta functions.

Apart from that: Yes. You can easily check the result by use of the divergence theorem and integration over a sphere/circle. In fact, that is one way to find the constant factors in front.

I'm quoting the first expression from Introductions to Electrodynamics (David J. Griffiths) Section 1.5.3, Eqn 1.99
$$\triangle \cdot \frac{\hat{r}}{r^2}=4π\delta^3(r)$$

Trollfaz said:

I'm quoting the first expression from Introductions to Electrodynamics (David J. Griffiths) Section 1.5.3, Eqn 1.99
$$\triangle \cdot \frac{\hat{r}}{r^2}=4π\delta^3(r)$$

You are missing the point (and hopefully misquoting Griffiths as the expression as written here makes no sense). The expression in your OP was correct. My comment was that it was not really something dependent on using spherical coordinates.

right, integral in ##R^3-B_\epsilon(0)##, and use divergence thm

if your are curious about the equation everypoint (specially point 0), that is Dirac delta, try to understand it as a functional like ##<\delta, \varphi>##, that is ## \int \limits_{R^3} \delta \varphi ##