- #1

Trollfaz

- 140

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$$\triangledown \cdot \frac{\hat{\textbf{r}}}{r^2}=4π\delta^3(\textbf{r})$$

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?