Relation between an indefinte and definite integral

Question

How is the following relation established? $$ \int e^{-p r^2} d\mathbf{r}=4 \pi\int_0^\infty r^2 e^{-pr^2} dr $$ where $p$ is a real and positive number.

Answer 1

I think that $d\mathbf{r}$ means here the volume integral and that the integral on the left is over all $\mathbb{R}^3$.

If we pass to spherical polars $(r,\theta,\phi)$, and remember the formula for the volume element, then the integral will become $$ \int_{r=0}^{\infty}\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi} e^{-pr^2}\ r^2\ \sin\phi\ dr \ d\theta\ d\phi $$ which evaluates to the RHS.