# Integral of d3p

1. Oct 5, 2013

### Ang Han Wei

I was recently told that the integral for a 3-momentum d^3p = 4∏∫p^2 dp

But I dunno what how is this integtal done.

Any help?

2. Oct 5, 2013

### jfizzix

This integral was done in spherical coordinates
$\int dxdydz = \int r^{2}sin(\theta)dr d\theta d\phi$
if what you're integrating over does not depend on $\theta$ or $\phi$, then you can integrate over those variables giving you an additional factor of $4 \pi$. Thus
$\int dxdydz = 4\pi \int r^{2}d r$.

I expect d^3p is shorthand for dp_{x}dp_{y}dp_{z}.