# Integral of d3p

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?

$\int dxdydz = \int r^{2}sin(\theta)dr d\theta d\phi$
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
$\int dxdydz = 4\pi \int r^{2}d r$.