I'd like to understand why $ \int{\rho}dV = \frac{1}{c} \int{j^0}dV = \frac{1}{c} \int{j^i}dS_i $ (the second equality), where
$j^i = \rho \frac{dx^i}{dt} $ is the current density 4-vector
$\mathbf{j} = \rho \mathbf{v}$ is the current density 3-vector
$ j^i = (c\rho, \mathbf{j}) $
$\rho$ is the charge density
$dS_i$ is the element $-dx-dy-dz+cdt$
Are you able to explain me this equality?
Thank you very much!