Integral form

Differential form

(1)	$\begin{array}{l}\displaystyle \frac{d}{dt} \iiint_\Omega \rho \, dV = \frac{dm_\Omega}{dt}\end{array}$

(2)	$\begin{array}{l}\displaystyle \frac{\partial \rho}{\partial t} + \nabla (\rho \, {\bf u}) = \frac{dm (t, {\bf r})}{dt}\end{array}$

where

$\begin{array}{l}t\end{array}$	time
$\begin{array}{l}{\bf r }\end{array}$	position vector
$\begin{array}{l}\Omega\end{array}$	space volume (could be finite or infinite)
$\begin{array}{l}\rho(t, {\bf r})\end{array}$	continuum body spatial density distribution
$\begin{array}{l}{\bf u}(t, {\bf r)\end{array}$	continuum body spatial velocity distribution
$\begin{array}{l}\displaystyle \frac{dm_\Omega}{dt}\end{array}$	mass generation rate with the space volume $\begin{array}{l}\Omega\end{array}$
$\begin{array}{l}\displaystyle \frac{dm(t, {\bf r})}{dt}\end{array}$	mass generation rate at a given point in space $\begin{array}{l}{\bf r }\end{array}$

For the specific case of stationary process when density is not explicitly dependent on time:

(3)	$\begin{array}{l}\displaystyle \frac{\partial \rho}{\partial t} = 0 \rightarrow \nabla (\rho \, {\bf u}) = 0\end{array}$

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See also