Let the vector field $\vec{U}$ of $\mathbb{R}^3$ be interpreted, as in the remark of the parent entry, as the velocity field of a stationary flow of a liquid. Then the flux$$ \oint_a\vec{U}\cdot d\vec{a}$$ of $\vec{U}$ through a closed surface $a$ expresses how much more liquid per time-unit it comes from inside of $a$ to outside than contrarily. Since for a usual incompressible liquid, the outwards flow and the inwards flow are equal, we must think in the case that the flux differs from 0 either that the flowing liquid is suitably compressible or that there are inside the surface some sources creating liquid and sinks annihilating liquid. Ordinarily, one uses the latter idea. Both the sources and the sinks may be called sources, when the sinks are negative sources. The flux of the vector $\vec{U}$ through $a$ is called the productivity or the strength of the sources inside $a$ .

For example, the sources and sinks of an electric field ($\vec{E}$ ) are the locations containing positive and negative charges, respectively. The gravitational field has only sinks, which are the locations containing mass.

The expression$$ \frac{1}{\Delta v}\oint_{\partial\Delta v}\vec{U}\cdot d\vec{a},$$ where $\Delta v$ means a region in the vector field and also its volume, is the productivity of the sources in $\Delta v$ per a volume-unit. When we let $\Delta v$ to shrink towards a point $P$ in it, to an infinitesimal volume-element $dv$ , we get the limiting value

(1)

The Gauss's theorem$$ \int_v\nabla\cdot\vec{U}\,dv \;=\; \oint_a\vec{U}\cdot d\vec{a}$$ applied to $dv$ says that

(2)

(3)

(4)