$\textbf{My understanding of divergence:}$ Consider any vector field $\textbf{u}$, then $\operatorname{div}(u) = \nabla \cdot u$. More conceptually, if I place an arbitrarily small sphere around any point of the vector field $\textbf{u}$, divergence measures the amount of "particles" exiting the sphere, i.e. positive divergence represent a vector field which is "moving faster" as we move to the right. However, how do I interpret $$ \int_U \operatorname{div}(u) \, dx$$ where $U$ is any bounded open subset of $\mathbb{R}^n$.
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You would like to explain divergence theorem (Gauss's law)? – kennytm May 01 '15 at 17:53
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Rather than being the "amount of 'particles' exiting the sphere", the divergence is the amount per unit of volume, i.e. the ratio of quantity of stuff exiting to volume. The integral is then the total amount exiting.
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One definition of the divergence is
$$\nabla \cdot \vec F(\vec r) = \lim_{\Delta V \to 0} \left(\frac{\oint_S \vec F(\vec r') \cdot \hat n' dS'}{\Delta V}\right)$$
where the surface $S$ is the boundary for the volume $\Delta V$, which includes the point $\vec r$. This clearly shows that the divergence is the net outward flux (of $F$) per unit volume at a point.
Mark Viola
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