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This is about continuity equation. What does the last integral mean? $$\frac{\mathrm{d}Q_V}{\mathrm{d}t}=\iiint_V \mathrm{d}^3x \,\frac{\partial\rho}{\partial t}=-\iiint_V\! \mathrm{d}^3x\,\operatorname{div}\,\mathbf{j}=-\iint\limits_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf j\;\cdot\mathbf n\,{d}S\,,$$

EDIT: And what does $\partial V$ mean? Why not S like surface?

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$\partial V$ is just the standard notation for boundary of the region $V$. –  MBN Feb 26 '11 at 19:09
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BTW it is a very sensible definition: the boundary is a derivative of the volume, in some proper sense. Note that even dimensionally, the surface has the same dimension as the volume divided by $x$, just like the derivative. In homology theory, $\partial$ is the boundary operator. It squares to zero: $\partial^2=0$. It means the fact that the boundary is a closed manifold - the boundary of a boundary is nothing. –  Luboš Motl Feb 26 '11 at 20:34

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up vote 8 down vote accepted

The last integral is a surface integral. It is the fluid (current) flux crossing the volume surface integrated over the whole surface. It is how much charge is leaving the volume per second. ($\partial V$ means boundary of V and you sum all local $j_\bot dS$.)

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dV is a slice of a volume right? But why is partial V the boundary? –  kame Feb 26 '11 at 19:08
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It is not a partial derivative; it is a mathematical notation that reads "boundary of V". In other words, it is $S$. –  Vladimir Kalitvianski Feb 26 '11 at 19:09

Note that the range of integration has changed from $V$ meaning over the volume to $\partial{V}$ and that the variable of integration has changed from $\text{d}^3x$ to $\text{d}S$.

It means an integral over the surface that bounds the volume.

The closed curve over the integration signs implies that the surface must be closed.


This expression is an application of what physicist tend to call Gauss's Law, and mathematicians tend to know as the divergence theorem.

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