Gauss divergence theorem for vectors can be easily explained by mass balance. But I can't think about one example for scalar gauss divergence theorem.

Gauss Divergence Theorem for scalars:

$$\int\limits_\Omega\nabla f(x)~\mathrm d\Omega=\int\limits_\Gamma f(x) \boldsymbol{n}~\mathrm d\Gamma,$$



$\Omega$ is an open set, such that $\Omega \subset \mathbb{R}^2$ (or $\mathbb{R}^3$)

$\Gamma$ is the boundary of $\Omega$ and $\boldsymbol n$ is the normal vector to $\Gamma$.


closed as unclear what you're asking by ACuriousMind, Gert, CuriousOne, Cosmas Zachos, knzhou Jul 14 '16 at 22:47

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  • $\begingroup$ This question (v8) seems like a list question. $\endgroup$ – Qmechanic Jul 14 '16 at 2:58
  • $\begingroup$ @Qmechanic Yes, it can be interpreted as a list question, but can you understand its importance? If yes, how could it be stated without being a list question like? I was looking for just one good example and I got one. $\endgroup$ – Vitor Abella Jul 14 '16 at 3:06

This theorem can be used to prove Archimede's Principle in a region with a non-uniform gravitational field.

The weight of the displaced fluid is $$\vec W=\int_\Omega \rho \vec g(\vec r)~\mathrm d\Omega.$$ Let us consider a body fully immersed. Then the buoyancy force is given by $$\vec B=-\oint_\Gamma p(\vec r)~\mathrm d\vec \Gamma =-\int_\Omega\vec\nabla p~\mathrm d\Omega,$$ where $p(\vec r)$ is the pressure, $d\vec \Gamma$ is the oriented area element and the given theorem is used.

A static fluid satisfy $$\vec\nabla p=\rho\vec g.$$ Plugging this into the integral for $\vec B$ and comparing with the integral for $\vec W$ we see that the buoyancy force equals the weight of the displaced fluid even when the gravitational field is non-uniform.


Proof of the result claimed by the OP:

Consider a vector field $f(x)\vec v$, where $f$ is a scalar field and $\vec v$ is a constant non vanishing vector. By the divergence theorem: $$\oint_\Gamma f\vec v\cdot ~\mathrm d\vec \Gamma=\int_\Omega \vec\nabla\cdot (f\vec v)~\mathrm d\Omega.$$ Since $\vec\nabla (f\vec v)=\vec v\cdot\vec\nabla f$, you get $$\vec v\cdot \oint_\Gamma f~\mathrm d\vec \Gamma=\vec v\cdot\int_\Omega\vec\nabla f~\mathrm d\Omega.$$ Since this holds for any constant vector $\vec v$ we get the result $$\oint_\Gamma f~\mathrm d\vec \Gamma=\int_\Omega\vec\nabla f~\mathrm d\Omega.$$ In fact I have seen people calling this the "gradient theorem".


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