# 3.4 Gauss' theorem

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A simple exposition of Gauss's theorem or the divergence theorem.

## Gauss' theorem

Consider the following volume enclosed by a surface we will call $S$ .

Now we will embed $S$ in a vector field:

We will cut the the object into two volumes that are enclosed by surfaces we will call ${S}_{1}$ and ${S}_{2}$ .

Again we embed it in the same vectorfield. It is clear that flux through ${S}_{1}$ + ${S}_{2}$ is equal to flux through $S\text{.}$ This is because the flux through one side of the plane is exactly opposite to theflux through the other side of the plane: So we see that ${\oint }_{S}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{a}={\oint }_{{S}_{1}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{1}}+{\oint }_{{S}_{2}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{2}}\text{.}$ We could subdivide the surface as much as we want and so for $n$ subdivisions the integral becomes:

${\oint }_{S}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{a}=\sum _{i=1}^{n}{\oint }_{{S}_{i}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{i}}\text{.}$ What is ${\oint }_{{S}_{i}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{i}}$ .? We can subdivide the volume into a bunch of littlecubes:

To first order (which is all that matters since we will take the limit of a smallvolume) the field at a point at the bottom of the box is ${F}_{z}+\frac{\Delta x}{2}\frac{\partial {F}_{z}}{\partial x}+\frac{\Delta y}{2}\frac{\partial {F}_{z}}{\partial y}$ where we have assumed the middle of the bottom of the box is the point $\left(x+\frac{\Delta x}{2},y+\frac{\Delta y}{2},z\right)$ . Through the top of the box $\left(x+\frac{\Delta x}{2},y+\frac{\Delta y}{2},z+\Delta z\right)$ you get ${F}_{z}+\frac{\Delta x}{2}\frac{\partial {F}_{z}}{\partial x}+\frac{\Delta y}{2}\frac{\partial {F}_{z}}{\partial y}+\Delta z\frac{\partial {F}_{z}}{\partial z}$ Through the top and bottom surfaces you get Flux Top - Flux bottom 

Which is $\Delta x\Delta y\Delta z\frac{\partial {F}_{z}}{\partial z}=\Delta V\frac{\partial {F}_{z}}{\partial z}$

Likewise you get the same result in the other dimensionsHence ${\oint }_{{S}_{i}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{i}}=\Delta {V}_{i}\left[\frac{\partial {F}_{x}}{\partial x}+\frac{\partial {F}_{y}}{\partial y}+\frac{\partial {F}_{z}}{\partial z}\right]$

or ${\oint }_{{S}_{i}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{i}}=\stackrel{⃗}{\nabla }\cdot \stackrel{⃗}{F}\Delta {V}_{i}$ $\begin{array}{c}{\oint }_{S}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{a}=\sum _{i=1}^{n}{\oint }_{{S}_{i}}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{{a}_{i}}\\ =\sum _{i=1}^{n}\stackrel{⃗}{\nabla }\cdot \stackrel{⃗}{F}\Delta {V}_{i}\end{array}$

So in the limit that $\Delta {V}_{i}\to 0$ and $n\to \infty$ ${\oint }_{S}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{a}={\oint }_{V}\stackrel{⃗}{\nabla }\cdot \stackrel{⃗}{F}dV$

This result is intimately connected to the fundamental definition of the divergence which is $\stackrel{⃗}{\nabla }\cdot \stackrel{⃗}{F}\equiv \underset{V\to 0}{{lim}}\frac{1}{V}{\oint }_{S}\stackrel{⃗}{F}\cdot d\stackrel{⃗}{a}$ where the integral is taken over the surface enclosing the volume $V$ . The divergence is the flux out of a volume, per unit volume, in the limit ofan infinitely small volume. By our proof of Gauss' theorem, we have shown that the del operator acting on a vector field captures this definition.

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