# Calculating the divergence of a vector field using non-cubic volumes

The divergence of a vector field is defined is formally defined as:

$\operatorname{div}\,\mathbf{F}(p) = \lim_{V \rightarrow \{p\}} \iint_{S(V)} {\mathbf{F}\cdot\mathbf{n} \over |V| } \; dS$

http://en.wikipedia.org/wiki/Divergence#Definition_of_divergence

Then by carrying out the integration over a box and taking the limit as the size of the box approach zero, we reach the well-known formula for calculating the divergence:

$\operatorname{div}\,\mathbf{F} = \nabla\cdot\mathbf{F} =\frac{\partial U}{\partial x} +\frac{\partial V}{\partial y} +\frac{\partial W}{\partial z }$

Now my question is, is there any formal proof that whatever volume we use, we get the same formula?

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As far as I know, there is only a really good elementary proof for shapes like cubes, spheres, or polyhedra. Proving this for general regions requires a bit more work.

The simplest way to prove the statement you are asking in the general case is to first prove Gauss's Divergence Theorem. Then for any volume $V$ (with piecewise-smooth boundary), we have $$\int\!\!\!\!\int_{S(V)} (\textbf{F}\cdot\textbf{n})\,dS \;=\; \int\!\!\!\!\int\!\!\!\!\int_V (\mathrm{div}\;\textbf{F})\,dV$$ As $V$ shrinks toward the point $p$, the integral on the right approaches $\mathrm{div}\;\textbf{F}(p)$ multiplied by the volume of $V$. We can make this precise using the following lemma:

Lemma. Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be a continuous function, and let $p\in\mathbb{R}^n$. Then for every $\epsilon>0$, there exists a $\delta>0$ with the following property: if $V$ is a region in $\mathbb{R}^n$ with positive volume and $V$ lies entirely within a distance $\delta$ of the point $p$, then $$\left|f(p) - \frac{1}{\mathrm{vol}(V)}\int\!\!\!\!\int\!\!\!\!\int_V f\,dV\right| < \epsilon.$$ Proof: Let $\epsilon>0$. Since $f$ is continuous, there exists a $\delta > 0$ so that $$\|x-p\|<\delta \qquad\Rightarrow\qquad |f(x)-f(p)|<\epsilon$$ for all $x\in\mathbb{R}^n$. Let $V$ be a region in $\mathbb{R}^n$, and suppose that $V$ lies entirely within $\delta$ of the point $p$. Then $f(p) - \epsilon < f(x) < f(p)+\epsilon$ for all $x\in V$, so $$[f(p) - \epsilon]\,\mathrm{vol}(V) \;\leq\; \int\!\!\!\!\int\!\!\!\!\int_V f\,dV \;\leq\; [f(p) + \epsilon]\,\mathrm{vol}(V)$$ and the result follows.$\qquad\square$

Using this theorem and Gauss's Divergence Theorem, we immediately get:

Theorem. Let $\textbf{F}\colon\mathbb{R}^n\to\mathbb{R}^n$ be a continuously differentiable vector field, and let $p\in\mathbb{R}^n$. Then for every $\epsilon>0$, there exists a $\delta>0$ with the following property: if $V$ is a region in $\mathbb{R}^n$ with positive volume and a piecewise-smooth boundary and $V$ lies entirely within a distance $\delta$ of the point $p$, then $$\left|\mathrm{div}\;\textbf{F}(p) - \frac{1}{\mathrm{vol}(V)}\int\!\!\!\!\int_{S(V)} (\textbf{F}\cdot\textbf{n})\,dS\right| < \epsilon.\tag*{\square}$$

It is possible to restate this theorem using sequences and limits. For the following theorem, if $V$ is a region in $\mathbb{R}^n$, define $$\mathrm{rad}(V,p) \;=\; \sup\{\|x-p\| \mid x\in V\}.$$ That is, $\mathrm{rad}(V,p)$ is the radius of the smallest sphere centered at $p$ that contains $V$. Then

Theorem. Let $\textbf{F}\colon\mathbb{R}^n\to\mathbb{R}^n$ be a continuously differentiable vector field, and let $p\in\mathbb{R}^n$. Let $V_1,V_2,V_3,\ldots$ be a sequence of regions in $\mathbb{R}^n$, all with positive volume and piecewise-smooth boundary, and suppose that $$\lim_{n\to\infty} \mathrm{rad}(V,p) \;=\; 0.$$ Then $$\lim_{n\to\infty} \frac{1}{\mathrm{vol}(V_n)}\int\!\!\!\!\int_{S(V_n)} (\textbf{F}\cdot\textbf{n})\,dS \;=\; \mathrm{div}\;\textbf{F}(p).\tag*{\square}$$

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Thanks for the answer, and for the time you spent writing it :-) – Rafid May 22 '12 at 5:26