# How do I prove that $\text{div }\textbf{F}=\nabla\cdot\textbf{F}$ using the formal definition of divergence in 3D?

The formal definition of divergence is, $$\text{div }\color{#08F}{\textbf{F}} = \lim_{\color{#F00}{|V|}\to0} \frac{1}{\color{#F00}{|V|}}\iint_{\partial\color{#F00}{V}}\color{#08F}{\textbf{F}}\cdot\color{#080}{\hat{\mathrm n}\,}\color{#F00}{\mathrm{d}\Sigma}$$

Where $\color{#08F}{\textbf{F}}$ is a 3D vector field, $\color{#F00}{V}$ is a 3D region containing a specific point $\color{#080}{(x,y,z)}$ and $\color{#F00}{|V|}$ is its volume. $\partial \color{#F00}{V}$ is the surface of $\color{#F00}{V}$.

How to show that the $\text{div }\color{#08F}{\textbf{F}}$ is independent of the region $\color{#F00}{V}$?

It was easy to prove, $\text{div }\color{#08F}{\textbf{F}}=\frac{\partial \color{#08F}{P}}{\partial x}+\frac{\partial \color{#08F}{Q}}{\partial y}+\frac{\partial \color{#08F}{R}}{\partial z}$ using a spherical or a cubical region. But not for any wierd region.(Where $\color{#08F}{\textbf{F}}=\begin{bmatrix} \color{#08F}{P} \\ \color{#08F}{Q} \\ \color{#08F}{R}\end{bmatrix}$)

If I show the independence from the region, I am done. Also if I first show that $\text{div }\textbf{F}=\nabla\cdot\textbf{F}$, then its trivial that it is independent from the region.

In any case your region $V$ is variable. If the limit exist for "all" regions then will exits for "nice" regions and it will be equal.
• @DevanshSehta, OK, the concrete procedure is irrelevant. The important fact is that considering only "nice" regions is similar to taking a subsequence in a convergent sequence. Technically, your limit is the limit of a net (en.wikipedia.org/wiki/Net_(mathematics)) where the elements of the directed set are the regions $V$ (again: $V$ is variable and "independent of $V$" does not makes sense) ordered by reverse inclusion. Considering only "nice" $V$ is taking a subnet. – Martín-Blas Pérez Pinilla Mar 7 '18 at 13:41