(adsbygoogle = window.adsbygoogle || []).push({}); The Background:

I'm trying to construct a rigorous proof for the divergence theorem, but I'm far from my goal. So far, I have constructed a basic proof, but it is filled with errors, assumptions, non-rigorousness, etc.

I want to make it rigorous; in so doing, I will learn how to construct rigorous proofs.

The Question:

My question is - can I use this line of reasoning to rigorously prove the divergence theorem for all cases? I realize that this is a time-consuming request, but I will greatly appreciate any assistance. I want to finally prove something rigorously, unlike my regular high school "let's prove that [tex]\vec{F} \equiv q\vec{E}[/tex]!" type of proofs.

The "Proof"

Given: Vector function [itex]\bf{F}(x, y, z)[/itex], a closed surface [itex]S[/itex], and an enclosed volume [itex]V[/itex].

We can split up [itex]S[/itex] into many small rectangular parallelipipeds (henceforth ``boxes'') as an approximation. We can now approximate the flux of the vector function over [itex]S[/itex] using these boxes.

[tex]

\Phi \approx \sum \bf{F} \cdot \bf{\hat{n}} \Delta S

[/tex]

We divide and multiply the flux expression by [itex]\Delta V[/itex].

[tex]

\Delta V \left [ \sum \bf{F} \cdot \bf{\hat{n}} \Delta S \right ] = \sum \bf{F} \cdot \bf{\hat{n}} \Delta S

[/tex]

Now we limit the expression as [itex]\Delta[/itex] approaches zero. This means that the width, height, and length of the boxes will approach infinitesimal values. By definition, the bracketed quantity will equal the divergence of [itex]\bf{F}[/itex]. In addition, the area for the flux will approach the infinitesimal value [itex]dS[/itex].

[tex]

\lim_{\Delta \to 0} \Delta V \left [ \sum \bf{F} \cdot \bf{\hat{n}} \Delta S \right ] = \nabla \cdot \bf{F} \,dV = \sum \bf{F} \cdot \bf{\hat{n}} \,dS

[/tex]

If we integrate both sides (one side as a volume, the other as a surface), we get the following expression:

[tex]

\int \!\!\! \int \!\!\! \int_V \nabla \cdot \bf{F} \,dV = \int \!\!\! \int_S \bf{F} \cdot \bf{\hat{n}} \,dS

[/tex]

If [itex]\bf{F}[/itex] is not a continuous function on [itex]V[/itex] and [itex]S[/itex], then the infinitesimal form cannot be integrated. Therefore, [itex]\bf{F}[/itex] must be continuous on [itex]V[/itex] and [itex]S[/itex].

If \[itex]\bf{F}[/itex] is not differentiable at any point on S, then the dot product of [itex]\bf{F}[/itex] and the unit normal vector is indeterminate. Therefore, the flux would be indeterminate. For the same reason, it follows that the spatial derivative of [itex]\bf{F}[/itex] must be continuous on S.

Thus, we have the divergence theorem: If [itex]\bf{F}(x, y, z)[/itex] is a vector function differentiable on closed surface [itex]S[/itex] and continuous on enclosed volume [itex]V[/itex], then

[tex]

\int \!\!\! \int \!\!\! \int_V \nabla \cdot \bf{F} \,dV = \int \!\!\! \int_S \bf{F} \cdot \bf{\hat{n}} \,dS

[/tex]

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Rigorous Divergence Theorem Proof

**Physics Forums | Science Articles, Homework Help, Discussion**