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I was reading Introduction to Electrodynamics by Griffiths and I wanted to check if I understood Gauss' Divergence Theorem correctly. The theorem states:

$$\int \int \int_V \vec{\nabla} \cdot \vec{C} \hspace{0.1cm} dV = \oint\oint_S \vec{C} \cdot d\vec{a}$$

where the LHS is integrating over a volume and the RHS is integrating over a closed surface.

Here is my analogy. Suppose you have some pipes, like this:

enter image description here

Imagine that all pipes are outputting water at an equal rate.

Now, I say that the LHS of the theorem says you can calculate the flux of water by adding up all the divergences$^{\dagger}$ across the volume that the water creates. In contrast the RHS says that you can calculate the total flux by adding the individual fluxes at the boundaries (indicated in red).

Is a correct way of thinking about the theorem?

$^{\dagger}$ Where the divergence is a measure of the spread of a vector (of a vector function) at a given point.

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    $\begingroup$ I don't believe this analogy holds up well. Your 2D-surfaces aren't closed, for one; your dimensionality is off. This is nearly akin to Stoke's Theorem, but not quite. $\endgroup$ – BMS Apr 4 '14 at 17:47
  • $\begingroup$ @BMS When I say boundaries I mean the area forms the exit hole; does that make any difference? $\endgroup$ – turnip Apr 4 '14 at 17:56
  • $\begingroup$ The boundary must completely cover the hose exit, entrance and sides. A boundary of something can't have a boundary itself. $\endgroup$ – George G Apr 4 '14 at 18:58
  • $\begingroup$ I don't understand your picture at all, but your words sound like they might be correct, but they are unclear. For example, I don't know what "the volume that the water creates" means, and that misunderstanding might change everything. $\endgroup$ – garyp Apr 4 '14 at 18:59
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The boundaries in your example aren't of the correct dimensionality. Actually, the geometry of what you've chosen would be convenient for Stokes' theorem, but that involves curls and line integrals instead.

Here is a similar but more accurate analogy. Imagine a free cylindrical pipe you can hold in your hand. The pipe has two open ends, and water is gushing out of both ends. (How is this possible? I don't know: apparently water is being created out of nothing. But let's ignore this impossibility, since it will be convenient.)

Now you want to add up all the infinitesimal amounts of divergences $\iiint \vec{\nabla}\cdot\vec{C}\,dV$ for your water flow vector field $\vec{C}$. The straightforward way to do this is to actually perform the volume integral. I'm imagining some sort of Dirac-delta sources as contributing to this volume integral, but as long as you imagine water appearing out of nowhere, this volume integral will be non-zero since "sources" of such fields have positive divergences.

Now you want to use the divergence theorem instead. This states that, instead of evaluating the volume integral above, you evaluate the flux through closed surface integral $$\iint_\text{closed} \vec{C}\cdot d\vec{A}.$$ This should make intuitive sense, since the water that comes out of the magical "source" inside the pipe must flow out. The area you're integrating over is the imaginary closed area that surrounds the entire pipe. So it takes the shape of the closed cylinder, which has a cylindrical tube and two circular end caps.

Almost there. The surface integral around the curved portion would be zero since there is no flow here; no water's coming out. The only contribution would come from the the integral over the (imaginary) circular end caps.

That's it. Now contrast this with your own analogy. You were using the edges of the circular end caps as being relevant for the RHS of the divergence theorem. Instead, it's the entire imaginary end caps that play the important role. The cylindrical surface is also important, but that contribution happened to be zero.

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  • $\begingroup$ Ah so, in your example the "surface" that we integrate over in the RHS is the imaginary cross sectional area of the exit holes? $\endgroup$ – turnip Apr 4 '14 at 20:00
  • $\begingroup$ The exit holes are only part of the area you integrate over. You need to integrate over a closed surface, meaning a surface can be trapped inside of, and you can only exit by going through that surface. If you just had the exit holes, you could go through the sides of the cylinder. So in this case, you need to integrate over the sides of the cylinder as well. $\endgroup$ – BMS Apr 4 '14 at 20:13
  • $\begingroup$ But I thought you said the integral around the curved portion would be zero? $\endgroup$ – turnip Apr 4 '14 at 20:25
  • $\begingroup$ It is zero, but for the divergence theorem requires us to choose a closed surface, which necessitated the curved surface here. It just so happened that its contribution was zero in this case. $\endgroup$ – BMS Apr 4 '14 at 21:09
  • $\begingroup$ Ok, thanks for clearing that up. I guess my analogy was halfway there; by boundaries I did mean the cross sectional area but I now I see what was missing. Thanks! $\endgroup$ – turnip Apr 4 '14 at 21:16

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