# Help me understand this proof, so I can present it

1. Feb 16, 2005

### semidevil

so I googled up a proof of te divergence theorem, and plan to explain it....I want to make sure I am understanding the whole proof before I make a fool of myself.

here is a link to the proof I found. It is section 7.2

http://www.enm.bris.ac.uk/admin/courses/EMa2/Lecture Notes 04-05/vcalc7.pdf

ok, so basically, I"m gonna start out and give the formula for divergence, and I will try to prove it.

-so let F = Fi + Fj + Fk, and I"m gonna divde this solid into N small cubes, of volume del(V). so the volume of each cube is devl(V)

-so if we can show that it holds for each little cube, then the proof is done.

because: the volume is equal to the sum, and (1)each little cube will be either inside, or (2) each will be part of the surface.

(in the pdf, it states that in (1), the flux integeral is exactly canceled by the flux integeral of the neighboring cub. I dont know what that means actually...can someone simplify that meaning for me....because it seems important, and if I dont know what it means, then I dont know how to explain it to the class. And in (2), all these contributions sum to the toal flux of intergral of S. again, dunno what that means....)

-so I consider a cube. I will label each face S1, 2, 3, 4, 5, S6.

so for S1, ndS = idydz, and double integeral of F * ndA = integeral from (zi-del(z) to zi +del(z) and integeral (yi-del(y) + yi+del(y) of F1(xi + del x, y, z)dydz.

now, first, I dont know what this integeral is saying. what are we computing here? and where is all the liimits coming from? it's been a while for me since calc class, so I really forgot what n and the del's are.
and then, we take S2 and do the same, w/ the corresponding limits, and different function.

so first of all, where are the functions coming from?

anyways, once we get S1 and S2, we add them up, and by FTC, we get triple integeral partial F/partial x, dV.

and we do the same for S3 + S4, and S5 + S6.

and then we add them all up and get the divergence theorem.

so my final question is, when we did the integeration, what did we integerate exactly...and why when we added them all up, did it prove the divergence theorem.

I really wan't to undersand this, so if someone can help..that would be great.

thanx

2. Feb 16, 2005

### learningphysics

I'll try to help. We have this object A... we divide it up into a bunch of cubes... The first thing to understand is that the flux through the surface of A, is the sum of the fluxes through all the little cubes.

The proof doesn't exactly make this clear. That picture right under where it says "Proof of the divergence theorem"... where it has one cube on top of the other... Suppose we want the total flux through that 2 cuboid object.

The flux through the bottom surface of the top cube, is cancelled by the flux through the top surface of the bottom cube. This happens because the normal vector is in the opposite direction. The force vector is the same and dA is the same.

So the total flux of the 2 cube object is the flux over the surface area (5 sides of the one cube and 5 sides of the other cube).

So when two cubes have a surface in common (they are neighbors)... their fluxes through that side cancel each other out... In a similar manner the fluxes through all the cubes inside the object A cancel each other out.... until all that's left is the flux through the surface....

Think of it like this... when adding up the fluxes of all the little cubes, you're adding up the fluxes through the square surfaces of the cubes... for each cubical surface that's within the object, two cubes that have that surface. So those two fluxes cancel. The only cubical surfaces that are left are those that are on the surface (outer part) of the object.

I hope this helps... Let me know, and I'll try to help with the rest.

3. Feb 16, 2005

### Galileo

Hmm... I may sound picky, but no finite union of cubes together will be equal to, say, a sphere. So the 'sum of the cubes' is in general not equal to the interior of a closed surface. (The idea is to take the limit ofcourse).

(1)Consider a side of a small cube. This side belongs to exactly two cubes. So the flux through a side of a cube is the same as the flux through this side when it belongs to the neighbouring cube, but with opposite sign. Hence, they cancel.

(2)Since all the internal sides of the cubes cancel out against each other (a cube that is surrounded by 6 other cubes doesn't add to the total flux, because it all cancels against the flux through the sides belonging to the 6 neighbouring cubes), the only flux that matters is the one going through the cubes at the surface, since they don't have neighbouring cubes all around them.

Haven't read the rest of the proof. Instead, try to prove it in the following way.
Assume the region is of the following type:
$$V=\{(x,y,z)|(x,y) \in D, u_1(x,y) \leq z \leq u_2(x,y)\}$$
where D is the projection of V onto the xy-plane.
Then show that:
$$\iint_S F_1\vec i \cdot \vec n dS = \iiint_V \frac{\partial F_1}{\partial x}dV$$
$$\iint_S F_2\vec j \cdot \vec n dS = \iiint_V \frac{\partial F_2}{\partial y}dV$$
$$\iint_S F_3\vec k \cdot \vec n dS = \iiint_V \frac{\partial F_3}{\partial z}dV$$
(Does require calculus dealing with integration over general regions).
Note that the DT is proven for this class of regions if the above is shown.

To complete the proof, show that the theorem holds for a finite union of V type regions.

The proof in the pdf is more insightful intuitively, but the above is a little more rigorous. Try doing both.

4. Feb 17, 2005

### semidevil

ok, thanx guys...I Understand the flux part.....now, where it says ndS = idydyz, and it does that double integeral for S1......what does that mean actually?

what is it?

and why did we consider a cuboid w/ length 2del(x), 2del(y), and 2del(z)?

what does 2(del)x mean?

Last edited: Feb 17, 2005
5. Feb 17, 2005

### learningphysics

They're getting the flux through the surface S1. An infintesimal area on S1, is dS=dxdy. The unit normal to the surface S1 is the vector i.

Over an infintesimal area on S1, the flux through the surface is F.ndS=F.idS=F.idxdy (I'm using . for dot product... F and i are vectors. i've still to learn latex)

So the total area on S1 is that integral of F.idxdy over the surface S1, which becomes a double integral, as you have to integrate over all the x and all the y, on that surface.

In the proof they seem to sometimes use dS, and sometimes dA but they're the same thing.

It's just arbitrary... the idea is to eventually let the dimensions of the cube approach 0(in the limit). They could have made the dimensions, del(x), 5(del)y, 8(del)z... it doesn't matter. Only advantage is to make the math look neater.

Last edited: Feb 17, 2005
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