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Divergence theorem proof (part 5)

Home stretch. Proving the Type I part. Created by Sal Khan.

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  • aqualine ultimate style avatar for user Aaron Williams
    Sal's "proof" of the divergence theorem is straightforward and fairly simple owing to the shape of the surface over which he was integrating, but does anyone know of a way to apply this method to more general surfaces? My intuition suggests that any closed region in space can be broken into many Type !, II, and III regions(which type we choose depends on which component of F we choose to focus on), and from there one can apply sal's method to each individual region. We then add all the regions to get the final result. This approach may not seem very rigorous, but at least it comes much closer to generalizing the divergence theorem.
    (5 votes)
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    • spunky sam blue style avatar for user Laserbeamcool
      Yes, that's how you do it. so we take the sum of all the little surfaces/regions (depending on whether you're looking at the left hand side or the right of the divergence theorem) and the flux inside the region cancels out. Similar thing can be said about Sal's stokes theorem proof, he said it was a special case where z=z(x,y) and the reason why the inside stuff cancel out is because of the positive and negative orientations.
      (6 votes)
  • leaf orange style avatar for user Mario S
    Most people's mathematical journey ends here. It was a fun adventure and I hope you (person who stops studying mathematics here) enjoy your life to the fullest.

    BTW, how do I solve the following exercise (I found this exercise on a multivariable calculus book)?

    . Suppose S is a surface with boundary C and F is a vector function such that ∇ ×F is
    tangent to S at each point of S. Prove that ∫ ⋅ =
    F dr 0 .
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

Now we can work on the triple integral part of our problem or our proof, this right over here. And so I can rewrite that. So this is the triple integral over our region, which we're assuming is a type I region, of the partial of R with respect to Z, which we can write like this. It doesn't matter. Partial of r with respect to do z. And I'll dV. And we can rewrite this as-- we can assume we're going to integrate with respect to Z first. So I'm going to integrate with respect to Z. Let me do that in another color. I'm going integrate with respect to Z first. The lower bound on Z in our type I region, the lower bound is f1. The upper bound is f2. So we're going to integrate from f1 of x, y to f2 of x, y. And I'm going to integrate the partial of R with respect to Z. So let me do that in that same yellow color-- partial of R with respect to Z. And then I have dZ, and then I'll have to integrate with respect to y and x or with respect to x and y. So it's dx dy or dy dz. I can just write that as dA. So what you could think of it-- we can evaluate the yellow part. And then we're just going to take the double integral over the x, y domain. So this is just going to be over the x, y domain. Let me put some brackets here just to make it clear what we're going to do. So all we're doing is we're integrating with respect to Z first, and we have the bounds there. Well, this is pretty straightforward. This is all going to be equal to-- I'll write the outside first-- the double integral over the domain. And I have the dA right over here. Actually, let me give myself some real estate-- dA. Well, what's the antiderivative of this? This is just R, and this is just R, or R of x, y, z evaluated when Z is f1-- or when Z is f2. And from that, we evaluate when Z is f1. So this is just going to be R of x, y and z, and we evaluate when Z is equal to that. And from that, we subtract when Z is equal to that. So that's going to be equal to-- so R of x, y z evaluated when Z is equal to that is R of x, y, f2 of x, y. And from that, we need to subtract R when Z is this-- minus R of x, y f1 of x, y, and then make sure that we got our parentheses. Now, this is exactly what we saw in the last video. It is exactly that, which shows that this is exactly this. So when we assumed it was a type I region, we got that this is exactly equal to this. You do the exact same argument with the type II region to show that this is equal to this, type III region to show this is equal to that, and you have your divergence theorem proved. And we can consider ourselves done.