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Divergence theorem proof (part 2)

Breaking up the surface integral. Created by Sal Khan.

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  • spunky sam blue style avatar for user Ethan Dlugie
    At , I understand why Sal crosses out the third integral for his situation (the cylinder). But isn't that taking out some generalization? What if the sides of the cylinder bulged out? Sort of like a barrel. The normal vectors on the third surface would not be orthogonal to the k vectors, right?
    (9 votes)
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    • male robot donald style avatar for user John
      Actually, if the cylinder bulged out like a barrel it would still be a type 1, 2 and 3 region. The (x,y) domain would be a slightly larger circle to represent the maximum diameter of the barrel. S1 would be the entire lower half of the barrel and S2 would be the entire upper half of the barrel. As S1 and S2 meet at the middle of the barrel, there would be no S3 to consider.
      (5 votes)
  • male robot hal style avatar for user Aiman
    Why is K dot N going to be zero for the third integral at ?
    (2 votes)
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  • blobby green style avatar for user max
    So if "flux" would be like mass flow per unit area (kg/s*m^2), then the "flux across the boundary" is the change in momentum (kg*m/s)?
    (1 vote)
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    • aqualine ultimate style avatar for user Aaron Williams
      If your aim is to calculate the rate at which mass flows through a surface, then the vector function along the surface is density times velocity. Its units are( kg/(s*m^2). Along each infinitesimal surface area, you multiply a component of the vector function in the direction of the normal vector by the area(with units m^2) to get kg/s. When you add all the contributions to the flux, you end up with the rate at which mass flows(kg/s), so the flux across the boundary is not a change in momentum.
      (2 votes)

Video transcript

I think we're now ready to get into the meat of the proof. In the last video, we said if we can just prove that each of these parts are equal to each other, we essentially have proven that that is equal to that. Because here in yellow is another way of writing the flux across the surface and here in green is another way of writing the triple integral over our region of the divergence of f. And what I'm going to do in this video and probably in the next video is prove that these two are equivalent to each other. And I'm going to prove it using the fact that our original region is a simple solid region, in particular is a Type 1 region. And then that's essentially going to be it because you can use the exact same argument and the fact that it is a Type 2 region to prove this, and the exact same argument and the fact that it is a Type 3 region to prove that. So I'm going to assume it's a Type 1, which I can. It's a Type 1, 2, and 3 region. So given the fact that it's Type 1, I'm now going to prove this relationship right over here. And then I'll leave it up to you to do the exact same argument with a Type 2 and a Type 3 region. So let's get going. So Type 1 region, just to remind ourselves, a Type 1 region is a region that is equal to the set of all x, y's and z's such that the xy pairs are a member of a domain in the xy plane, and z is bounded by two functions. z's lower bound is f1 of x and y. And that's going to be less than or equal to z. And z's upper bound, we can call it f2 of xy. And then let me close the set notation right over here. And let me just draw a general version of a Type 1 region. So let me draw my x, y, and z-axes. So this is my z-axis. This is my x-axis. And there is my y-axis. And so we might have a region D. So our region, I'll draw it as a little circle right over here. This is our region D. And for any xy in our region D, you can evaluate the function f. You can figure out an f1, I should say. So this might define an f1, which we can kind of imagine as a surface or a little thing that's at the bottom of a cylinder if you want. So every xy there, when you evaluate or when you figure out what the corresponding f1 of those points in this domain would be, you might get a surface that looks something like this. It doesn't have to be flat, but hopefully this gives the idea. It doesn't have to be completely flat. It can be curved or whatever else. But this just shows you that every xy, when you evaluate it right over here, it gets associated with a point, this lower bound surface right over here. And I'll draw a dotted line to show that it's only for the xy's in this domain. And then we have an upper bound surface that might be up here. Give me any xy. When I evaluate f2, I get this surface up here. And once again, they don't have to look the same. This could be like a dome or it could be slanted or who knows what it might be. But this will give you the general idea. And then z fills up the region. Remember, the region isn't just the surface of the figure, it's the entire volume inside of it. So when z varies between that surface and that surface, for any given xy in our domain, we fill up the entire region. So we can fill up this entire region. And this is the way I drew it. It looks like a cylinder, but it doesn't have to be a cylinder like this. And these two surfaces might touch each other, in which case there would be no side of the cylinder. They could be lumpier than this. They might be inclined in some way. But hopefully, this is a good generalization of a Type 1 region. Now, a Type 1 region, you can kind of think of it, it can be broken up into three parts. It can be broken up into surface-- or the surfaces of a Type 1 region, I should say, can be broken up into three parts. Let's call that surface one. Let's call this right over here surface two, the top of the cylinder or whatever kind of lumpy top it might be. And let's call the side, if these two surfaces don't touch each other, let's call that surface three. There might not necessarily even be a surface three if these two touch each other as in the case of the sphere. But let's just assume that there actually is a surface three. So if we're evaluating the surface integral-- we'll think about this triple integral in a second. But let's think about how we can rewrite this surface integral right over here. This entire surface is S1 plus S2 plus S3. So we can essentially break this up into three separate surface integrals. So let's do that. Just remember we're just focusing on this part right over here. So the surface integral of R times k dot n, the dot product of k and n, ds can be rewritten as the-- let me write it this way-- can be rewritten as the surface integral over S2 of R times k dot n ds, plus the surface integral-- I'm just breaking up the surface here-- plus the surface integral over S1 of R times k dot n ds plus the surface integral over surface three of the same thing, R times k dot n ds. Now, the way I've drawn it, and this is actually the case, S3 is if those surfaces don't touch each other. And for Type 1 situation right over here, the normal vector at any given point on the side of the cylinder for this Type 1 region-- if there is this in between region. There always isn't. In a sphere, there wouldn't be this surface, in which case this would be 0. But if there is this surface in a Type 1 region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component. The normal vector will always be pointing flat out. It will only have an i and j component. So if you take this normal vector right over here does not have a k component and you're dotting it with a k vector, then the dot product of two things that are orthogonal, the k vector goes like that, you're going to get 0. So this thing is going to be 0 because k dot n is going to be 0 in this situation for this surface. k dot n is going to be equal to 0. So this part right over here simplifies to this right over here. Now in the next video what we can do is essentially express these in terms of surface integrals, but in terms of double integrals over this domain right over here. We'll kind of evaluate these surface integrals.