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Divergence theorem proof (part 4)

More evaluation of the surface integral. Created by Sal Khan.

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Video transcript

Sometimes when you're doing a large multipart proof like this, it's easy to lose your bearings. We're trying to prove the divergence theorem. Well, we started off just rewriting the flux across the surface and rewriting the triple integral of the divergence. And we said, well, if we can prove that each of these components are equal to each other, we will have our proof done. And what I said is that we're going to use the fact this is a Type 1 region to prove this part, the fact that it's a Type 2 region to prove this part, and the fact that it's a Type 3 region to prove this part. In particular, I'm going to show the fact that, if it is a Type 1 region, we can prove this. And you can use the exact same argument to prove the other two-- in which case, the divergence theorem would be correct. So we're focused on this part right here. And in particular, we're focused on evaluating this surface integral. And to evaluate that surface integral, we broke up the entire surface into three surfaces-- that upper bound on z, the lower bound on z, and kind of the side of-- if you imagine some type of a funky cylinder. These don't have to be flat tops and bottoms. And for a Type 1 region, you don't even have to have this side surface. This is only for the case where the two surfaces don't touch each other within this domain right over here. So we broke it up into three surfaces. But we said, look, the normal vector along the side right over here is never going to have a k component. And so when you take the dot product with the k unit vector, this is just going to cross out. So our surface integral simplified to the surface integral over s2 and the surface integral of s1. And in the last video, we evaluated the surface integral of s2-- or at least, we turned it into a double integral over the domain. Now, I'm going to do the same exact thing with s1. So let's just remind ourselves what we are concerned with. We want to re-express s1. This surface integral-- I should say we want to re-express the surface integral over s1. It was r of x, y, and z. Up here we just wrote r, but this is making it explicit that r is a function of x, y, and z. Times k dot n ds. And the way we evaluate any surface integral is we want to have a parameterization for our actual surface. So let's introduce a parameterization for our surface. Let's say that s1-- and I'll use-- the letter O is kind of a weird one to use. It looks like a 0, and we're already using it for our normal vector. Let's use e. So the parameterization of our surface could be x times i plus y times j plus-- and since it's a Type 2 region, our surface is a function of x and y-- plus f1-- which is a function of xy-- times k. We see f1 right over here. That's our lower bound on our region. And then, this is for all of the xy pairs that are a member of our domain in question. So we have our parameterization. Now, we can think about how we can write this right over here, the nds. And so n times ds-- so we've done this multiple, multiple, multiple times-- which is actually the same thing as ds. We've done this multiple times. This is equal to the cross product of the parameterization in one direction with respect to one parameter. And then, that cross with respect to the other parameter. And then, that times da. But we want to make sure we get the order right. So I'm going to claim that this is going to be the partial of our parameterization with respect to y crossed with the partial of our parameterization with respect to x. And then, we have times da. But we need to make sure this has the right orientation. Because, for this bottom surface, remember, we need to be pointed straight down, outward from the region. So if we're going in the y direction, the partial with respect to y is like that. The partial with respect to x is like that. And if you use the right-hand rule, your thumb will point downward, like that. I could draw my right hand. My index finger could go like that. My middle finger would bend like this. I don't really care what my other two fingers do. And then, my thumb would go downwards. So this is the right order, which is a different order, or it's the opposite ordering, as we did last time. You'll see, we'll just get a negative value, but let's just work it through just to be a little bit more convincing. So this business right over here is going to be equal to-- let's write our i, j, and k unit vectors. It's going to be all of that times da. So the partial of e with respect to y is 0, 1, partial of f1 with respect to y. Partial with respect to x is going to be 0, 1, partial of f1 with respect to x. And then, when you evaluate this whole thing-- and so let me draw a little dotted line here-- this is going to be equal to some business times i minus some other business times j. And on our k component, you're going to have 0 times 0 minus 1 times k. So minus k, and then, all of that times da. And now, the reason why I didn't take even worry about what these things are going to be is I'm going to have to take the dropped product with k. So this whole thing-- all of this business right over here-- I can now express in the xy domain. I can now write is going to be equal to the double integral. And let me do it in that same purple color because that was the original color for that surface integral. It's going to be the double integral over our parameter's domain, in the xy plane of r of x-- let me write this a little bit cleaner-- r of xy. Instead of z, I'm going to write z over that surface is f1 of x and y, so that we have everything in terms of our parameters. Times all of this business k dotted this. Well, what is k dotted this? Well, the dot product of k and negative k is negative 1. So we're just going to be left with negative da. So we'll put the da out front here. And we have a negative da right over here. So we have now expressed this surface integral as the sum of two double integrals. Let me find it. The sum of this, the sum of that and that right over there. And actually, let me just rewrite it just so that we can make everything clear. So this surface integral. So let me write it right over here. The surface integral over our entire surface of r times k dot n ds is equal to the double integral-- and I'll do this in a new color-- is equal to the double integral in the domain d of this thing minus this thing da. So I'll write r of xy and f2 of xy, and that's that minus this-- minus r of xy and f1-- be careful here-- f1 of xy. That's this thing right over here. All of that times da. Now, we just showed this is equal to this. All we have to show now is that this is also equal to that same expression. And we will have proven this for the Type 1 case. And we can use the exact same argument for the Type 2 and Type 3 cases and feel good that the divergence theorem is about to be proven for a simple solid region.