Divergence theorem proof (part 3)
Evaluating the surface integral. Created by Sal Khan.
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- why did we take the S3 such that it's normal vector has no k^ component where as there can be the case when it's not so, for example section of a cone above xy-plane. Also is it necessary that we should be taking surface such that it's domain for f1 and f2 are same, as in this video we have taken for both f1 ans f2 as D.(3 votes)
- if it was a cone, the pointy part would be S2, which would touch S1, so there would not be an S3 (and thus no k^ component). As for having the domains be the same, yes, to be a type I, II, or III, the two functions must both be applied to the same region.(3 votes)
- What video(s) can I watch to understand the intuition for the equation at1:40?(2 votes)
- Why for the t(x,y) function the K component only incorporates the f2 surface function and not the f1?(2 votes)
- At1:30Sal takes two partial derivatives: t with respect to x and t with respect to y, crosses them and multiples with dA. I understand why we would multiply with dA, just not where the partials of t come from and why we are even crossing them(1 vote)
- Why isn't the normal vector, n, a unit vector? Shouldn't we divide tx X ty over its magnitude?(1 vote)
Let's now evaluate this green surface integral right over here. And to do that, we need to parameterize surface 2. So let's say surface 2 can be represented by the vector position function. I'll just call it t-- t for two-- so t-- which is a vector-- it's going to be a function of x and y. Those are going are going to be our parameters. And it's going to be equal to-- and we can do this, once again, because our surface is a function of x and y. So it's going to be equal to x times i plus y times j plus f2 of xy times k for all the xy's that are a member of our domain. Now, with that out of the way, we can re-express what k dot n ds is. So let me write this over here. k dot n ds is equal to-- and we could put parentheses here that we're going to take this dot product. That's, at least, how I like to think about it. This is the exact same thing as k dotted with the cross product of the partial of t-- let me make this clear. I'm going to do the magenta color. The partial of t with respect to x crossed with the partial of t with respect to y times a little chunk of our area-- times da. A little chunk of our area in the xy domain. We've done this multiple times as we evaluated surface integrals. And we got the intuition for why this works. And so then, we're essentially just evaluating the surface integral. And the one thing we want to make sure is make sure this has the right orientation. Because, remember, in order for the divergence theorem to be true, the way we've defined it is, all the normal vectors have to be outward-facing. So for this top surface, the normal vector has to be pointing straight up. Not necessarily straight up. At least, upwards. If this is a curve, it wouldn't be necessarily straight up. But it needs to be up kind of outward-facing, like that. On the sides, it would be outward-facing like that. And down here, it would be outward-facing going in the general downwards direction. So let's just make sure that this is upwards-facing. If we're changing with respect to x, we're going in this direction. Changing with respect to y, we're going in that direction. Take the right-hand rule with the cross product-- index finger there, middle finger there, your right thumb, I should say, will go straight up. So it goes in the right direction. So this would be an upward-pointing vector. So we got the right orientation for our surface. Now, let's think about what this is. And it's important to realize we could calculate all of the components of this, but then we're just going to take the dot product of that with k. So really, we care about the k component only, but I'll work it out. This is equal to k times a matrix i, j, k of the partial of t with respect to x. Well, the partial of t with respect to x-- I'll do this in blue-- is going to be 1, 0, and the partial of f2 with respect to x. And then the partial of t with respect to y is just going to be 0, 1, and the partial of f2 with respect to y. And then, of course, we have to multiply times da. And this is all going to be equal to unit vector k dotted with-- and I don't even have to work it all out. It's going to be something times the i unit vector minus something-- checkerboard pattern-- minus something times something else necessarily-- times our j unit vector. And now, we can think about the k unit vector. So the k unit vector is going to be 1 times 1 minus 0. So it's just going to be plus the k unit vector. We know it's 1 times the k unit vector. And of course, we have our da out right here. But when you take this dot product, you only are left with the k components. And it's essentially just 1 times 1. You end up with a scalar quantity of 1. All of this business just simplified to da. So now, we can rewrite our surface integral. And we're going to rewrite it in the xy domain now-- in our parameters domain. So our surface integral right up here-- so this will be good for this video. And then, we'll do the same thing with this surface here. Just making sure that we get the orientation right. So this surface integral, s2-- and I'll even rewrite a little bit-- s2, which is a function. r is a function of x, y, and z times k dot n ds-- I just rewrote all of this right over here-- is equivalent to the double integral over our parameters domain-- which is just D-- of r of x, y, z times all of this business. All of this business just simplified to da. And since I want to write it in terms of my parameters, I'll write it as R of xy. And while we're on that surface, z is equal to f2. So it's xy and f2 of xy. And then, all of this business we just saw simplified to da. So you might be saying, hey, Sal. It didn't look like you simplified it a lot. But at least now put it in terms of a double integral instead of a surface integral. So at least, in my mind, that is a simplification. In the next video, we're going to the exact same thing with this-- just making sure that our vectors are oriented properly. And we could just introduce a negative sign to make sure that they are. And then, we're going to think about the triple integrals and try to simplify those.