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Studying for a test? Prepare with these 11 lessons on Green's, Stokes', and the divergence theorems.
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Video transskription
Where we left off in the last video, we'd expressed our line integral around the boundary of our surface. So this is f.dr where our path is this boundary right over here-- this path c. We had expressed it in terms of a line integral around this boundary-- around path c1-- which is the boundary of region R. And the reason why this is going to be valuable to us is now we can directly apply Green's Theorem to this to turn this into a double integral over this region right over here-- the region that it is actually bounding. So let's actually do that. We're just applying Green's Theorem here. So Green's Theorem. And actually, let me box this off right over here. This was just something that we wrote to remind ourselves in the last video. So let me just box it off right over here. But when you apply Green's Theorem, you get that this is going to be the exact same thing as the double integral over the region that c1 bounds-- that region, R, that's in our xy plane-- of the partial of this with respect to x of this business. So I'll do that in that same green color. Q plus R times the partial of z with respect to y minus the partial with respect to y of P plus R times the partial of z with respect to x. And then, dA, little differential of our region. dA. And so let's actually calculate this right here. You take the partials of each of these expressions. And we'll see that if we expand them and then simplify, we'll get something very similar-- or actually, hopefully, identical-- to this right over here. And that'll show that this line integral for this special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case. So let's do it. So let's just apply the partial derivative operator. So first we want to take the partial derivative with respect to Q. And we need to remind ourselves-- and we did this way up here where we first thought about it-- we saw that P, Q, and R, they're each functions of x, y, and z. So we're assuming that's how they're represented. And if z was not a function of x, then, if we were to take the partial of Q with respect to x, we would just write it as a partial of Q with respect to x. But we know that we've assumed that z is itself a function of x and y. So if we're taking the partial with respect to x over here, we have to think first-- Well, how can Q directly change with respect to x? And then, how can it change due to something else changing due to x? And so that other thing that could change due to x is z. y is independent of x, but z is a function of x. So let's keep that in mind. So we're really going to do the multi-variable chain rule. So when we try to take the derivative of this part, of this whole function with respect to x, we have to think about-- How will Q change directly with respect to x? And to that, we have to add how Q could change due to changes in other variables due to changes in x. And the only other variable that Q is a function of that could change due to a change in x is z. So Q could also change to z because z has changed due to x. So the operator here is the partial with respect to x plus the partial of Q with respect to z times the partial of z with respect to x. If we rewrote Q so that z was substituted with x's and y's-- because it is a function of x and y's-- then we would just have to write this first term right over here. But we're assuming this is expressed as a function of x, y, and z. And z itself is a function of x. And so that's why we had to use the multi-variable chain rule. Now, let's move to the next part. And both of these might have some x's in them. So we have to use the product rule right over here. So first, we can take the derivative of r with respect to x. And then, multiply that times z sub y. Then, we have to take the derivative of z sub y with respect to x and multiply that times R. So this is going to be plus. If we take the derivative of this with respect to x, same exact logic-- R can change directly due to x, and it can change due to y. And that could change due to z and multiply that times how z could change due to x. Once again, you could view this as the multi-variable chain rule in action here. But of course, we take the derivative of the first term times the second term. And I'll do the second term in magenta. So the partial of z with respect to y plus the derivative of the second term-- which is the partial of z with respect to y, and then taking the partial of that with respect to x, which we could just write as that-- times the first term. So that's the partial with respect to x of all of this business right over here. And then, we need to subtract the partial of this with respect to y. And we're going to use the exact same logic. So then, we're going to subtract-- and I'll put it in parentheses like that. So P could change directly due to y. I will circle P. Let me do it with a color that I haven't used yet. P could change directly due to y. So we could say the partial of P with respect to y, but it could also change due to z changing because of y. So plus the partial with respect to z times the partial of z with respect to y-- and I'll do R maybe in that same color-- plus the derivative of R. Well, we already figured that, but actually now it's with respect to x, not with respect to y. You have to be careful. So it's going to be the partial of R with respect to y plus the partial of R with respect to z times the partial of z with respect to y times z sub x plus-- now, we take the derivative of the second term times the first term. The derivative of the partial of z with respect to x, then with respect to y, is going to be z. And then we're going to multiply that times R. And now, let's see if we can expand this out. And hopefully, things simplify. And just a reminder, I'm just working on the inside of this double integral. I'll rewrite the double integral and the dA once I get this all cleaned up a little bit. So let me rewrite it a little bit. So this is equal to the partial of Q. I'll try to color-code it the same way. And this is really just algebra at this point. The partial of Q with respect to x plus the partial of Q with respect to z times the partial of z with respect to x plus the partial of R with respect to x times the partial of z with respect to y. And then plus the partial of R with respect to z times the partial of z with respect to x times the partial of z with respect to y. And then we have this term right over here-- which I'll just do in purple-- plus the partial of z with respect to y, and then respect to x, times R. And now, we're going to subtract all of this business right over here. I'll do it in the blue. Minus the partial of P with respect to y minus the partial of P with respect to z times the partial of z with respect to y. And then we're going to subtract from that. Minus the partial of r with respect to y times-- we want to distribute this-- times the partial of z with respect to x minus the partial of R. This gets a little bit tedious. But hopefully, it'll get us to where we need to go. The partial of R with respect to z times the partial of z with respect to y times the partial of z with respect to x. And then, finally, this term right over here minus this-- because we have that negative out there-- minus the partial of z with respect to y, then with respect to x, and then with respect to yR. Now, let's see if we can simplify things. So the first thing-- this and this look to be the same. We just can commute the order in which we actually multiply. But these are the exact same term. So that is going to cancel out with that. And because we assumed way up here that we have continuous second derivatives of the function z-- z is a function of x and y, that that is equal to that-- we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out. And so this simplified things a good bit. And let's see if I can group terms in a way that might start to make sense. And actually, I'm going to try to see if I can make them similar to this. So I have all the terms that have a z sub x and the z sub y and then the rest of them. So z sub x I'll do in blue. So you have the terms that have a z sub x in it. So you have this term right over here and this term right over here. And we can factor out the z sub x. And we get the partial of z with respect to x times the partial of Q with respect to z minus the partial of R with respect to y. And then, let's do the-- and I want to get the colors the same way, too. I did yellow next. Plus, I have all the terms of the partial of z with respect to y. So it's that term and that term right over here. So it becomes plus the partial of z with respect to y times the partial of R with respect to x minus the partial of p with respect to z. And then we have these last two terms. And I used the color green up there, so I'll use the color green again. So for these two terms, I'll just write plus the partial of Q with respect to x minus the partial of P with respect to y. So our double integrals have kind of-- I can't really say simplified-- but we can rewrite it like this. And we don't want to forget this was all a simplification of our double integral over the region dA. This is what we have been able to using Green's theorem and the multi-variable chain rule and whatever else. We've been able to say that that line integral around the boundary of our surface is the same thing as this. And now, we can compare that to what our surface integral was. So let's see if I have space. So copy. And then, let me see if I have space up here to paste it. Well, it doesn't look like I actually have much space to paste it, although I'll try anyway. So if I paste it, you see that they are identical. They are identical. Our line integral is identical to this. We get the exact same thing. So our line integral, f.dr, around this path c simplified to this and our surface integral simplified to this. So using the assumptions we had, they both simplified to the same thing. So now we know, for this special case, our line integral is equal to our surface integral. And we are done.