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### Course: Multivariable calculus>Unit 5

Lesson 9: Proof of Stokes' theorem

# Stokes' theorem proof part 4

Starting to work on the line integral about the surface. Created by Sal Khan.

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## Video transcript

Let's continue with our proof of Stokes' theorem. And this time we're going to focus on the other side of Stokes' theorem. We're going to try to figure out what is the line integral over the boundary C-- where this is C right over here, the boundary of our surface of f dot dr. And what we're going to see is that we're going to get the exact same result as we have right up here. Before we do that, I'm going to take a little bit of a detour to kind of build up to this. So let's just take this and put it to the side for now. Actually, let me actually just delete it right now. And what I'm actually going to do is I'm going to focus on this region down here-- this region in the xy plane. This is path C, which is the boundary of our surface. I'm going to focus on the path that is the boundary of this region. This path that sits in the xy plane. I will call that path C1. And so, one, we can think about a parameterization of just that path in the xy plane. We could say that C1 could be parameterized as x is equal to x of t and y is also a function of t. And t is obviously our parameter. And it can go between A and B. So maybe when t is equal to A, it sits right here. And then, as t gets larger and larger and larger, it goes all the way around. And eventually, when t is equal to B, it gets to that exact same point. So that's our parameterization right there. And now, just to make the rest of this proof a little bit more understandable, I'm going to give you a little bit of a review of something. Imagine that we have some vector field G. And G, at minimum, is defined on the xy plane. And it could be defined other places, but let's say that G is equal to m of x, y i plus N of x, y j. And this is all a review. We've seen this a long time ago. What would be the line integral over the path C1? Not C, but this path that sits in the xy plane. What would be the line integral over the path C1? I like to write that sometimes. And I'm using G so I don't get confused with F-- our original vector field-- of G, our vector field here, along that path. G dot dr. Well, dr is just going to be equal to dxi plus dyj. So if you take the dot product of these two things right over here, you're going to get the line integral over our path C1. Remember, C1 is this path down here. Let me do it in that same color so you don't think I'm changing colors on you. The line integral over our path C1. But when you take this dot product, you multiply the x components and then add that to the product of the y component. So you have m times dx plus n times dy. I just took the dot product of G and dr. n times dy. And when you evaluate these things, one way to think about it is that dx is the same thing as-- let me write it up here in a different color-- dx is the same thing as the derivative of x with respect to t, dt. And same logic for y. dy is equal to the derivative of y with respect to t, dt. One way to think about it, these dt's cancel out. And you are just left with dx. And this is an important thing to think about because then this allows us to take this line integral into the domain of our parameter. So then, this will be equal to the integral in the domain of our parameter. So now we are in the t domain. And t is going to vary between a and b. We are in the t domain between a and b. This is going to be equal to m times-- instead of writing dx, I'm going to write dx/dt dt. So it's going to be dx-- let me write it this way. dx-- the derivative of x with respect to t-- dt-- that's the first expression-- plus n-- and then, the exact same thing-- times dy dt. n times dy dt. These are all equivalent statements. Now, with all of that out of the way-- and all of this is really just a reminder so that the rest of this proof becomes a little bit intuitive. With that out of the way, let's come up with the parameterization for this path up here, for C. Remember, we just did C1 down in the xy plane. Now we're going to do C that sits up here that kind of rises above the xy plane. Well, for C, the parameterizations for x and y can still be the exact same thing because the x and y values are going to be the exact same thing. The x and y value there is the exact same thing as the x and y value there. The only difference is we now have a z component. We defined it way up here. Our z component is going to be a function of x and y. It tells us how high to go. We can parameterize C as-- maybe I'll write it as a vector. So let me parameterize. So I'll write C as a vector. Actually, no. Let me write it this way. Let me write C. Let me do that purple color. C, we can say is x is x of t-- actually, let me write this as a vector. And I'll use a vector r, not to be confused with this r right over here. So these are two different r's, but I'll just use r because that tends to be the convention. So in order to parameterize C, it's going to be the position vector r, which is going to be a function of t. And x is still just going to be x of t i plus y of tj. And now we're going to have a z component. And z is going to be a function of x and y, which are, in turn, functions of t. So z is a function of x-- which is a function of t-- and a function of y-- which is a function of t k-- that tells us how high above to, essentially, get each of those points. And then, once again, we know that t is between a and b. t is greater than or equal to a and less than or equal to b. So we have that parameterization right there. And now we can start to think about the line integral of f dot dr along this path. Before, we did dr along this path. Now we're going to do dr on this path right over here. This is now our parameterization r. So I'll leave you there, and I'll see you in the next video.