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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.

## 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.