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.