# Stokes' theorem proof part 6 | Multivariable Calculus | Khan Academy

over the boundary of our surface in terms of dt,
in the dt domain.
We expressed what f dot dr is going to be equal to.
What I'm going to do in this video
is do a little bit of algebraic manipulation,
and then we will actually apply Green's theorem.
And whether or not we have time in this video,
we'll then manipulate that to show
that it's the exact same thing that we saw in the earlier
part, where we evaluated the surface integral.
And so let's do that.
So this integral right over here is the same thing
as the integral from a to b.
And then what I'm going to do is I'm
going to group the things that are being multiplied by dx dt.
So if I group them and then factor out a dx dt,
so if I take this part right here and this part right
over here, I'm essentially going to distribute
the R. Let me make it clear.
I'm going to distribute the R. And so I'm
going to group that and that right over there,
I will be left with-- and I'll factor out the dx dt.
I will be left with P plus R times the partial of z
with respect to x times dx dt.
And I'll do the same thing for the dy dt's.
So that's that part right over there.
And then the R is going to get distributed
in this thing right over here.
So it's going to be plus Q plus R-- the R is going
to get distributed-- times the partial of z with respect to y.
And I'm going to factor out a dy dt.
And then we can't forget all of that
is going to be multiplied by dt.
Now, this right over here is interesting,
because it's starting to look very similar to what
field.
In fact, let me copy and paste it.
So actually, I don't know if I'm on the right a layer of my work
right over here.
So let me copy and paste.
No, that didn't work.
Let me try it one more time.
So if I try to copy, I'll go a layer down.
I'm using an art program for this.
Copy-- and then I think this will work-- and paste.
There we go.
So this is a result that we had before.
This is kind of a template to look at.
But what is going on over here if we just
look at this template?
We see that we're in the t domain.
We're integrating over t right over here.
But then we have these things.
We have some function that's a function of x and y times dx
dt, and then some function, the function of x and y times
dy dt, and we're integrating with respect to dt.
Well, that's exactly what we're doing right over here.
We can distribute the dt, and we have
something that looks exactly like this right over here,
where M is analogous to this piece right over here.
Let me make it clear.
M, this piece right over here, looks a lot like M
in our example right over here.
It's being multiplied by dx dt, and then this dt,
which you can distribute, and this piece right over here,
looks a lot like N.
And so, we can say that, well, we
have something that looks like this.
We can rewrite it like this and go back
into the-- kind of deparametrize it.
So this thing is going to be equal to now the line
integral of C1.
We are in the xy plane.
We started with the curve C, but now we're going to go to C1.
It's completely analogous.
These are only functions of x's and y's.
Everything here is.
So now, this is going to be the line integral over C1--
and I could even draw it as that, if I like-- of M dx,
and that makes sense.
Because if you multiply dt times dx dt, the dt's cancel out,
and you're just left with dx.
So M times dx, so let me write it that way.
So it's going to be P plus R times the partial of z
with respect to x dx plus n.
Let me scroll to the right a little bit.
Plus n, which is Q plus R times the partial
of z with respect to y dy.
And then this is really interesting,
because this path that we are now concerned with,
it's completely analogous.
I hope you don't think I'm doing some voodoo here.
This statement is completely analogous to this statement,
where M could be this and N could be this.
And so, we can revert it back to now the path
C1 that sits in the xy plane.
Not our original boundary C, but now we're
just dealing with a boundary in the xy plane.
So it reverts to this.
But what's powerful about getting it to this point,
is we can now apply Green's theorem to this to essentially
turn this into a double integral over the region
that this path surrounds, the original region
over this region R. And when we change
how we manipulate that, when we play around with it,
we're going to see that we get the exact same result that we
got in earlier videos.
And I'll leave you there, and I'll
see if I can do that in the next video.