In the last video, we began to explore Stokes' theorem.
And what I want to do in this video
is to see whether it's consistent with some
of what we have already seen.
And to do that, let's imagine-- so let me draw my axes.
So that's my z-axis.
That is my x-axis.
And then that is my y-axis.
And now let's imagine a region in the xy plane.
So let me draw it like this.
So let's say this is my region in the xy plane.
I will call that region R. And I also
have a boundary of that region.
And let's say we care about the direction
that we traverse the boundary.
And let's say we're going to traverse it
in a counterclockwise direction.
So we have this path that goes around this region.
We can call that c.
So we'll call that c, and we're going
to traverse it in the counterclockwise direction.
And let's say that we also have a vector field f.
That essentially its i component is just
going to be a function of x and y.
And its j component is only going
to be a function of x and y.
And let's say it has no k component.
So the vector field on this region,
it might look something like this.
I'm just drawing random things.
And then if you go off that region,
if you go in the z direction, it's
just going to look the same as you go higher and higher.
So that vector, it wouldn't change
as you change your z component.
And all of the vectors would essentially
be parallel to, or if z is 0, actually
sitting on the xy plane.
Now given this, let's think about what
Stokes' theorem would tell us about the value of the line
integral over the contour-- let me
draw that a little bit neater-- the line
integral over the contour c of f dot
dr, f dot lowercase dr, Where dr is obviously
going along the contour.
So if we take Stokes' theorem, then this quantity
right over here should be equal to this quantity
right over here.
It should be equal to the double integral over the surface.
Well this region is really just a surface
that's sitting in the xy plane.
So it should really just be the double integral--
let me write that in that same-- it'll
be the double integral over our region, which is really
just the same thing as our surface,
of the curl of f dot n.
So let's just think about what the curl of f dot n is.
And then d of s would just be a little chunk
of our region, a little chunk of our flattened surface
right over there.
So instead of ds, I'll just write da.
But let's think of what curl of f dot n would actually be.
So let's work on curl of f first.
So the curl of f-- and the way I always
remember it is we're going to take
the determinant of this ijk partial
with respect to x, partial with respect to y,
partial with respect to z.
This is just the definition of taking the curl.
We're figuring out how much this vector field
would cause something to spin.
And then we want the i component, which
is our function p, which is just a function of x and y,
j component, which is just the function q.
And there was no z component over here, so 0.
And so this is going to be equal to-- well
if we look at the i component, it's
going to be the partial of y of 0.
That's just going to be 0, minus the partial of q with respect
to z.
Well what's the partial of q with respect to z?
Well q isn't a function of z at all.
So that's also going to be 0-- let me write this out
just so it's not too confusing.
So our i component, it's going to be partial of 0
with respect to y.
Well that's just going to be 0 minus the partial of q
with respect to z.
Well the partial of q with respect to z
is just going to be 0.
So we have a 0 i component.
And then we want to subtract the j component.
And then the j component partial of 0 with respect to x is 0.
And then from that you're going to subtract the partial of p
with respect to z.
Well once again, p is not a function of z at all.
So that's going to be 0 again.
And then you have plus k times the partial of q with respect
to x.
Remember this is just the partial derivative operator.
So the partial of q with respect to x.
And from that we're going to subtract the partial of p
with respect to y.
So the curl of f just simplifies to this right over here.
Now what is n?
What is the unit normal vector.
Well we're in the xy plane.
So the unit normal vector is just
going to be straight up in the z direction.
It's going to have a magnitude of 1.
So in this case, our unit normal vector
is just going to be the k vector.
So we're essentially just going to take-- so curl of f is this.
And our unit normal vector is just
going to be equal to the k.
It's just going to be the k unit vector.
It's going to go straight up.
So what happens if we take the curl of f dot k?
If we just dot this with k.
We're just dotting this with this.
Well, we're just going to end up with this part right over here.
So curl of f dot the unit normal vector is just
going to be equal to this business.
It's just going to be equal to the partial of q with respect
to x minus the partial of p with respect to y.
And this is neat because using Stokes' theorem
in the special case, where we're dealing
with a flattened-out surface in the xy plane,
in this situation, this just boiled down to Green's theorem.
This thing right over here just boiled down to Green's theorem.
So we see that Green's theorem is really just a special case--
let me write theorem a little bit neater.
We see that Green's theorem is really
just a special case of Stokes' theorem,
where our surface is flattened out, and it's in the xy plane.
So that should make us feel pretty good, although we still
have not proven Stokes' theorem.
But the one thing that I do like about this is seeing
that Green's theorem and Stokes' theorem is consistent
is now it starts to make sense of this right over here.
When we first learned Green's theorem, we were like,
what is this?
what's going on over here?
But now this is telling us this is just
taking the curl in this region along this surface.
And now starts to make a lot of sense based on the intuition
that we saw in the last video.