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# Green's and Stokes' theorem relationship

Video transcript

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.