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In the last video, we began
to explore Stokes' theorem.

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And what I want to
do in this video

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is to see whether it's
consistent with some

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of what we have already seen.

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And to do that, let's imagine--
so let me draw my axes.

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So that's my z-axis.

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That is my x-axis.

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And then that is my y-axis.

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And now let's imagine a
region in the xy plane.

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So let me draw it like this.

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So let's say this is my
region in the xy plane.

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I will call that
region R. And I also

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have a boundary of that region.

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And let's say we care
about the direction

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that we traverse the boundary.

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And let's say we're
going to traverse it

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in a counterclockwise direction.

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So we have this path that
goes around this region.

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We can call that c.

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So we'll call that
c, and we're going

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to traverse it in the
counterclockwise direction.

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And let's say that we also
have a vector field f.

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That essentially its
i component is just

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going to be a
function of x and y.

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And its j component
is only going

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to be a function of x and y.

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And let's say it
has no k component.

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So the vector field
on this region,

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it might look
something like this.

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I'm just drawing random things.

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And then if you go
off that region,

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if you go in the
z direction, it's

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just going to look the same
as you go higher and higher.

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So that vector,
it wouldn't change

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as you change your z component.

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And all of the vectors
would essentially

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be parallel to, or
if z is 0, actually

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sitting on the xy plane.

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Now given this, let's
think about what

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Stokes' theorem would tell us
about the value of the line

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integral over the
contour-- let me

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draw that a little
bit neater-- the line

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integral over the
contour c of f dot

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dr, f dot lowercase dr,
Where dr is obviously

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going along the contour.

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So if we take Stokes'
theorem, then this quantity

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right over here should
be equal to this quantity

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right over here.

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It should be equal to the double
integral over the surface.

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Well this region is
really just a surface

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that's sitting in the xy plane.

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So it should really just
be the double integral--

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let me write that
in that same-- it'll

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be the double integral over
our region, which is really

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just the same thing
as our surface,

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of the curl of f dot n.

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So let's just think about
what the curl of f dot n is.

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And then d of s would
just be a little chunk

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of our region, a little chunk
of our flattened surface

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right over there.

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So instead of ds,
I'll just write da.

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But let's think of what curl
of f dot n would actually be.

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So let's work on
curl of f first.

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So the curl of f--
and the way I always

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remember it is
we're going to take

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the determinant of
this ijk partial

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with respect to x,
partial with respect to y,

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partial with respect to z.

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This is just the definition
of taking the curl.

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We're figuring out how
much this vector field

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would cause something to spin.

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And then we want the
i component, which

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is our function p, which is
just a function of x and y,

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j component, which is
just the function q.

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And there was no z
component over here, so 0.

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And so this is going
to be equal to-- well

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if we look at the
i component, it's

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going to be the
partial of y of 0.

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That's just going to be 0, minus
the partial of q with respect

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to z.

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Well what's the partial
of q with respect to z?

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Well q isn't a
function of z at all.

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So that's also going to be
0-- let me write this out

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just so it's not too confusing.

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So our i component, it's
going to be partial of 0

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with respect to y.

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Well that's just going to
be 0 minus the partial of q

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with respect to z.

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Well the partial of
q with respect to z

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is just going to be 0.

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So we have a 0 i component.

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And then we want to
subtract the j component.

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And then the j component partial
of 0 with respect to x is 0.

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And then from that you're going
to subtract the partial of p

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with respect to z.

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Well once again, p is not
a function of z at all.

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So that's going to be 0 again.

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And then you have plus k times
the partial of q with respect

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to x.

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Remember this is just the
partial derivative operator.

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So the partial of q
with respect to x.

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And from that we're going
to subtract the partial of p

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with respect to y.

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So the curl of f just simplifies
to this right over here.

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Now what is n?

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What is the unit normal vector.

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Well we're in the xy plane.

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So the unit normal
vector is just

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going to be straight
up in the z direction.

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It's going to have
a magnitude of 1.

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So in this case, our
unit normal vector

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is just going to
be the k vector.

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So we're essentially just going
to take-- so curl of f is this.

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And our unit normal
vector is just

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going to be equal to the k.

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It's just going to
be the k unit vector.

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It's going to go straight up.

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So what happens if we
take the curl of f dot k?

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If we just dot this with k.

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We're just dotting
this with this.

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Well, we're just going to end up
with this part right over here.

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So curl of f dot the unit
normal vector is just

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going to be equal
to this business.

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It's just going to be equal to
the partial of q with respect

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to x minus the partial
of p with respect to y.

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And this is neat because
using Stokes' theorem

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in the special case,
where we're dealing

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with a flattened-out
surface in the xy plane,

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in this situation, this just
boiled down to Green's theorem.

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This thing right over here just
boiled down to Green's theorem.

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So we see that Green's theorem
is really just a special case--

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let me write theorem
a little bit neater.

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We see that Green's
theorem is really

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just a special case
of Stokes' theorem,

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where our surface is flattened
out, and it's in the xy plane.

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So that should make us feel
pretty good, although we still

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have not proven Stokes' theorem.

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But the one thing that I do
like about this is seeing

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that Green's theorem and
Stokes' theorem is consistent

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is now it starts to make
sense of this right over here.

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When we first learned Green's
theorem, we were like,

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what is this?

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what's going on over here?

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But now this is
telling us this is just

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taking the curl in this
region along this surface.

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And now starts to make a lot
of sense based on the intuition

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that we saw in the last video.