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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been learning
about two-dimensional

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flux and the normal form of
Green's theorem, so normal

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here meaning perpendicular.

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So I want to give you a
problem about that.

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So what I'd like you to do is
to verify Green's theorem in

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normal form for this particular
field, the field F

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equals xi hat plus yj hat and
the curve C that consists of

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the upper half of the unit
circle and the x-axis

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interval (-1, 1).

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So first of all, let me say what
I mean by this C. So C,

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it's the usual unit circle,
circle of radius 1 centered at

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the origin, so just its top
half, And the x-axis interval

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(-1, 1), I mean, the line
segment that connects the

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points (-1, 0) and (1, 0),
so the diameter of that

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semicircle.

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So that's the curve
C and the field F.

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What I mean by verify Green's
theorem is I'd like you to

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compute both the double integral
that appears in

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Green's theorem and the line
integral that appears in

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Green's theorem and check
that they're really

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equal to each other.

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So that'll confirm Green's
theorem in this particular

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instance and hopefully help
give us a feel for how it

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works a little bit.

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So why don't you pause the
video, have a go, compute both

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of those integrals, come
back and we can

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work them out together.

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Hopefully you had some
luck with this.

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Let's have a go at it.

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So what Green's theorem tells
you is that the flux across a

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curve, which we usually compute
as a line integral, is

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also equal to an integral of the
divergence over the region

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bounded by that curve.

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So here the curve has to be a
closed curve so that it bounds

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a region of the plane.

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So in particular, let's draw
the picture here so we know

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what we're talking about.

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So the curve C, so we've got the
segment from -1 to 1 along

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the x-axis, and then
we have the top

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half of the unit circle.

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That's the curve C. And I didn't
specify an orientation

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but in any context like this,
when you don't specify an

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orientation, what you mean
is positively oriented.

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So it's a positively
oriented curve.

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So that's our curve C.

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And so the region that it bounds
is this half of the

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circle, the semicircle.

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I keep saying half of
the semicircle.

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Sorry about that.

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This upper half of the circle,
of the disc, in fact.

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That region of the plane.

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OK, so what Green's theorem
tells us is that when we

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compute the surface integral--

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sorry, the double integral--

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of the divergence of F over this
region, that should be

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the same as what we do if we
compute F dot n around the

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boundary of the curve.

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And now we're going
to check it.

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So let's do the double integral
first. The double

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integral in this case, so it's
the double integral over--

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let me call that region R--

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so it's the double integral
over R of the

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divergence of F dA.

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So what is the divergence
of F?

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Well, here's F. It's
xi hat plus yj hat.

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So it's divergence is the
partial of x with respect to

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x, plus the partial of
y with respect to y.

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So that's 1 plus 1.

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So the divergence of F is
just 2, in this case.

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So it's equal to the double
integral over the

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semicircle of 2 dA.

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And of course when you integrate
a constant over a

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region, what you get is just
that constant times the area

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of the region.

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So dA here is the area of the
semicircle, it's half of a

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circle of radius 1.

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So circle of radius
1 has area pi.

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So this is 2 times 1/2 pi.

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So this is pi.

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OK.

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So that's the double integral
that we get from Green's

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theorem in normal form.

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And what Green's theorem says
is that this is equal to a

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particular line integral.

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So what is the line integral?

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Well, it's the integral
around C of F dot n.

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So let's write down what
it is, the line

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integral part now.

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So it's an integral.

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It's a closed curve.

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So it's in and around
C of F dot n ds.

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So that is the line integral
that we're looking to compute.

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So how do we compute this?

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Well, usually we compute it by
using the coordinates of F. So

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this is equal to, and we know
that this is always equal to,

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the integral around C of--

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and let me make sure I'm getting
this right before I

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screw anything up.

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Yes.

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OK, I am.

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Good.

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M dy minus N dx, where M and N
are the coordinates of F-- or

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the components of F. M and N are
the components of F. So M

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is the first component and N
is the second component.

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So in our case, F has this
fairly simple form.

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So in our case this is equal to
the integral around C of--

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so M, the first component
of F is x.

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So this is xdy minus the second
component is ydx.

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So this is the line integral
we're interested in computing.

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But, of course, this curve
is not easy to

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parameterize as a single go.

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So we want to split it
into two pieces.

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So let's look.

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So the first piece we want
to split it into

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is that line segment.

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So let's call that maybe--

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well, I'm not even going to
bother giving them names.

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We want to split it into the
integral over the line segment

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plus the integral over
the semicircle.

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The boundary of that--

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yes, the semicircle.

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So this is equal to--

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so it's the integral over
the line segment.

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So let's see.

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So that integral--

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well, OK, I will give
them names.

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I will give them names.

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I take it back.

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We'll call the line segment
C1 and we'll call

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the semicircle C2.

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So it's equal to the
integral over c1.

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OK, well, what is the
integral over c1?

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What are x and dy and y
and dx in this case?

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So in this case, well, x is
what we're integrating.

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But dy, we're on this line
segment, y isn't changing.

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y is constant.

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So dy is just 0.

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So it's 0 minus--

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OK, and now on this line
segment y is 0 also.

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So it's 0dx.

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So the first integral, the
integral over C1, is the

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integral of 0.

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Plus we have to integrate
over C2.

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Of OK, so x dy minus y dx.

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All right.

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So this one's just going
to be 0, that's easy.

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So now we just have to work
with this second one.

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OK, so for the second one, we're
integrating over the

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semicircle so we want
to parameterize it.

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And we're going to use our
usual parameterization.

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x equals cosine t,
y equals sine t.

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And in this case, we just want
to do this semicircle.

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So we just want to go from
(1, 0) all the way

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around to (-1, 0).

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So that is from going
from 0 to pi.

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So this is equal to--

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so this first integral
is just 0.

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It's the integral of
0 and that's 0.

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So it's the integral--

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OK, so t is going to
go from 0 to pi.

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So now x dy.

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So x is cosine t.

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y is sine t.

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So dy is sine t dt.

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Sorry, it's cosine t dt.

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y is sine t.

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dy is cosine t dt.

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So it's cosine t times
cosine t dt minus.

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All right, now y is
sine t again.

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And x is cosine t.

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So dx is minus sine t dt.

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So this is times minus
sine t dt.

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OK, well what happens here?

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So this becomes cosine squared
t dt minus minus

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sine squared t.

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Minus minus is plus.

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So it's cosine squared t dt
plus sine squared t dt.

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But, of course, cosine squared
plus sine squared is just 1.

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So we can write this even more
simply as the integral from 0

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to pi of 1 dt or just dt, and
the integral dt is just t.

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So this is t between 0
and pi, which is pi.

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Whew!

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OK.

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Pi.

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Good.

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And what did we get before?

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We also got pi.

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Great.

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So we have successfully verified
Green's theorem in

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normal form in this particular
instance.

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So let's just recap
again what we did.

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We had this field F and this
curve C. This closed curve C

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that bounded some region.

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And so what we've done is we
computed the double integral

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over the region of d of F dA.

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So that's what we did first.
Double integral over the

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region bounded by the curve.

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And then second we computed the
line integral around the

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boundary of F dot n ds.

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So that was what-- this is
always a useful form in which

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to write this F dot n dA.

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OK, and then we substituted
and computed it.

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And Green's theorem tells us
that the two intervals have to

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be equal to each other.

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And indeed, for this particular
F and this

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particular C, we verified
that in this case they

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both give us pi.

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And, of course, Green's theorem
tells us that that

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would have been true, that they
would have come out the

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same, regardless of what the
choice of F and the choice of

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C that we made were.

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So I'll stop there.

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