WEBVTT
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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 flux
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and the normal form of
Green's theorem-- so normal
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here meaning perpendicular--
so I want to give you
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a problem about that.
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So what I'd like you
to do is to verify
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Green's theorem in normal form
for this particular field,
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the field F equals x i hat
plus y j hat and the curve C
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that consists of the upper
half of the unit circle
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and the x-axis
interval minus 1 to 1.
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So first of all, let me
say what I mean by this C.
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So C, it's the usual unit
circle, circle of radius 1
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centered at the origin,
so just its top half,
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And the x-axis
interval minus 1 to 1,
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I mean the line segment that
connects the points (-1, 0)
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and (1, 0), so the diameter
of that 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
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is I'd like you to compute
both the double integral that
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appears in Green's theorem and
the line integral that appears
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in Green's theorem
and check that they're
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really equal to each other.
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So that'll confirm
Green's theorem
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in this particular
instance and hopefully
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help give us a feel for
how it works a little bit.
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So why don't you pause
the video, have a go,
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compute both of those
integrals, come back
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and we can 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
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is that the flux across
a curve, which we usually
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compute as a line
integral, is also
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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
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bounds a region of the plane.
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So in particular, let's
draw the picture here
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so we know what
we're talking about.
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So the curve C, so we've got
the segment from minus 1 to 1
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along the x-axis, and
then we have the top half
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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
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you don't specify an
orientation, what you
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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
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of the 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
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us is that when we compute
the surface integral-- sorry,
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the double integral--
of the divergence of F
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over this region, that
should be the same as what
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we do if we compute F
dot n around the boundary
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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.
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The double integral
in this case,
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so it's the double
integral over--
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let me call that region R-- so
it's the double integral over R
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of the 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
x i hat plus y j hat.
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So its divergence
is the partial of x
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with respect to 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
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over the semicircle of 2 dA.
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And of course when you integrate
a constant over a region, what
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you get is just that constant
times the area of the region.
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So dA here is the area
of the semicircle,
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it's half of a
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
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get from Green's
theorem in normal form.
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And what Green's theorem
says is that this
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is equal to a 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 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 integral
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.
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So this is equal to, and we know
that this is always equal to,
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the integral around
C of-- and let
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me make sure I'm getting this
right before I screw anything
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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--
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or the components of F. M and
N are the components of F.
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So M 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
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around C of-- so M, the
first component of F is x.
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So this is x*dy minus-- the
second component is y*dx.
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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
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easy to 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--
well, I'm not even going
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to bother giving them names.
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We want to split it into
the integral over the line
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segment plus the integral
over the semicircle.
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The boundary of that--
yes, the semicircle.
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So this is equal to-- so it's
the integral over the line
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segment.
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So let's see.
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So that integral-- 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 C_1 and we'll
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call the semicircle C_2.
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So it's equal to the
integral over C_1--
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OK, well, what is the
integral over C_1?
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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-- OK, and now on
this line segment y is 0 also.
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So it's 0*dx.
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So the first integral,
the integral over C_1,
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is the integral of 0.
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Plus-- we have to
integrate over C_2 of-- OK,
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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
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over the 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 around to
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(-1, 0).
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So that is from
going from 0 to pi.
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So this is equal to-- 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-- OK, so
t is going to go from 0 to pi.
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So now x*dy 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-- all right,
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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 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
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in 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
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C. This closed curve C
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 div F dA.
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So that's what we did first.
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Double integral over the
region bounded by the curve.
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And then second, we
computed the line
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integral, around the
boundary, of F dot n ds.
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So that was what--
this is always
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a useful form in which
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 integrals
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have to be equal to each other.
00:10:45.600 --> 00:10:48.450
And indeed, for this particular
F and this particular C,
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we verified that in this
case they both give us pi.
00:10:52.084 --> 00:10:53.750
And, of course, Green's
theorem tells us
00:10:53.750 --> 00:10:56.125
that that would have been
true, that they would have come
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out the same, regardless
of what the choice of F
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and the choice of C
that we made were.
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So I'll stop there.