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CHRISTINE BREINER: Welcome
back to recitation.

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In this video I would like us
to use Green's theorem to

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compute the following integral
where it's the integral over

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the curve c, where c is
the circle drawn here.

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So the circle is oriented so
the interior is on the left

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and it's centered at the point
where x equals a, y equals 0.

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So the integral--

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you can read it but I will read
it also for you-- it's

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the integral of 3x squared y
squared dx plus 2x squared

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times the quantity
1 plus xy dy.

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So you are supposed
to use Green's

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theorem to compute that.

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And why don't you pause the
video, give it a shot, and

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then when you're ready to see my
solution you can bring the

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video back up.

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OK, welcome back.

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Well, what we're going to do,
obviously to use Green's

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theorem to compute this--
because I gave you a big hint

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that you're supposed to use
Green's theorem because we

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have an integral over
a closed curve.

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And what we're going to do is
now take the integral of

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another certain integrand--

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obviously related in
some way to this--

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over the region that
I will shade here.

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So we're interested in this
shaded region now.

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So let me write down what
Green's theorem is and then

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we'll put in the important
parts.

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OK, so just to remind you, I'm
not going to write down all

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the hypotheses of Green's
theorem that we need, but the

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point that I want to make is
that if we start the integral

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over this closed curve c of
mdx plus ndy we can also

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integrate over the region that
C encloses of N sub x minus M

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sub y dy dx.

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So that is ultimately what
we're going to do.

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And so in this case again, as
always, m is going to be the

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function associated with
the dx portion.

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It's the ice component
of the vector field.

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And n is going to be the
function associated

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here with the dy.

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That's standard obviously.

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So now what we want to do is
transform what we have there

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into something that
looks like this.

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So the region--

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I'm just going to keep calling
it R for the moment--

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but the region R you'll notice
is the thing that I shaded in

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the drawing.

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And so now let's compute
what this is--

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well you have the capacity
to do that.

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That's just some straight
taking derivative.

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So I'm just going to write down
what it is and not show

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you all the individual pieces.

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So I'll just write down what
you get and then the

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simplified version.

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So you get 6x squared y
plus 4x is N sub x.

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And then M sub y is negative
6x squared y.

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And I'm going to call dy dx, I'm
just going to refer to it

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now as dA, because that makes
it a lot easier to write.

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Oh, I guess I should
call this--

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well, dA, this is the area.

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The volume form there.

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So this simplifies and I just
get the integral of 4x over

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this region dA.

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Now this-- you saw an example
of this in lecture of how to

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deal with these types of
problems. At this point, if I

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really wanted to I could figure
out what the bounds are

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in that region in terms of x and
y, and I could do a lot of

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work and integrate it
all, or I could

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remember one simple fact.

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Which is that if I have--

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I think you see it in class as
x bar-- the center of mass

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should be equal to 1 over the
volume of the region times the

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integral of x dA over
the region.

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That's what we know.

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This should be-- this maybe
doesn't look like a v. But so

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in this case volume
is area, isn't it?

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Maybe I should write area, that
might make you nervous.

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So if I take the area--

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I could just say A of R-- if
I take the area, 1 over the

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area, and then I multiply by the
integral of x over R with

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dy dx with respect to dA, then
I get the center of mass.

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Well, let's look at what in
this picture-- what is the

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center of mass here?

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If I want to balance this thing
on a pencil tip, if I

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want to balance this disc--
assuming the density is

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everywhere the same on a pencil
tip-- where am I going

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to put the pencil?

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I'm going to put it right
at the center.

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The x value there is a, the
y value there is 0.

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So if I had the y center of
mass, I would want 0.

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But I want the x center
of mass, so I want a.

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This x bar is actually
equal to, from the

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picture, is equal to a.

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And now let's notice
what I've done.

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I have taken--

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I had this quantity here--

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I've taken this quantity except
for the 4 and I have a

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way of writing explicitly what
this quantity is without

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actually doing any of
the integration.

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I haven't done any sophisticated
things at this

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point except know what the
center of mass is.

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So also what is the area
of the region?

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I'm going to need that.

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The area of the region-- it's
a circle of radius a.

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So the area is pi a squared.

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So this quantity is
pi a squared.

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And all this together tells me
that the integral of x over R

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dA, if I solve for this part,
I get a times pi a squared.

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So I get pi a cubed.

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And that only differs from
our answer by one thing.

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There's a scalar--

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you multiply by 4 and that
gives us what we want.

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I said differs from
our answer.

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Differs from what we
want by one thing.

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We just multiplied by 4 there,
so we multiply by 4 here.

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So in fact--

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maybe there looks like there
was a little magic here, so

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let me point out some of the
key points at the end.

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I started off knowing I was
trying to integrate 4x over

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the region R. And R in this case
was a circle centered at

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(a, 0) and of radius a.

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And then I said, well,
I don't want to do a

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lot of work for this.

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So I'm going to not cheat, but
I'm going to use my knowledge

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of the center of mass
to make this easier.

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So the center of mass is equal
to 1 divided by the area of

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the region times the integral
of x over the region.

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So I want to find the integral
of x over the region, I just

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solve for the integral
of x over the region.

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I just solve for that part.

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The center of mass-- from just
looking at the picture and

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understanding what the center of
mass means-- the center of

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mass in the x component is a.

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The area is pi a squared.

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So I end up with a pi a
cubed when I solve for

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the integral of x.

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And then because I wanted the
integral of 4x, I just

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multiply by 4.

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And so the final answer is
actually 4 pi a cubed.

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So what I was trying to find,
if you remember, was I was

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trying to find the value when
I integrated over this curve

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of a certain vector field.

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And that one was going
to be a little messy

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to do it that way.

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But Green's theorem, actually
there's a lot of cancellation

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which makes it much easier, and
then the center of mass is

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a nice little trick
to use at the end.

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And then the calculation
is quite simple.

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So I think that's
where I'll stop.

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