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Welcome back to recitation.

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In this video, what I'd like
you to do is use Green's theorem

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to find the area between
one arch of the cycloid that

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is described by x equals
a times the quantity

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theta minus sine theta, y equals
a times the quantity 1 minus

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cosine theta, and the x-axis.

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So probably it's
helpful for you to draw

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a picture of this first.

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So you have one arch
of the cycloid defined

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in terms of theta, the x of
theta and y of theta are here,

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and then the x-axis
will be the other bound.

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And the big thing I
want you to notice

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is you're going to use Green's
theorem to find the area.

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So why don't you pause the
video, work on the problem,

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when you're ready to
see my solution you

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can bring the video back up.

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OK, welcome back.

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So again, what we want to do is
we want to use Green's theorem

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to find an area.

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OK, and what I'm
going to do first

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is I'm going to try and figure
out what the picture looks like

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and then I'm going to
figure out how to use

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Green's theorem in this region.

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So lets first draw a sketch of
the picture of what's going on.

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And because I want the area
between one arch of the cycloid

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and the x-axis, what I
really want to do first

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is figure out what
theta values will

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make y positive, or
greater than or equal to 0,

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and go from there.

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So if I want y to be
greater than or equal to 0,

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then what I need is
1 minus cosine theta

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to be greater than
or equal to 0.

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And that's going to happen for
every theta from 0 to 2*pi,

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and at 0 and 2*pi I get 0.

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So we know we want theta
to go between 0 and 2*pi.

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Those are good parameters
to have for theta,

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because this statement
is always true,

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but I get 0 when theta is
0 and when theta is 2*pi.

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So I'm going to let theta
run between 0 and 2*pi,

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as I mentioned.

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So now let's think about
what the picture is.

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When theta is 0, if I come over
to what x is, x is a times 0

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minus 0, So x is 0.

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And y we said was 0.

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I know that I'll
achieve a maximum height

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at theta equal pi because that's
where cosine theta is minus 1.

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So the 1 minus
cosine theta is 2.

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And so if you look at
what the y-value will be,

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the y-value when
theta is pi is 2a.

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And the value is going to
be a times pi minus sine pi.

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Well, sine of pi is 0.

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So I just get a*pi.

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So I should go over a*pi.

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And I know there's a point at
a*pi comma-- I think I said 2a.

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And then it's actually going
to come back down the same way,

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so let me draw the first part.

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This is not a perfect
drawing of this.

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I don't have lots of points.

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I'm having the smallest
amount of points

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possible to figure
out what's happening.

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So then when theta is 2*pi,
1 minus cosine theta is 0.

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So I get y equals 0.

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And then here I get 2
pi minus sine of 2*pi.

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Sine of 2 pi is 0,
so I get 2*pi*a.

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So I come all the way over
here and I get 2*pi*a.

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Yeah, that's not maybe the
most beautiful picture,

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but it's a good start.

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So I'm interested in finding
the area of this region.

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What I want to do then is, to
find the area of that region

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I'm going to use
Green's theorem.

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That's what we asked you to do.

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So what I want to do is I
want to pick a curve that's

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going to have 2 components.

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It's going to have C_1,
will be this bottom part,

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and C2 will be this top part.

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And we need to
keep it oriented so

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when I'm walking along the
curve the region is on the left.

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And then I have to figure out
how to use Green's theorem.

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So to find the area of this
region, all I actually need

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is the integral over
the region of dA.

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That would give me the area.

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And so what quantities--
what functions would be good?

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Sorry.

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In order to find this.

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Well, I believe you
actually saw in class

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that you have two options.

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You have minus y*dx
over the closed curve,

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or you have x*dy over
the closed curve.

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You have one of
those two options.

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Now which one of those
is better in this case?

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We have a choice, so let's
make the best choice.

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Which one of those is better?

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Well, if you notice that
on these two curves,

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x is changing on both of them
but y is fixed on this one.

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So it'd be nicer for
us if we could just

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integrate on the curve
on this part, which

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we would be able to do if
we were choosing this one.

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Because y is constant and
it is equal to 0 along here.

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So what you need to do to
solve this problem is really,

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you only need to integrate
on C_2, minus y*dx.

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That's going to give
you the entire area.

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And so now we have to figure
out how to put everything in.

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Well, C_2-- I'm
parameterized in theta--

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and C_2 parameterized in theta,
if I start here I'm at 2*pi

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for theta.

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And if I go this direction
I'm at 0 for theta.

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So what I'm going
to do is I'm going

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to do something that might seem
tricky, but it's quite natural.

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I'm going to replace this minus
by a plus and then integrate

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from 0 to 2*pi.

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Because I was going to be
integrating from 2*pi to 0

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and so I just flipped the order
and made that sign a plus.

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So that should be fine.

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And then y, I already
know what it is.

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It's a times 1
minus cosine theta.

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And now I just have to
figure out what is dx.

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So we have to come back
over, for one second

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we're going to come over here
and we're going to look at dx.

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If x is equal to a times
theta minus sine theta,

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then dx is just a times
1 minus cosine theta.

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So that's nice because that's
actually what we already have.

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So we come back
over here-- oh, I

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should say also with a d
theta-- come back over here,

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dx gives you a squared a,
it gives you a squared this,

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and it gives you a d theta.

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So that actually gives
you everything you need.

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Now I am not going to actually
write out all the steps

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in order to do this.

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But the way you
could easily do this,

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you can pull out the
a squared and you're

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left with a 1 minus
cosine theta squared.

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You might think about
squaring all the terms

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and then you have 1 cosine
squared theta to deal with.

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But ultimately, when
you get your answer--

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I'm just going to
write down what it is--

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you get 3*pi a squared.

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And you'll see you get a
little bit of cancellation

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and it's not too hard
to solve at this point.

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Since it's single
variable I'm going

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to assume I don't need to do it.

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But you can check
your final answer.

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You should get-- I think, yes.

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I got 3*pi a squared.

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So what we did was we
took a problem, again,

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where we had a curve
defined in terms of theta.

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We had a curve, in x and y,
it was parameterized in theta.

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And we wanted to find an
area between that curve

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and the x-axis.

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And the reason we could use
Green's theorem so easily is

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because area of a region-- of
this connected region here--

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is equal to the integral over
the closed loop surrounding it

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of minus y*dx, or alternatively
you could have done x*dy.

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And because I got to choose,
I picked the easier one.

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And I picked the
easier one in this case

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because y is equal to 0
on one part of the curve.

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So I just ignored the C_1
because that's where y is 0,

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and I looked along C_2.

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I noticed C_2 is parameterized
in this direction from theta

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equals 2*pi to theta equals 0.

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So that's why I dropped the
minus sign and changed it from

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0 to 2*pi.

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That's where that came from.

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And then you end up having
to just determine what y is

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and what dx are,
in terms of theta.

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Which we had them already
in terms of theta,

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so we could just explicitly
determine everything,

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do all the substitution and
then evaluate the integral.

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So then, that's all
there is to this problem.

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So hopefully that
taught you a little more

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about Green's theorem.

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And actually, now
you've done problems

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going from the left to the right
and from the right to the left.

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So you've done a
little bit of both.

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So hopefully this
revealed a little bit more

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to you than you knew before.

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That's where I'll stop.