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DAVID JORDAN: Hello, and welcome
back to recitation.

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So today I want to practice
doing some computings of

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integrals with you at
polar coordinates.

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So we have these three integrals
set up here.

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And the way the integrals are
given to you, they're given to

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you in rectangular
coordinates.

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So the first thing you need to
do is to re-express these in

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polar coordinates.

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And so for the first one, part
a, I want you to completely

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compute the integral.

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For part b and c, we have this
function f, which we haven't

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specified yet.

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So the exercise is to
set up the integral.

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We won't actually compute
it, we'll just set it up

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completely.

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So rewrite it in terms
of r and theta.

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So why don't you pause the video
and get started on that.

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And check back with me and we'll
work it out together.

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Welcome back.

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Let's get started.

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So for all of these, when
we're transferring from

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rectangular to polar
coordinates, the most

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difficult part is understanding
what region

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we're integrating over.

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And if we can understand that,
then the rest of it is just

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straightforward calculation.

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So in part a, let's try to think
about what this region

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is that we're given.

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So we're given the region from
x equals 1 to x equals 2.

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So let's draw those lines. x
equals 1 and x equals 2.

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OK, so that's how x varies
in between this band.

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And now the y varies from
0 to the function x.

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So we'd better draw the
line y equals x.

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OK, so this is the
line y equals x.

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And the region that we're after
is in between this band

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and in between these two lines,
so it's this region

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right here.

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That's the region
that we want.

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All right.

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So now that we have that, what
we need to do to express

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something in polar coordinates
is we need to sweep out rays

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of constant angle, and
then measure the

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radius along the ray.

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So let's see what
I mean by that.

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So we can kind of see that if
we're inside this region here,

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the smallest value that theta
can be is just theta equals 0.

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Because, sprint is right here is
a point at theta equals 0.

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And then theta runs
all the way up.

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And this whole line here is
where theta reaches its

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maximum, and that's at
theta is pi over 4.

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So this tells us that it's
easiest to write the theta

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integral on the outside.

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And we're going to have theta
running from 0 to pi over 4.

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OK, very good.

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Now, in order to get the r
ranges, it's going to be a

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little bit more subtle, because
what we need to do is

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we need to fix some arbitrary
theta in the middle here.

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So I'll just draw kind of
a representative theta.

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It would be like
this one here.

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So there is a representative
theta.

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And what we need to do is, we
need to describe how r varies

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along this pink line here.

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So that's the line that
we really want.

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OK.

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So in order to do that, we just
need to do some simple

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trigonometry now.

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So let's see.

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So r turns on at this
line right here.

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At the line x equals 1: that's
where r turns on.

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So let's see what the value
of r is at that line here.

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So, the pont is that we
have --so let me draw

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this triangle here--.

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Let me blow this triangle up.

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OK.

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So this is our line x equals
1, and this is

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this triangle here.

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We just put a magnifying
glass on it.

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So we have this angle theta.

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OK.

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So this x-value is 1, because
we're on the line x equals 1.

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So the length of this leg
is 1, and this is

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r that we're after.

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OK?

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So in basic trigonometry
we know that cos

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theta is equal to--

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so cosine is adjacent
over hypotenuse--

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is equal to 1 over r.

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OK, so that tells us that
r is equal to sec theta.

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OK?

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So what that means is that this,
for any given theta,

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this pink band turns on at
precisely r equals sec theta.

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And so that's what we need to
write in the bottom here.

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OK?

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Now, you can see that it turns
off at r equals 2 sec theta.

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Because now instead of going
over length 1, we

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go over length 2.

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And so our integral becomes
the integral from

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0 to pi over 4.

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And r equals sec theta
to 2 sec theta.

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OK, and now we need to
re-express our original

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function in terms
of r and theta.

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So our original function was,
so let's go back over to the

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formula here.

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So we have x squared plus
y squared to the 3/2.

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So x squared plus y squared
to the 1/2 that's r.

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And so x squared plus
y squared to the

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3/2 that's r cubed.

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So this is 1 over r cubed.

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So this is 1 over
r cubed in here.

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And as we know, dy dx becomes
r dr d theta.

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OK.

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So this is the integral that
we need to compute, and now

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that we've done all the
geometry, this is going to be

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a straightforward integral
to compute.

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So let's do it together.

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I'll just rewrite
it over here.

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So we have theta running from 0
to pi over 4, and r running

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from sec theta to 2 sec theta.

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And we have 1 over r squared,
dr d theta.

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OK.

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So that equals--

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taking the inside
integral first--

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we get negative 1 over r from
2 sec theta to sec theta.

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OK?

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And now this is nice,
because sec theta--

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if we plug it in for 1 over r--
it's just going to turn

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back into a cosine, right?

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So this is just the integral
from theta equals

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0 to pi over 4.

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OK, because we have this minus
sign here, we're going to get

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cos theta minus--

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1 over 2 sec theta becomes--

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1/2 cos theta, d theta.

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And this we can compute to be
the square root of 2 over 4.

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So this is an elementary
integral that we can compute:

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the square root of 2, over 4.

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OK.

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And that's it.

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That's all there is to a.

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So notice that the difficult
part is figuring out how the

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boundary curves of your original
region re-express in

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the r and theta coordinates.

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So let's see if we can get more
practice with that on

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parts b and c.

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So in part b--

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let me just recall--

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we're taking the integral
x equals 0 to 1.

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And y equals x squared
to x, f dy dx.

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So let's draw this
region again.

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So the bottom curve is y equals
x squared and the top

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curve is y equals x, and
x runs from 0 to 1.

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So this is the region
that we're after.

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OK.

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So once again in this case, it's
pretty easy to figure out

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the range for theta.

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Because you see this parabola
here, as it approaches the

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origin, it becomes flat.

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So the theta, as we get
closer and closer to

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the origin, is 0.

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So the initial bound
for theta is 0.

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It doesn't matter that we have
this curve here, it's still 0.

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And then the top bound,
again, is pi over 4.

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OK?

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Now, for this curve y equals x
squared, what we just have to

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do is we just have to use our
polar change of coordinates

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and re-express this curve.

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So y equals x squared.

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Well, y is r sine theta.

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And x is r cos theta.

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So this is altogether r squared
cos squared theta.

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OK?

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And now all we need to do is
just solve for r here.

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So we get r equals-- so I can
cancel that r with that one--

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and it looks to me like we get
r equals tan theta secant

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theta by solving.

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OK.

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So that's the r-value
at each of these

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points along the curve.

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So that is going to
be our top bound.

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So if we draw little segments
of constant theta, they all

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start at r equals 0, because we
have this sharp point here

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at the origin.

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And they all stop at this
curve: r equals

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tan theta sec theta.

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All right?

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And so now f we're just
going to leave put.

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And then finally we
have r dr d theta.

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OK.

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So that's our answer to b.

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All right.

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Now c is going to be a
little bit tricky.

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What's tricky about c is drawing
a picture, but we'll

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see what we can do about that.

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So let me just remind you. y
ranges from 0 to 2, and x

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ranges from 0 to the square root
of 2y minus y squared.

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And then we have f dx dy.

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OK.

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Well, the y bounds are pretty
straightforward, 0 to 2.

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This function, when I first saw
this, I didn't recognize

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this function right away.

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So let's see what we can
make out of this curve.

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So the top curve for x is
going to be x equals the

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square root of 2y
minus y squared.

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And this looks like
it wants to be the

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equation of a circle.

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So let's see if we can
turn this into an

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equation of a circle.

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So if we square both sides of
this equation, then we get x

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squared equals 2y
minus y squared.

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And now this over here, 2y minus
y squared, so if we add

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this over to the other side,
we have x squared plus y

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squared minus 2y equals 0.

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And now the thing that I notice
about this expression

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here is it really looks
almost like the

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quantity y minus 1 squared.

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Yeah?

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If there were a plus 1 here,
then that would be the case.

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So this is actually x squared,
plus y minus 1 squared--

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except the left-hand side is not
quite equal to that-- it's

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equal to that, minus 1.

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So this equation
just equals 0.

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So this equation is just
equivalent to that one.

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And that's good news, because
this says that, this is the

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equation if I add this 1 over
to the other side--

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let me go over here--

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we have x squared, plus y minus
1 squared, equals 1.

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And this, we know
what this is.

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00:14:21,270 --> 00:14:25,650

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This is a circle which is
centered not at the origin,

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but at the point 0 comma
1 in the y-direction.

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So if we draw this, well,
this is the equation

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for the entire circle.

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But we only want the positive
half, because we were told

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that x starts at 0 and goes up
to this positive number.

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So we only want the positive
half here.

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OK.

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So this is the region that
we're integrating.

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And now that we know this, we
can again figure out the

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values of theta and
the values of r.

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00:15:10,300 --> 00:15:13,090

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00:15:13,090 --> 00:15:19,880
So first of all, theta is going
to range again from 0,

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because this bottom curve, as
it comes into the origin, it

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comes in flat.

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So we have points of arbitrarily
small angles.

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So theta is going to start a
0, and theta is going to go

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all the way up until pi over
2, because theta can be

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00:15:35,770 --> 00:15:37,020
pointing straight up.

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00:15:37,020 --> 00:15:39,630

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00:15:39,630 --> 00:15:40,210
OK?

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00:15:40,210 --> 00:15:51,790
And now we have to think about
these lines of constant angle,

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and we need to think about
what r does in there.

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So we can see from the picture
that r always starts at 0 and

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it stops at this curve.

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00:16:04,280 --> 00:16:07,900
So we need to figure out what
the r-value is along this

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curve in terms of theta.

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So what we want to do
is use the equation.

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00:16:15,660 --> 00:16:16,720
So we had it over here.

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00:16:16,720 --> 00:16:25,380
So x squared plus y squared
minus 2y equals 0.

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00:16:25,380 --> 00:16:27,950
So that was one of the
equivalent equations that we

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00:16:27,950 --> 00:16:29,420
got for our curve.

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00:16:29,420 --> 00:16:32,360
And now this is nice, because
this here is just r

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squared, isn't it?

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00:16:33,935 --> 00:16:36,630
So this is r squared.

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00:16:36,630 --> 00:16:42,190
And this is minus 2y, and
y is r sine theta.

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00:16:42,190 --> 00:16:46,360

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OK.

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00:16:47,040 --> 00:16:51,410
So we get r squared minus
2r sine theta equals 0.

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00:16:51,410 --> 00:16:55,610
Solving this for r, we get that
r is just 2 sine theta.

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00:16:55,610 --> 00:16:59,260

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00:16:59,260 --> 00:17:00,510
OK.

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00:17:00,510 --> 00:17:04,420

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00:17:04,420 --> 00:17:06,310
So r equals 2 sine theta.

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00:17:06,310 --> 00:17:09,060
That is the equation in r,
theta coordinates, which

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00:17:09,060 --> 00:17:11,550
describes this semicircle
here.

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00:17:11,550 --> 00:17:17,930
And so altogether, we get the
range is from r equals 0 to r

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00:17:17,930 --> 00:17:22,580
equals 2 sine theta.

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00:17:22,580 --> 00:17:25,300
And f just comes along
for the ride, and

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00:17:25,300 --> 00:17:30,716
we have r dr d theta.

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00:17:30,716 --> 00:17:32,830
And I'll leave it at that.

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00:17:32,830 --> 00:17:33,053