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DAVID JORDAN: Hello, and
welcome back to recitation.

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So today I want to practice
doing, computing some integrals

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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,

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they're given to you in
rectangular coordinates.

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So the first thing
you need to do

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is to re-express these
in polar coordinates.

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And so for the
first one, part a,

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I want you to completely
compute the integral.

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For part b and c, we
have this function f,

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which we haven't 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

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just set it up 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 rectangular

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to polar coordinates,
the most difficult part

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is understanding what region
we're integrating over.

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

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is just straightforward
calculation.

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

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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 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

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need to do to express something
in polar coordinates is

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we need to sweep out
rays of constant angle,

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and then measure the
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, for instance 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

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theta reaches its 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,

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it's going to be a
little bit more subtle,

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because what we need to do is
we need to fix some arbitrary

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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

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to describe how r varies
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

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to do some simple
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 point is
that we have-- so

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let me draw 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 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 r that we're after.

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OK?

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So in basic trigonometry
we know that cos theta is

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equal to-- 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,

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we go over length 2.

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And so our integral becomes the
integral from 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

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our original function
in terms of r and theta.

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So our original
function was-- so let's

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go back over to
the 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 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,

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and now that we've
done all the geometry,

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this is going to be a
straightforward integral

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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,

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and r running 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-- 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 back

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into a cosine, right?

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So this is just the
integral from theta

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equals 0 to pi over 4.

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OK, because we have
this minus sign here,

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we're going to get cos theta
minus-- 1 over 2 sec theta

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becomes 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

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how the boundary curves of
your original region re-express

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in the r and theta coordinates.

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

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and c.

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So in part b-- 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 curve is

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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

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to figure out the
range for theta.

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Because you see
this parabola here,

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as it approaches the
origin, it becomes flat.

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So the theta, as we get closer
and closer to 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

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have to do is we
just have to use

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our polar change of coordinates
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 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,

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they all start at r
equals 0, because we

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have this sharp point
here at the origin.

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And they all stop at this curve:
r equals 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,

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but we'll see what
we can do about that.

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

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and x ranges from 0
to the square root

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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,

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I didn't recognize 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

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x equals the square root
of 2y minus y squared.

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And this looks like it wants
to be the equation of a circle.

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So let's see if we can turn this
into an equation of a circle.

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So if we square both
sides of this equation,

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then we get x squared
equals 2y minus y squared.

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And now this over
here, 2y minus y

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squared, so if we add this
over to the other side,

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we have x squared plus y
squared minus 2y equals 0.

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And now the thing that I notice
about this expression here

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is it really looks almost like
the 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

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squared-- except the
left-hand side is not

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quite equal to that, it's
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

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says that, this is
the equation if I add

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this 1 over to the other
side-- 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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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,

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because we were told
that x starts at 0

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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

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figure out the values of
theta and the values of r.

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So first of all, theta
is going to range again

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from 0, because
this bottom curve,

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as it comes into the
origin, it 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

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is going to go all the
way up until pi over 2,

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because theta can be
pointing straight up.

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OK?

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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

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and it stops at this curve.

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So we need to figure out
what the r-value is along

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this curve in terms of theta.

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00:16:10,440 --> 00:16:15,660
So what we want to do
is use the equation.

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So we had it over here.

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So x squared plus y
squared minus 2y equals 0.

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So that was one of the
equivalent equations

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00:16:27,720 --> 00:16:29,420
that we got for our curve.

246
00:16:29,420 --> 00:16:33,040
And now this is nice, because
this here is just r squared,

247
00:16:33,040 --> 00:16:33,935
isn't it?

248
00:16:33,935 --> 00:16:36,630
So this is r squared.

249
00:16:36,630 --> 00:16:42,190
And this is minus 2y,
and y is r sine theta.

250
00:16:46,360 --> 00:16:47,040
OK.

251
00:16:47,040 --> 00:16:51,410
So we get r squared minus
2r sine theta equals 0.

252
00:16:51,410 --> 00:16:55,610
Solving this for r, we get
that r is just 2 sine theta.

253
00:16:59,041 --> 00:16:59,540
OK.

254
00:17:04,420 --> 00:17:06,310
So r equals 2 sine theta.

255
00:17:06,310 --> 00:17:09,060
That is the equation in r,
theta coordinates, which

256
00:17:09,060 --> 00:17:11,550
describes this semicircle here.

257
00:17:11,550 --> 00:17:15,360
And so altogether,
we get the range

258
00:17:15,360 --> 00:17:22,580
is from r equals 0 to
r equals 2 sine theta.

259
00:17:22,580 --> 00:17:30,716
And f just comes along for the
ride, and we have r dr d theta.

260
00:17:30,716 --> 00:17:32,553
And I'll leave it at that.