WEBVTT
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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
00:00:23.805 --> 00:00:27.850
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
00:01:09.830 --> 00:01:13.310
to polar coordinates,
the most difficult part
00:01:13.310 --> 00:01:17.400
is understanding what region
we're integrating over.
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And if we can understand
that, then the rest of it
00:01:19.560 --> 00:01:22.340
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
00:09:19.170 --> 00:09:22.980
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
00:09:55.640 --> 00:09:59.430
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
00:12:05.760 --> 00:12:08.730
of 2y minus y squared.
00:12:13.060 --> 00:12:18.740
And then we have f dx dy.
00:12:18.740 --> 00:12:21.130
OK.
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Well, the y bounds are pretty
straightforward, 0 to 2.
00:12:24.170 --> 00:12:26.887
This function, when
I first saw this,
00:12:26.887 --> 00:12:28.720
I didn't recognize this
function right away.
00:12:28.720 --> 00:12:30.750
So let's see what we can
make out of this curve.
00:12:33.940 --> 00:12:39.580
So the top curve
for x is going to be
00:12:39.580 --> 00:12:46.490
x equals the square root
of 2y minus y squared.
00:12:46.490 --> 00:12:53.987
And this looks like it wants
to be the equation of a circle.
00:12:53.987 --> 00:12:56.570
So let's see if we can turn this
into an equation of a circle.
00:12:56.570 --> 00:13:00.100
So if we square both
sides of this equation,
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then we get x squared
equals 2y minus y squared.
00:13:11.260 --> 00:13:15.990
And now this over
here, 2y minus y
00:13:15.990 --> 00:13:19.810
squared, so if we add this
over to the other side,
00:13:19.810 --> 00:13:28.610
we have x squared plus y
squared minus 2y equals 0.
00:13:28.610 --> 00:13:32.900
And now the thing that I notice
about this expression here
00:13:32.900 --> 00:13:37.530
is it really looks almost like
the quantity y minus 1 squared.
00:13:37.530 --> 00:13:38.030
Yeah?
00:13:38.030 --> 00:13:41.680
If there were a plus 1 here,
then that would be the case.
00:13:41.680 --> 00:13:49.190
So this is actually x
squared, plus y minus 1
00:13:49.190 --> 00:13:53.100
squared-- except the
left-hand side is not
00:13:53.100 --> 00:13:56.190
quite equal to that, it's
equal to that, minus 1.
00:13:56.190 --> 00:13:59.150
So this equation just equals 0.
00:13:59.150 --> 00:14:02.020
So this equation is just
equivalent to that one.
00:14:02.020 --> 00:14:04.624
And that's good
news, because this
00:14:04.624 --> 00:14:06.290
says that, this is
the equation if I add
00:14:06.290 --> 00:14:10.760
this 1 over to the other
side-- let me go over here--
00:14:10.760 --> 00:14:19.570
we have x squared, plus y
minus 1 squared, equals 1.
00:14:19.570 --> 00:14:21.270
And this, we know what this is.
00:14:25.650 --> 00:14:28.310
This is a circle which is
centered not at the origin,
00:14:28.310 --> 00:14:33.370
but at the point 0 comma
1 in the y-direction.
00:14:33.370 --> 00:14:46.990
So if we draw this, well,
this is the equation
00:14:46.990 --> 00:14:50.420
for the entire circle.
00:14:50.420 --> 00:14:53.720
But we only want
the positive half,
00:14:53.720 --> 00:14:56.530
because we were told
that x starts at 0
00:14:56.530 --> 00:14:58.280
and goes up to this
positive number.
00:14:58.280 --> 00:15:00.730
So we only want the
positive half here.
00:15:00.730 --> 00:15:03.390
OK.
00:15:03.390 --> 00:15:05.290
So this is the region
that we're integrating.
00:15:05.290 --> 00:15:07.880
And now that we know
this, we can again
00:15:07.880 --> 00:15:10.300
figure out the values of
theta and the values of r.
00:15:13.090 --> 00:15:19.250
So first of all, theta
is going to range again
00:15:19.250 --> 00:15:22.520
from 0, because
this bottom curve,
00:15:22.520 --> 00:15:25.060
as it comes into the
origin, it comes in flat.
00:15:25.060 --> 00:15:28.380
So we have points of
arbitrarily small angles.
00:15:28.380 --> 00:15:31.150
So theta is going to
start a 0, and theta
00:15:31.150 --> 00:15:34.831
is going to go all the
way up until pi over 2,
00:15:34.831 --> 00:15:36.580
because theta can be
pointing straight up.
00:15:39.630 --> 00:15:40.210
OK?
00:15:40.210 --> 00:15:51.790
And now we have to think about
these lines of constant angle,
00:15:51.790 --> 00:15:55.960
and we need to think about
what r does in there.
00:15:55.960 --> 00:16:02.300
So we can see from the picture
that r always starts at 0
00:16:02.300 --> 00:16:04.280
and it stops at this curve.
00:16:04.280 --> 00:16:07.740
So we need to figure out
what the r-value is along
00:16:07.740 --> 00:16:10.440
this curve in terms of theta.
00:16:10.440 --> 00:16:15.660
So what we want to do
is use the equation.
00:16:15.660 --> 00:16:16.720
So we had it over here.
00:16:16.720 --> 00:16:25.380
So x squared plus y
squared minus 2y equals 0.
00:16:25.380 --> 00:16:27.720
So that was one of the
equivalent equations
00:16:27.720 --> 00:16:29.420
that we got for our curve.
00:16:29.420 --> 00:16:33.040
And now this is nice, because
this here is just r squared,
00:16:33.040 --> 00:16:33.935
isn't it?
00:16:33.935 --> 00:16:36.630
So this is r squared.
00:16:36.630 --> 00:16:42.190
And this is minus 2y,
and y is r sine theta.
00:16:46.360 --> 00:16:47.040
OK.
00:16:47.040 --> 00:16:51.410
So we get r squared minus
2r sine theta equals 0.
00:16:51.410 --> 00:16:55.610
Solving this for r, we get
that r is just 2 sine theta.
00:16:59.041 --> 00:16:59.540
OK.
00:17:04.420 --> 00:17:06.310
So r equals 2 sine theta.
00:17:06.310 --> 00:17:09.060
That is the equation in r,
theta coordinates, which
00:17:09.060 --> 00:17:11.550
describes this semicircle here.
00:17:11.550 --> 00:17:15.360
And so altogether,
we get the range
00:17:15.360 --> 00:17:22.580
is from r equals 0 to
r equals 2 sine theta.
00:17:22.580 --> 00:17:30.716
And f just comes along for the
ride, and we have r dr d theta.
00:17:30.716 --> 00:17:32.553
And I'll leave it at that.