WEBVTT

00:00:06.910 --> 00:00:09.300
DAVID JORDAN: Hello, and
welcome back to recitation.

00:00:09.300 --> 00:00:12.030
So today I want to practice
doing, computing some integrals

00:00:12.030 --> 00:00:13.280
with you at polar coordinates.

00:00:13.280 --> 00:00:17.880
So we have these three
integrals set up here.

00:00:17.880 --> 00:00:20.431
And the way the integrals
are given to you,

00:00:20.431 --> 00:00:22.430
they're given to you in
rectangular coordinates.

00:00:22.430 --> 00:00:23.805
So the first thing
you need to do

00:00:23.805 --> 00:00:27.850
is to re-express these
in polar coordinates.

00:00:27.850 --> 00:00:31.000
And so for the
first one, part a,

00:00:31.000 --> 00:00:34.870
I want you to completely
compute the integral.

00:00:34.870 --> 00:00:37.509
For part b and c, we
have this function f,

00:00:37.509 --> 00:00:38.800
which we haven't specified yet.

00:00:38.800 --> 00:00:40.910
So the exercise is to
set up the integral.

00:00:40.910 --> 00:00:42.830
We won't actually
compute it, we'll

00:00:42.830 --> 00:00:44.490
just set it up completely.

00:00:44.490 --> 00:00:47.410
So rewrite it in
terms of r and theta.

00:00:47.410 --> 00:00:50.370
So why don't you pause the
video and get started on that.

00:00:50.370 --> 00:00:52.620
And check back with me and
we'll work it out together.

00:01:00.439 --> 00:01:00.980
Welcome back.

00:01:00.980 --> 00:01:01.750
Let's get started.

00:01:05.120 --> 00:01:09.830
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.

00:01:17.400 --> 00:01:19.560
And if we can understand
that, then the rest of it

00:01:19.560 --> 00:01:22.340
is just straightforward
calculation.

00:01:22.340 --> 00:01:26.647
So in part a, let's try to
think about what this region is

00:01:26.647 --> 00:01:27.355
that we're given.

00:01:32.580 --> 00:01:37.790
So we're given the region
from x equals 1 to x equals 2.

00:01:37.790 --> 00:01:45.110
So let's draw those lines.
x equals 1 and x equals 2.

00:01:45.110 --> 00:01:49.950
OK, so that's how x varies,
in between this band.

00:01:49.950 --> 00:01:54.350
And now the y varies
from 0 to the function x.

00:01:54.350 --> 00:01:56.630
So we'd better draw
the line y equals x.

00:01:59.720 --> 00:02:03.960
OK, so this is the
line y equals x.

00:02:03.960 --> 00:02:07.110
And the region that we're
after is in between this band

00:02:07.110 --> 00:02:10.870
and in between these two lines,
so it's this region right here.

00:02:13.760 --> 00:02:15.590
That's the region that we want.

00:02:15.590 --> 00:02:16.270
All right.

00:02:16.270 --> 00:02:18.370
So now that we
have that, what we

00:02:18.370 --> 00:02:22.510
need to do to express something
in polar coordinates is

00:02:22.510 --> 00:02:30.360
we need to sweep out
rays of constant angle,

00:02:30.360 --> 00:02:32.605
and then measure the
radius along the ray.

00:02:32.605 --> 00:02:33.980
So let's see what
I mean by that.

00:02:38.570 --> 00:02:42.610
So we can kind of see that if
we're inside this region here,

00:02:42.610 --> 00:02:46.940
the smallest value that theta
can be is just theta equals 0.

00:02:46.940 --> 00:02:51.350
Because, for instance right here
is a point at theta equals 0.

00:02:51.350 --> 00:02:53.920
And then theta runs
all the way up.

00:02:53.920 --> 00:02:56.660
And this whole
line here is where

00:02:56.660 --> 00:02:59.990
theta reaches its maximum, and
that's at theta is pi over 4.

00:03:04.050 --> 00:03:07.370
So this tells us that it's
easiest to write the theta

00:03:07.370 --> 00:03:08.950
integral on the outside.

00:03:08.950 --> 00:03:16.690
And we're going to have theta
running from 0 to pi over 4.

00:03:16.690 --> 00:03:17.850
OK, very good.

00:03:17.850 --> 00:03:24.760
Now, in order to
get the r ranges,

00:03:24.760 --> 00:03:26.510
it's going to be a
little bit more subtle,

00:03:26.510 --> 00:03:29.360
because what we need to do is
we need to fix some arbitrary

00:03:29.360 --> 00:03:30.970
theta in the middle here.

00:03:30.970 --> 00:03:34.185
So I'll just draw kind of
a representative theta.

00:03:34.185 --> 00:03:36.330
It would be like this one here.

00:03:38.910 --> 00:03:40.580
So there is a
representative theta.

00:03:40.580 --> 00:03:48.370
And what we need
to do is, we need

00:03:48.370 --> 00:03:51.970
to describe how r varies
along this pink line here.

00:03:51.970 --> 00:03:55.500
So that's the line
that we really want.

00:03:55.500 --> 00:03:56.470
OK.

00:03:56.470 --> 00:03:58.620
So in order to do
that, we just need

00:03:58.620 --> 00:04:02.300
to do some simple
trigonometry now.

00:04:02.300 --> 00:04:03.000
So let's see.

00:04:03.000 --> 00:04:06.260
So r turns on at
this line right here.

00:04:06.260 --> 00:04:09.070
At the line x equals 1;
that's where r turns on.

00:04:09.070 --> 00:04:14.101
So let's see what the value
of r is at that line here.

00:04:14.101 --> 00:04:17.045
So, the point is
that we have-- so

00:04:17.045 --> 00:04:18.560
let me draw this triangle here.

00:04:21.210 --> 00:04:22.730
Let me blow this triangle up.

00:04:30.850 --> 00:04:31.870
OK.

00:04:31.870 --> 00:04:40.610
So this is our line x equals 1,
and this is this triangle here.

00:04:40.610 --> 00:04:43.310
We just put a
magnifying glass on it.

00:04:43.310 --> 00:04:46.736
So we have this angle theta.

00:04:46.736 --> 00:04:47.235
OK.

00:04:50.770 --> 00:04:56.000
So this x-value is 1, because
we're on the line x equals 1.

00:04:56.000 --> 00:05:02.260
So the length of this leg is 1,
and this is r that we're after.

00:05:02.260 --> 00:05:02.760
OK?

00:05:02.760 --> 00:05:08.230
So in basic trigonometry
we know that cos theta is

00:05:08.230 --> 00:05:14.400
equal to-- so cosine is
adjacent over hypotenuse--

00:05:14.400 --> 00:05:17.560
is equal to 1 over r.

00:05:17.560 --> 00:05:23.260
OK, so that tells us that
r is equal to sec theta.

00:05:23.260 --> 00:05:24.020
OK?

00:05:24.020 --> 00:05:26.940
So what that means is that
this, for any given theta,

00:05:26.940 --> 00:05:30.940
this pink band turns on at
precisely r equals sec theta.

00:05:30.940 --> 00:05:33.620
And so that's what we need
to write in the bottom here.

00:05:36.560 --> 00:05:37.470
OK?

00:05:37.470 --> 00:05:43.670
Now, you can see that it turns
off at r equals 2 sec theta.

00:05:46.470 --> 00:05:49.570
Because now, instead
of going over length 1,

00:05:49.570 --> 00:05:51.110
we go over length 2.

00:05:51.110 --> 00:05:56.270
And so our integral becomes the
integral from 0 to pi over 4.

00:05:56.270 --> 00:05:59.540
And r equals sec
theta to 2 sec theta.

00:05:59.540 --> 00:06:03.220
OK, and now we
need to re-express

00:06:03.220 --> 00:06:06.180
our original function
in terms of r and theta.

00:06:06.180 --> 00:06:09.800
So our original
function was-- so let's

00:06:09.800 --> 00:06:14.580
go back over to
the formula here.

00:06:14.580 --> 00:06:18.310
So we have x squared plus
y squared to the 3/2.

00:06:18.310 --> 00:06:21.650
So x squared plus y squared
to the 1/2, that's r.

00:06:21.650 --> 00:06:24.500
And so x squared plus y squared
to the 3/2, that's r cubed.

00:06:24.500 --> 00:06:26.360
So this is 1 over r cubed.

00:06:26.360 --> 00:06:31.270
So this is 1 over
r cubed in here.

00:06:31.270 --> 00:06:34.600
And as we know, dy dx
becomes r dr d theta.

00:06:38.760 --> 00:06:39.480
OK.

00:06:39.480 --> 00:06:42.340
So this is the integral
that we need to compute,

00:06:42.340 --> 00:06:44.074
and now that we've
done all the geometry,

00:06:44.074 --> 00:06:45.990
this is going to be a
straightforward integral

00:06:45.990 --> 00:06:46.490
to compute.

00:06:46.490 --> 00:06:47.510
So let's do it together.

00:06:50.580 --> 00:06:52.265
I'll just rewrite it over here.

00:06:52.265 --> 00:06:56.390
So we have theta running
from 0 to pi over 4,

00:06:56.390 --> 00:07:03.360
and r running from sec
theta to 2 sec theta.

00:07:03.360 --> 00:07:09.710
And we have 1 over r
squared, dr d theta.

00:07:09.710 --> 00:07:10.770
OK.

00:07:10.770 --> 00:07:21.030
So that equals-- taking
the inside integral first,

00:07:21.030 --> 00:07:29.030
we get negative 1 over r from
2 sec theta to sec theta.

00:07:32.940 --> 00:07:33.900
OK?

00:07:33.900 --> 00:07:37.090
And now this is nice,
because sec theta--

00:07:37.090 --> 00:07:39.890
if we plug it in for 1 over r--
it's just going to turn back

00:07:39.890 --> 00:07:41.350
into a cosine, right?

00:07:41.350 --> 00:07:45.680
So this is just the
integral from theta

00:07:45.680 --> 00:07:48.820
equals 0 to pi over 4.

00:07:48.820 --> 00:07:50.570
OK, because we have
this minus sign here,

00:07:50.570 --> 00:07:56.830
we're going to get cos theta
minus-- 1 over 2 sec theta

00:07:56.830 --> 00:08:05.430
becomes 1/2 cos theta-- d theta.

00:08:05.430 --> 00:08:13.750
And this we can compute to be
the square root of 2 over 4.

00:08:13.750 --> 00:08:17.105
So this is an elementary
integral that we can compute:

00:08:17.105 --> 00:08:19.580
the square root of 2 over 4.

00:08:19.580 --> 00:08:20.640
OK.

00:08:20.640 --> 00:08:21.240
And that's it.

00:08:21.240 --> 00:08:22.281
That's all there is to a.

00:08:22.281 --> 00:08:26.970
So notice that the difficult
part is figuring out

00:08:26.970 --> 00:08:30.460
how the boundary curves of
your original region re-express

00:08:30.460 --> 00:08:32.010
in the r and theta coordinates.

00:08:32.010 --> 00:08:36.110
So let's see if we can get more
practice with that on parts b

00:08:36.110 --> 00:08:36.610
and c.

00:08:39.670 --> 00:08:42.170
So in part b-- let
me just recall--

00:08:42.170 --> 00:08:47.006
we're taking the
integral x equals 0 to 1.

00:08:47.006 --> 00:08:53.850
And y equals x
squared to x, f dy dx.

00:08:53.850 --> 00:08:55.670
So let's draw this region again.

00:09:00.320 --> 00:09:07.530
So the bottom curve is y equals
x squared and the top curve is

00:09:07.530 --> 00:09:12.430
y equals x, and x
runs from 0 to 1.

00:09:12.430 --> 00:09:16.090
So this is the region
that we're after.

00:09:16.090 --> 00:09:16.790
OK.

00:09:16.790 --> 00:09:19.170
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.

00:09:22.980 --> 00:09:25.180
Because you see
this parabola here,

00:09:25.180 --> 00:09:27.980
as it approaches the
origin, it becomes flat.

00:09:27.980 --> 00:09:32.910
So the theta, as we get closer
and closer to the origin, is 0.

00:09:32.910 --> 00:09:39.800
So the initial bound
for theta is 0.

00:09:39.800 --> 00:09:42.420
It doesn't matter that we have
this curve here, it's still 0.

00:09:42.420 --> 00:09:44.850
And then the top bound,
again, is pi over 4.

00:09:47.640 --> 00:09:49.040
OK?

00:09:49.040 --> 00:09:55.640
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

00:09:59.430 --> 00:10:02.860
our polar change of coordinates
and re-express this curve.

00:10:02.860 --> 00:10:07.860
So y equals x squared.

00:10:07.860 --> 00:10:11.130
Well, y is r sine theta.

00:10:14.510 --> 00:10:16.870
And x is r cos theta.

00:10:16.870 --> 00:10:23.330
So this is altogether r
squared cos squared theta.

00:10:23.330 --> 00:10:24.020
OK?

00:10:24.020 --> 00:10:27.380
And now all we need to do
is just solve for r here.

00:10:27.380 --> 00:10:32.360
So we get r equals-- so I can
cancel that r with that one--

00:10:32.360 --> 00:10:39.890
and it looks to me like we
get r equals tan theta secant

00:10:39.890 --> 00:10:42.030
theta by solving.

00:10:42.030 --> 00:10:43.980
OK.

00:10:43.980 --> 00:10:49.000
So that's the r-value at each
of these points along the curve.

00:10:49.000 --> 00:10:51.050
So that is going to
be our top bound.

00:10:51.050 --> 00:10:55.710
So if we draw little
segments of constant theta,

00:10:55.710 --> 00:10:58.820
they all start at r
equals 0, because we

00:10:58.820 --> 00:11:01.510
have this sharp point
here at the origin.

00:11:01.510 --> 00:11:07.140
And they all stop at this curve:
r equals tan theta sec theta.

00:11:18.550 --> 00:11:19.580
All right?

00:11:19.580 --> 00:11:24.160
And so now f we're just
going to leave put.

00:11:24.160 --> 00:11:26.960
And then finally we
have r dr d theta.

00:11:34.450 --> 00:11:34.950
OK.

00:11:34.950 --> 00:11:37.870
So that's our answer to b.

00:11:37.870 --> 00:11:38.370
All right.

00:11:42.060 --> 00:11:46.460
Now c is going to be
a little bit tricky.

00:11:46.460 --> 00:11:49.270
What's tricky about c
is drawing a picture,

00:11:49.270 --> 00:11:53.590
but we'll see what
we can do about that.

00:11:53.590 --> 00:11:59.070
So let me just remind
you. y ranges from 0 to 2,

00:11:59.070 --> 00:12:05.760
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.

00:12:21.130 --> 00:12:24.170
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,

00:13:00.100 --> 00:13:11.260
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.