WEBVTT

00:00:07.450 --> 00:00:07.930
JOEL LEWIS: Hi.

00:00:07.930 --> 00:00:09.560
Welcome back to recitation.

00:00:09.560 --> 00:00:12.090
In lecture, you've been learning
about line integrals

00:00:12.090 --> 00:00:13.210
of vector fields.

00:00:13.210 --> 00:00:16.200
And I have a couple of nice
questions on that subject for

00:00:16.200 --> 00:00:17.260
you right here.

00:00:17.260 --> 00:00:21.550
So I want F to be the vector
field whose first coordinate

00:00:21.550 --> 00:00:23.980
is xy and whose second
coordinate is x

00:00:23.980 --> 00:00:25.120
squared plus y squared.

00:00:25.120 --> 00:00:29.270
And so what I'd like you do is
compute the line integral of F

00:00:29.270 --> 00:00:33.010
around two different curves C.
So both curves start at the

00:00:33.010 --> 00:00:36.620
point 1, 1 and they end
at the point 2, 4.

00:00:36.620 --> 00:00:39.600
So in part a, the curve is just
the straight line that

00:00:39.600 --> 00:00:42.190
connects the point
1, 1 to 2, 4.

00:00:42.190 --> 00:00:46.780
And in part b, the curve is
this sort of piecewise.

00:00:46.780 --> 00:00:48.580
It's two sides of a
rectangle, right?

00:00:48.580 --> 00:00:53.020
It goes straight up until it
gets to the point 1, 4, and

00:00:53.020 --> 00:00:54.990
then it goes across
to the point 2, 4.

00:00:54.990 --> 00:01:00.500
So it's a piecewise, smooth,
curved path that connects

00:01:00.500 --> 00:01:02.320
those two points.

00:01:02.320 --> 00:01:05.460
So I'd like you to compute the
integral over each of these

00:01:05.460 --> 00:01:08.830
curves of F dot dr. So why don't
you pause the video,

00:01:08.830 --> 00:01:10.460
have a go at that,
come back, and we

00:01:10.460 --> 00:01:11.710
can work it out together.

00:01:19.710 --> 00:01:22.540
So, when you're computing a line
integral over a curve,

00:01:22.540 --> 00:01:24.460
really the thing that you want
to do is you want to

00:01:24.460 --> 00:01:29.570
parametrize the curve, and then
that gives you stuff that

00:01:29.570 --> 00:01:30.300
you can plug in.

00:01:30.300 --> 00:01:32.480
You'll have expressions
for x and y in

00:01:32.480 --> 00:01:33.650
terms of your parameter.

00:01:33.650 --> 00:01:36.180
So you can plug it in and you
just turn this integral right

00:01:36.180 --> 00:01:38.650
into a nice single variable
integral, and then you can

00:01:38.650 --> 00:01:39.330
compute it.

00:01:39.330 --> 00:01:43.020
So that's our basic strategy
for computing integrals of

00:01:43.020 --> 00:01:44.620
this form line integrals
of vector fields.

00:01:44.620 --> 00:01:45.910
So let's have a go.

00:01:45.910 --> 00:01:47.700
Let's start with part a.

00:01:47.700 --> 00:01:52.490
So in part a, what we need to
do to apply this method is

00:01:52.490 --> 00:01:54.990
that we need to parametrize
the curve in question.

00:01:54.990 --> 00:01:56.670
So this is a straight line.

00:01:56.670 --> 00:01:59.010
And if you look at it, it's the
line through the points 1,

00:01:59.010 --> 00:02:00.670
1 and 2, 4.

00:02:00.670 --> 00:02:07.120
So this line has equation
y equals 3x minus 2.

00:02:07.120 --> 00:02:08.760
That's our line, and OK.

00:02:08.760 --> 00:02:13.630
So we need to choose some
parameter that will give us

00:02:13.630 --> 00:02:15.551
this segment of this line.

00:02:15.551 --> 00:02:15.740
So a natural thing to
do on this case

00:02:15.740 --> 00:02:16.990
is --it's easy enough--

00:02:19.550 --> 00:02:22.860
y is already written in terms
of x, so it's natural enough

00:02:22.860 --> 00:02:25.090
just to take a parameter
that's equal to x.

00:02:25.090 --> 00:02:27.820
So it's up to you whether you
introduce the letter t in

00:02:27.820 --> 00:02:29.300
order to do this, or not.

00:02:29.300 --> 00:02:34.220
I'm going to do it with the
letter t here in part a, but

00:02:34.220 --> 00:02:36.430
you could do this problem
exactly the same way just

00:02:36.430 --> 00:02:37.810
using the letter x.

00:02:37.810 --> 00:02:44.510
So what I'm going to do is I'm
going to let x equals t so

00:02:44.510 --> 00:02:49.840
that y is equal to 3t minus 2.

00:02:49.840 --> 00:02:53.390
OK, so that is the parametric
equation for the entire line,

00:02:53.390 --> 00:02:55.160
but we only want the
part between the

00:02:55.160 --> 00:02:57.150
points 1, 1 and 2, 4.

00:02:57.150 --> 00:03:05.860
So the part between the lines
1, 1 and 2, 4 is the part

00:03:05.860 --> 00:03:07.760
where t is between 1 and 2.

00:03:07.760 --> 00:03:09.620
Where x is between 1 and 2.

00:03:09.620 --> 00:03:10.350
OK.

00:03:10.350 --> 00:03:12.700
So this is our parametrization.

00:03:12.700 --> 00:03:18.850
So now we need to figure out
what is the field F in this

00:03:18.850 --> 00:03:22.590
parametrization, and what is
dr. And then after we have

00:03:22.590 --> 00:03:25.320
those, we can just put them into
our integral and compute.

00:03:25.320 --> 00:03:33.190
So F. In this parametrization,
well, we take the equation for

00:03:33.190 --> 00:03:36.780
F, which is xy comma x squared
plus y squared, and

00:03:36.780 --> 00:03:38.170
we just plug in.

00:03:38.170 --> 00:03:43.100
So in this case, xy is going to
be 3t minus 2 times t, so

00:03:43.100 --> 00:03:46.390
that's 3t squared minus 2t.

00:03:46.390 --> 00:03:51.840
And x squared plus y squared,
well that's t squared, plus 3t

00:03:51.840 --> 00:03:54.250
minus 2 quantity squared.

00:03:57.170 --> 00:03:58.630
So that's what F is.

00:03:58.630 --> 00:04:02.370
And also we have that dr--

00:04:02.370 --> 00:04:06.850
well, we just take the
differentials of x and y-- so

00:04:06.850 --> 00:04:12.490
this is going to be
dt comma 3dt.

00:04:12.490 --> 00:04:16.800
Or if you like, 1 comma 3 times
dt if you like to factor

00:04:16.800 --> 00:04:17.720
out your dt.

00:04:17.720 --> 00:04:20.090
So that's what F and dr is.

00:04:20.090 --> 00:04:23.130
So now we need to compute
our integral.

00:04:23.130 --> 00:04:28.810
So the integral over C
of F dot dr, well,

00:04:28.810 --> 00:04:29.600
you just plug in.

00:04:29.600 --> 00:04:33.410
So this is the integral
over C now.

00:04:33.410 --> 00:04:33.790
So OK.

00:04:33.790 --> 00:04:35.910
So now we need to look
at our bounds.

00:04:35.910 --> 00:04:39.370
So the integral over C means
the integral as t varies in

00:04:39.370 --> 00:04:42.330
the range that we
need to cover.

00:04:42.330 --> 00:04:43.090
That whole curve.

00:04:43.090 --> 00:04:46.310
So in this case, we said that
was from t equals 1 to 2.

00:04:46.310 --> 00:04:50.690
So it's the integral as t goes
from 1 to 2 of F dot dr.

00:04:50.690 --> 00:04:54.860
So in the first coordinates,
let me factor out

00:04:54.860 --> 00:04:56.610
the dt at the end.

00:04:56.610 --> 00:05:05.710
So that's going to be 3t squared
minus 2t, times 1,

00:05:05.710 --> 00:05:09.430
plus --OK well, let's expand
this out now--.

00:05:09.430 --> 00:05:14.100
3t minus 2 quantity squared
that's going to give me a 9t

00:05:14.100 --> 00:05:18.052
squared minus 12t plus 4
--so this is 9t squared

00:05:18.052 --> 00:05:23.250
minus 12t plus 4--

00:05:23.250 --> 00:05:25.470
and then we have to add
t squared to it.

00:05:25.470 --> 00:05:38.310
So this is plus 10t squared
minus 12t plus 4, times 3, and

00:05:38.310 --> 00:05:40.000
then this whole thing is dt.

00:05:40.000 --> 00:05:43.105
dt is the whole integrand,
there.

00:05:43.105 --> 00:05:46.900
I could even put in another pair
of parentheses just to

00:05:46.900 --> 00:05:48.380
emphasize that, perhaps.

00:05:48.380 --> 00:05:49.190
OK.

00:05:49.190 --> 00:05:52.310
Now this is straightforward.

00:05:52.310 --> 00:05:55.840
I mean, it's a little
complicated looking, but it's

00:05:55.840 --> 00:05:58.130
just an integral of
a polynomial.

00:05:58.130 --> 00:06:01.310
Easy enough to do.

00:06:01.310 --> 00:06:04.970
Let's first just combine terms.
OK, so let's look at

00:06:04.970 --> 00:06:06.240
the t squareds.

00:06:06.240 --> 00:06:08.350
We have a 10t squared times 3.

00:06:08.350 --> 00:06:11.670
So 30t squared, and then
another 3 times 1.

00:06:11.670 --> 00:06:25.150
So 33t squared minus 2t minus
36t is minus 38t plus 12--

00:06:25.150 --> 00:06:27.220
4 times 3--

00:06:27.220 --> 00:06:27.770
dt.

00:06:27.770 --> 00:06:29.400
OK, and now we integrate.

00:06:29.400 --> 00:06:33.620
So this is equal
to 11t cubed--

00:06:33.620 --> 00:06:34.980
that's a 3--

00:06:37.580 --> 00:06:47.810
minus 19t squared plus 12t as
t varies between 1 and 2.

00:06:47.810 --> 00:06:48.930
And all right.

00:06:48.930 --> 00:06:51.430
OK, so now we've got to plug
in and evaluate and so on.

00:06:51.430 --> 00:07:03.380
So at 2, this is 88 minus
76 plus 24, minus 11

00:07:03.380 --> 00:07:09.100
minus 19 plus 12.

00:07:09.100 --> 00:07:12.820
So you do some arithmetic
and this is going

00:07:12.820 --> 00:07:15.860
to work out to 32.

00:07:15.860 --> 00:07:19.500
OK, so there's part a.

00:07:19.500 --> 00:07:21.670
It's a nice, simple curve,
so we had a nice, simple

00:07:21.670 --> 00:07:23.025
parametrization.

00:07:23.025 --> 00:07:29.040
We computed F and dr, then we
dotted them, and integrated.

00:07:29.040 --> 00:07:32.410
OK, so now we're going to do the
same exact thing for part

00:07:32.410 --> 00:07:34.940
b, but in part b, the curve is
a little more complicated.

00:07:34.940 --> 00:07:38.620
Let's come over here where we've
got some empty space.

00:07:38.620 --> 00:07:42.730
So in part b, our curve
looks like this.

00:07:42.730 --> 00:07:49.070
So it starts at the point 1, 1,
and then it goes up to the

00:07:49.070 --> 00:07:57.090
point 1, 4, and then it goes
over to the point 2, 4.

00:07:57.090 --> 00:07:58.120
All right?

00:07:58.120 --> 00:08:02.040
So it's hard to parametrize in
one fell swoop something that

00:08:02.040 --> 00:08:03.890
makes a sharp right
angle like that.

00:08:03.890 --> 00:08:06.890
So a natural thing to do is to
split the integral over this

00:08:06.890 --> 00:08:09.900
whole curve into the integrals
over the two different pieces.

00:08:09.900 --> 00:08:13.990
So let's call this vertical
part C1 and this

00:08:13.990 --> 00:08:16.480
horizontal part C2.

00:08:16.480 --> 00:08:22.570
And so we know that the integral
over C of F dot dr is

00:08:22.570 --> 00:08:28.910
equal to the integral over C1 of
F dot dr plus the integral

00:08:28.910 --> 00:08:34.560
over C2 of F dot dr. And so
now, it's easy enough to

00:08:34.560 --> 00:08:37.930
parametrize these two separate
curves separately.

00:08:37.930 --> 00:08:42.020
C1, for example, is the straight
line segment that

00:08:42.020 --> 00:08:45.050
goes from 1, 1 to 1, 4.

00:08:45.050 --> 00:08:46.530
So C1.

00:08:46.530 --> 00:08:52.660
So that means we have x equal to
1, and 1 less than or equal

00:08:52.660 --> 00:08:55.280
to y less than or equal to 4.

00:08:55.280 --> 00:08:58.090
So a natural parametrization
here is just the

00:08:58.090 --> 00:09:01.830
parametrization that uses
the parameter y.

00:09:01.830 --> 00:09:02.190
Right?

00:09:02.190 --> 00:09:04.920
So in this one, I'm not going
to bother introducing

00:09:04.920 --> 00:09:05.730
a new letter t.

00:09:05.730 --> 00:09:07.510
I'm just going to stick
with x and y.

00:09:07.510 --> 00:09:11.460
So we have x equals 1, and y is
our parameter and it goes

00:09:11.460 --> 00:09:13.000
from 1 to 4.

00:09:13.000 --> 00:09:17.050
So now let's look at
what F and dr are.

00:09:17.050 --> 00:09:19.290
So in this case,
F is equal to--

00:09:19.290 --> 00:09:23.410
its first coordinate is xy,
and x is just 1 here--

00:09:23.410 --> 00:09:25.260
so this is y.

00:09:25.260 --> 00:09:28.560
And its second coordinate was x
squared plus y squared, and

00:09:28.560 --> 00:09:33.510
so that's going to be
1 plus y squared.

00:09:33.510 --> 00:09:36.480
And dr--

00:09:36.480 --> 00:09:39.950
well, r here is 1 comma y--

00:09:39.950 --> 00:09:45.610
so dr is equal to 0 comma dy.

00:09:45.610 --> 00:09:50.070
Or 0, 1 times dy, if you wanted
to factor that dy out

00:09:50.070 --> 00:09:51.060
to the end.

00:09:51.060 --> 00:09:52.910
OK.

00:09:52.910 --> 00:09:53.280
Good.

00:09:53.280 --> 00:09:55.510
So we're all set to do
that first integral.

00:09:55.510 --> 00:09:57.980
So let's do that.

00:09:57.980 --> 00:10:07.692
So we have the integral over
C1 of F dot dr is equal to

00:10:07.692 --> 00:10:09.490
what we dot these two
things together.

00:10:09.490 --> 00:10:13.780
And the first term gives me y
times 0, and that's just 0.

00:10:13.780 --> 00:10:16.510
So that's going to die, and
all we're left with is the

00:10:16.510 --> 00:10:17.510
second term.

00:10:17.510 --> 00:10:21.600
So it's the integral
of 1 plus y squared

00:10:21.600 --> 00:10:24.120
dy, but we need bounds.

00:10:24.120 --> 00:10:24.560
Right?

00:10:24.560 --> 00:10:27.975
OK, so y was going from 1
to 4 in this integral.

00:10:27.975 --> 00:10:34.620
So it's the integral from 1 to
4 of 1 plus y squared dy.

00:10:34.620 --> 00:10:34.940
OK.

00:10:34.940 --> 00:10:37.880
So we can either continue and
evaluate this now, or we could

00:10:37.880 --> 00:10:39.920
go and do the second one.

00:10:43.210 --> 00:10:45.840
Let's finish evaluating it since
we've already got it

00:10:45.840 --> 00:10:46.830
written up here.

00:10:46.830 --> 00:10:56.220
So this is equal to y, plus y
cubed over 3, between 1 and 4.

00:10:56.220 --> 00:10:56.750
So what is this?

00:10:56.750 --> 00:11:08.080
This is 4 plus 64/3,
minus 1 plus 1/3.

00:11:08.080 --> 00:11:11.880
So that looks like
it's 24 to me.

00:11:11.880 --> 00:11:14.540
OK, so we get 24 for
the first part.

00:11:14.540 --> 00:11:16.465
Now, let's do the second part.

00:11:16.465 --> 00:11:18.415
So it's C2 here.

00:11:18.415 --> 00:11:22.640
I'll draw a little line there
to separate them.

00:11:22.640 --> 00:11:24.500
Now on curve C2 --let's go
back and look at it--

00:11:24.500 --> 00:11:28.960
OK, so curve C2 is the segment
connecting the points 1, 4 and

00:11:28.960 --> 00:11:31.230
the point 2, 4.

00:11:31.230 --> 00:11:36.690
OK, so y is always 4 on this
curve, and x goes from 1 to 2.

00:11:36.690 --> 00:11:41.305
So 1 is less than or equal to
x less than or equal to 2, y

00:11:41.305 --> 00:11:43.820
is equal to 4.

00:11:43.820 --> 00:11:47.060
So a natural parametrization
here again, is just to take x

00:11:47.060 --> 00:11:48.230
to be our parameter.

00:11:48.230 --> 00:11:50.260
And again, I'm not going to
introduce a letter t.

00:11:50.260 --> 00:11:52.330
We're just using x
as our parameter.

00:11:52.330 --> 00:11:56.340
So in this case, F--

00:11:56.340 --> 00:12:01.366
well, it's xy, so x is
just x and y is 4--

00:12:01.366 --> 00:12:04.160
so that's 4x comma--

00:12:04.160 --> 00:12:07.050
and the second coordinate is x
squared plus y squared-- so

00:12:07.050 --> 00:12:10.980
that's x squared plus 16.

00:12:10.980 --> 00:12:22.210
And dr is equal to dx comma 0.

00:12:22.210 --> 00:12:24.440
OK, so that's F and dr.

00:12:24.440 --> 00:12:30.320
So the integral that I want now
is the integral over C2 of

00:12:30.320 --> 00:12:36.900
F dot dr. OK, so we just plug
in here what we've got.

00:12:36.900 --> 00:12:41.330
So this is equal to
the integral of--

00:12:41.330 --> 00:12:44.530
well, the first coordinates
are 4x dx and the second

00:12:44.530 --> 00:12:47.875
coordinates just give me
0-- so it's 4x dx.

00:12:47.875 --> 00:12:49.230
And again, I need my bounds.

00:12:49.230 --> 00:12:50.320
Well, I had-- over here--

00:12:50.320 --> 00:12:53.215
I had 1 less than or equal to x
is less than or equal to 2.

00:12:53.215 --> 00:12:55.620
So that's the integral
between 1 and 2.

00:12:58.350 --> 00:12:59.200
4x--

00:12:59.200 --> 00:13:04.490
integrate that-- and I get 2x
squared between 1 and 2, which

00:13:04.490 --> 00:13:09.295
is equal to 8 minus 2, or 6.

00:13:11.870 --> 00:13:12.650
All right.

00:13:12.650 --> 00:13:16.080
So let's see what we've got.

00:13:16.080 --> 00:13:17.820
So we had--

00:13:17.820 --> 00:13:18.990
back here--

00:13:18.990 --> 00:13:22.050
we had our curve C, which
we split into the two

00:13:22.050 --> 00:13:24.040
parts, C1 and C2.

00:13:24.040 --> 00:13:27.360
And we wanted to know what the
integral over C was, and we've

00:13:27.360 --> 00:13:32.965
separately computed the
integral over C1.

00:13:32.965 --> 00:13:35.790
And we computed that to be 24.

00:13:35.790 --> 00:13:39.075
And we computed the integral
over C2, and that was 6.

00:13:42.380 --> 00:13:49.190
So the integral over the whole
curve of F dot dr is equal to

00:13:49.190 --> 00:13:55.170
24 plus 6, which is 30.

00:13:55.170 --> 00:13:56.350
OK.

00:13:56.350 --> 00:13:58.250
So there's your answer
for the second part.

00:13:58.250 --> 00:14:01.120
Now one thing I'd like you to
notice is that over this curve

00:14:01.120 --> 00:14:03.150
C in part b--

00:14:03.150 --> 00:14:05.060
over the whole curve
in part b--

00:14:05.060 --> 00:14:09.645
we got that the integral
of this field F was 30.

00:14:09.645 --> 00:14:14.400
And now if you remember, right
here, in the first part, in

00:14:14.400 --> 00:14:17.640
part a, we computed the integral
over a different

00:14:17.640 --> 00:14:21.340
curve that connected the
two same endpoints.

00:14:21.340 --> 00:14:24.300
And we found that the integral
came out to 32.

00:14:24.300 --> 00:14:26.770
So one thing you should take
away from this is that the

00:14:26.770 --> 00:14:30.310
integral over a curve joining
two points can depend on which

00:14:30.310 --> 00:14:32.200
curve you choose, right?

00:14:32.200 --> 00:14:34.440
So we had two different curves
and we got two different

00:14:34.440 --> 00:14:36.150
answers, even though
the two curves

00:14:36.150 --> 00:14:37.920
connected the same points.

00:14:37.920 --> 00:14:39.190
So that's interesting.

00:14:39.190 --> 00:14:40.890
And the other thing to take
away from this is just the

00:14:40.890 --> 00:14:42.190
general approach.

00:14:42.190 --> 00:14:48.080
Which is that whenever you have
a problem like this, what

00:14:48.080 --> 00:14:50.430
you want to do is you want
to take your curve--

00:14:50.430 --> 00:14:56.360
so whether it be well, in part
a we had this straight line,

00:14:56.360 --> 00:14:57.640
slanted line.

00:14:57.640 --> 00:15:00.820
In part b where we had this
nice piecewise linear with

00:15:00.820 --> 00:15:03.060
these vertical and horizontal
parts--

00:15:03.060 --> 00:15:07.280
you want to break it into nice
pieces and parametrize them.

00:15:07.280 --> 00:15:09.870
You know, sometimes you only
need one piece when it's an

00:15:09.870 --> 00:15:12.770
easy-to-parametrize
curve like that.

00:15:12.770 --> 00:15:15.060
Sometimes, if it has corners
or so on, you

00:15:15.060 --> 00:15:16.700
might want more pieces.

00:15:16.700 --> 00:15:19.380
Break it into pieces, choose a
nice parametrization, and that

00:15:19.380 --> 00:15:21.910
reduces your problem just to
computing integrals, just like

00:15:21.910 --> 00:15:26.370
we've done in Calculus
I-- in 18.01--

00:15:26.370 --> 00:15:28.620
and then you just integrate.

00:15:28.620 --> 00:15:29.400
All right.

00:15:29.400 --> 00:15:31.020
I'll end there.