WEBVTT
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JOEL LEWIS: Hi.
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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.
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And I have a couple
of nice questions
00:00:14.870 --> 00:00:17.260
on that subject
for you right here.
00:00:17.260 --> 00:00:21.750
So I want F to be the vector
field whose first coordinate is
00:00:21.750 --> 00:00:25.120
x*y and whose second coordinate
is x squared plus y squared.
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And so what I'd like you do is
compute the line integral of F
00:00:29.270 --> 00:00:32.350
around two different
curves C. So both curves
00:00:32.350 --> 00:00:36.620
start at the point (1, 1) and
they end at the point (2, 4).
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So in part a, the curve is just
the straight line that connects
00:00:40.130 --> 00:00:42.190
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 --> 00:00:52.890
It goes straight up until
it gets to the point (1, 4),
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and then it goes across
to the point (2, 4).
00:00:54.990 --> 00:00:59.170
So it's a piecewise
smooth curve, path,
00:00:59.170 --> 00:01:02.320
that connects those two points.
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So I'd like you to compute
the integral over each
00:01:05.140 --> 00:01:08.804
of these curves of F dot dr. So
why don't you pause the video,
00:01:08.804 --> 00:01:10.970
have a go at that, come
back, and we can work it out
00:01:10.970 --> 00:01:11.470
together.
00:01:19.710 --> 00:01:22.540
So, when you're computing a
line integral over a curve,
00:01:22.540 --> 00:01:24.100
really the thing
that you want to do
00:01:24.100 --> 00:01:25.960
is you want to
parametrize the curve,
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and then that gives you
stuff that you can plug in.
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You'll have
expressions for x and y
00:01:32.380 --> 00:01:33.650
in terms of your parameter.
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So you can plug it in and you
just turn this integral right
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into a nice single
variable integral,
00:01:38.164 --> 00:01:39.330
and then you can compute it.
00:01:39.330 --> 00:01:42.704
So that's our basic strategy
for computing integrals
00:01:42.704 --> 00:01:44.620
of this form, line
integrals of vector fields.
00:01:44.620 --> 00:01:45.910
So let's have a go.
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Let's start with part a.
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So in part a, what we need
to do to apply this method
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is that we need to parametrize
the curve in question.
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So this is a straight line.
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And if you look at it, it's the
line through the points (1, 1)
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and (2, 4).
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So this line has equation
y equals 3x minus 2.
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That's our line, and OK.
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So we need to choose some
parameter that will give us
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this segment of this line.
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So a natural thing to do in
this case is-- it's easy enough,
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y is already written
in terms of x,
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so it's natural
enough just to take
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a parameter that's equal to x.
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So it's up to you whether you
introduce the letter t in order
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to do this, or not.
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I'm going to do it with the
letter t here in part a,
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but you could do this problem
exactly the same way just using
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the letter x.
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So what I'm going to do is
I'm going to let x equals t
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so that y is equal
to 3t minus 2.
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OK, so that is the parametric
equation for the entire line,
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but we only want the part
between the points (1, 1) and
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(2, 4).
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So the part between the
lines (1, 1) and (2, 4)
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is the part where t
is between 1 and 2.
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Where x is between 1 and 2.
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OK.
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So this is our parametrization.
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So now we need to
figure out what
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is the field F in
this parametrization,
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and what is dr. And then
after we have those,
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we can just put them into
our integral and compute.
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So F, in this
parametrization, well,
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we take the equation for F,
which is x*y comma x squared
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plus y squared, and
we just plug in.
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So in this case, x*y is going
to be 3t minus 2 times t,
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so that's 3 t squared minus 2t.
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And x squared plus y squared,
well that's t squared plus 3t
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minus 2 quantity squared.
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So that's what F is.
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And also we have that dr-- well,
we just take the differentials
00:04:05.680 --> 00:04:12.490
of x and y-- so this is
going to be dt comma 3*dt.
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Or if you like, 1
comma 3 times dt
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if you like to
factor out your dt.
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So that's what F and dr is.
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So now we need to
compute our integral.
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So the integral over C of F
dot dr, well, you just plug in.
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So this is the
integral over C now.
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So OK.
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So now we need to
look at our bounds.
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So the integral over
C means the integral
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as t varies in the range
that we need to cover.
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That whole curve.
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So in this case, we said that
was from t equals 1 to 2.
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So it's the integral as t
goes from 1 to 2 of F dot dr.
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So in the first coordinates--
let me factor out
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the dt at the end.
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So that's going to be 3 t
squared minus 2t, times 1,
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plus-- OK, well, let's
expand this out now.
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3t minus 2 quantity
squared, that's
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going to give me a 9 t
squared minus 12t plus 4--
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so this is 9 t squared
minus 12t plus 4,
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and then we have to
add t squared to it.
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So this is plus 10 t squared
minus 12t plus 4, times 3,
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and then this whole thing is dt.
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dt is the whole
integrand, there.
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I could even put in another
pair of parentheses just
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to emphasize that, perhaps.
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OK.
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Now this is a
straightforward-- I mean,
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it's a little
complicated looking,
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but it's just an
integral of a polynomial.
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Easy enough to do.
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Let's first just combine terms.
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OK, so let's look
at the t squareds.
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We have a 10 t squared times 3.
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So 30 t squared, and
then another 3 times 1.
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So 33 t squared.
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And I've got minus 2t
minus 36t is minus 38t,
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plus 12-- 4 times 3-- dt.
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OK, and now we integrate.
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So this is equal to
11 t cubed-- that's
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a 3-- minus 19 t
squared plus 12t as t
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varies between 1 and 2.
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And all right.
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OK, so now we've got to plug
in and evaluate and so on.
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So at 2, this is 88 minus 76
plus 24, minus 11 minus 19
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plus 12.
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So you do some
arithmetic and this
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is going to work out to 32.
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OK, so there's part a.
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It's a nice, simple
curve, so we had
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a nice, simple parametrization.
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We computed F and dr, then we
dotted them, and integrated.
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OK, so now we're going to do
the same exact thing for part b,
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but in part b, the curve is
a little more complicated.
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Let's come over here where
we've got some empty space.
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So in part b, our
curve looks like this.
00:07:42.730 --> 00:07:46.796
So it starts at
the point (1, 1),
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and then it goes up
to the point (1, 4),
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and then it goes over
to the point (2, 4).
00:07:57.090 --> 00:07:58.120
All right?
00:07:58.120 --> 00:08:01.440
So it's hard to parametrize
in one fell swoop
00:08:01.440 --> 00:08:03.890
something that makes a
sharp right angle like that.
00:08:03.890 --> 00:08:06.450
So a natural thing to do
is to split the integral
00:08:06.450 --> 00:08:08.940
over this whole curve into
the integrals over the two
00:08:08.940 --> 00:08:09.900
different pieces.
00:08:09.900 --> 00:08:15.080
So let's call this vertical part
C_1 and this horizontal part
00:08:15.080 --> 00:08:16.480
C_2.
00:08:16.480 --> 00:08:20.640
And so we know that the
integral over C of F
00:08:20.640 --> 00:08:25.124
dot dr is equal to
the integral over C_1
00:08:25.124 --> 00:08:33.380
of F dot dr plus the integral
over C_2 of F dot dr.
00:08:33.380 --> 00:08:35.720
And so now, it's easy enough
to parametrize these two
00:08:35.720 --> 00:08:37.930
separate curves separately.
00:08:37.930 --> 00:08:42.020
C_1, 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 C_1.
00:08:46.530 --> 00:08:52.500
So that means we have x
equal to 1, and 1 less than
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or equal to y less
than or equal to 4.
00:08:55.280 --> 00:08:57.690
So a natural
parametrization here
00:08:57.690 --> 00:09:01.690
is just the parametrization
that uses the parameter y.
00:09:01.690 --> 00:09:02.190
Right?
00:09:02.190 --> 00:09:05.231
So in this one, I'm not going to
bother introducing a new letter
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t.
00:09:05.730 --> 00:09:07.510
I'm just going to
stick with x and y.
00:09:07.510 --> 00:09:10.935
So we have x equals 1,
and y is our parameter
00:09:10.935 --> 00:09:13.000
and it goes 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 --> 00:09:21.340
So in this case, F is equal to--
its first coordinate is x*y,
00:09:21.340 --> 00:09:25.260
and x is just 1
here, so this is y.
00:09:25.260 --> 00:09:28.215
And its second coordinate
was x squared plus y squared,
00:09:28.215 --> 00:09:33.510
and so that's going to
be 1 plus y squared.
00:09:33.510 --> 00:09:39.950
And dr-- 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 --> 00:09:49.380
Or (0, 1) times dy, if
you wanted to factor that
00:09:49.380 --> 00:09:51.060
dy out to the end.
00:09:51.060 --> 00:09:52.780
OK.
00:09:52.780 --> 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 --> 00:10:07.820
So we have the integral over C_1
of F dot dr is equal to-- well,
00:10:07.820 --> 00:10:09.490
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 --> 00:10:15.780
So that's going to
die, and all we're
00:10:15.780 --> 00:10:17.510
left with is the second term.
00:10:17.510 --> 00:10:24.060
So it's the integral of 1 plus y
squared dy, but we need bounds.
00:10:24.060 --> 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 --> 00:10:34.440
So it's the integral from 1
to 4 of 1 plus y squared dy.
00:10:34.440 --> 00:10:34.940
OK.
00:10:34.940 --> 00:10:37.430
So we can either continue
and evaluate this now,
00:10:37.430 --> 00:10:39.920
or we could go and
do the second one.
00:10:43.210 --> 00:10:45.560
Let's finish evaluating
it since we've already
00:10:45.560 --> 00:10:46.830
got it written up here.
00:10:46.830 --> 00:10:56.084
So this is equal to y, plus y
cubed over 3, between 1 and 4.
00:10:56.084 --> 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 C_2 here.
00:11:18.415 --> 00:11:21.270
I'll draw a little line
there to separate them.
00:11:21.270 --> 00:11:24.770
Now, curve C_2-- let's go
back and look at it-- OK,
00:11:24.770 --> 00:11:28.820
so curve C_2 is the segment
connecting the points (1, 4)
00:11:28.820 --> 00:11:31.230
and 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 --> 00:11:41.070
So 1 is less than or equal
to x less than or equal to 2,
00:11:41.070 --> 00:11:43.820
y is equal to 4.
00:11:43.820 --> 00:11:45.550
So a natural
parametrization here,
00:11:45.550 --> 00:11:48.219
again, is just to take
x to be our parameter.
00:11:48.219 --> 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 --> 00:11:58.280
So in this case,
F-- well, it's x*y,
00:11:58.280 --> 00:12:04.160
so x is just x and y is
4-- so that's 4x comma--
00:12:04.160 --> 00:12:06.645
and the second coordinate is
x squared plus y squared--
00:12:06.645 --> 00:12:10.980
so 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 --> 00:12:30.000
So the integral that I want
now is the integral over C_2
00:12:30.000 --> 00:12:36.900
of F dot dr. OK, so we just
plug in here what we've got.
00:12:36.900 --> 00:12:41.920
So this is equal to
the integral of-- well,
00:12:41.920 --> 00:12:45.120
the first coordinates are 4x dx
and the second coordinates just
00:12:45.120 --> 00:12:47.875
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 --> 00:12:51.507
Well, I had-- over
here, I had 1 less than
00:12:51.507 --> 00:12:53.215
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 --> 00:13:04.300
4x-- integrate that-- and I get
2 x squared between 1 and 2,
00:13:04.300 --> 00:13:09.295
which 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 --> 00:13:21.610
So we had-- back here, we had
our curve C, which we split
00:13:21.610 --> 00:13:24.040
into the two parts C_1 and C_2.
00:13:24.040 --> 00:13:26.830
And we wanted to know what
the integral over C was,
00:13:26.830 --> 00:13:32.965
and we've separately computed
the integral over C_1.
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 C_2, and that was 6.
00:13:42.380 --> 00:13:45.910
So the integral over
the whole curve of F
00:13:45.910 --> 00:13:55.170
dot dr is equal to 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 --> 00:13:59.930
Now one thing I'd
like you to notice
00:13:59.930 --> 00:14:03.150
is that over this
curve C in part b--
00:14:03.150 --> 00:14:05.210
over the whole
curve in part b-- we
00:14:05.210 --> 00:14:09.645
got that the integral
of this field F was 30.
00:14:09.645 --> 00:14:13.550
And now if you
remember, right here,
00:14:13.550 --> 00:14:16.950
in the first part, in part
a, we computed the integral
00:14:16.950 --> 00:14:20.130
over a different curve
that connected the two
00:14:20.130 --> 00:14:21.340
same endpoints.
00:14:21.340 --> 00:14:24.300
And we found that the
integral came out to 32.
00:14:24.300 --> 00:14:26.160
So one thing you should
take away from this
00:14:26.160 --> 00:14:29.190
is that the integral over
a curve joining two points
00:14:29.190 --> 00:14:32.200
can depend on which
curve you choose, right?
00:14:32.200 --> 00:14:33.770
So we had two
different curves and we
00:14:33.770 --> 00:14:36.620
got two different answers, even
though the two curves connected
00:14:36.620 --> 00:14:37.920
the same points.
00:14:37.920 --> 00:14:39.190
So that's interesting.
00:14:39.190 --> 00:14:40.939
And the other thing
to take away from this
00:14:40.939 --> 00:14:42.190
is just the 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 --> 00:15:00.460
In part b where we had
this nice piecewise
00:15:00.460 --> 00:15:03.060
linear with these vertical
and horizontal parts--
00:15:03.060 --> 00:15:06.925
you want to break it into
nice pieces, parametrize them.
00:15:06.925 --> 00:15:08.300
You know, sometimes
you only need
00:15:08.300 --> 00:15:11.620
one piece when it's an
easy-to-parametrize curve
00:15:11.620 --> 00:15:12.770
like that.
00:15:12.770 --> 00:15:14.990
Sometimes, if it has
corners or so on,
00:15:14.990 --> 00:15:16.700
you might want more pieces.
00:15:16.700 --> 00:15:19.000
Break it into pieces, choose
a nice parametrization,
00:15:19.000 --> 00:15:21.450
and that reduces your problem
just to computing integrals,
00:15:21.450 --> 00:15:26.370
just like 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.