WEBVTT
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JOEL LEWIS: Hi.
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Welcome back to recitation.
00:00:09.750 --> 00:00:12.490
In lecture, you've been
learning about Stokes' Theorem,
00:00:12.490 --> 00:00:15.320
and I have a nice exercise on
Stokes' Theorem for you here.
00:00:15.320 --> 00:00:19.370
So I'm going to
let F be this field
00:00:19.370 --> 00:00:21.270
that I've written just above me.
00:00:21.270 --> 00:00:27.600
So it's 2x*z minus 2y comma 2y*z
plus 2x comma x square plus y
00:00:27.600 --> 00:00:29.390
square plus z square.
00:00:29.390 --> 00:00:31.080
And I've got C.
00:00:31.080 --> 00:00:34.380
So C is this
complicated-looking curve here.
00:00:34.380 --> 00:00:38.320
So it sort of dips up
and down and back around.
00:00:38.320 --> 00:00:40.370
But the thing I'm
really going to tell you
00:00:40.370 --> 00:00:49.020
about it is that it all lies
on this cylinder of radius b.
00:00:49.020 --> 00:00:52.120
So C is this curve in
the cylinder of radius b
00:00:52.120 --> 00:00:56.580
that wraps around it once,
but behaves kind of oddly
00:00:56.580 --> 00:00:58.090
while it's wrapping around.
00:00:58.090 --> 00:00:59.680
So what I'd like
you to do is I'd
00:00:59.680 --> 00:01:02.140
like you to use Stokes'
Theorem to compute
00:01:02.140 --> 00:01:05.740
the integral around this
curve of F dot dr. Now,
00:01:05.740 --> 00:01:08.910
my hint to you is that
for Stokes' Theorem,
00:01:08.910 --> 00:01:11.404
you can use-- just like you
have for Green's Theorem
00:01:11.404 --> 00:01:13.820
and for Divergence Theorem
that we've talked about before,
00:01:13.820 --> 00:01:15.640
you have these
extended versions that
00:01:15.640 --> 00:01:18.000
let you consider more
than one boundary piece.
00:01:18.000 --> 00:01:21.300
So the same thing works
for Stokes' Theorem.
00:01:21.300 --> 00:01:24.027
So Stokes' Theorem
works perfectly well
00:01:24.027 --> 00:01:25.860
when you have a piece
of a surface with more
00:01:25.860 --> 00:01:29.370
than one boundary curve,
provided you orient everything
00:01:29.370 --> 00:01:30.170
correctly.
00:01:30.170 --> 00:01:33.390
So you might think about how
you can use Stokes' Theorem
00:01:33.390 --> 00:01:37.680
to replace this complicated
curve with a surface integral
00:01:37.680 --> 00:01:39.780
and an easier to
understand curve.
00:01:39.780 --> 00:01:42.200
And if you can do that,
then computing the other two
00:01:42.200 --> 00:01:43.260
gives you the third one.
00:01:43.260 --> 00:01:44.050
All right.
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So that's my hint to you
for computing this integral.
00:01:48.120 --> 00:01:50.741
So why don't you pause the
video, have a go at that,
00:01:50.741 --> 00:01:52.490
come back, and we can
work on it together.
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Hopefully, you had some luck
working on this problem.
00:02:03.600 --> 00:02:05.000
Let's talk about it.
00:02:05.000 --> 00:02:07.120
So before I left, I
gave you this hint
00:02:07.120 --> 00:02:10.210
that maybe the thing to do here
isn't to try and parametrize
00:02:10.210 --> 00:02:13.500
this curve directly and compute
the line integral directly
00:02:13.500 --> 00:02:15.520
since it's a
complicated-looking curve,
00:02:15.520 --> 00:02:18.170
and also since I haven't really
given you enough information
00:02:18.170 --> 00:02:22.540
to do that, and instead to think
about applying Stokes' Theorem.
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So to think about applying
Stokes' Theorem, what we'd like
00:02:25.880 --> 00:02:31.570
is a nice surface, with this
curve as part of its boundary.
00:02:31.570 --> 00:02:32.830
Well, what is such a surface?
00:02:32.830 --> 00:02:36.950
Well, this curve lies all
on the cylinder of radius b.
00:02:36.950 --> 00:02:39.000
So a natural choice
for a surface
00:02:39.000 --> 00:02:41.240
is to use some piece
of this cylinder.
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So maybe we could use the
piece of this cylinder
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with this as its upper boundary.
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So then what might be a natural
lower boundary to choose?
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Well, we just want to choose
something nice and simple.
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Right?
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So, what's nice and simple?
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Well, maybe we can
choose this bottom circle
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that's in the plane y equals x.
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All right.
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So I'm going to call
that circle C_1.
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So that's the circle of
radius b in the xy-plane.
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Sorry, not the plane y equals x.
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The xy-plane.
00:03:09.730 --> 00:03:12.730
The plane z equals 0.
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So we've got the top curve C
and we've got this bottom curve
00:03:17.540 --> 00:03:18.460
C_1.
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Now, the way I've
oriented them, I've
00:03:20.550 --> 00:03:24.150
oriented them both so that
they're going counterclockwise
00:03:24.150 --> 00:03:27.030
as you look down
from the z-axis.
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So in that case, what
does Stokes' Theorem say?
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Well, Stokes' Theorem
says that the integral
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over the piece of the
surface between them--
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let's call it S-- of curl F
dot n with respect to surface
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area is equal to-- OK.
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So let's say we can give it the
outward pointing normal, say.
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In which case, C_1 will
be positively oriented
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and C will be
negatively oriented.
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So this is equal to
the line integral
00:04:00.760 --> 00:04:12.150
over C_1 of F dot dr minus
the line integral over C
00:04:12.150 --> 00:04:15.660
of F dot dr.
00:04:15.660 --> 00:04:18.780
And so what's nice
about this formula
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is that it replaces computing
the integral that we want.
00:04:24.340 --> 00:04:27.160
Instead of computing that,
we can try and compute
00:04:27.160 --> 00:04:29.770
this other line integral
and this surface integral.
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And if these are
easier to compute,
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then computing the two of them
gives us what the value of this
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is just by subtracting, or
by adding and subtracting,
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or whatever.
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By arithmetic, right?
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So if these integrals
are easy to compute,
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then that makes this one
easy without actually having
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to parametrize and compute it.
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So let's take a look at
what these integrals are.
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Let's do the surface integral
first since it's on the left.
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So in order to compute
the surface integral,
00:04:59.390 --> 00:05:02.350
we're going to need to
compute the curl of F.
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So OK.
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So F is this kind of
messy-looking thing here.
00:05:05.590 --> 00:05:13.660
So curl of F, well,
what have we got?
00:05:13.660 --> 00:05:17.410
So it's going to be
big thing times i hat.
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So it's going to be i hat
times this determinant, right?
00:05:20.890 --> 00:05:22.790
So let me write the determinant.
00:05:22.790 --> 00:05:26.250
So on top we've got
i hat, j hat, k hat,
00:05:26.250 --> 00:05:28.307
then we have the partial
x, partial y, partial z,
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and then we have the components.
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So these are 2x*z minus
2y, and 2y*z plus 2x,
00:05:40.740 --> 00:05:45.250
and x square plus y
square plus z square.
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All right.
00:05:45.750 --> 00:05:51.075
So that's i hat,
j hat, and k hat.
00:05:51.075 --> 00:05:55.900
And then we've got
partial over partial x,
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partial over partial y,
and partial over partial z.
00:06:01.360 --> 00:06:03.690
So this is what the
curl is, and so now we
00:06:03.690 --> 00:06:04.860
have to expand this out.
00:06:04.860 --> 00:06:10.426
So for i, it's going to be
partial y of x squared plus y
00:06:10.426 --> 00:06:14.890
squared plus z
squared-- so that's 2y--
00:06:14.890 --> 00:06:23.860
minus partial z of 2y*z plus
2x, so that's minus 2y, i hat.
00:06:23.860 --> 00:06:31.820
Plus-- for j hat, it's going to
be partial z of 2x*z minus 2y,
00:06:31.820 --> 00:06:40.240
so that's 2x-- minus partial
x of x squared plus y squared
00:06:40.240 --> 00:06:45.500
plus z squared, so
that's minus 2x, j hat.
00:06:45.500 --> 00:06:52.380
Plus-- for k hat, we want
partial x of 2y*z plus 2x,
00:06:52.380 --> 00:06:59.750
so that's 2-- minus
partial y of 2x*z minus 2y,
00:06:59.750 --> 00:07:07.300
so that's minus minus 2,
so that's plus 2, k hat.
00:07:07.300 --> 00:07:07.800
Oh.
00:07:07.800 --> 00:07:08.410
All right.
00:07:08.410 --> 00:07:08.910
OK.
00:07:08.910 --> 00:07:12.060
So the i-component is 0
and the j-component is 0.
00:07:12.060 --> 00:07:13.400
So this is a nice, simple one.
00:07:13.400 --> 00:07:16.170
So the curl here, the
k-component is just 4.
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So this is equal to 4k.
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OK.
00:07:20.680 --> 00:07:23.460
So that's what the curl of F is.
00:07:23.460 --> 00:07:25.040
Now what do we need to compute?
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We need to compute curl of F dot
the normal vector with respect
00:07:29.880 --> 00:07:30.770
to surface area.
00:07:30.770 --> 00:07:32.960
Now let's look at
what our surface is.
00:07:32.960 --> 00:07:38.680
Our surface is right here, and
it's this vertical cylinder.
00:07:38.680 --> 00:07:41.410
Well, what is the normal
vector of a vertical cylinder?
00:07:41.410 --> 00:07:46.220
Well, it's pointing
straight away from the axis.
00:07:46.220 --> 00:07:46.820
Right?
00:07:46.820 --> 00:07:48.903
It's perpendicular to the
surface of the cylinder,
00:07:48.903 --> 00:07:51.600
so it's parallel
to the xy-plane.
00:07:54.320 --> 00:07:56.630
It rotates as you go
around the cylinder,
00:07:56.630 --> 00:07:58.690
but it's always in the xy-plane.
00:07:58.690 --> 00:07:59.810
So what does that mean?
00:07:59.810 --> 00:08:02.050
Well, that means
in particular, it's
00:08:02.050 --> 00:08:06.190
perpendicular to things
in the z-direction.
00:08:06.190 --> 00:08:06.690
Right?
00:08:06.690 --> 00:08:10.200
So if we look, we see our curl
here is just straight upward
00:08:10.200 --> 00:08:11.550
in the z-direction.
00:08:11.550 --> 00:08:14.730
And our normal vector
has no z-component.
00:08:14.730 --> 00:08:17.480
It's only in the xy-plane.
00:08:17.480 --> 00:08:23.080
So this k hat is
orthogonal to n, OK?
00:08:23.080 --> 00:08:26.460
So the curl and
n are orthogonal.
00:08:26.460 --> 00:08:28.330
So their dot product is 0.
00:08:28.330 --> 00:08:31.790
So this surface integral
is a surface integral of 0.
00:08:31.790 --> 00:08:33.340
So it just gives you 0.
00:08:33.340 --> 00:08:33.850
OK.
00:08:33.850 --> 00:08:34.350
So great.
00:08:34.350 --> 00:08:35.830
So that's really nice.
00:08:35.830 --> 00:08:38.050
That simplifies
our life very much.
00:08:38.050 --> 00:08:42.380
Now, our line
integral that we want.
00:08:42.380 --> 00:08:45.350
We just have it in terms of
this one other line integral.
00:08:45.350 --> 00:08:45.850
Right?
00:08:45.850 --> 00:08:48.352
So the surface integral is 0.
00:08:48.352 --> 00:08:49.400
And let me see.
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Where should I put this?
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OK, the curl is 4k, so the
surface integral curl F dot n
00:09:01.320 --> 00:09:05.440
dS is also equal to 0.
00:09:05.440 --> 00:09:07.670
So having made that
simplification,
00:09:07.670 --> 00:09:10.070
now we just need
this other integral.
00:09:10.070 --> 00:09:13.280
We need this line
integral over C_1.
00:09:13.280 --> 00:09:16.860
And that'll give
us what we need.
00:09:16.860 --> 00:09:18.350
So let's have a go at that.
00:09:18.350 --> 00:09:30.280
So C_1 is the circle
of radius b centered
00:09:30.280 --> 00:09:32.840
at the origin in the xy-plane.
00:09:32.840 --> 00:09:33.410
OK.
00:09:33.410 --> 00:09:34.460
I'm not going to
write that down.
00:09:34.460 --> 00:09:35.501
I'm just going to say it.
00:09:35.501 --> 00:09:38.510
The circle of radius b centered
at the origin in the xy-plane.
00:09:38.510 --> 00:09:39.010
So OK.
00:09:39.010 --> 00:09:40.551
So it's not that
hard to parametrize.
00:09:42.850 --> 00:09:49.010
So it's parametrized by
x equals b cosine theta,
00:09:49.010 --> 00:09:52.030
y equals b sine theta.
00:09:52.030 --> 00:09:52.906
We should check.
00:09:52.906 --> 00:09:54.280
We should double-check
that we're
00:09:54.280 --> 00:09:57.070
doing the right direction
of parametrization.
00:09:57.070 --> 00:09:59.050
Let's go have a look.
00:09:59.050 --> 00:10:00.580
Let's see.
00:10:00.580 --> 00:10:01.080
Yes.
00:10:01.080 --> 00:10:01.580
OK.
00:10:01.580 --> 00:10:04.670
So we parametrized
this circle going
00:10:04.670 --> 00:10:06.660
counterclockwise
in the xy-plane.
00:10:06.660 --> 00:10:07.160
So good.
00:10:07.160 --> 00:10:09.490
So this parametrization is
going the right direction.
00:10:09.490 --> 00:10:12.940
Otherwise, we'd have to change
the sign of theta or something.
00:10:12.940 --> 00:10:16.060
So it's x is b cosine
theta, y is b sine theta.
00:10:16.060 --> 00:10:17.800
And we're going once
around the circle,
00:10:17.800 --> 00:10:20.770
so we want 0 less than or equal
to theta less than or equal
00:10:20.770 --> 00:10:22.600
to 2*pi.
00:10:22.600 --> 00:10:24.410
And so what do we have?
00:10:24.410 --> 00:10:28.250
So now, we want to
compute the integral
00:10:28.250 --> 00:10:37.500
over the circle of F dot
dr. So let's see what
00:10:37.500 --> 00:10:41.060
F looks like in this situation.
00:10:41.060 --> 00:10:46.150
So let's go back and look at
the expression for F over here.
00:10:46.150 --> 00:10:51.740
So in this plane, we
have z is equal to 0.
00:10:51.740 --> 00:10:59.130
So F is minus 2y, plus 2x,
x squared plus y squared.
00:10:59.130 --> 00:11:01.300
OK?
00:11:01.300 --> 00:11:03.600
OK, so let's come back then.
00:11:03.600 --> 00:11:07.590
So F is what I
just said, so this
00:11:07.590 --> 00:11:17.210
is equal to the integral
over C of minus 2y dx,
00:11:17.210 --> 00:11:23.830
plus 2x dy-- plus x
squared plus y squared dz,
00:11:23.830 --> 00:11:25.610
but we're in the
plane z equals 0,
00:11:25.610 --> 00:11:28.490
so dz is always
0 in that plane--
00:11:28.490 --> 00:11:30.250
so we don't have a
third term there.
00:11:30.250 --> 00:11:32.510
Great.
00:11:32.510 --> 00:11:34.490
So this is our
integral, and now we
00:11:34.490 --> 00:11:37.270
can substitute from our
parametrization here.
00:11:37.270 --> 00:11:43.800
So this is equal to the
integral from 0 to 2*pi.
00:11:43.800 --> 00:11:47.510
So minus 2y dx.
00:11:47.510 --> 00:11:56.360
So that's minus 2b
sine theta, times-- dx
00:11:56.360 --> 00:12:08.380
is minus b sine theta
d theta-- plus 2x-- so
00:12:08.380 --> 00:12:17.861
that's 2b cosine theta-- times
dy, which is b cosine theta d
00:12:17.861 --> 00:12:18.360
theta.
00:12:23.350 --> 00:12:23.850
Whew.
00:12:23.850 --> 00:12:26.665
This is quite a long
equation, isn't it?
00:12:26.665 --> 00:12:29.050
Or a long expression, I guess.
00:12:29.050 --> 00:12:34.640
So our line integral around
C of F dot dr is equal
00:12:34.640 --> 00:12:38.560
to the integral from 0 to 2*pi
of minus 2b sine theta times
00:12:38.560 --> 00:12:39.960
minus b sine theta d theta.
00:12:39.960 --> 00:12:44.230
So this is 2b squared sine
squared theta d theta.
00:12:44.230 --> 00:12:47.820
And this is 2b cosine
squared theta d theta.
00:12:47.820 --> 00:12:48.320
So OK.
00:12:48.320 --> 00:12:50.660
So that 2 b squared
is a constant.
00:12:50.660 --> 00:12:52.010
We can just factor it out.
00:12:52.010 --> 00:12:54.840
And we're left with sine squared
theta plus cosine squared theta
00:12:54.840 --> 00:12:55.610
d theta.
00:12:55.610 --> 00:12:56.380
All right.
00:12:56.380 --> 00:12:58.150
OK.
00:12:58.150 --> 00:12:59.370
That's great.
00:12:59.370 --> 00:13:00.500
I'm happy to have that.
00:13:00.500 --> 00:13:01.000
Right?
00:13:01.000 --> 00:13:02.583
Sine squared theta
plus cosine squared
00:13:02.583 --> 00:13:03.950
theta, that's going to be 1.
00:13:03.950 --> 00:13:04.610
OK.
00:13:04.610 --> 00:13:05.990
So we can rewrite this.
00:13:05.990 --> 00:13:07.650
I'm going to bring
it back up here.
00:13:07.650 --> 00:13:16.250
So that's equal to the integral
from 0 to 2*pi of 2 b squared d
00:13:16.250 --> 00:13:22.650
theta, which is 4*pi b squared.
00:13:22.650 --> 00:13:23.330
Great.
00:13:23.330 --> 00:13:29.401
OK, so that's our line integral
around this bottom curve C.
00:13:29.401 --> 00:13:29.900
Oh, dear.
00:13:29.900 --> 00:13:33.630
I've been writing C, but this
is not our original curve C,
00:13:33.630 --> 00:13:36.761
this is our new curve
C_1, like I wrote there.
00:13:36.761 --> 00:13:37.260
Sorry.
00:13:37.260 --> 00:13:39.557
So everywhere I wrote the
line integral over C--
00:13:39.557 --> 00:13:41.140
both of these places--
it was supposed
00:13:41.140 --> 00:13:43.300
to be a line integral over C_1.
00:13:43.300 --> 00:13:45.260
Sorry about that.
00:13:45.260 --> 00:13:48.020
So we've got this line
integral over C_1,
00:13:48.020 --> 00:13:51.590
and it worked out
to 4*pi b squared,
00:13:51.590 --> 00:13:56.040
just using our usual
parametrize-and-compute
00:13:56.040 --> 00:13:58.110
technique for computing
line integrals.
00:13:58.110 --> 00:13:58.610
So OK.
00:13:58.610 --> 00:14:00.020
So now, let's see
where we're at.
00:14:00.020 --> 00:14:03.640
Let's go back over here
to when we wrote down
00:14:03.640 --> 00:14:06.270
what the extended Stokes'
Theorem says in our case.
00:14:06.270 --> 00:14:08.962
So Stokes' Theorem
told us that the thing
00:14:08.962 --> 00:14:10.670
we were interested
in-- this is the thing
00:14:10.670 --> 00:14:12.160
we're trying to compute, right?
00:14:12.160 --> 00:14:15.340
The problem asked us to
compute the line integral
00:14:15.340 --> 00:14:19.460
over C of F dot
dr. Well, extended
00:14:19.460 --> 00:14:21.830
Stokes' Theorem said, in
order to compute this line
00:14:21.830 --> 00:14:24.564
integral, what you can do is
you can compute this surface
00:14:24.564 --> 00:14:26.480
integral over S, and you
can compute this line
00:14:26.480 --> 00:14:28.780
integral over this
other curve C_1,
00:14:28.780 --> 00:14:31.970
and then these three things have
to satisfy this relationship.
00:14:31.970 --> 00:14:33.780
That's what's
Stokes' Theorem says.
00:14:33.780 --> 00:14:34.800
And now we've computed.
00:14:34.800 --> 00:14:36.300
We've computed the
surface integral,
00:14:36.300 --> 00:14:40.684
and we found it was equal to 0
by a simple geometric argument
00:14:40.684 --> 00:14:42.850
that didn't require us to
actually compute a surface
00:14:42.850 --> 00:14:44.560
integral.
00:14:44.560 --> 00:14:48.110
And we computed this
line integral, just now,
00:14:48.110 --> 00:14:49.760
by parametrizing
and computing it.
00:14:49.760 --> 00:14:50.260
So OK.
00:14:50.260 --> 00:14:55.020
So this was 0 and this
was 4*pi b squared.
00:14:55.020 --> 00:14:58.570
So if we just add our
integral in question
00:14:58.570 --> 00:15:00.510
to the other side,
what we find--
00:15:00.510 --> 00:15:04.360
I'm going to go find some empty
board space to write it down--
00:15:04.360 --> 00:15:14.780
so our integral, the
integral over C of F
00:15:14.780 --> 00:15:22.785
dot dr is equal to this other
line integral minus the surface
00:15:22.785 --> 00:15:23.285
integral.
00:15:23.285 --> 00:15:27.864
So it's equal to 4*pi
b squared minus 0.
00:15:27.864 --> 00:15:30.280
Just rearranging that equation
we were looking at a second
00:15:30.280 --> 00:15:31.550
ago from Stokes' Theorem.
00:15:31.550 --> 00:15:36.390
So it's just 4*pi b squared.
00:15:36.390 --> 00:15:39.160
So that's the answer,
and I'll end there.