WEBVTT
00:00:07.065 --> 00:00:07.690
JOEL LEWIS: Hi.
00:00:07.690 --> 00:00:09.309
Welcome back to recitation.
00:00:09.309 --> 00:00:11.600
In lecture, you've been
learning about Stokes' Theorem.
00:00:11.600 --> 00:00:13.480
And I have a nice
question here for you
00:00:13.480 --> 00:00:15.466
that can put Stokes'
Theorem to the test.
00:00:15.466 --> 00:00:16.840
So what I'd like
you to do is I'd
00:00:16.840 --> 00:00:19.300
like you to consider
this field F.
00:00:19.300 --> 00:00:22.910
So its components
are 2z, x, and y.
00:00:22.910 --> 00:00:27.450
And the surface S that is the
top half of the unit sphere.
00:00:27.450 --> 00:00:29.930
So it's the sphere
of radius 1 centered
00:00:29.930 --> 00:00:32.080
at the origin, but
only its top half.
00:00:32.080 --> 00:00:34.540
Only the part where z is
greater than or equal to 0.
00:00:34.540 --> 00:00:37.480
So what I'd like you to do
is to verify Stokes' Theorem
00:00:37.480 --> 00:00:38.540
for this surface.
00:00:38.540 --> 00:00:41.650
So that is, I'd
like you to compute
00:00:41.650 --> 00:00:44.870
the surface integral
that comes from Stokes'
00:00:44.870 --> 00:00:47.630
Theorem for this surface,
and the line integral that
00:00:47.630 --> 00:00:49.650
comes from Stokes'
Theorem for the surface,
00:00:49.650 --> 00:00:52.030
and check that they're
really equal to each other.
00:00:52.030 --> 00:00:54.310
Now, before we
start, we should just
00:00:54.310 --> 00:00:57.610
say one brief thing about
compatible orientation.
00:00:57.610 --> 00:00:59.380
So I didn't give you
any orientations,
00:00:59.380 --> 00:01:01.350
but of course, it doesn't
matter as long as you
00:01:01.350 --> 00:01:03.170
choose ones that are compatible.
00:01:03.170 --> 00:01:06.620
So if you think about your rules
that you have for finding them.
00:01:06.620 --> 00:01:09.540
So if you imagine yourself
walking along this boundary
00:01:09.540 --> 00:01:12.940
circle with your left
hand out over that sphere.
00:01:12.940 --> 00:01:18.270
So you'll be walking in this
counterclockwise direction
00:01:18.270 --> 00:01:21.970
when your head is sticking
out of the sphere.
00:01:21.970 --> 00:01:22.470
All right?
00:01:22.470 --> 00:01:26.680
So in other words, the outward
orientation on the sphere
00:01:26.680 --> 00:01:30.640
is compatible with the
counterclockwise orientation
00:01:30.640 --> 00:01:33.130
on the circle that
is the boundary.
00:01:33.130 --> 00:01:35.340
So let's actually put
in a little arrow here
00:01:35.340 --> 00:01:38.835
to just indicate that is our
orientation for the circle.
00:01:41.730 --> 00:01:44.010
And our normal is an
outward-pointing normal.
00:01:44.010 --> 00:01:46.685
And let's call our
circle C, and our S
00:01:46.685 --> 00:01:48.620
is our sphere is our surface.
00:01:48.620 --> 00:01:49.230
OK.
00:01:49.230 --> 00:01:52.701
So just so we have
the same notation.
00:01:52.701 --> 00:01:53.200
Good.
00:01:53.200 --> 00:01:56.630
So why don't you work this
out, compute the line integral,
00:01:56.630 --> 00:01:58.354
compute the surface
integral, come back,
00:01:58.354 --> 00:01:59.770
and we can work
them out together.
00:02:08.134 --> 00:02:10.300
Hopefully you had some luck
working on this problem.
00:02:10.300 --> 00:02:12.260
We have two things to compute.
00:02:12.260 --> 00:02:16.500
I think I'm going to start
with the line integral.
00:02:16.500 --> 00:02:18.960
So let me write that
down: line integral.
00:02:23.240 --> 00:02:25.240
So what I need to do
to compute the line
00:02:25.240 --> 00:02:31.050
integral is I need to compute
the integral over the curve
00:02:31.050 --> 00:02:41.290
C of F dot dr. And so I know
what F is on that circle.
00:02:41.290 --> 00:02:43.444
So I need to know what dr is.
00:02:43.444 --> 00:02:44.610
So I need to know what r is.
00:02:44.610 --> 00:02:46.400
I need a parametrization
of that circle.
00:02:46.400 --> 00:02:49.440
Well, you know, that is a pretty
easy circle to parametrize.
00:02:49.440 --> 00:02:51.800
It's the unit circle
in the xy-plane.
00:02:51.800 --> 00:02:58.690
So we have-- for C, we have--
and we're wandering around it
00:02:58.690 --> 00:02:59.760
counterclockwise.
00:02:59.760 --> 00:03:01.485
So it's our usual
parametrization.
00:03:01.485 --> 00:03:02.360
It's the one we like.
00:03:02.360 --> 00:03:10.030
So we have x equals cosine
t, y equals sine t--
00:03:10.030 --> 00:03:16.450
where t goes from 0 to
2*pi-- and this is in three
00:03:16.450 --> 00:03:19.960
dimensions, so the other part of
the parametrization is z equals
00:03:19.960 --> 00:03:20.730
0.
00:03:20.730 --> 00:03:26.310
So this is my parametrization
of this circle.
00:03:26.310 --> 00:03:29.370
OK, so let's go ahead
and put that in.
00:03:29.370 --> 00:03:44.160
So the integral over C of F
dot dr is the integral from 0
00:03:44.160 --> 00:03:45.370
to 2 pi.
00:03:45.370 --> 00:03:46.670
So we've got three parts.
00:03:46.670 --> 00:03:53.030
So the first part--
so F is 2z, x, y.
00:03:53.030 --> 00:03:57.690
So it's 2z*dx plus
x*dy plus y*dz.
00:03:57.690 --> 00:04:00.900
But z is 0 on this whole circle.
00:04:00.900 --> 00:04:03.190
So that piece just dies.
00:04:03.190 --> 00:04:05.470
And dz is also 0, so
that piece just dies.
00:04:05.470 --> 00:04:07.490
So we're just left with x*dy.
00:04:07.490 --> 00:04:11.521
So this is equal to
the integral x dy.
00:04:11.521 --> 00:04:12.020
Oh.
00:04:12.020 --> 00:04:14.620
So I guess this is
not from 0 to 2*pi.
00:04:14.620 --> 00:04:18.411
This is still over
C. Sorry about that.
00:04:18.411 --> 00:04:18.910
OK.
00:04:18.910 --> 00:04:25.670
And now I change to
my parametrization.
00:04:25.670 --> 00:04:26.170
OK.
00:04:26.170 --> 00:04:26.730
Yes.
00:04:26.730 --> 00:04:27.230
Right.
00:04:27.230 --> 00:04:32.430
So this is still in dx, dy,
dz form, so it's still over C.
00:04:32.430 --> 00:04:36.230
Now we switch to the dt form, so
now t is going from 0 to 2*pi.
00:04:36.230 --> 00:04:37.550
OK, so now we have x*dy.
00:04:37.550 --> 00:04:42.430
So x is cosine t, and
dy-- so y is sine t,
00:04:42.430 --> 00:04:44.680
so dy is cosine t dt.
00:04:44.680 --> 00:04:51.090
So this is cosine t times
cosine t, is cosine squared t.
00:04:51.090 --> 00:04:51.820
dt, gosh.
00:04:51.820 --> 00:04:54.545
So now you have to
remember way back in 18.01
00:04:54.545 --> 00:04:56.920
when you learned how to compute
trig integrals like this.
00:04:56.920 --> 00:04:58.952
So I think the thing
that we do, when
00:04:58.952 --> 00:05:01.410
we have a cosine squared t, is
we use a half-angle formula.
00:05:01.410 --> 00:05:04.390
So let me come back down
here just to finish this off
00:05:04.390 --> 00:05:06.190
in one board.
00:05:06.190 --> 00:05:11.570
OK, so cosine squared t is
the integral from 0 to 2*pi.
00:05:11.570 --> 00:05:20.410
So cosine squared t is 1
plus cosine 2t over 2, dt.
00:05:20.410 --> 00:05:24.960
And now cosine 2t, as t goes
between 0 and 2*pi, well,
00:05:24.960 --> 00:05:27.381
that's two whole loops of it.
00:05:27.381 --> 00:05:27.880
Right?
00:05:27.880 --> 00:05:30.650
Two whole periods of cosine 2t.
00:05:30.650 --> 00:05:32.280
And it's a trig function.
00:05:32.280 --> 00:05:33.504
It's a nice cosine function.
00:05:33.504 --> 00:05:35.670
So the positive parts and
the negative parts cancel.
00:05:35.670 --> 00:05:39.200
The cosine 2t part, when we
integrate it from 0 to 2*pi,
00:05:39.200 --> 00:05:40.830
that gives us 0.
00:05:40.830 --> 00:05:45.340
So we're left with 1/2
integrated from 0 to 2*pi,
00:05:45.340 --> 00:05:49.202
and that's just going to give
us 1/2 of 2*pi, so that's pi.
00:05:49.202 --> 00:05:50.980
All right.
00:05:50.980 --> 00:05:51.480
So good.
00:05:51.480 --> 00:05:52.730
So that was the line integral.
00:05:52.730 --> 00:05:54.540
A very straightforward thing.
00:05:54.540 --> 00:05:57.910
We had our circle back here.
00:05:57.910 --> 00:05:59.150
We had our field.
00:05:59.150 --> 00:06:02.170
So we parametrized
the curve that
00:06:02.170 --> 00:06:04.500
is the circle, that
is the boundary.
00:06:04.500 --> 00:06:06.690
And then we just computed
the line integral,
00:06:06.690 --> 00:06:08.750
and it was a nice,
easy one to do.
00:06:08.750 --> 00:06:11.090
You had to remember one
little trig identity in order
00:06:11.090 --> 00:06:12.170
to do it.
00:06:12.170 --> 00:06:13.230
All right.
00:06:13.230 --> 00:06:14.360
That's the first one.
00:06:14.360 --> 00:06:18.290
So let's go on to
the surface integral.
00:06:25.760 --> 00:06:27.570
So the surface
integral that you have
00:06:27.570 --> 00:06:29.870
to compute in Stokes'
Theorem is you
00:06:29.870 --> 00:06:33.720
have to compute
the double integral
00:06:33.720 --> 00:06:42.340
over your surface of the
curl of F dot n with respect
00:06:42.340 --> 00:06:43.090
to surface area.
00:06:43.090 --> 00:06:47.610
So this is the integral
we want to compute here.
00:06:47.610 --> 00:06:48.210
So OK.
00:06:48.210 --> 00:06:49.793
So the first thing
we're going to need
00:06:49.793 --> 00:06:52.440
is we're going to need
to find the curl of F.
00:06:52.440 --> 00:06:55.754
So F-- let me just write it
here so we don't have to walk
00:06:55.754 --> 00:06:56.920
all the way back over there.
00:06:56.920 --> 00:07:04.030
So F is [2z, x, y].
00:07:04.030 --> 00:07:07.950
So curl of F-- OK, you should
have lots of experience
00:07:07.950 --> 00:07:10.755
computing curls by
now-- So it's going
00:07:10.755 --> 00:07:13.690
to be this-- I
always think of it,
00:07:13.690 --> 00:07:16.400
so you've got these
little 2 by 2 determinants
00:07:16.400 --> 00:07:19.160
with the partial derivatives
in them, but most of those
00:07:19.160 --> 00:07:20.360
are going to be 0.
00:07:20.360 --> 00:07:23.940
We've got a d_x x term
that's coming up in k,
00:07:23.940 --> 00:07:27.410
and a d_y y term
that's coming up in i,
00:07:27.410 --> 00:07:32.471
and a d_z 2z term
that's coming up in j.
00:07:32.471 --> 00:07:32.970
So OK.
00:07:32.970 --> 00:07:35.370
So almost half the terms are 0.
00:07:35.370 --> 00:07:37.160
The others are really
easy to compute.
00:07:37.160 --> 00:07:41.790
I trust that you can
also compute and get
00:07:41.790 --> 00:07:44.300
that the curl is [1, 2, 1] here.
00:07:44.300 --> 00:07:46.891
OK, so this is F. This
is curl of F. Great.
00:07:46.891 --> 00:07:47.390
So OK.
00:07:47.390 --> 00:07:48.260
So that's curl of F.
00:07:48.260 --> 00:07:51.250
So now we need n.
00:07:51.250 --> 00:07:52.000
Well, let's think.
00:07:52.000 --> 00:07:55.880
So we need the unit
normal to our surface.
00:07:55.880 --> 00:07:58.260
So back at the beginning
before we started,
00:07:58.260 --> 00:08:00.780
we said it was the
outward-pointing normal.
00:08:00.780 --> 00:08:02.460
So we need the
outward-pointing normal.
00:08:02.460 --> 00:08:04.360
Well, this is a sphere, right?
00:08:04.360 --> 00:08:07.790
So the normal is parallel
to the position vector.
00:08:07.790 --> 00:08:13.270
So that means n
should be parallel
00:08:13.270 --> 00:08:17.850
to the vector [x, y, z].
00:08:17.850 --> 00:08:20.270
So n should be parallel
to this vector [x, y, z],
00:08:20.270 --> 00:08:22.590
but in fact, we're
even better than that.
00:08:22.590 --> 00:08:24.380
We're on a unit sphere.
00:08:24.380 --> 00:08:27.230
So the position vector
has length of 1.
00:08:27.230 --> 00:08:30.132
So n should be pointing in the
same direction as this vector,
00:08:30.132 --> 00:08:32.090
and they both have length
1, so they had better
00:08:32.090 --> 00:08:34.460
be equal to each other.
00:08:34.460 --> 00:08:36.520
Great.
00:08:36.520 --> 00:08:40.320
So this unit normal n is
just this very simple vector,
00:08:40.320 --> 00:08:41.570
[x, y, z].
00:08:41.570 --> 00:08:43.980
If it had been a
bigger sphere, then you
00:08:43.980 --> 00:08:46.700
would have to divide
this by the radius
00:08:46.700 --> 00:08:47.910
to scale it appropriately.
00:08:50.770 --> 00:08:51.280
All right.
00:08:51.280 --> 00:08:54.770
So we've got curl
F. We've got n.
00:08:54.770 --> 00:09:02.140
So the integral that we
want is this double integral
00:09:02.140 --> 00:09:05.400
over the surface
of curl F dot n.
00:09:05.400 --> 00:09:14.290
So that's x plus 2y plus z,
with respect to surface area.
00:09:14.290 --> 00:09:14.840
OK.
00:09:14.840 --> 00:09:16.790
Well, now we've just
got a surface integral.
00:09:16.790 --> 00:09:19.800
It's over a hemisphere.
00:09:19.800 --> 00:09:21.574
Not a terrible thing
to parametrize.
00:09:21.574 --> 00:09:22.740
So that's what we should do.
00:09:22.740 --> 00:09:24.850
We should go in, we
should parametrize it,
00:09:24.850 --> 00:09:28.050
and then we should just compute
it like a surface integral,
00:09:28.050 --> 00:09:29.140
like we know how to do.
00:09:29.140 --> 00:09:30.905
So before we start
though, I want
00:09:30.905 --> 00:09:32.640
to make one little observation.
00:09:32.640 --> 00:09:34.430
Well, maybe two
little observations.
00:09:34.430 --> 00:09:36.370
We can simplify this.
00:09:36.370 --> 00:09:37.040
All right?
00:09:37.040 --> 00:09:39.180
x.
00:09:39.180 --> 00:09:42.880
We're integrating x over
the surface of a hemisphere
00:09:42.880 --> 00:09:44.820
centered at the origin.
00:09:44.820 --> 00:09:47.150
This hemisphere is
really symmetric.
00:09:47.150 --> 00:09:50.320
And on the back
side-- the part where
00:09:50.320 --> 00:09:54.690
x is negative-- we're getting
negative contributions from x.
00:09:54.690 --> 00:09:56.210
And on the front
side-- where x is
00:09:56.210 --> 00:09:59.310
positive-- we're getting
positive contributions from x.
00:09:59.310 --> 00:10:01.770
And because this sphere
is totally symmetric,
00:10:01.770 --> 00:10:04.440
those just cancel
out completely.
00:10:04.440 --> 00:10:12.850
So when we integrate x
over the whole hemisphere,
00:10:12.850 --> 00:10:14.060
it just kills itself.
00:10:14.060 --> 00:10:16.185
I mean, the negative parts
kill the positive parts.
00:10:16.185 --> 00:10:16.880
We just get 0.
00:10:16.880 --> 00:10:20.690
Similarly, this hemisphere is
symmetric between its left side
00:10:20.690 --> 00:10:25.270
and its right side, and so
the parts where y are negative
00:10:25.270 --> 00:10:28.770
cancel out exactly the
parts where y are positive.
00:10:28.770 --> 00:10:31.010
So as a simplifying
step, we can realize,
00:10:31.010 --> 00:10:33.120
right at the beginning,
that this is actually
00:10:33.120 --> 00:10:39.040
just the integral over S of z
with respect to surface area.
00:10:39.040 --> 00:10:42.289
Now, if you didn't
realize that, that's OK.
00:10:42.289 --> 00:10:43.830
What you would have
done is you would
00:10:43.830 --> 00:10:46.280
have done the parametrization
that we're about to do.
00:10:46.280 --> 00:10:48.770
And in doing that
parametrization,
00:10:48.770 --> 00:10:51.320
you would have found that you
were integrating something like
00:10:51.320 --> 00:10:55.600
cosine theta between 0 and
2*pi, or something like this.
00:10:55.600 --> 00:10:57.110
And that would have given you 0.
00:10:57.110 --> 00:11:01.250
So you would have
found this symmetry,
00:11:01.250 --> 00:11:03.290
even if you didn't
realize it right now,
00:11:03.290 --> 00:11:05.850
you would have found it in
the process of computing
00:11:05.850 --> 00:11:08.640
this integral, but it's
a little bit easier on us
00:11:08.640 --> 00:11:10.910
if we can recognize
that symmetry first.
00:11:10.910 --> 00:11:13.960
Now, notice that z doesn't
cancel, because this is just
00:11:13.960 --> 00:11:16.850
the top hemisphere, so it
doesn't have a bottom half
00:11:16.850 --> 00:11:17.750
to cancel out with.
00:11:17.750 --> 00:11:18.250
Right?
00:11:18.250 --> 00:11:21.930
So the z part we can't
use this easy analysis on.
00:11:21.930 --> 00:11:24.204
If we integrated this z
over the whole sphere--
00:11:24.204 --> 00:11:26.120
if we had the other half
of the sphere-- well,
00:11:26.120 --> 00:11:28.550
then that would also give us 0.
00:11:28.550 --> 00:11:31.980
But we only have the
top half of the sphere.
00:11:31.980 --> 00:11:34.315
So it's going to give us
something positive, because z
00:11:34.315 --> 00:11:35.940
is always positive up there.
00:11:35.940 --> 00:11:39.470
OK, so let's actually set
about parametrizing it.
00:11:39.470 --> 00:11:41.800
We want to parametrize
the unit sphere.
00:11:41.800 --> 00:11:42.440
Well, OK.
00:11:42.440 --> 00:11:44.430
So we have our standard
parametrization
00:11:44.430 --> 00:11:46.220
that comes from
spherical coordinates.
00:11:46.220 --> 00:11:48.120
So rho is just 1.
00:11:48.120 --> 00:11:48.670
Right?
00:11:48.670 --> 00:11:54.917
So x is equal to, it's
going to be cosine--
00:11:54.917 --> 00:11:55.500
You know what?
00:11:55.500 --> 00:11:57.430
I always get a little
confused, so I'm just
00:11:57.430 --> 00:12:01.320
going to check, carefully, that
I'm doing this perfectly right.
00:12:01.320 --> 00:12:06.660
x is going to be
cosine theta sine phi.
00:12:06.660 --> 00:12:07.520
Good.
00:12:07.520 --> 00:12:14.370
y is going to be
sine theta sine phi.
00:12:14.370 --> 00:12:20.250
And z is going to be cosine phi.
00:12:20.250 --> 00:12:22.140
So that's our parametrization.
00:12:22.140 --> 00:12:25.820
But we need bounds, of
course, on theta and phi
00:12:25.820 --> 00:12:28.376
in order to properly describe
just this hemisphere.
00:12:28.376 --> 00:12:29.000
So let's think.
00:12:29.000 --> 00:12:31.600
So for phi, we
want the hemisphere
00:12:31.600 --> 00:12:36.200
that goes from the z-axis
down to the xy-plane.
00:12:36.200 --> 00:12:40.360
So that means we
want 0 to be less
00:12:40.360 --> 00:12:45.931
than or equal to phi to be less
than or equal to pi over 2.
00:12:45.931 --> 00:12:46.430
Right?
00:12:46.430 --> 00:12:48.594
That will give us
just that top half.
00:12:48.594 --> 00:12:49.760
And we want the whole thing.
00:12:49.760 --> 00:12:51.134
We want to go all
the way around.
00:12:51.134 --> 00:12:55.850
So we want 0 less than or equal
to theta less than or equal
00:12:55.850 --> 00:12:58.970
to 2*pi.
00:12:58.970 --> 00:13:01.760
OK, so this is what
x, y, and z are.
00:13:01.760 --> 00:13:06.320
These are the bounds for our
parameters phi and theta.
00:13:06.320 --> 00:13:07.740
Now, the only
other thing we need
00:13:07.740 --> 00:13:10.150
is we need to know what dS is.
00:13:10.150 --> 00:13:12.540
So in spherical
coordinates, we know
00:13:12.540 --> 00:13:19.270
that dS-- I'll put it
right above here-- so dS
00:13:19.270 --> 00:13:26.530
is equal to sine
phi d phi d theta.
00:13:26.530 --> 00:13:28.966
Let me again just double-check
that, that I'm not
00:13:28.966 --> 00:13:29.840
doing anything silly.
00:13:32.580 --> 00:13:39.120
So dS is equal to sine
phi d phi d theta.
00:13:39.120 --> 00:13:41.750
So we've got our
parametrization.
00:13:41.750 --> 00:13:43.450
We've got our bounds
on our parameters.
00:13:43.450 --> 00:13:44.709
We know what dS is.
00:13:44.709 --> 00:13:46.750
And we have the integral
that we want to compute.
00:13:46.750 --> 00:13:48.583
So now we just have to
substitute everything
00:13:48.583 --> 00:13:50.760
in and actually compute it
as an iterated integral.
00:13:50.760 --> 00:13:51.380
Great.
00:13:51.380 --> 00:13:52.420
So let's do that.
00:13:52.420 --> 00:13:55.100
So, this integral
that we want, I'm
00:13:55.100 --> 00:13:57.510
going to write a big
equal sign that's
00:13:57.510 --> 00:14:00.870
going to carry me
all the way up here.
00:14:00.870 --> 00:14:02.080
That's an equal sign.
00:14:02.080 --> 00:14:02.580
All right.
00:14:02.580 --> 00:14:06.480
So our integral, the
integral over S of z
00:14:06.480 --> 00:14:08.010
with respect to surface area.
00:14:08.010 --> 00:14:12.700
So z becomes cosine phi.
00:14:12.700 --> 00:14:16.070
So we've got our double integral
becomes an iterated integral.
00:14:16.070 --> 00:14:20.670
z becomes cosine phi.
00:14:20.670 --> 00:14:23.940
dS becomes sine
phi d phi d theta.
00:14:31.380 --> 00:14:32.430
And our bounds.
00:14:32.430 --> 00:14:38.510
So let's see: phi we said is
going from 0 to pi over 2.
00:14:38.510 --> 00:14:41.450
Zero, pi over 2.
00:14:41.450 --> 00:14:46.480
And theta is going
from 0 to 2*pi.
00:14:46.480 --> 00:14:47.200
OK.
00:14:47.200 --> 00:14:49.110
So now we just have a
nice, straightforward
00:14:49.110 --> 00:14:50.750
iterated integral
here to compute.
00:14:50.750 --> 00:14:54.460
So let's do the inner one first.
00:14:54.460 --> 00:14:57.940
So we're computing--
the inner integral
00:14:57.940 --> 00:15:07.060
is the integral from 0 to pi
over 2, of cosine phi sine phi
00:15:07.060 --> 00:15:08.001
d phi.
00:15:08.001 --> 00:15:08.500
And OK.
00:15:08.500 --> 00:15:10.770
So there are a bunch
of different ways
00:15:10.770 --> 00:15:12.060
you could do this.
00:15:12.060 --> 00:15:14.839
If you wanted to get fancy, you
could do a double-angle formula
00:15:14.839 --> 00:15:16.880
here, but that's really
more fancy than you need.
00:15:16.880 --> 00:15:22.970
Because this is like sine
phi times d sine phi, right?
00:15:22.970 --> 00:15:25.670
So this is equal
to-- another way
00:15:25.670 --> 00:15:28.040
of saying that is you can
make the substitution u equals
00:15:28.040 --> 00:15:28.990
sine phi.
00:15:28.990 --> 00:15:33.470
Anyhow, this is all Calc I
stuff that hopefully you're
00:15:33.470 --> 00:15:34.551
pretty familiar with.
00:15:34.551 --> 00:15:35.050
So OK.
00:15:35.050 --> 00:15:37.400
So this is equal
to-- in the end,
00:15:37.400 --> 00:15:44.020
we get sine squared phi over
2, between 0 and pi over 2.
00:15:44.020 --> 00:15:44.520
OK.
00:15:44.520 --> 00:15:45.311
So we plug this in.
00:15:45.311 --> 00:15:48.670
So sine squared pi
over 2, that's 1/2,
00:15:48.670 --> 00:15:52.280
minus-- sine squared
0 over 2 is 0 over 2.
00:15:52.280 --> 00:15:54.200
So it's just 1/2.
00:15:54.200 --> 00:15:56.180
So the inner integral is 1/2.
00:15:56.180 --> 00:15:58.906
So let's see about
the outer one.
00:15:58.906 --> 00:16:06.080
The outer integral is just the
integral from 0 to 2*pi d theta
00:16:06.080 --> 00:16:08.100
of whatever the
inner integral was.
00:16:08.100 --> 00:16:10.370
Well, the inner
integral was 1/2.
00:16:10.370 --> 00:16:14.660
So the integral from 0
to 2*pi of 1/2 is pi.
00:16:14.660 --> 00:16:15.500
Straightforward.
00:16:15.500 --> 00:16:16.000
Good.
00:16:16.000 --> 00:16:16.500
So OK.
00:16:16.500 --> 00:16:19.490
So that's what the
surface integral gives us.
00:16:19.490 --> 00:16:22.090
So let's go back
here and compare.
00:16:22.090 --> 00:16:27.450
So way back at the beginning
of this recitation,
00:16:27.450 --> 00:16:32.120
we did the line
integral for this circle
00:16:32.120 --> 00:16:35.360
that's the boundary of this
hemisphere, and we got pi.
00:16:35.360 --> 00:16:38.667
And just now what we did is
we had the surface integral--
00:16:38.667 --> 00:16:41.000
the associated surface integral
that we get from Stokes'
00:16:41.000 --> 00:16:43.570
Theorem, this curl F dot n dS.
00:16:43.570 --> 00:16:47.320
So we computed F
and curl F and n.
00:16:47.320 --> 00:16:50.455
And then we'd noticed a
little nice symmetry here.
00:16:50.455 --> 00:16:51.830
Although if you
didn't notice it,
00:16:51.830 --> 00:16:55.060
you should have had no trouble
computing the extra terms
00:16:55.060 --> 00:16:57.240
in the integral that you
actually ended up with it.
00:16:57.240 --> 00:17:00.140
It would've been another
couple of trig terms
00:17:00.140 --> 00:17:02.230
there after you made
the substitution.
00:17:02.230 --> 00:17:04.320
So we parametrized
our surface nicely.
00:17:04.320 --> 00:17:07.650
Because it's a sphere,
it's easy to do.
00:17:07.650 --> 00:17:09.880
And then we computed
the double integral
00:17:09.880 --> 00:17:11.682
and we also came out with pi.
00:17:11.682 --> 00:17:13.390
And we better have
also come out with pi,
00:17:13.390 --> 00:17:15.180
because Stokes' Theorem
tells us that the line
00:17:15.180 --> 00:17:16.554
integral and the
surface integral
00:17:16.554 --> 00:17:18.624
have to give us the same value.
00:17:18.624 --> 00:17:19.290
So that's great.
00:17:19.290 --> 00:17:21.790
So that's exactly what we
were hoping would happen.
00:17:21.790 --> 00:17:25.020
And now we've sort of
convinced ourselves, hopefully,
00:17:25.020 --> 00:17:27.350
that through an
example now, we have
00:17:27.350 --> 00:17:30.766
a feel for what sorts of things
Stokes' Theorem can do for us.
00:17:30.766 --> 00:17:32.360
I'll end there.