WEBVTT

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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
learning about Stokes' Theorem.

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And I have a nice
question here for you

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that can put Stokes'
Theorem to the test.

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So what I'd like
you to do is I'd

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like you to consider
this field F.

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So its components
are 2z, x, and y.

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And the surface S that is the
top half of the unit sphere.

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So it's the sphere
of radius 1 centered

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at the origin, but
only its top half.

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Only the part where z is
greater than or equal to 0.

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So what I'd like you to do
is to verify Stokes' Theorem

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for this surface.

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So that is, I'd
like you to compute

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the surface integral
that comes from Stokes'

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Theorem for this surface,
and the line integral that

00:00:47.630 --> 00:00:49.650
comes from Stokes'
Theorem for the surface,

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and check that they're
really equal to each other.

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Now, before we
start, we should just

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say one brief thing about
compatible orientation.

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So I didn't give you
any orientations,

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but of course, it doesn't
matter as long as you

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choose ones that are compatible.

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So if you think about your rules
that you have for finding them.

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So if you imagine yourself
walking along this boundary

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circle with your left
hand out over that sphere.

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So you'll be walking in this
counterclockwise direction

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when your head is sticking
out of the sphere.

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All right?

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So in other words, the outward
orientation on the sphere

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is compatible with the
counterclockwise orientation

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on the circle that
is the boundary.

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So let's actually put
in a little arrow here

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to just indicate that is our
orientation for the circle.

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And our normal is an
outward-pointing normal.

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And let's call our
circle C, and our S

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is our sphere is our surface.

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OK.

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So just so we have
the same notation.

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Good.

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So why don't you work this
out, compute the line integral,

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compute the surface
integral, come back,

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and we can work
them out together.

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Hopefully you had some luck
working on this problem.

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We have two things to compute.

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I think I'm going to start
with the line integral.

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So let me write that
down: line integral.

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So what I need to do
to compute the line

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integral is I need to compute
the integral over the curve

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C of F dot dr. And so I know
what F is on that circle.

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So I need to know what dr is.

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So I need to know what r is.

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I need a parametrization
of that circle.

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Well, you know, that is a pretty
easy circle to parametrize.

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It's the unit circle
in the xy-plane.

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So we have-- for C, we have--
and we're wandering around it

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counterclockwise.

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So it's our usual
parametrization.

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It's the one we like.

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So we have x equals cosine
t, y equals sine t--

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where t goes from 0 to
2*pi-- and this is in three

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dimensions, so the other part of
the parametrization is z equals

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0.

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So this is my parametrization
of this circle.

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OK, so let's go ahead
and put that in.

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So the integral over C of F
dot dr is the integral from 0

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to 2 pi.

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So we've got three parts.

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So the first part--
so F is 2z, x, y.

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So it's 2z*dx plus
x*dy plus y*dz.

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But z is 0 on this whole circle.

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So that piece just dies.

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And dz is also 0, so
that piece just dies.

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So we're just left with x*dy.

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So this is equal to
the integral x dy.

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Oh.

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So I guess this is
not from 0 to 2*pi.

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This is still over
C. Sorry about that.

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OK.

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And now I change to
my parametrization.

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OK.

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Yes.

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Right.

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So this is still in dx, dy,
dz form, so it's still over C.

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Now we switch to the dt form, so
now t is going from 0 to 2*pi.

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OK, so now we have x*dy.

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So x is cosine t, and
dy-- so y is sine t,

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so dy is cosine t dt.

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So this is cosine t times
cosine t, is cosine squared t.

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dt, gosh.

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So now you have to
remember way back in 18.01

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when you learned how to compute
trig integrals like this.

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So I think the thing
that we do, when

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we have a cosine squared t, is
we use a half-angle formula.

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So let me come back down
here just to finish this off

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in one board.

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OK, so cosine squared t is
the integral from 0 to 2*pi.

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So cosine squared t is 1
plus cosine 2t over 2, dt.

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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.

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Right?

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Two whole periods of cosine 2t.

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And it's a trig function.

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It's a nice cosine function.

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So the positive parts and
the negative parts cancel.

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The cosine 2t part, when we
integrate it from 0 to 2*pi,

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that gives us 0.

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So we're left with 1/2
integrated from 0 to 2*pi,

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and that's just going to give
us 1/2 of 2*pi, so that's pi.

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All right.

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So good.

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So that was the line integral.

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A very straightforward thing.

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We had our circle back here.

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We had our field.

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So we parametrized
the curve that

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is the circle, that
is the boundary.

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And then we just computed
the line integral,

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and it was a nice,
easy one to do.

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You had to remember one
little trig identity in order

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to do it.

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All right.

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That's the first one.

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So let's go on to
the surface integral.

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So the surface
integral that you have

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to compute in Stokes'
Theorem is you

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have to compute
the double integral

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over your surface of the
curl of F dot n with respect

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to surface area.

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So this is the integral
we want to compute here.

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So OK.

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So the first thing
we're going to need

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is we're going to need
to find the curl of F.

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So F-- let me just write it
here so we don't have to walk

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all the way back over there.

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So F is [2z, x, y].

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So curl of F-- OK, you should
have lots of experience

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computing curls by
now-- So it's going

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to be this-- I
always think of it,

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so you've got these
little 2 by 2 determinants

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with the partial derivatives
in them, but most of those

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are going to be 0.

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We've got a d_x x term
that's coming up in k,

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and a d_y y term
that's coming up in i,

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and a d_z 2z term
that's coming up in j.

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So OK.

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So almost half the terms are 0.

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The others are really
easy to compute.

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I trust that you can
also compute and get

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that the curl is [1, 2, 1] here.

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OK, so this is F. This
is curl of F. Great.

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So OK.

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So that's curl of F.

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So now we need n.

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Well, let's think.

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So we need the unit
normal to our surface.

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So back at the beginning
before we started,

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we said it was the
outward-pointing normal.

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So we need the
outward-pointing normal.

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Well, this is a sphere, right?

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So the normal is parallel
to the position vector.

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So that means n
should be parallel

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to the vector [x, y, z].

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So n should be parallel
to this vector [x, y, z],

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but in fact, we're
even better than that.

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We're on a unit sphere.

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So the position vector
has length of 1.

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So n should be pointing in the
same direction as this vector,

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and they both have length
1, so they had better

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be equal to each other.

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Great.

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So this unit normal n is
just this very simple vector,

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[x, y, z].

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If it had been a
bigger sphere, then you

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would have to divide
this by the radius

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to scale it appropriately.

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All right.

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So we've got curl
F. We've got n.

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So the integral that we
want is this double integral

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over the surface
of curl F dot n.

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So that's x plus 2y plus z,
with respect to surface area.

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OK.

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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.

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Well, maybe two
little observations.

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We can simplify this.

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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.