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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

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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,

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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.

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It's over a hemisphere.

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Not a terrible thing
to parametrize.

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So that's what we should do.

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We should go in, we
should parametrize it,

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and then we should just compute
it like a surface integral,

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like we know how to do.

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So before we start
though, I want

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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?

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x.

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We're integrating x over
the surface of a hemisphere

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centered at the origin.

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This hemisphere is
really symmetric.

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And on the back
side-- the part where

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x is negative-- we're getting
negative contributions from x.

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And on the front
side-- where x is

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positive-- we're getting
positive contributions from x.

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And because this sphere
is totally symmetric,

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those just cancel
out completely.

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So when we integrate x
over the whole hemisphere,

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it just kills itself.

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I mean, the negative parts
kill the positive parts.

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We just get 0.

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Similarly, this hemisphere is
symmetric between its left side

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and its right side, and so
the parts where y are negative

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cancel out exactly the
parts where y are positive.

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So as a simplifying
step, we can realize,

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right at the beginning,
that this is actually

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just the integral over S of z
with respect to surface area.

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00:10:39,040 --> 00:10:42,289
Now, if you didn't
realize that, that's OK.

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What you would have
done is you would

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have done the parametrization
that we're about to do.

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And in doing that
parametrization,

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you would have found that you
were integrating something like

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cosine theta between 0 and
2*pi, or something like this.

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And that would have given you 0.

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So you would have
found this symmetry,

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even if you didn't
realize it right now,

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you would have found it in
the process of computing

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this integral, but it's
a little bit easier on us

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if we can recognize
that symmetry first.

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Now, notice that z doesn't
cancel, because this is just

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the top hemisphere, so it
doesn't have a bottom half

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to cancel out with.

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00:11:17,750 --> 00:11:18,250
Right?

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00:11:18,250 --> 00:11:21,930
So the z part we can't
use this easy analysis on.

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00:11:21,930 --> 00:11:24,204
If we integrated this z
over the whole sphere--

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if we had the other half
of the sphere-- well,

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then that would also give us 0.

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But we only have the
top half of the sphere.

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So it's going to give us
something positive, because z

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is always positive up there.

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00:11:35,940 --> 00:11:39,470
OK, so let's actually set
about parametrizing it.

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00:11:39,470 --> 00:11:41,800
We want to parametrize
the unit sphere.

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00:11:41,800 --> 00:11:42,440
Well, OK.

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00:11:42,440 --> 00:11:44,430
So we have our standard
parametrization

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00:11:44,430 --> 00:11:46,220
that comes from
spherical coordinates.

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So rho is just 1.

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Right?

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00:11:48,670 --> 00:11:54,917
So x is equal to, it's
going to be cosine--

253
00:11:54,917 --> 00:11:55,500
You know what?

254
00:11:55,500 --> 00:11:57,430
I always get a little
confused, so I'm just

255
00:11:57,430 --> 00:12:01,320
going to check, carefully, that
I'm doing this perfectly right.

256
00:12:01,320 --> 00:12:06,660
x is going to be
cosine theta sine phi.

257
00:12:06,660 --> 00:12:07,520
Good.

258
00:12:07,520 --> 00:12:14,370
y is going to be
sine theta sine phi.

259
00:12:14,370 --> 00:12:20,250
And z is going to be cosine phi.

260
00:12:20,250 --> 00:12:22,140
So that's our parametrization.

261
00:12:22,140 --> 00:12:25,820
But we need bounds, of
course, on theta and phi

262
00:12:25,820 --> 00:12:28,376
in order to properly describe
just this hemisphere.

263
00:12:28,376 --> 00:12:29,000
So let's think.

264
00:12:29,000 --> 00:12:31,600
So for phi, we
want the hemisphere

265
00:12:31,600 --> 00:12:36,200
that goes from the z-axis
down to the xy-plane.

266
00:12:36,200 --> 00:12:40,360
So that means we
want 0 to be less

267
00:12:40,360 --> 00:12:45,931
than or equal to phi to be less
than or equal to pi over 2.

268
00:12:45,931 --> 00:12:46,430
Right?

269
00:12:46,430 --> 00:12:48,594
That will give us
just that top half.

270
00:12:48,594 --> 00:12:49,760
And we want the whole thing.

271
00:12:49,760 --> 00:12:51,134
We want to go all
the way around.

272
00:12:51,134 --> 00:12:55,850
So we want 0 less than or equal
to theta less than or equal

273
00:12:55,850 --> 00:12:58,970
to 2*pi.

274
00:12:58,970 --> 00:13:01,760
OK, so this is what
x, y, and z are.

275
00:13:01,760 --> 00:13:06,320
These are the bounds for our
parameters phi and theta.

276
00:13:06,320 --> 00:13:07,740
Now, the only
other thing we need

277
00:13:07,740 --> 00:13:10,150
is we need to know what dS is.

278
00:13:10,150 --> 00:13:12,540
So in spherical
coordinates, we know

279
00:13:12,540 --> 00:13:19,270
that dS-- I'll put it
right above here-- so dS

280
00:13:19,270 --> 00:13:26,530
is equal to sine
phi d phi d theta.

281
00:13:26,530 --> 00:13:28,966
Let me again just double-check
that, that I'm not

282
00:13:28,966 --> 00:13:29,840
doing anything silly.

283
00:13:29,840 --> 00:13:32,580


284
00:13:32,580 --> 00:13:39,120
So dS is equal to sine
phi d phi d theta.

285
00:13:39,120 --> 00:13:41,750
So we've got our
parametrization.

286
00:13:41,750 --> 00:13:43,450
We've got our bounds
on our parameters.

287
00:13:43,450 --> 00:13:44,709
We know what dS is.

288
00:13:44,709 --> 00:13:46,750
And we have the integral
that we want to compute.

289
00:13:46,750 --> 00:13:48,583
So now we just have to
substitute everything

290
00:13:48,583 --> 00:13:50,760
in and actually compute it
as an iterated integral.

291
00:13:50,760 --> 00:13:51,380
Great.

292
00:13:51,380 --> 00:13:52,420
So let's do that.

293
00:13:52,420 --> 00:13:55,100
So, this integral
that we want, I'm

294
00:13:55,100 --> 00:13:57,510
going to write a big
equal sign that's

295
00:13:57,510 --> 00:14:00,870
going to carry me
all the way up here.

296
00:14:00,870 --> 00:14:02,080
That's an equal sign.

297
00:14:02,080 --> 00:14:02,580
All right.

298
00:14:02,580 --> 00:14:06,480
So our integral, the
integral over S of z

299
00:14:06,480 --> 00:14:08,010
with respect to surface area.

300
00:14:08,010 --> 00:14:12,700
So z becomes cosine phi.

301
00:14:12,700 --> 00:14:16,070
So we've got our double integral
becomes an iterated integral.

302
00:14:16,070 --> 00:14:20,670
z becomes cosine phi.

303
00:14:20,670 --> 00:14:23,940
dS becomes sine
phi d phi d theta.

304
00:14:23,940 --> 00:14:31,380


305
00:14:31,380 --> 00:14:32,430
And our bounds.

306
00:14:32,430 --> 00:14:38,510
So let's see: phi we said is
going from 0 to pi over 2.

307
00:14:38,510 --> 00:14:41,450
Zero, pi over 2.

308
00:14:41,450 --> 00:14:46,480
And theta is going
from 0 to 2*pi.

309
00:14:46,480 --> 00:14:47,200
OK.

310
00:14:47,200 --> 00:14:49,110
So now we just have a
nice, straightforward

311
00:14:49,110 --> 00:14:50,750
iterated integral
here to compute.

312
00:14:50,750 --> 00:14:54,460
So let's do the inner one first.

313
00:14:54,460 --> 00:14:57,940
So we're computing--
the inner integral

314
00:14:57,940 --> 00:15:07,060
is the integral from 0 to pi
over 2, of cosine phi sine phi

315
00:15:07,060 --> 00:15:08,001
d phi.

316
00:15:08,001 --> 00:15:08,500
And OK.

317
00:15:08,500 --> 00:15:10,770
So there are a bunch
of different ways

318
00:15:10,770 --> 00:15:12,060
you could do this.

319
00:15:12,060 --> 00:15:14,839
If you wanted to get fancy, you
could do a double-angle formula

320
00:15:14,839 --> 00:15:16,880
here, but that's really
more fancy than you need.

321
00:15:16,880 --> 00:15:22,970
Because this is like sine
phi times d sine phi, right?

322
00:15:22,970 --> 00:15:25,670
So this is equal
to-- another way

323
00:15:25,670 --> 00:15:28,040
of saying that is you can
make the substitution u equals

324
00:15:28,040 --> 00:15:28,990
sine phi.

325
00:15:28,990 --> 00:15:33,470
Anyhow, this is all Calc I
stuff that hopefully you're

326
00:15:33,470 --> 00:15:34,551
pretty familiar with.

327
00:15:34,551 --> 00:15:35,050
So OK.

328
00:15:35,050 --> 00:15:37,400
So this is equal
to-- in the end,

329
00:15:37,400 --> 00:15:44,020
we get sine squared phi over
2, between 0 and pi over 2.

330
00:15:44,020 --> 00:15:44,520
OK.

331
00:15:44,520 --> 00:15:45,311
So we plug this in.

332
00:15:45,311 --> 00:15:48,670
So sine squared pi
over 2, that's 1/2,

333
00:15:48,670 --> 00:15:52,280
minus-- sine squared
0 over 2 is 0 over 2.

334
00:15:52,280 --> 00:15:54,200
So it's just 1/2.

335
00:15:54,200 --> 00:15:56,180
So the inner integral is 1/2.

336
00:15:56,180 --> 00:15:58,906
So let's see about
the outer one.

337
00:15:58,906 --> 00:16:06,080
The outer integral is just the
integral from 0 to 2*pi d theta

338
00:16:06,080 --> 00:16:08,100
of whatever the
inner integral was.

339
00:16:08,100 --> 00:16:10,370
Well, the inner
integral was 1/2.

340
00:16:10,370 --> 00:16:14,660
So the integral from 0
to 2*pi of 1/2 is pi.

341
00:16:14,660 --> 00:16:15,500
Straightforward.

342
00:16:15,500 --> 00:16:16,000
Good.

343
00:16:16,000 --> 00:16:16,500
So OK.

344
00:16:16,500 --> 00:16:19,490
So that's what the
surface integral gives us.

345
00:16:19,490 --> 00:16:22,090
So let's go back
here and compare.

346
00:16:22,090 --> 00:16:27,450
So way back at the beginning
of this recitation,

347
00:16:27,450 --> 00:16:32,120
we did the line
integral for this circle

348
00:16:32,120 --> 00:16:35,360
that's the boundary of this
hemisphere, and we got pi.

349
00:16:35,360 --> 00:16:38,667
And just now what we did is
we had the surface integral--

350
00:16:38,667 --> 00:16:41,000
the associated surface integral
that we get from Stokes'

351
00:16:41,000 --> 00:16:43,570
Theorem, this curl F dot n dS.

352
00:16:43,570 --> 00:16:47,320
So we computed F
and curl F and n.

353
00:16:47,320 --> 00:16:50,455
And then we'd noticed a
little nice symmetry here.

354
00:16:50,455 --> 00:16:51,830
Although if you
didn't notice it,

355
00:16:51,830 --> 00:16:55,060
you should have had no trouble
computing the extra terms

356
00:16:55,060 --> 00:16:57,240
in the integral that you
actually ended up with it.

357
00:16:57,240 --> 00:17:00,140
It would've been another
couple of trig terms

358
00:17:00,140 --> 00:17:02,230
there after you made
the substitution.

359
00:17:02,230 --> 00:17:04,320
So we parametrized
our surface nicely.

360
00:17:04,320 --> 00:17:07,650
Because it's a sphere,
it's easy to do.

361
00:17:07,650 --> 00:17:09,880
And then we computed
the double integral

362
00:17:09,880 --> 00:17:11,682
and we also came out with pi.

363
00:17:11,682 --> 00:17:13,390
And we better have
also come out with pi,

364
00:17:13,390 --> 00:17:15,180
because Stokes' Theorem
tells us that the line

365
00:17:15,180 --> 00:17:16,554
integral and the
surface integral

366
00:17:16,554 --> 00:17:18,624
have to give us the same value.

367
00:17:18,624 --> 00:17:19,290
So that's great.

368
00:17:19,290 --> 00:17:21,790
So that's exactly what we
were hoping would happen.

369
00:17:21,790 --> 00:17:25,020
And now we've sort of
convinced ourselves, hopefully,

370
00:17:25,020 --> 00:17:27,350
that through an
example now, we have

371
00:17:27,350 --> 00:17:30,766
a feel for what sorts of things
Stokes' Theorem can do for us.

372
00:17:30,766 --> 00:17:32,360
I'll end there.

373
00:17:32,360 --> 00:17:32,986