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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

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about Stoke's Theorem.

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And I have a nice question
here for you that can put

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Stoke's Theorem to the test.

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

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field F. 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 at the origin, but

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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 Stoke's Theorem for

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this surface.

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So that is, I'd like you to
compute the surface integral

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that comes from Stoke's Theorem
for this surface, and

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the line integral that comes
from Stoke's Theorem for the

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surface, and check that
they're really

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equal to each other.

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Now, before we start, we should
just say one brief

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thing about compatible
orientation.

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So I didn't give you any
orientations, but of course,

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it doesn't matter as long as
you choose ones that are

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compatible.

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So if you think about your
rules that you have for

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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 is

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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 to just

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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 is our

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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

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integral, compute the surface
integral, come back, and we

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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 integral is I

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need to compute the integral
over the curve C of F dot dr.

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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 circle.

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

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parametrize.

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It's the unit circle
in the xy plane.

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So we have C. 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--

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and this is in three dimensions,
so the other part

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of the parametrization
is z equals 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

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integral from 0 to 2 pi.

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So we've got three parts.

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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

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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--

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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 when you

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learned how to compute trig
integrals like this.

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So I think the thing that we
do when we have a cosine

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squared t is we use a
half-angle formula.

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So let me come back down
here just to finish

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this off 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--

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as t goes between 0 and 2 pi--

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well, 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, and

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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 is the

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circle that is the boundary.

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And then we just computed the
line integral, and it was a

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nice, easy one to do.

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You had to remember one
little trig identity

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in order 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 to compute in Stoke's

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Theorem is you have to compute
the double integral over your

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surface of the curl of F dot n
with respect 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 is we're going to need

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to find the curl of F. Let me
just write it here so we don't

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have to walk all the way
back over there.

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So F is 2z, x, y.

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So curl of F--

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OK, you should have
lots of experience

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computing curls by now--

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is going to be--

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I always think of it as these
little 2 by 2 determinants

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with the partial derivatives in
them, but most of those are

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going to be 0.

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We've got a dx x term that's
coming up in k, and a dy y

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term that's coming up in i,
and a dz 2z term that's

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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 that the curl

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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, we said it

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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 to the

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vector x, y, z.

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So n should be parallel to this
vector x, y, z, but in

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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 be equal to

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each other.

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Great.

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So this unit normal n is
just this very simple

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vector x, y, z.

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If it had been a bigger sphere,
then you would have to

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divide this by the radius 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 over

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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, and then we

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should just compute it like
a surface integral, like

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we know how to do.

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So before we start though,
I want to make one little

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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 x is negative--

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we're getting negative
contributions from x.

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And on the front side--
where x is positive--

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we're getting positive
contributions from x.

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And because this sphere is
totally symmetric, those just

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cancel out completely.

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So when we integrate x over the
whole hemisphere, it just

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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

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

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negative cancel out exactly the

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parts where y are positive.

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So as a simplifying step, we
can realize right at the

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beginning, that this is actually
just the integral

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over S of z with respect
to surface area.

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Now, if you didn't realize
that, that's OK.

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What you would have done is
you would have done the

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parametrization that
we're about to do.

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And in doing that
parametrization, you would

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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 even if you don't

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realize it right now.

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You would have found it in the
process of computing this

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integral, but it's a little
bit easier on us if we can

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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 to

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cancel out with.

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Right?

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So the z part we can't use
this easy analysis on.

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If we integrated this z over
the whole sphere--

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

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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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OK, so let's actually set
about parametrizing it.

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We want to parametrize
the unit sphere.

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Well, OK.

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So we have our standard
parametrization that comes

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from spherical coordinates.

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So rho is just 1.

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Right?

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00:11:49,370 --> 00:11:54,990

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You know what?

257
00:11:55,500 --> 00:11:57,610
I always get a little confused,
so I'm just going to

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00:11:57,610 --> 00:12:01,320
check carefully that I'm doing
this perfectly right.

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

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

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

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

263
00:12:20,250 --> 00:12:22,140
So that's our parametrization.

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00:12:22,140 --> 00:12:26,240
But we need bounds, of course,
on theta and phi in order to

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00:12:26,240 --> 00:12:28,470
properly describe just
this hemisphere.

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00:12:28,470 --> 00:12:29,000
So let's think.

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00:12:29,000 --> 00:12:33,430
So for phi, we want the
hemisphere that goes from the

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00:12:33,430 --> 00:12:36,200
z-axis down to the xy plane.

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00:12:36,200 --> 00:12:42,130
So that means we want 0 to be
less than or equal to phi to

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00:12:42,130 --> 00:12:46,150
be less than or equal
to pi over 2.

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00:12:46,150 --> 00:12:46,430
Right?

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00:12:46,430 --> 00:12:48,650
That will give us just
that top half.

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00:12:48,650 --> 00:12:49,760
And we want the whole thing.

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00:12:49,760 --> 00:12:51,060
We want to go all
the way around.

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00:12:51,060 --> 00:12:55,510
So we want 0 less than or equal
to theta less than or

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00:12:55,510 --> 00:12:58,970
equal to 2 pi.

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

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

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00:13:06,320 --> 00:13:08,290
Now, the only other thing
we need is we need to

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00:13:08,290 --> 00:13:10,150
know what dS is.

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00:13:10,150 --> 00:13:15,340
So in spherical coordinates,
we know that dS--

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00:13:15,340 --> 00:13:17,420
I'll put it right above here--

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00:13:17,420 --> 00:13:26,530
so dS is equal to sine
phi d phi d theta.

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00:13:26,530 --> 00:13:29,080
Let me again just double-check
that I'm not

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00:13:29,080 --> 00:13:30,330
doing anything silly.

286
00:13:30,330 --> 00:13:32,580

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

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

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

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00:13:43,450 --> 00:13:44,980
We know what dS is.

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00:13:44,980 --> 00:13:46,750
And we have the integral that
we want to compute.

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00:13:46,750 --> 00:13:48,450
So now we just have to
substitute everything in and

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00:13:48,450 --> 00:13:50,760
actually compute it as
an iterated integral.

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

295
00:13:51,380 --> 00:13:52,420
So let's do that.

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00:13:52,420 --> 00:13:55,870
So, this integral that we want,
I'm going to write a big

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00:13:55,870 --> 00:14:00,870
equal sign that's going to carry
me all the way up here.

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00:14:00,870 --> 00:14:02,380
That's an equal sign.

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00:14:02,380 --> 00:14:02,580
All right.

300
00:14:02,580 --> 00:14:03,565
So our integral.

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00:14:03,565 --> 00:14:08,010
The integral over S of z with
respect to surface area.

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

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00:14:12,700 --> 00:14:14,100
So we've got our
double integral

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00:14:14,100 --> 00:14:16,070
becomes an iterated integral.

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

306
00:14:20,670 --> 00:14:23,940
dS becomes sine phi
d phi d theta.

307
00:14:23,940 --> 00:14:31,380

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

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00:14:32,430 --> 00:14:36,720
So let's see: phi we said is
going from 0 to pi over 2.

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00:14:36,720 --> 00:14:41,450

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

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

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00:14:47,200 --> 00:14:49,560
So now we just have a nice,
straightforward iterated

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00:14:49,560 --> 00:14:50,750
integral here to compute.

315
00:14:50,750 --> 00:14:56,282
So let's do the inner one first.
So we're computing.

316
00:14:56,282 --> 00:15:02,340
The inner integral is the
integral from 0 to pi over 2,

317
00:15:02,340 --> 00:15:08,050
of cosine phi sine phi d phi.

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00:15:08,050 --> 00:15:08,500
And OK.

319
00:15:08,500 --> 00:15:10,770
So there are a bunch
of different ways

320
00:15:10,770 --> 00:15:12,060
you could do this.

321
00:15:12,060 --> 00:15:14,500
If you wanted to get fancy, you
could do a double-angle

322
00:15:14,500 --> 00:15:16,040
formula here, but that's
really more

323
00:15:16,040 --> 00:15:16,880
fancy than you need.

324
00:15:16,880 --> 00:15:22,510
Because this is like sine phi
times d sine phi, right?

325
00:15:22,510 --> 00:15:25,130

326
00:15:25,130 --> 00:15:27,550
Another way of saying that is
you can make the substitution

327
00:15:27,550 --> 00:15:28,990
u equals sine phi.

328
00:15:28,990 --> 00:15:33,470
Anyhow, this is all CALC I stuff
that hopefully you're

329
00:15:33,470 --> 00:15:34,650
pretty familiar with.

330
00:15:34,650 --> 00:15:35,050
So OK.

331
00:15:35,050 --> 00:15:36,660
So this is equal to--

332
00:15:36,660 --> 00:15:43,670
in the end-- we get sine squared
phi over 2, between 0

333
00:15:43,670 --> 00:15:44,310
and pi over 2.

334
00:15:44,310 --> 00:15:44,520
OK.

335
00:15:44,520 --> 00:15:45,280
So we plug this in.

336
00:15:45,280 --> 00:15:49,660
So sine squared pi over
2, that's 1/2, minus--

337
00:15:49,660 --> 00:15:52,280
sine squared 0 over
2 is 0 over 2.

338
00:15:52,280 --> 00:15:54,200
So it's just 1/2.

339
00:15:54,200 --> 00:15:56,180
So the inner integral is 1/2.

340
00:15:56,180 --> 00:15:58,906
So let's see about
the outer one.

341
00:15:58,906 --> 00:16:05,410
The outer integral is just the
integral from 0 to 2 pi d

342
00:16:05,410 --> 00:16:08,100
theta of whatever the
inner integral was.

343
00:16:08,100 --> 00:16:10,370
Well, the inner integral
was 1/2.

344
00:16:10,370 --> 00:16:14,660
So the integral from 0
to 2 pi of 1/2 is pi.

345
00:16:14,660 --> 00:16:15,650
Straightforward.

346
00:16:15,650 --> 00:16:16,000
Good.

347
00:16:16,000 --> 00:16:16,490
So OK.

348
00:16:16,490 --> 00:16:19,490
So that's what the surface
integral gives us.

349
00:16:19,490 --> 00:16:22,090
So let's go back here
and compare.

350
00:16:22,090 --> 00:16:28,530
So way back at the beginning of
this recitation, we did the

351
00:16:28,530 --> 00:16:33,170
line integral for this circle
that's the boundary of this

352
00:16:33,170 --> 00:16:35,360
hemisphere, and we got pi.

353
00:16:35,360 --> 00:16:38,850
And just now what we did is we
had the surface integral--

354
00:16:38,850 --> 00:16:40,650
the associated surface integral
that we get from

355
00:16:40,650 --> 00:16:43,570
Stoke's Theorem-- this
curl F dot n dS.

356
00:16:43,570 --> 00:16:47,320
So we computed F and
curl F and n.

357
00:16:47,320 --> 00:16:50,580
And then we'd noticed a little
nice symmetry here.

358
00:16:50,580 --> 00:16:53,070
Although if you didn't notice
it, you should have had no

359
00:16:53,070 --> 00:16:56,180
trouble computing the extra
terms in the integral that you

360
00:16:56,180 --> 00:16:57,240
actually ended up with it.

361
00:16:57,240 --> 00:17:00,680
It would've been another couple
of trig terms there

362
00:17:00,680 --> 00:17:02,230
after you made the
substitution.

363
00:17:02,230 --> 00:17:04,320
So we parametrized our
surface nicely.

364
00:17:04,320 --> 00:17:07,650
Because it's a sphere,
it's easy to do.

365
00:17:07,650 --> 00:17:10,510
And then we computed the double
integral and we also

366
00:17:10,510 --> 00:17:11,780
came out with pi.

367
00:17:11,780 --> 00:17:13,970
And we had better of also come
out with pi, because Stoke's

368
00:17:13,970 --> 00:17:15,990
Theorem tells us that the line
integral and the surface

369
00:17:15,990 --> 00:17:18,670
integral have to give
us the same value.

370
00:17:18,670 --> 00:17:19,290
So that's great.

371
00:17:19,290 --> 00:17:21,790
So that's exactly what we were
hoping would happen.

372
00:17:21,790 --> 00:17:24,060
And now we've sort of
convinced ourselves,

373
00:17:24,060 --> 00:17:28,010
hopefully, that through an
example now, we have a feel

374
00:17:28,010 --> 00:17:30,766
for what sorts of things Stoke's
Theorem can do for us.

375
00:17:30,766 --> 00:17:32,016
I'll end there.

376
00:17:32,016 --> 00:17:32,986