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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like to do
two problems that ask us to

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determine the flux of a vector
field along a surface.

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So the first one is I'd like you
to find the outward flux

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of the vector z comma x comma
y through the piece of the

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cylinder that's shown.

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So it's just shading the
cylindrical part.

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So it's a cylinder of radius a,
and you're taking the piece

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in the first octant
up to height h.

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So that's the first question
I'd like you to answer.

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Is the flux of this vector field
through that piece of

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the cylinder the outward flux?

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And then I'd also like you to
find the outward flux of this

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vector field xz comma yz comma z
squared through the piece of

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the sphere of radius a
in the first octant.

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So again, it will be in the
first octant like this one is

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in the first octant.

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It'll be the piece of the sphere
that sits in the same

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part of three-dimensional space
as the piece of the

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cylinder we're looking
at here.

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So what I'd like you to do,
again, is just find the flux--

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in both cases, the outward
flux-- of the vector listed

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through the surface
that is listed.

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And then when you feel
comfortable and confident with

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your answer, you can bring the
video back up and I'll show

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you how I did it.

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OK, welcome back.

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So again, what we're trying to
do is determine flux of a

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vector field through
a surface.

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So I am going to determine first
what the normal is here,

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and then what F dot n is here,
and then using the fact that I

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know F dot n and I know dS in
a good parametrization, I'm

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going to be able to calculate
the integral quite simply.

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So let's point out first, that
the normal at any point on the

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surface is going to be equal to
x comma y comma 0 divided

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by a, where a is, again,
the radius.

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Because in the normal, I know
there's no z component.

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And I actually know it's the
same as the normal would be on

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a circle of radius a.

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And so this vector x comma y
has length a obviously, so

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that's why I'm dividing by a.

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So when I look at the normal and
I dot it with F, let's see

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what I get.

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F dotted with the normal is
going to be xz plus xy divided

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by a, right?

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So I just took x dotted with
x times z, y times z, and 0

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times y, I add those up, and I
still have to divide by my a.

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So that's actually the vector
dotted with the normal.

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Now, what a natural
parametrization to use here is

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obviously the cylindrical
coordinates

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because I'm on a cylinder.

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The radius is fixed.

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But what I'm interested in is
changes in theta and changes

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in the height.

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So I'm going to be interested in
d theta and dz, and I need

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to understand what dS
in this case is.

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And it's just a d theta dz.

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So let me write down what I'm
going to need to completely

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determine the rest
of number one.

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I'm integrating over
the surface.

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And I'll put the bounds
momentarily.

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Actually, I should also point
out, I'm going to write x and

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y in terms of theta, because
now I know what they are.

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x in terms of theta is a cosine
theta, and y in terms

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of theta is a sine theta.

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So when I simplify the
expression F dot n, I get z

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times cosine theta for
my first component.

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And I get a times cosine
theta sine theta

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for my second component.

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So let me maybe point out
again how I got that.

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Let me come over here.

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x is a cosine theta.

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So I get z times a cosine
theta divided by a.

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So I get z times cosine theta.

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x again is a cosine theta.
y is a sine theta.

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So I get a squared cosine
theta sine theta.

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And I divide by a, so
I get a single a

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cosine theta sine theta.

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And then my dS, as I mentioned
before, is

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a d theta dz, right?

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And so now I just have to figure
out the bounds in theta

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and the bounds in z.

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Well, the bounds in
z are very easy.

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The bounds in z are simply 0 to
h, and so I'm going to put

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those here on the outside.

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And then the bounds in theta.

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Maybe it's helpful to come over
and look at my picture.

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This direction, in the
x-direction, is

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theta equals 0.

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When I swing around to the
y-direction, I'm at theta

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equals pi over 2.

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So I need to go from 0
to pi over 2, right?

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So let me come back over here.

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OK, so now, really, I have a
couple of constants that are

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letters, but everything now,
I'm ready to integrate.

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So a is just a constant, and h
is a constant, and z and theta

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are my variables, and I can
actually do this integration

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quite simply.

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I'm not going to do it because
I know this is at this point

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something we already know
how to do, but I'll

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give you the answer.

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So let me write down
the answer.

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You get ah squared over 2,
plus a squared h over 2.

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So that's the solution
you actually get.

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So again, I mean you have to
integrate in d theta first. So

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you have to deal with this, you
have to deal with this,

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and this uses a good
trig identity.

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You can use the fact that 2
sine theta cosine theta is

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sine 2 theta.

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I'll give you that little hint,
and then you can figure

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it out from there.

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And then you integrate in z, and
you evaluate from 0 to h.

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So that's the solution
to number one.

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So now, let's look
at number two.

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Number two we have to
again figure out.

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We have our vector field and we
know our surface is a piece

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of a sphere.

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And so if we're going to
parametrize the surface of a

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sphere, we know we want to use
d theta and d phi, OK?

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And let me point out first,
that again, the normal is

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going to be in some ways similar
to what we saw on the

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cylinder, but now instead of x
comma y comma 0, because it's

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a sphere, it's going to be x
comma y comma z divided by a.

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So again, as before, let me just
point out the normal that

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I'm going to be using is
x comma y comma z,

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all divided by a.

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So I'm going to dot F with
n and look at the surface

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integral with respect to dS--

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d capital S there--

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and I'll see what I get.

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So again, I know how I'm going
to parametrize this sphere.

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I already mentioned it, but
let me say it again.

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It's going to be in theta
and phi, right?

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Because we have a
constant radius.

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We're on a sphere of radius a.

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So I don't need to change rho.

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It's a two-dimensional thing.

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So theta is varying and
phi is varying.

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So let's see what we get first.
Let me do a little work

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and see what we get when we
look at F dotted with n.

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So let me first point out that
F dotted with n looks like

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it's x squared z plus y squared
z plus-- just to make

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this obvious--

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z squared z, all divided by a.

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x squared plus y squared plus z
squared is a squared, right?

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So it's actually a squared
times z divided by a.

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So it's just a times z, right?

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So far, all I've done was I
dotted F with the normal, and

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I knew the fact that x squared
plus y squared plus z squared

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was a squared, because I was
on a sphere of radius a.

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So a squared times z divided
by a is a times z.

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And now if I want to use the
right coordinates, think about

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theta and phi.

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z is a cosine phi.

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So F dot n in the coordinates
I'm interested in is going to

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be a squared cosine phi.

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So let me get out of the way
so you can see that.

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So that's what our
F dot n will be.

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And now, if we're going to
figure out the flux, of

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course, it's the integral
of F dot n dS.

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And let's remind ourselves
what dS is.

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dS is going to be a squared
sine phi d theta d phi.

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You saw this in lecture,
actually, also.

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So this should look familiar.

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And so now I just have to
integrate F dot n dS over the

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right bounds for
theta and phi.

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So let's determine
what those are.

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I'll put everything together,
and we'll

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determine what those are.

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So I've got F dot n dS.

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That's going to be a to the
fourth sine phi cosine phi d

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theta d phi.

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And let's think about what is
the picture that I need in

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terms of the first octant
of a sphere.

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Maybe I should draw a quick
picture over here so we can

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remember what that looks like.

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So it's going to be--

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this is not going to be the
greatest drawing ever-- but

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it's something like this.

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And so I've got pieces of
a circle at each level.

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I've got a piece of
a circle, right?

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If this is the x-direction, this
is the y-direction, and

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this is the z-direction,
theta is going from 0--

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again--

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to pi over 2, and phi is going
from 0 to pi over 2, right?

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So they're both going
from 0 to pi over 2.

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So hopefully, you were able to
get this far at least, in

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terms of figuring out the
flux of F through that

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piece of the sphere.

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And I'm, again, just going to
write down the solution, and

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then you can check your answer
against the solution.

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And I got a to the fourth
over 4 times pi.

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So this whole solution
is a to the fourth

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divided by 4 times pi.

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So you can check your
solution there.

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Again, I want to point out, what
we did in both of these

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problems is we were trying to
compute the flux of a certain

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vector field through
a surface.

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And if you'll notice, what I
actually did in this case is I

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kept the parametrization in
terms of the x, y, z variables

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first, and then I put it in the
parametrization of theta

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and phi after.

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And that made it a little
easier to hang on to and

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figure out what it was.

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Because notice, that x, y, and
z become very complicated in

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theta and phi.

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I have to write a lot
more down, I guess.

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And then simplify things
more carefully.

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This way, it was very obvious
that I got an a squared times

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z in the numerator.

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So sometimes it's a little
easier to compute the F dot n

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in the initial x, y, z
variables, and then change it

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to the appropriate
parametrization for the

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surface you're looking at.

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So that's what we
did, actually.

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In both cases we computed F dot
n, we put it in the right

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parametrization, and then we had
to figure out what dS was.

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We had to make sure we knew dS,
and then we just had to

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integrate over the appropriate
bounds for our parameters.

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And that's giving us the flux
across the surface.

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So I think that's
where I'll stop.

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