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Welcome back to recitation.

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What I'd like us to do in
these two problems is to

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understand how to compute the
flux in three dimensions-- the

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flux of a vector in
three dimensions--

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across a surface, without
maybe doing a lot of

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calculations.

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So we're going to see if we can
figure out how to do these

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problems without doing
a lot of computation.

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So the first one is to find
the flux of the vector k

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through the infinite cylinder
x squared plus y

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squared equals 1.

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So notice this doesn't depend
on z, but in fact at every

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height it is a unit circle.

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So it's an infinite cylinder.

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And then the second problem I'd
like you to think about

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and to try is to to find the
flux of the vector j through

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one square that has side length
1 in the x-z plane.

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So you pick any square in the
x-z plane of side length 1 and

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find the flux of j through
that square.

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And so I think that that's
enough information.

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So why don't you try both of
those problems, pause the

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video, and then when you're
ready to see my explanation of

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how they work, bring
the video back up.

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OK, welcome back.

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Again, what we're trying to do
is to understand the flux of a

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vector field across a surface.

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And we're hoping to do it
with the least amount of

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calculation possible for these
particular problems. So

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obviously it's going to be
helpful if you can draw a

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picture to draw a picture.

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So I'm going to draw the surface
that is the infinite

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cylinder first and then I'm
going to look at the

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vector field k.

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So let me draw my picture
first. If these are my

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coordinate axes, I get--

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I think you have something
usually like this.

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This is x, this is
y, and this is z.

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This is how you do them
in your class.

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And so it's not going to be a
great picture but I'm going to

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try and make it look like a
cylinder up here coming down.

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So this is in fact going to be
infinitely long, going down

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forever, but I'll stop it
somewhere down here.

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And then so every slice in the
z at a fixed height for z is

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going to be a circle.

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I should think about these are
intersecting the x and y axes

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at (1, 0) and (0, 1).

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So there's actually a sort of
unit circle down here as well.

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Now that's the surface.

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Now what does the vector
field look like?

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Well, in general what does the
vector field look like? k is

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just a constant vector field
that points in this direction.

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This is k.

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So k at every point on the
surface is just the vector

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that's pointing straight
up in this direction.

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And the normal to the surface,
if you think about it, the

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normal to the surface
is independent of z.

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It doesn't depend on z at all.

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It is always going to be a
vector that is in the x-y

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direction only.

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It's going to be--

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essentially at every point it's
going to be sitting in

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the plane z equals
the constant.

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Because if you think about what
you have, you have a unit

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circle at every height.

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And it doesn't vary in the z
direction at all, with the

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bending of that unit circle.

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So in fact the normal is always
going to point straight

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out from the unit circle.

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There's going to be
no z component.

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Or I should say the
z component's 0.

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Maybe that's the best
way to say it.

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So that means that the
normal dotted with k

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is going to be 0.

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And so the answer to the first
question is the flux of k

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through the infinite cylinder
is actually 0.

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So the answer to part a is 0.

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Been getting a lot of zeroes
in my video so far.

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The next one is to find the flux
of the vector j through

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one square in the x-z plane
where the squares

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have side length 1.

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So I can draw any
square I want.

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That seems to imply that maybe
it'll be the same answer

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everywhere.

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So let's see what we get.

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Let me draw a picture
for part b.

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Label this first maybe.

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Let me label my axes.

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And I'm going to draw the
simplest one I can.

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That does not look like
a square but I'm

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not great at this.

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There we go.

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So this is my surface sitting
in the x-z plane.

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This side length is 1 and
this side length is 1.

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So what is the normal
to that surface?

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Well, we have two choices and
so we will actually have a

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possibility of two answers.

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So let me point out that the
normal to the surface-- well,

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what direction does
it point in?

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Because this plane is in the x-z
plane, the normal to the

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surface is either j
or it's minus j.

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And so if I'm integrating j
dotted with the normal over

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the surface--

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I'll just call this surface
capital R--

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if I'm integrating over R j
dotted with the normal dS, j

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dotted with the normal
is going to be

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either 1 or minus 1.

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And hopefully that makes
sense because j--

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let me draw this--
j is pointing

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exactly in the y direction.

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And the normal is either in
this direction or in the

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opposite direction.

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Up to how I choose to
orient the surface.

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And so j dotted with n is either
plus or minus 1, and so

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I just get the area of R with
a plus or minus-- ooh, that

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doesn't look like a plus
or minus-- with a

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plus or minus in front.

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Depending on whether j dotted
with n is 1 or whether j

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dotted with n is minus 1.

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So the solution for this
computation is just the area

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of R or minus the area
of R. Well, what's

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the area of the region?

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The area, it's a square
of side length 1,

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so it has area 1.

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So the final answer is
just plus or minus 1.

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So again, let me remind you
what we're trying to do.

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We're trying to determine these
fluxes of vector fields

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across surfaces without doing
a lot of calculation.

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And in the first case we had a
vector field that pointed in

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the z direction only, and
the normal was only

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in the x and y direction.

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And so the flux was 0, even
though it was a vector field

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on an infinite cylinder,
the flux was still 0.

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And in the other case, I had
actually here, I had a surface

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that was exactly in
the x-z plane.

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And so its normal was exactly
either in the same direction

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as j or 180 degrees
around from j.

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So j dotted with the normal was
either plus or minus 1.

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And so I only had to know
the area of the region.

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Which is why it didn't
matter where I

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moved this unit square.

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I didn't tell you where the
unit square had to sit so

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that's where you can see
why it didn't matter.

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Because it's just the area.

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OK, I think that's
where I'll stop.

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