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Welcome back to recitation.

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

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

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the flux of a vector in three
dimensions-- across a surface,

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without maybe doing a
lot of calculations.

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So we're going to see
if we can figure out

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how to do these 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

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plus y squared equals 1.

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

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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 find
the flux of the vector j

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through one square that has
side length 1 in the xz-plane.

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So you pick any square, in
the xz-plane, of side length 1

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and 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,

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pause the video,
and then when you're

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ready to see my explanation
of how they work,

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bring the video back up.

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OK, welcome back.

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Again, what we're
trying to do is

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to understand the flux of a
vector field across a surface.

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And we're hoping to
do it with the least

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amount of calculation possible,
for these particular problems.

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So obviously it's
going to be helpful

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if you can draw a picture,
to draw a picture.

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So I'm going to draw
the surface that

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is the infinite cylinder
first and then I'm

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going to look at
the vector field k.

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So let me draw my picture first.

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If these are my
coordinate axes, I

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get-- 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

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but I'm going to try and make
it look like a cylinder up

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here coming down.

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So this is in fact going
to be infinitely long,

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going down forever, but I'll
stop at somewhere down here.

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

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is 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?

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k is just a constant
vector field

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that points in this direction.

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This is k.

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So k at every point
on the surface

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is just the vector
that's pointing straight

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up in this direction.

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

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the 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

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is in the x-y direction only.

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It's going to be--
essentially at every point,

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it's going to be sitting in
the plane z equals a constant.

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Because if you think
about what you have,

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you have a unit circle
at every height.

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And it doesn't vary
in the z-direction

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at all, the bending
of that unit circle.

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So in fact the normal is always
going to point straight out

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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 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

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through one square
in the xz-plane

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where the squares
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

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be the same answer 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 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

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a possibility of two answers.

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

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well, what direction
does it point in?

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Because this plane
is in the xz-plane,

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the normal to the surface
is either j or it's minus j.

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And so if I'm
integrating j dotted

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with the normal over
the surface-- I'll just

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call this surface capital R--
if I'm integrating over R j

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dotted with the
normal dS, j dotted

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with the normal is going
to be 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
exactly in the y-direction.

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And the normal is
either in this direction

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or in the 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,

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

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

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with a plus or minus in front.

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Depending on whether
j dotted with n is 1

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or whether j dotted
with n is minus 1.

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So the solution for
this computation

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is just the area of R or
minus the area of R. Well,

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what's the area of the region?

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

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of vector fields across
surfaces without doing

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a lot of calculation.

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And in the first case
we had a vector field

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that pointed in the
z-direction only,

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and the normal was only
in the x and y direction.

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And so the flux was
0, even though it

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was a vector field on
an infinite cylinder,

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the flux was still 0.

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

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had a surface that was
exactly in the xz-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 moved this unit square.

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

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so 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.