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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 flux and

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surface integrals in the
divergence theorem, and I have

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a nice problem about
that for you here.

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So I've got this field F, and
it's a little bit ugly right?

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All right.

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So its coordinates are x to the
fourth y, minus 2x cubed y

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squared, and z squared.

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And it's passing through the
surface of a solid that's

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bounded by the plane z equals
0, by the plane z equals h,

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and by the surface
x squared plus y

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squared equals R squared.

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So often we call this
solid a cylinder.

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So it's got its bottom surface
in the plane z equals 0, and

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its top surface in the plane
z equals h, and it's got a

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circular base with
radius R there.

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So what I'd like you to do is
to compute the flux of this

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field F through this cylinder.

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So I'll point out before I let
you at it, that to compute

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this as a surface integral,
you could do it.

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You could do it.

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If you really want an exercise
in nasty arithmetic, I invite

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you to do it.

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But you might be able to think
of a way to do this that

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requires less effort than
parametrizing the three

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surfaces and integrating
and so on.

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So I'll leave you with that.

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Why don't you pause the video,
work this one out, come back,

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and we can work on
it together.

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Hopefully, you had some luck
working on this problem.

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Right before I left, I mentioned
that you were

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computing a flux through a
surface here, but that doing

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it as a surface integral is
maybe not the best way to go.

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And so, even without that hint,
probably many of you

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realized that really the way
that we want to go about this

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problem is with the divergence
theorem.

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OK.

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So in our case, the divergence
theorem--

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I'm just abbreviating
it div T-H-M here--

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says start with the double
integral over the surface of F

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dot n d surface area.

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So S here is the surface
of this solid.

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So the divergence theorem says
that this surface integral,

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which is the flux that we're
interested in, is equal to the

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triple integral over the solid
region D-- so that's bounded

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by the surface, and so that's
the solid cylinder here--

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of div F dV. OK.

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So in our case, this is nice,
because in fact, this solid

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region D is an easier to
understand, or easier to

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grapple with region than
the surface that we

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started with, right?

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It's just one solid piece.

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It's easy to parametrize,
in fact.

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It's easy to describe especially
in cylindrical

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coordinates, but also in
rectangular coordinates.

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Whereas this surface S, if we
wanted to talk about it, we'd

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need to split it up into three
pieces, and we'd need to

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parametrize it.

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And it's kind of a hassle,
relatively speaking.

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Also, the divergence of this
field F is a lot simpler than

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the field itself.

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If we go and look at this field,
all of its components

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are polynomials.

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To compute its divergence,
we take

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derivatives of all of them.

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And so that makes their degrees
lower, and then we

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just add them.

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Life is a little bit simpler.

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So OK.

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So this process of using the
divergence theorem is going to

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make our lives easier.

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It's going to make this nasty
surface integral into an easy

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to compute triple integral.

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So let's see actually
how it does.

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So let's compute div F first.
So we know what

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the integrand is.

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All right.

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So we need to look at the
components of F, and so we

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need to take the partial
of the first one

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with respect to x.

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So that's x to the fourth
y with respect to x.

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So put that down over here.

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That's 4x cubed y.

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We just treat y as a constant.

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OK, so now we come back
and we need to look

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at the second one.

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So it's minus 2x cubed
y squared.

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And it's the second one.

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We take its partial
with respect to y.

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So OK.

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So that's going to be minus
2x cubed times 2y.

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So that's going to be
minus 4x cubed y.

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And then we come back and we
look at the last component.

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And that's z squared.

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And so we need to take its
partial with respect to z.

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So in this case, that's
just 2z, and so we

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add that on as well.

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Plus 2z.

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And in this case, not only are
these polynomials simpler than

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the coordinates of F that we
had, but in fact, we've got

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some simplification here.

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Life gets really,
really simple.

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So in fact, this is just going
to work out to 2z.

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So the divergence here is very
simple compared with the

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function F. More simple than we
have a right to expect, but

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in any case, good.

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It's nice to work with.

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OK.

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So that's the divergence.

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So I'm going to write,
this is the flux.

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These integrals that we're
interested in.

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This surface integral, and
then by the divergence

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theorem, it's the same as
this triple integral.

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So the divergence is this 2z.

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So the flux is what I get when
I just put that in here.

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So flux is equal to the triple
integral over our solid of 2z

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dV. OK, so I've left some
stuff out of this.

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Because I'm going to start
writing down the bounds and

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writing this down as an
iterated integral now.

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OK.

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So we have to choose some
coordinate system in which to

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integrate over this solid.

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And so we have three kinds of
natural choices that we always

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look back to.

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There are rectangular
coordinates and cylindrical

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coordinates and spherical
coordinates.

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So spherical coordinates seem
pretty clearly inappropriate.

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Rectangular and cylindrical?

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You know, you could try and
do it in rectangular.

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It's not horrible.

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But this is a cylinder, right?

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I mean, it's crying out for
us to use cylindrical

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coordinates.

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So let's use cylindrical
coordinates.

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So we're going to use
cylindrical coordinates.

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So to get dV we need a z, an r,
and a theta, but remember

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there's this extra
factor of r.

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So it's going to be 2z times
r dz dr d theta.

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Right?

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This is dV. This r dz
dr d theta part.

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So that's what dV is when we use
cylindrical coordinates.

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OK, so now let's figure out
what the bounds are.

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So let's go look
at the cylinder

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that we had over here.

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So it's bounded between z equals
0 at the bottom surface

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and z equals h at
the top surface.

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OK.

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So that's easy enough.

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That's what the bounds
on z are.

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So let's put those in.

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So z is the innermost one, so
that's going from 0 to h.

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OK.

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How about the next one?

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So the next one is r.

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So let's go back over here.

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So r is the radius here after
we project it down.

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And we just get the circle
of radius big R

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centered at the origin.

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So little r is going
from 0 to big R.

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And theta is the circle.

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It's the whole circle.

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So theta is going
from 0 to 2 pi.

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So cylinders are really easy
to describe what they look

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like in cylindrical
coordinates.

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So let's put those in.

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So little r is going from
0 to big R, and theta is

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going from 0 to 2 pi.

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OK.

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Wonderful.

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Now we just have to compute
this, right?

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We've got our flux is this
triple integral.

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So let's compute it.

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Let's walk over to this little
bit of empty board space.

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OK, so we have an iterated
integral.

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So let's do it.

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So the inner integral
is the integral from

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0 to h of 2zr dz.

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Well, that's not that bad.

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That's equal to r
as a constant.

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So it's equal to rz squared
as z goes between 0 and h.

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It's dz, so z is going
from 0 to h.

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So plug in, and we just get
h squared r minus 0.

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So just h squared r.

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OK.

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So now let's do the
middle integral.

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So the middle integral is the
integral from 0 to big R d

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little r of the inner
integral.

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So this is the integral from 0
to big R of the inner integral

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which was h squared little
r d little r.

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OK.

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And that's not that
bad either.

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So h is just a constant.

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So this is equal to 1/2 h
squared r squared from r

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equals 0 to big R. And
so that's 1/2 h

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squared big R squared.

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That's the middle integral.

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So the outer one now.

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OK.

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So let's go back and look.

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So we're doing d theta as theta
goes from 0 to 2 pi of

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whatever the middle
integral was.

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So it's the integral from 0 to
2 pi of whatever the value of

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the middle integral was.

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So this is 1/2 h squared
big R squared d theta.

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And this is all just constant
with respect to theta.

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So that's going to be just
pi h squared r squared.

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You're just multiplying
it by 2 pi.

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All right.

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So pi h squared r squared.

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So this is our final answer.

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Let's just quickly recap
what we did.

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We had to compute the flux of
this field F through the

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surface of a solid cylinder.

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And so we had options.

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We could do it directly by
trying to compute the surface

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integrals, but in this case,
life was a lot easier if we

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applied the divergence
theorem.

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So the divergence theorem
says that the flux--

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which is equal to this
surface integral--

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can also be written as the
triple integral over the solid

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region bounded by the
surface of the

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divergence of the field.

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All right.

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And so in our case, the
divergence was very nice and

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simple, and the solid region
D was relatively simpler to

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describe than-- its surface
that bounds it--

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S. So this is why we think of
the divergence theorem.

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Because the divergence of the
field is easy to understand,

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and the solid is easier to
describe than its surface.

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So those are both things that
make us think to use the

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divergence theorem for
a problem like this.

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So then by the divergence
theorem, the flux is just that

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triple integral, and so we wrote
it out here that we were

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integrating over a cylinder.

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So a natural thing to do is use
cylindrical coordinates.

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And then we computed the
triple integral just

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like we always do.

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I did it in three steps: inner,
middle, and outer.

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You don't have to do it
exactly this way if

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you don't want to.

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But it works for me.

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OK.

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And we got our final answer:
pi h squared r squared.

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I'll stop there.

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