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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 surface

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integrals in the
divergence theorem,

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and I have 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,

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minus 2 x cubed y
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

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z equals h, and by the surface
x squared plus y squared equals

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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,

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and its top surface in
the plane z equals h,

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and it's got a 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 field

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F through this cylinder.

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So I'll point out
before I let you

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at it, that to compute
this as a surface integral,

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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,

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I invite you to do it.

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But you might be able
to think of a way

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to do this that requires less
effort than parametrizing

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the three 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 computing a flux

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through a surface here, but that
doing it as a surface integral

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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

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want to go about this 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 that the double integral
over the surface of F dot

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n d surface area-- 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,

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is equal to the triple integral
over the solid region D--

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so that's bounded
by the surface,

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and so that's the
solid cylinder here--

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is equal to the triple
integral over D of div F dV.

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OK.

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

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this solid region D is
an easier to understand,

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or easier to grapple with
region than the surface

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that we 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

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in cylindrical
coordinates, but also

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in rectangular coordinates.

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

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

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and we'd need to 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

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is a lot simpler than
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,

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and then we 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, using
the divergence theorem,

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is going to make
our lives easier.

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It's going to make
this nasty surface

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integral into an easy 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.

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So we know what
the integrand is.

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All right.

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So we need to look at
the components of F,

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and so we 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 4 x 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 at the second one.

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So it's minus 2 x
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 2 x cubed times 2y.

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So that's going to
be minus 4 x 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 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

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than the coordinates of F
that we had, but in fact,

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we've got 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

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

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but 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 theorem,

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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

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of 2z dV.

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

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Because I'm going to start
writing down the bounds

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and 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

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to integrate over this solid.

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And so we have three
kinds of natural choices

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that we always 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 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,

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but remember 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.

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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 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 centered

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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

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what they look 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,

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and theta is 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 0 to h

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of 2*z*r*dz.

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Well, that's not that bad.

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That's equal to--
r is a constant.

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So it's equal to r z 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 we 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

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d little r of the
inner integral.

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So this is the integral
from 0 to big R

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of the inner integral, 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 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 whatever

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the middle integral was.

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

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00:10:03,225 --> 00:10:04,350
of 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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00:10:23,070 --> 00:10:25,010
So this is our final answer.

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00:10:25,010 --> 00:10:27,570
Let's just quickly
recap what we did.

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00:10:27,570 --> 00:10:31,490
We had to compute the
flux of this field F

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through the 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

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was a lot easier if we applied
the divergence theorem.

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So the divergence theorem
says that the flux-- which

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is equal to this
surface integral--

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00:10:52,170 --> 00:10:55,120
can also be written as
the triple integral,

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00:10:55,120 --> 00:10:59,020
over the solid region
bounded by the surface,

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of the divergence of the field.

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00:11:01,490 --> 00:11:02,050
All right.

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00:11:02,050 --> 00:11:03,540
And so in our case,
the divergence

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was very nice and simple,
and the solid region

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D was relatively simpler to
describe than its surface that

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bounds it, S. So this is why
we think of the divergence

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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

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us think to use the
divergence theorem

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for a problem like this.

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So then by the
divergence theorem,

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the flux is just
that triple integral,

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and so we wrote it out here.

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00:11:35,560 --> 00:11:37,340
We were integrating
over a cylinder.

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00:11:37,340 --> 00:11:40,530
So a natural thing to do is
use cylindrical coordinates.

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And then we computed
the triple integral

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just 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 you don't want to.

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00:11:49,280 --> 00:11:51,950
But it works for me.

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00:11:51,950 --> 00:11:52,450
OK.

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00:11:52,450 --> 00:11:56,220
And we got our final answer:
pi h squared r squared.

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I'll stop there.