WEBVTT
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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.
00:00:22.560 --> 00:00:27.210
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.
00:08:39.750 --> 00:08:42.690
That's equal to--
r is a constant.
00:08:42.690 --> 00:08:50.520
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.
00:08:52.950 --> 00:08:59.190
So we plug in, and we just
get h squared r minus 0.
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So just h squared r.
00:09:00.650 --> 00:09:01.150
OK.
00:09:01.150 --> 00:09:04.740
So now let's do the
middle integral.
00:09:04.740 --> 00:09:09.980
So the middle integral is
the integral from 0 to big R
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d little r of the
inner integral.
00:09:13.550 --> 00:09:19.570
So this is the integral
from 0 to big R
00:09:19.570 --> 00:09:25.260
of the inner integral, which was
h squared little r, d little r.
00:09:25.260 --> 00:09:25.930
OK.
00:09:25.930 --> 00:09:27.221
And that's not that bad either.
00:09:27.221 --> 00:09:28.600
So h is just a constant.
00:09:28.600 --> 00:09:35.350
So this is equal to 1/2 h
squared r squared from r
00:09:35.350 --> 00:09:43.810
equals 0 to big R. And so that's
1/2 h squared big R squared.
00:09:43.810 --> 00:09:45.230
That's the middle integral.
00:09:45.230 --> 00:09:48.230
So the outer one now.
00:09:48.230 --> 00:09:48.982
OK.
00:09:48.982 --> 00:09:50.290
So let's go back and look.
00:09:50.290 --> 00:09:55.480
So we're doing d theta as theta
goes from 0 to 2*pi of whatever
00:09:55.480 --> 00:09:57.340
the middle integral was.
00:09:57.340 --> 00:10:03.225
So it's the integral from 0
to 2*pi of whatever the value
00:10:03.225 --> 00:10:04.350
of the middle integral was.
00:10:04.350 --> 00:10:10.110
So this is 1/2 h squared
big R squared d theta.
00:10:10.110 --> 00:10:12.670
And this is all just constant
with respect to theta.
00:10:12.670 --> 00:10:18.950
So that's going to be just
pi h squared r squared.
00:10:18.950 --> 00:10:20.970
You're just
multiplying it by 2*pi.
00:10:20.970 --> 00:10:21.470
All right.
00:10:21.470 --> 00:10:23.070
So pi h squared r squared.
00:10:23.070 --> 00:10:25.010
So this is our final answer.
00:10:25.010 --> 00:10:27.570
Let's just quickly
recap what we did.
00:10:27.570 --> 00:10:31.490
We had to compute the
flux of this field F
00:10:31.490 --> 00:10:37.190
through the surface
of a solid cylinder.
00:10:37.190 --> 00:10:39.260
And so we had options.
00:10:39.260 --> 00:10:42.520
We could do it directly by
trying to compute the surface
00:10:42.520 --> 00:10:44.450
integrals, but in
this case, life
00:10:44.450 --> 00:10:47.560
was a lot easier if we applied
the divergence theorem.
00:10:47.560 --> 00:10:50.280
So the divergence theorem
says that the flux-- which
00:10:50.280 --> 00:10:52.170
is equal to this
surface integral--
00:10:52.170 --> 00:10:55.120
can also be written as
the triple integral,
00:10:55.120 --> 00:10:59.020
over the solid region
bounded by the surface,
00:10:59.020 --> 00:11:01.490
of the divergence of the field.
00:11:01.490 --> 00:11:02.050
All right.
00:11:02.050 --> 00:11:03.540
And so in our case,
the divergence
00:11:03.540 --> 00:11:07.220
was very nice and simple,
and the solid region
00:11:07.220 --> 00:11:12.010
D was relatively simpler to
describe than its surface that
00:11:12.010 --> 00:11:15.790
bounds it, S. So this is why
we think of the divergence
00:11:15.790 --> 00:11:16.290
theorem.
00:11:16.290 --> 00:11:20.890
Because the divergence of the
field is easy to understand,
00:11:20.890 --> 00:11:24.184
and the solid is easier to
describe than its surface.
00:11:24.184 --> 00:11:25.600
So those are both
things that make
00:11:25.600 --> 00:11:27.500
us think to use the
divergence theorem
00:11:27.500 --> 00:11:29.090
for a problem like this.
00:11:29.090 --> 00:11:30.950
So then by the
divergence theorem,
00:11:30.950 --> 00:11:34.394
the flux is just
that triple integral,
00:11:34.394 --> 00:11:35.560
and so we wrote it out here.
00:11:35.560 --> 00:11:37.340
We were integrating
over a cylinder.
00:11:37.340 --> 00:11:40.530
So a natural thing to do is
use cylindrical coordinates.
00:11:40.530 --> 00:11:42.750
And then we computed
the triple integral
00:11:42.750 --> 00:11:44.340
just like we always do.
00:11:44.340 --> 00:11:46.697
I did it in three steps:
inner, middle, and outer.
00:11:46.697 --> 00:11:49.280
You don't have to do it exactly
this way if you don't want to.
00:11:49.280 --> 00:11:51.950
But it works for me.
00:11:51.950 --> 00:11:52.450
OK.
00:11:52.450 --> 00:11:56.220
And we got our final answer:
pi h squared r squared.
00:11:56.220 --> 00:11:57.953
I'll stop there.