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 the divergence theorem,
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also known as Gauss's
theorem, and flux,
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and all that good stuff.
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So I have a nice exercise
on it for you here.
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So what I want-- so
I want you to take F,
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and I want it to be the
field whose components are
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x over rho cubed, y over rho
cubed, and z over rho cubed.
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So here, rho is your usual rho
from spherical coordinates.
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Rho is equal to the
square root of x squared
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plus y squared plus z squared.
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And I want S to be the
surface of the box whose
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vertices are plus or
minus 2, plus or minus 2,
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plus or minus 2.
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So it's a cubical box.
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So what I'd like you to
do is, first in part a,
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I'd like you to show
that the divergence of F
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is 0, wherever the
field F is defined.
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In part b, what I'd
like you to think about
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is whether we can
conclude from that,
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that the flux through the
surface of S is equal to 0.
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All right.
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And in part c, what
I'd like you to do
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is to use the extended
version of Gauss's theorem--
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or the extended version of
the divergence theorem--
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in order to actually
compute the flux through S
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by computing an integral.
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So why don't you pause the
video for a couple of minutes,
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work out this
problem, come back,
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and we can work it out together.
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Hopefully you had some
luck with this problem.
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Let's get started.
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Part a asks you to compute
the divergence of F.
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So in order to
compute that, we're
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going to need to take
the partial derivatives
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of the components of F. And in
order to do that, at some point
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I'm going to need to take a
partial derivative of rho.
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So let me first compute the
partial derivatives of rho,
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and that will save me a
tiny bit of work later.
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So rho is equal to the
square root of x squared
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plus y squared plus z squared.
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So partial rho partial
x-- well, you just
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apply your usual
chain rule here.
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And I guess we get
a half, but then we
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get a 2 that cancels
it, so I think
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this works out to x divided by
the square root of x squared
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plus y squared plus z squared,
so that's x divided by rho.
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All right.
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And I'm just going to
keep rho around here,
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because otherwise I have to
write out the square root of x
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squared plus y squared plus z
squared over and over again,
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and this is going to
save me some effort
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and would save you
some effort as well.
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So OK.
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So this is rho.
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So this is d rho dx.
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So we want to take the x partial
of the first component of F. So
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that's the x partial
of x over rho cubed.
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OK.
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And you just apply your
usual quotient rule,
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so what do we get?
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We get the derivative
of the top.
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So that's rho cubed
minus-- OK, so the top
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is x times the derivative
of the bottom, which
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is going to be 3 rho
squared times x over rho--
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so that's 3-- so we
have an x-- so it's
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3 x squared rho, divided
by the bottom squared,
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which is rho to the sixth.
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And I guess there's a
common factor of rho
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everywhere that
we can cancel out.
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So this is equal to rho squared
minus 3 x squared divided
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by rho to the fifth.
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OK.
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So that's the
x-partial derivative
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of the first component of F.
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Now we need the
y-partial derivative
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of the second component of F,
and the z-partial derivative
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of the third component of
F. But if you go and look
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back at what the
formula for F was,
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you see that this is a very,
very symmetric formula.
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So in order to get from
the first component
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to the second component,
we just change
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x to y, and to get from the
second component to the third,
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we just change y to z,
because of course rho
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treats x, y, and z the same.
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So what does that mean?
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Well, that means that
the partial derivatives
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are easy to compute.
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Having computed this
x-partial derivative,
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we also get that
partial over partial y
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of the second component--
which is y over rho cubed--
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is equal to rho squared
minus 3 y squared, over rho
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to the fifth.
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And the last one we get, partial
over partial z of z over rho
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cubed is equal to rho
squared minus-- I'm
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getting a little
cramped here-- 3 z
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squared, over rho to the fifth.
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And so adding these
up, we get that div F
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is equal to the sum
of those three things.
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So let's see what we've got.
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We've got a 3-- so
the denominators
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are all rho to the fifth.
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And we've got 3
rho squared minus 3
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x squared minus 3 y
squared minus 3 z squared.
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So this is equal
to 3 rho squared
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minus 3 x squared minus 3 y
squared minus 3 z squared, all
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over rho to the fifth.
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But of course, rho
squared is x squared
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plus y squared plus z squared,
so this numerator is just 0.
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So this is equal to 0.
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OK.
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Which is what we
thought it should be.
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All right.
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Good.
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So that's part a.
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We just computed the
partial derivatives of F,
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and then added them together
to get the divergence.
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And we found that, in fact, yes,
the divergence was equal to 0.
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Great.
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So that's part a.
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So let's go look
at what part b was.
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Part b asks, can we
conclude that the flux
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through the surface S is 0?
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All right.
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Now remember what the
divergence theorem says.
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The divergence theorem
says that the flux
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through a surface of a field
is equal to the triple integral
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of the divergence of that
field over the interior,
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provided the field is defined
and differentiable and nice,
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or whatever, Everywhere inside.
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OK?
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But this field has a problem.
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Almost everywhere, this
field is nicely behaved,
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but at 0, we have
a real problem.
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We're dividing by 0.
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Right?
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So this field is
not defined at 0.
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So there's a single point in
the middle of this cube where
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this field behaves badly.
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And that means we can't
apply the divergence
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theorem inside this cube.
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So since we can't apply
the divergence theorem,
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we aren't allowed to
conclude immediately
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that the flux through
this surface is 0.
00:07:04.230 --> 00:07:04.960
OK.
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So the answer is no.
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We can't conclude that
the flux through S
00:07:09.010 --> 00:07:11.765
is 0, because one of the
hypotheses of the divergence
00:07:11.765 --> 00:07:13.300
theorem isn't satisfied.
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Namely, the field isn't defined
everywhere inside the surface.
00:07:19.650 --> 00:07:20.590
OK.
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So the answer to b is no.
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OK, I'm just going
to write that.
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But it's no because the
hypotheses aren't satisfied.
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OK, so now let's look at part c.
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So part c suggests, we can't
conclude that the flux is 0.
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So we still want to
know what the flux is.
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That's still an
interesting question,
00:07:43.960 --> 00:07:46.790
so part c suggests,
maybe you can still
00:07:46.790 --> 00:07:49.460
use the divergence
theorem-- well,
00:07:49.460 --> 00:07:51.960
now we're calling it
extended Gauss's theorem--
00:07:51.960 --> 00:07:54.290
to compute what this flux is.
00:07:54.290 --> 00:07:56.610
So let's think about
how we could do that.
00:07:56.610 --> 00:07:59.250
So remember what extended
Gauss's theorem says?
00:07:59.250 --> 00:08:01.430
Or extended divergence theorem.
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I'm going to try and just say
Gauss's theorem from now on,
00:08:04.750 --> 00:08:06.000
so I stop having to say both.
00:08:06.000 --> 00:08:07.574
But I mean both.
00:08:07.574 --> 00:08:09.240
I mean, they're the
same theorem, right?
00:08:09.240 --> 00:08:09.740
OK.
00:08:13.660 --> 00:08:17.360
So Gauss's theorem says, when
you have a surface bounding
00:08:17.360 --> 00:08:20.387
a region, the flux
through the surface
00:08:20.387 --> 00:08:22.970
is equal to the triple integral
of divergence over the region,
00:08:22.970 --> 00:08:26.850
provided everything is
well-defined and nice.
00:08:26.850 --> 00:08:28.855
Extended Gauss's
theorem says, this
00:08:28.855 --> 00:08:35.120
is still true if your region
has more than one boundary.
00:08:35.120 --> 00:08:39.980
So for example, if your
region is a hollow something--
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so if it's a
spherical shell that
00:08:42.620 --> 00:08:47.430
has an outside sphere and an
inside sphere-- then extended
00:08:47.430 --> 00:08:49.970
Gauss' theorem says, OK,
so you do the same thing.
00:08:49.970 --> 00:08:52.280
You take the triple
integral of the divergence
00:08:52.280 --> 00:08:53.710
over the solid region.
00:08:53.710 --> 00:08:56.080
And then you take the
flux, but you add up
00:08:56.080 --> 00:08:57.970
the flux over all of
the boundary pieces.
00:08:57.970 --> 00:09:01.992
So you add up the flux over
the outside boundary surface,
00:09:01.992 --> 00:09:04.200
and also, if there is one,
through any other boundary
00:09:04.200 --> 00:09:05.340
surface.
00:09:05.340 --> 00:09:07.330
OK?
00:09:07.330 --> 00:09:09.420
And those two things are equal.
00:09:09.420 --> 00:09:13.750
So the total flux through
all of the boundary surface
00:09:13.750 --> 00:09:16.770
is equal to the
integral of divergence
00:09:16.770 --> 00:09:19.310
over the whole region
bounded by those surfaces.
00:09:19.310 --> 00:09:20.942
So how are we going to use this?
00:09:26.810 --> 00:09:29.790
We're trying to compute
the flux through a surface.
00:09:29.790 --> 00:09:31.970
OK, but we don't want to
compute a double integral
00:09:31.970 --> 00:09:33.220
if we can avoid it.
00:09:33.220 --> 00:09:35.200
We don't want to compute
the surface integral.
00:09:35.200 --> 00:09:36.950
So what we'd like
to do is we'd like
00:09:36.950 --> 00:09:40.420
to find a convenient
region over which
00:09:40.420 --> 00:09:44.940
to compute this integral,
to put us in a situation
00:09:44.940 --> 00:09:47.030
where we can apply
extended Gauss's theorem.
00:09:47.030 --> 00:09:49.400
We can't use just the
inside of the cube,
00:09:49.400 --> 00:09:50.892
so we want some other region.
00:09:50.892 --> 00:09:53.350
So what we're going to do is
we're going to walk over here,
00:09:53.350 --> 00:09:55.160
and we're going
to do-- there are
00:09:55.160 --> 00:09:57.970
many possible things you could
do, but this is a nice one.
00:09:57.970 --> 00:09:59.060
All right.
00:09:59.060 --> 00:10:01.530
One thing you could do is
you could take a big sphere.
00:10:01.530 --> 00:10:02.290
Take a big sphere.
00:10:02.290 --> 00:10:04.440
So we've got our cube here.
00:10:04.440 --> 00:10:07.200
This is the point (2, 2,
2), and this is the point
00:10:07.200 --> 00:10:09.900
2, 2, minus 2, and so on.
00:10:09.900 --> 00:10:12.710
So we've taken a big
sphere of radius R--
00:10:12.710 --> 00:10:18.750
for some big R-- that
contains our surface
00:10:18.750 --> 00:10:22.180
S that we're interested in, that
completely contains the cube.
00:10:22.180 --> 00:10:23.500
OK?
00:10:23.500 --> 00:10:24.547
So why have we done that?
00:10:24.547 --> 00:10:25.880
Well, extended Gauss's theorem--
00:10:39.750 --> 00:10:41.430
OK, so what does
extended Gauss' theorem
00:10:41.430 --> 00:10:45.610
say for the region between
the sphere and this cube.
00:10:45.610 --> 00:10:46.120
All right.
00:10:46.120 --> 00:10:51.710
So our cube is named S.
Let's call our sphere
00:10:51.710 --> 00:10:55.270
S_2, because why not?
00:10:55.270 --> 00:10:55.770
OK.
00:10:55.770 --> 00:11:11.930
And let's call the solid region
between them, between the cube
00:11:11.930 --> 00:11:14.450
and sphere-- just
for convenience,
00:11:14.450 --> 00:11:16.730
let's give it a name--
so, I don't know,
00:11:16.730 --> 00:11:20.100
we often call solid regions
D, so let's call it D.
00:11:20.100 --> 00:11:23.570
So it's this spherical region,
but it has a cubical hole
00:11:23.570 --> 00:11:24.890
in the middle of it.
00:11:24.890 --> 00:11:26.110
OK.
00:11:26.110 --> 00:11:29.070
So what does extended
Gauss's theorem say?
00:11:29.070 --> 00:11:34.820
So extended Gauss's theorem
says that the triple integral
00:11:34.820 --> 00:11:41.130
over D of the
divergence of F dV is
00:11:41.130 --> 00:11:48.650
equal to the sum of the fluxes
through each of the surfaces.
00:11:48.650 --> 00:11:52.990
But for this, we want the
flux out of the solid region.
00:11:55.910 --> 00:11:59.830
So for the sphere, the flux
out of the inside of the sphere
00:11:59.830 --> 00:12:01.290
is the flux out of the sphere.
00:12:01.290 --> 00:12:11.790
So that's integral over S_2
of F dot n, d surface area.
00:12:11.790 --> 00:12:15.160
But for the cube, the
flux out of this region
00:12:15.160 --> 00:12:18.220
is the flux into the cube.
00:12:18.220 --> 00:12:19.170
Right?
00:12:19.170 --> 00:12:21.750
Out here, you're living in
a region outside the cube,
00:12:21.750 --> 00:12:24.730
so when you leave that region,
you're going into the cube.
00:12:24.730 --> 00:12:28.550
So this is the negative of
the flux that we really want.
00:12:28.550 --> 00:12:39.290
So this is minus the flux
through the cube of F dot n,
00:12:39.290 --> 00:12:40.870
with respect to surface area.
00:12:40.870 --> 00:12:42.620
So remember, the signs
here are different,
00:12:42.620 --> 00:12:46.240
because I'm taking this normal
to be the outward pointing
00:12:46.240 --> 00:12:47.820
normal to both surfaces.
00:12:47.820 --> 00:12:49.900
The normal that points
away from the origin.
00:12:49.900 --> 00:12:54.700
But the normal pointing away
from the origin on the cube
00:12:54.700 --> 00:12:57.897
is the normal that points
into the solid region instead
00:12:57.897 --> 00:12:59.980
of the normal that points
out of the solid region.
00:12:59.980 --> 00:13:02.200
So that's why this
minus is here.
00:13:02.200 --> 00:13:03.040
OK.
00:13:03.040 --> 00:13:03.720
Whew.
00:13:03.720 --> 00:13:05.380
All right, so what
does this mean?
00:13:05.380 --> 00:13:06.930
Well, we've already
computed, in part
00:13:06.930 --> 00:13:09.340
a, that the divergence--
so first of all,
00:13:09.340 --> 00:13:11.400
F is well-defined
everywhere in this region
00:13:11.400 --> 00:13:13.890
D. The only place F was
badly behaved was the origin.
00:13:13.890 --> 00:13:15.620
And this region
doesn't contain it,
00:13:15.620 --> 00:13:17.920
which is why this trick works.
00:13:17.920 --> 00:13:20.955
So we've already computed
that the divergence of F
00:13:20.955 --> 00:13:21.870
is 0 everywhere.
00:13:21.870 --> 00:13:23.850
It's defined, so
it's 0 on all of D,
00:13:23.850 --> 00:13:26.470
and so this triple
integral is just 0.
00:13:26.470 --> 00:13:28.500
So if this triple
integral is 0, that
00:13:28.500 --> 00:13:32.260
means we can just add
the thing that we're
00:13:32.260 --> 00:13:34.500
interested in to
both sides, and we
00:13:34.500 --> 00:13:45.470
get that the surface
integral over the cube of F
00:13:45.470 --> 00:13:47.830
dot n, with respect
to surface area,
00:13:47.830 --> 00:13:55.380
is equal to the surface integral
over the sphere of F dot n,
00:13:55.380 --> 00:13:57.340
with respect to surface area.
00:13:57.340 --> 00:13:58.640
OK.
00:13:58.640 --> 00:14:03.905
So we've converted this original
integral-- our flux integral
00:14:03.905 --> 00:14:05.280
that we're interested
in-- and we
00:14:05.280 --> 00:14:08.640
found that it's equal
to this separate flux
00:14:08.640 --> 00:14:10.450
integral over a
different surface.
00:14:10.450 --> 00:14:12.740
This time over a big sphere.
00:14:12.740 --> 00:14:17.310
OK, so that's nice.
00:14:17.310 --> 00:14:20.000
Why do we want to do that?
00:14:20.000 --> 00:14:21.920
Well, we want to
do that because F
00:14:21.920 --> 00:14:25.980
is a really nicely behaved
field with respect to a sphere.
00:14:25.980 --> 00:14:28.150
F is a radial field.
00:14:28.150 --> 00:14:32.790
So F dot n is really
easy to understand.
00:14:32.790 --> 00:14:33.430
F dot
00:14:33.430 --> 00:14:39.630
n is just-- well, n is a unit
normal and F is a radial field.
00:14:39.630 --> 00:14:45.820
So on a sphere, the
normal is radial, right?
00:14:45.820 --> 00:14:48.590
It's parallel to
the position vector.
00:14:48.590 --> 00:14:49.800
And F is radial.
00:14:49.800 --> 00:14:52.260
So they're both pointing in
exactly the same direction.
00:14:52.260 --> 00:14:54.170
So when you take
that dot product,
00:14:54.170 --> 00:14:58.060
n is the unit vector in
the same direction as F,
00:14:58.060 --> 00:15:02.210
so when you dot that with F, you
just get the length of F. OK,
00:15:02.210 --> 00:15:03.440
so what does that mean?
00:15:03.440 --> 00:15:06.410
That means over
here, this integrand
00:15:06.410 --> 00:15:08.210
is really easy to understand.
00:15:08.210 --> 00:15:08.710
OK?
00:15:08.710 --> 00:15:13.640
This integrand F
dot n on the sphere
00:15:13.640 --> 00:15:19.960
is just equal to the
length of the vector F.
00:15:19.960 --> 00:15:22.120
Now what is the length
of the vector F?
00:15:22.120 --> 00:15:23.870
Well, we know what F is.
00:15:23.870 --> 00:15:28.140
It's x over rho cubed i hat,
plus y over rho cubed j hat,
00:15:28.140 --> 00:15:30.720
plus z over rho cubed k hat.
00:15:30.720 --> 00:15:33.610
So OK, so you compute the
length of that vector,
00:15:33.610 --> 00:15:34.480
and what do you get?
00:15:34.480 --> 00:15:38.020
Well, it's exactly
1 over rho squared.
00:15:38.020 --> 00:15:38.740
OK.
00:15:38.740 --> 00:15:40.350
But we said that
this is a sphere.
00:15:40.350 --> 00:15:42.650
I guess I didn't write it down.
00:15:42.650 --> 00:15:44.260
Let me write it down right here.
00:15:44.260 --> 00:15:47.910
This is a sphere whose radius
is big R. It doesn't really
00:15:47.910 --> 00:15:50.170
matter very much
what R we choose,
00:15:50.170 --> 00:15:52.030
we just want it to be
big enough so that it
00:15:52.030 --> 00:15:53.170
contains the whole cube.
00:15:53.170 --> 00:15:55.860
If you said this a
sphere of radius 10,
00:15:55.860 --> 00:15:57.420
that would completely
do the trick.
00:15:57.420 --> 00:15:59.300
That would be totally fine.
00:15:59.300 --> 00:16:03.550
OK, so the radius of
the sphere is big R,
00:16:03.550 --> 00:16:06.590
so the length of the field,
we said, back over here,
00:16:06.590 --> 00:16:09.900
is 1 over R squared.
00:16:09.900 --> 00:16:18.110
The length of the vector F.
So this flux integral then,
00:16:18.110 --> 00:16:21.525
is the integral over the
sphere S_2 of a constant.
00:16:25.100 --> 00:16:29.406
So it's the integral over the
sphere of 1 over R squared dS.
00:16:29.406 --> 00:16:30.780
But when you
integrate a constant
00:16:30.780 --> 00:16:34.360
over a surface, what you get
is just that constant times
00:16:34.360 --> 00:16:36.060
the surface area.
00:16:36.060 --> 00:16:37.370
Well, what's the surface area?
00:16:37.370 --> 00:16:38.090
This is a sphere.
00:16:38.090 --> 00:16:39.955
It's easy to understand
its surface area.
00:16:39.955 --> 00:16:45.690
Its surface area
is 4 pi R squared.
00:16:45.690 --> 00:16:46.300
Right?
00:16:46.300 --> 00:16:48.790
So this is equal to
the surface area,
00:16:48.790 --> 00:16:54.160
so that's 4 pi R squared, times
whatever that constant was.
00:16:54.160 --> 00:16:56.220
So the constant was
1 over R squared.
00:16:56.220 --> 00:16:58.220
And so the R squareds cancel.
00:16:58.220 --> 00:16:58.720
Right?
00:16:58.720 --> 00:17:00.370
This is why it
didn't matter what
00:17:00.370 --> 00:17:02.105
R we chose, because
they're just going
00:17:02.105 --> 00:17:03.810
to cancel at the end, anyhow.
00:17:06.630 --> 00:17:10.240
OK, so those cancel, and
we're left with 4*pi.
00:17:10.240 --> 00:17:15.880
So let's just quickly recap
what we did in this part c.
00:17:15.880 --> 00:17:18.630
We're looking to compute
the flux over the cube.
00:17:18.630 --> 00:17:22.390
But it's a kind of unpleasant
integral we'd have to compute,
00:17:22.390 --> 00:17:25.390
to total up the fluxes over
these various different faces
00:17:25.390 --> 00:17:26.150
and so on.
00:17:26.150 --> 00:17:28.690
So instead, we had
this clever idea
00:17:28.690 --> 00:17:32.110
that we'll apply the divergence
theorem to replace the cube
00:17:32.110 --> 00:17:34.195
with a more congenial surface.
00:17:34.195 --> 00:17:36.710
So the surface we
choose, because this
00:17:36.710 --> 00:17:40.020
is a nice radial vector
field-- that's our main hint.
00:17:40.020 --> 00:17:45.370
Because there was a rho involved
in the problem, if you will.
00:17:45.370 --> 00:17:49.420
So the surface that we
choose is some big sphere.
00:17:49.420 --> 00:17:52.540
And then we apply the
extended Gauss's theorem
00:17:52.540 --> 00:17:56.120
to the solid region between
the cube and the sphere.
00:17:56.120 --> 00:17:58.760
Outside the cube, but
inside the sphere.
00:17:58.760 --> 00:18:02.915
So because the divergence
of the field is 0,
00:18:02.915 --> 00:18:05.470
the extended Gauss's
theorem tells us
00:18:05.470 --> 00:18:10.290
that the two fluxes--
the flux out of the cube
00:18:10.290 --> 00:18:12.270
and the flux out of the
sphere-- are actually
00:18:12.270 --> 00:18:14.300
equal to each other.
00:18:14.300 --> 00:18:18.060
But since the fluxes are
actually equal to each other,
00:18:18.060 --> 00:18:20.270
in order to compute the
flux out of the cube,
00:18:20.270 --> 00:18:23.080
it's enough to compute the
flux out of the sphere.
00:18:23.080 --> 00:18:23.580
OK.
00:18:23.580 --> 00:18:25.525
But computing the
flux out of the sphere
00:18:25.525 --> 00:18:28.060
is relatively easy,
because on the sphere,
00:18:28.060 --> 00:18:31.870
the integrand F dot
n is just a constant.
00:18:31.870 --> 00:18:36.520
And so then we're integrating
a constant over the surface
00:18:36.520 --> 00:18:39.120
of a sphere, and that just
gives us the surface area
00:18:39.120 --> 00:18:40.700
of the sphere times
that constant,
00:18:40.700 --> 00:18:45.060
which is 4 pi R squared times 1
over R squared, which is 4*pi.
00:18:45.060 --> 00:18:50.220
So the flux out of the cube
then is also equal to 4*pi.
00:18:50.220 --> 00:18:51.699
I'll stop there.