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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.

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OK.

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So the answer is no.

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We can't conclude that
the flux through S

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is 0, because one of the
hypotheses of the divergence

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theorem isn't satisfied.

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Namely, the field isn't defined
everywhere inside the surface.

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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,

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so part c suggests,
maybe you can still

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use the divergence
theorem-- well,

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now we're calling it
extended Gauss's theorem--

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to compute what this flux is.

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So let's think about
how we could do that.

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So remember what extended
Gauss's theorem says?

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Or extended divergence theorem.

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I'm going to try and just say
Gauss's theorem from now on,

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so I stop having to say both.

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But I mean both.

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I mean, they're the
same theorem, right?

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OK.

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So Gauss's theorem says, when
you have a surface bounding

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a region, the flux
through the surface

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is equal to the triple integral
of divergence over the region,

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provided everything is
well-defined and nice.

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Extended Gauss's
theorem says, this

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is still true if your region
has more than one boundary.

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So for example, if your
region is a hollow something--

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so if it's a
spherical shell that

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has an outside sphere and an
inside sphere-- then extended

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Gauss' theorem says, OK,
so you do the same thing.

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You take the triple
integral of the divergence

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over the solid region.

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And then you take the
flux, but you add up

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the flux over all of
the boundary pieces.

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So you add up the flux over
the outside boundary surface,

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and also, if there is one,
through any other boundary

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surface.

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OK?

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And those two things are equal.

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So the total flux through
all of the boundary surface

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is equal to the
integral of divergence

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over the whole region
bounded by those surfaces.

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So how are we going to use this?

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We're trying to compute
the flux through a surface.

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OK, but we don't want to
compute a double integral

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if we can avoid it.

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We don't want to compute
the surface integral.

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So what we'd like
to do is we'd like

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to find a convenient
region over which

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to compute this integral,
to put us in a situation

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where we can apply
extended Gauss's theorem.

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We can't use just the
inside of the cube,

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so we want some other region.

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So what we're going to do is
we're going to walk over here,

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and we're going
to do-- there are

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many possible things you could
do, but this is a nice one.

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All right.

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One thing you could do is
you could take a big sphere.

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Take a big sphere.

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So we've got our cube here.

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This is the point (2, 2,
2), and this is the point

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2, 2, minus 2, and so on.

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So we've taken a big
sphere of radius R--

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for some big R-- that
contains our surface

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S that we're interested in, that
completely contains the cube.

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OK?

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So why have we done that?

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Well, extended Gauss's theorem--

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OK, so what does
extended Gauss' theorem

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say for the region between
the sphere and this cube.

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All right.

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So our cube is named S.
Let's call our sphere

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S_2, because why not?

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OK.

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And let's call the solid region
between them, between the cube

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and sphere-- just
for convenience,

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let's give it a name--
so, I don't know,

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we often call solid regions
D, so let's call it D.

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So it's this spherical region,
but it has a cubical hole

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in the middle of it.

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OK.

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So what does extended
Gauss's theorem say?

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So extended Gauss's theorem
says that the triple integral

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over D of the
divergence of F dV is

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equal to the sum of the fluxes
through each of the surfaces.

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But for this, we want the
flux out of the solid region.

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So for the sphere, the flux
out of the inside of the sphere

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is the flux out of the sphere.

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So that's integral over S_2
of F dot n, d surface area.

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But for the cube, the
flux out of this region

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is the flux into the cube.

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Right?

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00:12:19,170 --> 00:12:21,750
Out here, you're living in
a region outside the cube,

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so when you leave that region,
you're going into the cube.

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So this is the negative of
the flux that we really want.

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So this is minus the flux
through the cube of F dot n,

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with respect to surface area.

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So remember, the signs
here are different,

250
00:12:42,620 --> 00:12:46,240
because I'm taking this normal
to be the outward pointing

251
00:12:46,240 --> 00:12:47,820
normal to both surfaces.

252
00:12:47,820 --> 00:12:49,900
The normal that points
away from the origin.

253
00:12:49,900 --> 00:12:54,700
But the normal pointing away
from the origin on the cube

254
00:12:54,700 --> 00:12:57,897
is the normal that points
into the solid region instead

255
00:12:57,897 --> 00:12:59,980
of the normal that points
out of the solid region.

256
00:12:59,980 --> 00:13:02,200
So that's why this
minus is here.

257
00:13:02,200 --> 00:13:03,040
OK.

258
00:13:03,040 --> 00:13:03,720
Whew.

259
00:13:03,720 --> 00:13:05,380
All right, so what
does this mean?

260
00:13:05,380 --> 00:13:06,930
Well, we've already
computed, in part

261
00:13:06,930 --> 00:13:09,340
a, that the divergence--
so first of all,

262
00:13:09,340 --> 00:13:11,400
F is well-defined
everywhere in this region

263
00:13:11,400 --> 00:13:13,890
D. The only place F was
badly behaved was the origin.

264
00:13:13,890 --> 00:13:15,620
And this region
doesn't contain it,

265
00:13:15,620 --> 00:13:17,920
which is why this trick works.

266
00:13:17,920 --> 00:13:20,955
So we've already computed
that the divergence of F

267
00:13:20,955 --> 00:13:21,870
is 0 everywhere.

268
00:13:21,870 --> 00:13:23,850
It's defined, so
it's 0 on all of D,

269
00:13:23,850 --> 00:13:26,470
and so this triple
integral is just 0.

270
00:13:26,470 --> 00:13:28,500
So if this triple
integral is 0, that

271
00:13:28,500 --> 00:13:32,260
means we can just add
the thing that we're

272
00:13:32,260 --> 00:13:34,500
interested in to
both sides, and we

273
00:13:34,500 --> 00:13:45,470
get that the surface
integral over the cube of F

274
00:13:45,470 --> 00:13:47,830
dot n, with respect
to surface area,

275
00:13:47,830 --> 00:13:55,380
is equal to the surface integral
over the sphere of F dot n,

276
00:13:55,380 --> 00:13:57,340
with respect to surface area.

277
00:13:57,340 --> 00:13:58,640
OK.

278
00:13:58,640 --> 00:14:03,905
So we've converted this original
integral-- our flux integral

279
00:14:03,905 --> 00:14:05,280
that we're interested
in-- and we

280
00:14:05,280 --> 00:14:08,640
found that it's equal
to this separate flux

281
00:14:08,640 --> 00:14:10,450
integral over a
different surface.

282
00:14:10,450 --> 00:14:12,740
This time over a big sphere.

283
00:14:12,740 --> 00:14:17,310
OK, so that's nice.

284
00:14:17,310 --> 00:14:20,000
Why do we want to do that?

285
00:14:20,000 --> 00:14:21,920
Well, we want to
do that because F

286
00:14:21,920 --> 00:14:25,980
is a really nicely behaved
field with respect to a sphere.

287
00:14:25,980 --> 00:14:28,150
F is a radial field.

288
00:14:28,150 --> 00:14:32,790
So F dot n is really
easy to understand.

289
00:14:32,790 --> 00:14:33,430
F dot

290
00:14:33,430 --> 00:14:39,630
n is just-- well, n is a unit
normal and F is a radial field.

291
00:14:39,630 --> 00:14:45,820
So on a sphere, the
normal is radial, right?

292
00:14:45,820 --> 00:14:48,590
It's parallel to
the position vector.

293
00:14:48,590 --> 00:14:49,800
And F is radial.

294
00:14:49,800 --> 00:14:52,260
So they're both pointing in
exactly the same direction.

295
00:14:52,260 --> 00:14:54,170
So when you take
that dot product,

296
00:14:54,170 --> 00:14:58,060
n is the unit vector in
the same direction as F,

297
00:14:58,060 --> 00:15:02,210
so when you dot that with F, you
just get the length of F. OK,

298
00:15:02,210 --> 00:15:03,440
so what does that mean?

299
00:15:03,440 --> 00:15:06,410
That means over
here, this integrand

300
00:15:06,410 --> 00:15:08,210
is really easy to understand.

301
00:15:08,210 --> 00:15:08,710
OK?

302
00:15:08,710 --> 00:15:13,640
This integrand F
dot n on the sphere

303
00:15:13,640 --> 00:15:19,960
is just equal to the
length of the vector F.

304
00:15:19,960 --> 00:15:22,120
Now what is the length
of the vector F?

305
00:15:22,120 --> 00:15:23,870
Well, we know what F is.

306
00:15:23,870 --> 00:15:28,140
It's x over rho cubed i hat,
plus y over rho cubed j hat,

307
00:15:28,140 --> 00:15:30,720
plus z over rho cubed k hat.

308
00:15:30,720 --> 00:15:33,610
So OK, so you compute the
length of that vector,

309
00:15:33,610 --> 00:15:34,480
and what do you get?

310
00:15:34,480 --> 00:15:38,020
Well, it's exactly
1 over rho squared.

311
00:15:38,020 --> 00:15:38,740
OK.

312
00:15:38,740 --> 00:15:40,350
But we said that
this is a sphere.

313
00:15:40,350 --> 00:15:42,650
I guess I didn't write it down.

314
00:15:42,650 --> 00:15:44,260
Let me write it down right here.

315
00:15:44,260 --> 00:15:47,910
This is a sphere whose radius
is big R. It doesn't really

316
00:15:47,910 --> 00:15:50,170
matter very much
what R we choose,

317
00:15:50,170 --> 00:15:52,030
we just want it to be
big enough so that it

318
00:15:52,030 --> 00:15:53,170
contains the whole cube.

319
00:15:53,170 --> 00:15:55,860
If you said this a
sphere of radius 10,

320
00:15:55,860 --> 00:15:57,420
that would completely
do the trick.

321
00:15:57,420 --> 00:15:59,300
That would be totally fine.

322
00:15:59,300 --> 00:16:03,550
OK, so the radius of
the sphere is big R,

323
00:16:03,550 --> 00:16:06,590
so the length of the field,
we said, back over here,

324
00:16:06,590 --> 00:16:09,900
is 1 over R squared.

325
00:16:09,900 --> 00:16:18,110
The length of the vector F.
So this flux integral then,

326
00:16:18,110 --> 00:16:21,525
is the integral over the
sphere S_2 of a constant.

327
00:16:25,100 --> 00:16:29,406
So it's the integral over the
sphere of 1 over R squared dS.

328
00:16:29,406 --> 00:16:30,780
But when you
integrate a constant

329
00:16:30,780 --> 00:16:34,360
over a surface, what you get
is just that constant times

330
00:16:34,360 --> 00:16:36,060
the surface area.

331
00:16:36,060 --> 00:16:37,370
Well, what's the surface area?

332
00:16:37,370 --> 00:16:38,090
This is a sphere.

333
00:16:38,090 --> 00:16:39,955
It's easy to understand
its surface area.

334
00:16:39,955 --> 00:16:45,690
Its surface area
is 4 pi R squared.

335
00:16:45,690 --> 00:16:46,300
Right?

336
00:16:46,300 --> 00:16:48,790
So this is equal to
the surface area,

337
00:16:48,790 --> 00:16:54,160
so that's 4 pi R squared, times
whatever that constant was.

338
00:16:54,160 --> 00:16:56,220
So the constant was
1 over R squared.

339
00:16:56,220 --> 00:16:58,220
And so the R squareds cancel.

340
00:16:58,220 --> 00:16:58,720
Right?

341
00:16:58,720 --> 00:17:00,370
This is why it
didn't matter what

342
00:17:00,370 --> 00:17:02,105
R we chose, because
they're just going

343
00:17:02,105 --> 00:17:03,810
to cancel at the end, anyhow.

344
00:17:06,630 --> 00:17:10,240
OK, so those cancel, and
we're left with 4*pi.

345
00:17:10,240 --> 00:17:15,880
So let's just quickly recap
what we did in this part c.

346
00:17:15,880 --> 00:17:18,630
We're looking to compute
the flux over the cube.

347
00:17:18,630 --> 00:17:22,390
But it's a kind of unpleasant
integral we'd have to compute,

348
00:17:22,390 --> 00:17:25,390
to total up the fluxes over
these various different faces

349
00:17:25,390 --> 00:17:26,150
and so on.

350
00:17:26,150 --> 00:17:28,690
So instead, we had
this clever idea

351
00:17:28,690 --> 00:17:32,110
that we'll apply the divergence
theorem to replace the cube

352
00:17:32,110 --> 00:17:34,195
with a more congenial surface.

353
00:17:34,195 --> 00:17:36,710
So the surface we
choose, because this

354
00:17:36,710 --> 00:17:40,020
is a nice radial vector
field-- that's our main hint.

355
00:17:40,020 --> 00:17:45,370
Because there was a rho involved
in the problem, if you will.

356
00:17:45,370 --> 00:17:49,420
So the surface that we
choose is some big sphere.

357
00:17:49,420 --> 00:17:52,540
And then we apply the
extended Gauss's theorem

358
00:17:52,540 --> 00:17:56,120
to the solid region between
the cube and the sphere.

359
00:17:56,120 --> 00:17:58,760
Outside the cube, but
inside the sphere.

360
00:17:58,760 --> 00:18:02,915
So because the divergence
of the field is 0,

361
00:18:02,915 --> 00:18:05,470
the extended Gauss's
theorem tells us

362
00:18:05,470 --> 00:18:10,290
that the two fluxes--
the flux out of the cube

363
00:18:10,290 --> 00:18:12,270
and the flux out of the
sphere-- are actually

364
00:18:12,270 --> 00:18:14,300
equal to each other.

365
00:18:14,300 --> 00:18:18,060
But since the fluxes are
actually equal to each other,

366
00:18:18,060 --> 00:18:20,270
in order to compute the
flux out of the cube,

367
00:18:20,270 --> 00:18:23,080
it's enough to compute the
flux out of the sphere.

368
00:18:23,080 --> 00:18:23,580
OK.

369
00:18:23,580 --> 00:18:25,525
But computing the
flux out of the sphere

370
00:18:25,525 --> 00:18:28,060
is relatively easy,
because on the sphere,

371
00:18:28,060 --> 00:18:31,870
the integrand F dot
n is just a constant.

372
00:18:31,870 --> 00:18:36,520
And so then we're integrating
a constant over the surface

373
00:18:36,520 --> 00:18:39,120
of a sphere, and that just
gives us the surface area

374
00:18:39,120 --> 00:18:40,700
of the sphere times
that constant,

375
00:18:40,700 --> 00:18:45,060
which is 4 pi R squared times 1
over R squared, which is 4*pi.

376
00:18:45,060 --> 00:18:50,220
So the flux out of the cube
then is also equal to 4*pi.

377
00:18:50,220 --> 00:18:51,699
I'll stop there.