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.

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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

00:09:36.950 --> 00:09:40.420
to find a convenient
region over which

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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.

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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.

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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--

00:10:12.710 --> 00:10:18.750
for some big R-- that
contains our surface

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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.