JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been
learning about flux and surface
integrals in the
divergence theorem,
and I have a nice problem
about that for you here.
So I've got this field F, and
it's a little bit ugly, right?
All right.
So its coordinates
are x to the fourth y,
minus 2 x cubed y
squared, and z squared.
And it's passing through the
surface of a solid that's
bounded by the plane z
equals 0, by the plane
z equals h, and by the surface
x squared plus y squared equals
R squared.
So often we call this
solid a cylinder.
So it's got its bottom surface
in the plane z equals 0,
and its top surface in
the plane z equals h,
and it's got a circular
base with radius R there.
So what I'd like you to do is to
compute the flux of this field
F through this cylinder.
So I'll point out
before I let you
at it, that to compute
this as a surface integral,
you could do it.
You could do it.
If you really want an
exercise in nasty arithmetic,
I invite you to do it.
But you might be able
to think of a way
to do this that requires less
effort than parametrizing
the three surfaces and
integrating and so on.
So I'll leave you with that.
Why don't you pause the video,
work this one out, come back,
and we can work on it together.
Hopefully, you had some luck
working on this problem.
Right before I left, I mentioned
that you were computing a flux
through a surface here, but that
doing it as a surface integral
is maybe not the best way to go.
And so, even without that
hint, probably many of you
realized that really
the way that we
want to go about this problem
is with the divergence theorem.
OK.
So in our case, the
divergence theorem--
I'm just abbreviating
it div T-H-M here--
says that the double integral
over the surface of F dot
n d surface area-- so S here
is the surface of this solid.
So the divergence theorem says
that this surface integral,
which is the flux that
we're interested in,
is equal to the triple integral
over the solid region D--
so that's bounded
by the surface,
and so that's the
solid cylinder here--
is equal to the triple
integral over D of div F dV.
OK.
So in our case, this is
nice, because in fact,
this solid region D is
an easier to understand,
or easier to grapple with
region than the surface
that we started with, right?
It's just one solid piece.
It's easy to
parametrize, in fact.
It's easy to
describe, especially
in cylindrical
coordinates, but also
in rectangular coordinates.
Whereas this surface S, if
we wanted to talk about it,
we'd need to split it
up into three pieces,
and we'd need to parametrize it.
And it's kind of a hassle,
relatively speaking.
Also, the divergence
of this field F
is a lot simpler than
the field itself.
If we go and look at this
field, all of its components
are polynomials.
To compute its
divergence, we take
derivatives of all of them.
And so that makes
their degrees lower,
and then we just add them.
Life is a little bit simpler.
So OK.
So this process, using
the divergence theorem,
is going to make
our lives easier.
It's going to make
this nasty surface
integral into an easy to
compute triple integral.
So let's see
actually how it does.
So let's compute div F first.
So we know what
the integrand is.
All right.
So we need to look at
the components of F,
and so we need to take the
partial of the first one
with respect to x.
So that's x to the fourth
y with respect to x.
So put that down over here.
That's 4 x cubed y.
We just treat y as a constant.
OK, so now, we come back and we
need to look at the second one.
So it's minus 2 x
cubed y squared.
And it's the second one.
We take its partial
with respect to y.
So OK.
So that's going to be
minus 2 x cubed times 2y.
So that's going to
be minus 4 x cubed y.
And then we come back and we
look at the last component.
And that's z squared.
And so we need to take its
partial with respect to z.
So in this case, that's just 2z,
and so we add that on as well.
Plus 2z.
And in this case, not only
are these polynomials simpler
than the coordinates of F
that we had, but in fact,
we've got some
simplification here.
Life gets really, really simple.
So in fact, this is just
going to work out to 2z.
So the divergence here
is very simple compared
with the function F. More simple
than we have a right to expect,
but in any case, good.
It's nice to work with.
OK.
So that's the divergence.
So I'm going to write,
this is the flux.
These integrals that
we're interested in.
This surface integral, and
then by the divergence theorem,
it's the same as
this triple integral.
So the divergence is this 2z.
So the flux is what I get
when I just put that in here.
So flux is equal to the
triple integral over our solid
of 2z dV.
OK, so I've left some
stuff out of this.
Because I'm going to start
writing down the bounds
and writing this down as
an iterated integral now.
OK.
So we have to choose some
coordinate system in which
to integrate over this solid.
And so we have three
kinds of natural choices
that we always look back to.
There are rectangular
coordinates and cylindrical
coordinates and
spherical coordinates.
So spherical coordinates seem
pretty clearly inappropriate.
Rectangular and cylindrical?
You know, you could try
and do it in rectangular.
It's not horrible.
But this is a cylinder, right?
I mean, it's crying out for us
to use cylindrical coordinates.
So let's use
cylindrical coordinates.
So we're going to use
cylindrical coordinates.
So to get dV we need a
z, an r, and a theta,
but remember there's
this extra factor of r.
So it's going to be 2z
times r dz dr d theta.
Right?
This is dV.
This r dz dr d theta part.
So that's what dV is when we
use cylindrical coordinates.
OK, so now let's figure
out what the bounds are.
So let's go look at the
cylinder that we had over here.
So it's bounded between z
equals 0 at the bottom surface
and z equals h at
the top surface.
OK.
So that's easy enough.
That's what the bounds on z are.
So let's put those in.
So z is the innermost one,
so that's going from 0 to h.
OK.
How about the next one?
So the next one is r.
So let's go back over here.
So r is the radius here
after we project it down.
And we just get the circle
of radius big R centered
at the origin.
So little r is going
from 0 to big R.
And theta is the circle.
It's the whole circle.
So theta is going
from 0 to 2*pi.
So cylinders are
really easy to describe
what they look like in
cylindrical coordinates.
So let's put those in.
So little r is going
from 0 to big R,
and theta is going
from 0 to 2*pi.
OK.
Wonderful.
Now we just have to
compute this, right?
We've got our flux is
this triple integral.
So let's compute it.
Let's walk over to this little
bit of empty board space.
OK, so we have an
iterated integral.
So let's do it.
So the inner integral is
the integral from 0 to h
of 2*z*r*dz.
Well, that's not that bad.
That's equal to--
r is a constant.
So it's equal to r z squared
as z goes between 0 and h.
It's dz, so z is
going from 0 to h.
So we plug in, and we just
get h squared r minus 0.
So just h squared r.
OK.
So now let's do the
middle integral.
So the middle integral is
the integral from 0 to big R
d little r of the
inner integral.
So this is the integral
from 0 to big R
of the inner integral, which was
h squared little r, d little r.
OK.
And that's not that bad either.
So h is just a constant.
So this is equal to 1/2 h
squared r squared from r
equals 0 to big R. And so that's
1/2 h squared big R squared.
That's the middle integral.
So the outer one now.
OK.
So let's go back and look.
So we're doing d theta as theta
goes from 0 to 2*pi of whatever
the middle integral was.
So it's the integral from 0
to 2*pi of whatever the value
of the middle integral was.
So this is 1/2 h squared
big R squared d theta.
And this is all just constant
with respect to theta.
So that's going to be just
pi h squared r squared.
You're just
multiplying it by 2*pi.
All right.
So pi h squared r squared.
So this is our final answer.
Let's just quickly
recap what we did.
We had to compute the
flux of this field F
through the surface
of a solid cylinder.
And so we had options.
We could do it directly by
trying to compute the surface
integrals, but in
this case, life
was a lot easier if we applied
the divergence theorem.
So the divergence theorem
says that the flux-- which
is equal to this
surface integral--
can also be written as
the triple integral,
over the solid region
bounded by the surface,
of the divergence of the field.
All right.
And so in our case,
the divergence
was very nice and simple,
and the solid region
D was relatively simpler to
describe than its surface that
bounds it, S. So this is why
we think of the divergence
theorem.
Because the divergence of the
field is easy to understand,
and the solid is easier to
describe than its surface.
So those are both
things that make
us think to use the
divergence theorem
for a problem like this.
So then by the
divergence theorem,
the flux is just
that triple integral,
and so we wrote it out here.
We were integrating
over a cylinder.
So a natural thing to do is
use cylindrical coordinates.
And then we computed
the triple integral
just like we always do.
I did it in three steps:
inner, middle, and outer.
You don't have to do it exactly
this way if you don't want to.
But it works for me.
OK.
And we got our final answer:
pi h squared r squared.
I'll stop there.
