CHRISTINE BREINER: Welcome
back to recitation.
In this video, I'd like
to do two problems that
ask us to determine the flux of
a vector field along a surface.
So the first one is
I'd like you to find
the outward flux of the
vector z comma x comma
y through the piece of
the cylinder that's shown.
So it's just shading
the cylindrical part.
So it's a cylinder of
radius a, and you're
taking the piece in the
first octant up to height h.
So that's the first question
I'd like you to answer,
is the flux of this vector
field through that piece
of the cylinder,
the outward flux.
And then I'd also like you to
find the outward flux of this
vector field x*z comma y*z comma
z squared through the piece
of the sphere of radius
a in the first octant.
So again, it will be
in the first octant,
like this one is in
the first octant.
It'll be the piece
of the sphere that
sits in the same part of
three-dimensional space
as the piece of the cylinder
we're looking at here.
So what I'd like
you to do, again,
is just find the
flux-- in both cases,
the outward flux-- of the vector
listed through the surface that
is listed.
And then when you feel
comfortable and confident
with your answer, you can
bring the video back up
and I'll show you how I did it.
OK, welcome back.
So again, what
we're trying to do
is determine flux of a vector
field through a surface.
So I am going to determine
first what the normal is here,
and then what F dot
n is here, and then
using the fact
that I know F dot n
and I know dS in a
good parametrization,
I'm going to be
able to calculate
the integral quite simply.
So let's point out
first, that the normal
at any point on the
surface is going
to be equal to x comma
y comma 0 divided by a,
where a is, again, the radius.
Because in the normal, I
know there's no z-component.
And I actually know it's
the same as the normal would
be on a circle of radius a.
And so this vector x comma
y has length a obviously,
so that's why I'm dividing by a.
So when I look at the
normal and I dot it with F,
let's see what I get.
F dotted with the normal
is going to be x*z plus x*y
divided by a, right?
So I just took x dotted
with x times z, y times z,
and 0 times y, I add
those up, and I still
have to divide by my a.
So that's actually the vector
dotted with the normal.
Now, what's a natural
parametrization to use here
is obviously the
cylindrical coordinates,
because I'm on a cylinder.
The radius is fixed.
But what I'm interested in is
changes in theta and changes
in the height.
So I'm going to be
interested in d theta and dz,
and I need to understand
what dS in this case is.
And it's just a d theta dz.
So let me write
down what I'm going
to need to completely determine
the rest of number one
is I'm integrating
over the surface.
And I'll put the
bounds momentarily.
Actually, I should
also point out,
I'm going to write x
and y in terms of theta,
because now I know
what they are.
x in terms of theta is a cosine
theta, and y in terms of theta
is a sine theta.
So when I simplify the
expression F dot n,
I get z times cosine theta
for my first component.
And I get a times
cosine theta sine
theta for my second component.
So let me maybe point
out again how I got that.
Let me come over here.
x is a cosine theta.
So I get z times a cosine
theta divided by a.
So I get z times cosine theta.
x again is a cosine
theta. y is a sine theta.
So I get a squared
cosine theta sine theta.
And I divide by a, so I get
a single a cosine theta sine
theta.
And then my dS, as I mentioned
before, is a d theta dz, right?
And so now I just have to
figure out the bounds in theta
and the bounds in z.
Well, the bounds
in z are very easy.
The bounds in z
are simply 0 to h,
and so I'm going to put
those here on the outside.
And then the bounds in theta.
Maybe it's helpful to come
over and look at my picture.
This direction, in the
x-direction, is theta equals 0.
When I swing around
to the y-direction,
I'm at theta equals pi over 2.
So I need to go from
0 to pi over 2, right?
So let me come back over here.
OK, so now, really,
I have a couple
of constants that are letters,
but everything now, I'm
ready to integrate.
So a is just a constant,
and h is a constant,
and z and theta
are my variables,
and I can actually do this
integration quite simply.
I'm not going to do
it because I know
this is at this point something
we already know how to do,
but I'll give you the answer.
So let me write down the answer.
You get a h squared over
2, plus a squared h over 2.
So that's the solution
you actually get.
So again, I mean, you have to
integrate in d theta first.
So you have to deal with this,
you have to deal with this,
and this uses a
good trig identity.
You can use the fact
that 2 sine theta cosine
theta is sine 2 theta.
I'll give you that
little hint, and then you
can figure it out from there.
And then you integrate in z,
and you evaluate from 0 to h.
So that's the solution
to number one.
So now, let's look
at number two.
In number two, we
have to again figure
out-- we have our vector field
and we know our surface is
a piece of a sphere.
And so if we're
going to parametrize
the surface of a
sphere, we know we want
to use d theta and d phi, OK?
And let me point out first,
that again, the normal is going
to be in some ways similar to
what we saw on the cylinder,
but now instead of x comma y
comma 0, because it's a sphere,
it's going to be x comma
y comma z divided by a.
So again, as before, let me
just point out the normal
that I'm going to be using
is x comma y comma z,
all divided by a.
So I'm going to dot F with n
and look at the surface integral
with respect to dS--
d capital S there--
and I'll see what I get.
So again, I know how I'm going
to parametrize this sphere.
I already mentioned it,
but let me say it again.
It's going to be in
theta and phi, right?
Because we have a
constant radius.
We're on a sphere of radius a.
So I don't need to change rho.
It's a two-dimensional thing.
So theta is varying
and phi is varying.
So let's see what we get first.
Let me do a little work
and see what we get when
we look at F dotted with n.
So let me first point out
that F dotted with n looks
like it's x squared
z plus y squared
z plus-- just to
make this obvious-- z
squared z, all divided by a.
x squared plus y squared plus
z squared is a squared, right?
So it's actually a squared
times z divided by a.
So it's just a times z, right?
So far, all I've done was
I dotted F with the normal,
and I knew the fact that x
squared plus y squared plus z
squared was a squared, because
I was on a sphere of radius a.
So a squared times z
divided by a is a times z.
And now, if I want to use
the right coordinates,
think about theta and phi.
z is a cosine phi.
So F dot n in the
coordinates I'm interested in
is going to be a
squared cosine phi.
So let me get out of the
way so you can see that.
So that's what our
F dot n will be.
And now, if we're going to
figure out the flux, of course,
it's the integral of F dot n dS.
And let's remind
ourselves what dS is.
dS is going to be a squared
sine phi d theta d phi.
You saw this in
lecture, actually, also.
So this should look familiar.
And so now I just have
to integrate F dot n
dS over the right bounds
for theta and phi.
So let's determine
what those are.
I'll put everything
together, and we'll
determine what those are.
So I've got F dot n dS.
That's going to be a to the
fourth sine phi cosine phi
d theta d phi.
And let's think about
what is the picture
that I need in terms of the
first octant of a sphere.
Maybe I should draw a
quick picture over here
so we can remember
what that looks like.
So it's going to
be-- this is not
going to be the greatest
drawing ever-- but it's
something like this.
And so I've got pieces of
a circle at each level.
I've got a piece
of a circle, right?
If this is the x-direction,
this is the y-direction,
and this is the z-direction,
theta is going from 0-- again--
to pi over 2, and phi is going
from 0 to pi over 2, right?
So they're both going
from 0 to pi over 2.
So hopefully, you were able
to get this far at least,
in terms of figuring
out the flux of F
through that piece
of the sphere.
And I'm, again, just going
to write down the solution,
and then you can check your
answer against the solution.
And I got a to the
fourth over 4 times pi.
So this whole solution
is a to the fourth
divided by 4 times pi.
So you can check
your solution there.
Again, I want to
point out, what we
did in both of these
problems is we were trying
to compute the flux of a certain
vector field through a surface.
And if you'll notice, what
I actually did in this case
is I kept the parametrization in
terms of the x, y, z variables
first, and then I put it
in the parametrization
of theta and phi after.
And that made it a little easier
to hang on to and figure out
what it was.
Because notice, that x, y,
and z become very complicated
in theta and phi.
I have to write a lot
more down, I guess.
And then simplify
things more carefully.
This way, it was very obvious
that I got an a squared times
z in the numerator.
So sometimes it's
a little easier
to compute the F dot n in the
initial x, y, z variables,
and then change it to the
appropriate parametrization
for the surface
you're looking at.
So that's what we did, actually.
In both cases we
computed F dot n,
we put it in the
right parametrization,
and then we had to
figure out what dS was.
We had to make sure
we knew dS, and then
we just had to integrate
over the appropriate bounds
for our parameters.
And that's giving us the
flux across the surface.
So I think that's
where I'll stop.
