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.