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 xz comma yz 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 xz plus xy 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 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.
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 ah 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.
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.