JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been
learning about Stokes' Theorem.
And I have a nice
question here for you
that can put Stokes'
Theorem to the test.
So what I'd like
you to do is I'd
like you to consider
this field F.
So its components
are 2z, x, and y.
And the surface S that is the
top half of the unit sphere.
So it's the sphere
of radius 1 centered
at the origin, but
only its top half.
Only the part where z is
greater than or equal to 0.
So what I'd like you to do
is to verify Stokes' Theorem
for this surface.
So that is, I'd
like you to compute
the surface integral
that comes from Stokes'
Theorem for this surface,
and the line integral that
comes from Stokes'
Theorem for the surface,
and check that they're
really equal to each other.
Now, before we
start, we should just
say one brief thing about
compatible orientation.
So I didn't give you
any orientations,
but of course, it doesn't
matter as long as you
choose ones that are compatible.
So if you think about your rules
that you have for finding them.
So if you imagine yourself
walking along this boundary
circle with your left
hand out over that sphere.
So you'll be walking in this
counterclockwise direction
when your head is sticking
out of the sphere.
All right?
So in other words, the outward
orientation on the sphere
is compatible with the
counterclockwise orientation
on the circle that
is the boundary.
So let's actually put
in a little arrow here
to just indicate that is our
orientation for the circle.
And our normal is an
outward-pointing normal.
And let's call our
circle C, and our S
is our sphere is our surface.
OK.
So just so we have
the same notation.
Good.
So why don't you work this
out, compute the line integral,
compute the surface
integral, come back,
and we can work
them out together.
Hopefully you had some luck
working on this problem.
We have two things to compute.
I think I'm going to start
with the line integral.
So let me write that
down: line integral.
So what I need to do
to compute the line
integral is I need to compute
the integral over the curve
C of F dot dr. And so I know
what F is on that circle.
So I need to know what dr is.
So I need to know what r is.
I need a parametrization
of that circle.
Well, you know, that is a pretty
easy circle to parametrize.
It's the unit circle
in the xy-plane.
So we have-- for C, we have--
and we're wandering around it
counterclockwise.
So it's our usual
parametrization.
It's the one we like.
So we have x equals cosine
t, y equals sine t--
where t goes from 0 to
2*pi-- and this is in three
dimensions, so the other part of
the parametrization is z equals
0.
So this is my parametrization
of this circle.
OK, so let's go ahead
and put that in.
So the integral over C of F
dot dr is the integral from 0
to 2 pi.
So we've got three parts.
So the first part--
so F is 2z, x, y.
So it's 2z*dx plus
x*dy plus y*dz.
But z is 0 on this whole circle.
So that piece just dies.
And dz is also 0, so
that piece just dies.
So we're just left with x*dy.
So this is equal to
the integral x dy.
Oh.
So I guess this is
not from 0 to 2*pi.
This is still over
C. Sorry about that.
OK.
And now I change to
my parametrization.
OK.
Yes.
Right.
So this is still in dx, dy,
dz form, so it's still over C.
Now we switch to the dt form, so
now t is going from 0 to 2*pi.
OK, so now we have x*dy.
So x is cosine t, and
dy-- so y is sine t,
so dy is cosine t dt.
So this is cosine t times
cosine t, is cosine squared t.
dt, gosh.
So now you have to
remember way back in 18.01
when you learned how to compute
trig integrals like this.
So I think the thing
that we do, when
we have a cosine squared t, is
we use a half-angle formula.
So let me come back down
here just to finish this off
in one board.
OK, so cosine squared t is
the integral from 0 to 2*pi.
So cosine squared t is 1
plus cosine 2t over 2, dt.
And now cosine 2t, as t goes
between 0 and 2*pi, well,
that's two whole loops of it.
Right?
Two whole periods of cosine 2t.
And it's a trig function.
It's a nice cosine function.
So the positive parts and
the negative parts cancel.
The cosine 2t part, when we
integrate it from 0 to 2*pi,
that gives us 0.
So we're left with 1/2
integrated from 0 to 2*pi,
and that's just going to give
us 1/2 of 2*pi, so that's pi.
All right.
So good.
So that was the line integral.
A very straightforward thing.
We had our circle back here.
We had our field.
So we parametrized
the curve that
is the circle, that
is the boundary.
And then we just computed
the line integral,
and it was a nice,
easy one to do.
You had to remember one
little trig identity in order
to do it.
All right.
That's the first one.
So let's go on to
the surface integral.
So the surface
integral that you have
to compute in Stokes'
Theorem is you
have to compute
the double integral
over your surface of the
curl of F dot n with respect
to surface area.
So this is the integral
we want to compute here.
So OK.
So the first thing
we're going to need
is we're going to need
to find the curl of F.
So F-- let me just write it
here so we don't have to walk
all the way back over there.
So F is [2z, x, y].
So curl of F-- OK, you should
have lots of experience
computing curls by
now-- So it's going
to be this-- I
always think of it,
so you've got these
little 2 by 2 determinants
with the partial derivatives
in them, but most of those
are going to be 0.
We've got a d_x x term
that's coming up in k,
and a d_y y term
that's coming up in i,
and a d_z 2z term
that's coming up in j.
So OK.
So almost half the terms are 0.
The others are really
easy to compute.
I trust that you can
also compute and get
that the curl is [1, 2, 1] here.
OK, so this is F. This
is curl of F. Great.
So OK.
So that's curl of F.
So now we need n.
Well, let's think.
So we need the unit
normal to our surface.
So back at the beginning
before we started,
we said it was the
outward-pointing normal.
So we need the
outward-pointing normal.
Well, this is a sphere, right?
So the normal is parallel
to the position vector.
So that means n
should be parallel
to the vector [x, y, z].
So n should be parallel
to this vector [x, y, z],
but in fact, we're
even better than that.
We're on a unit sphere.
So the position vector
has length of 1.
So n should be pointing in the
same direction as this vector,
and they both have length
1, so they had better
be equal to each other.
Great.
So this unit normal n is
just this very simple vector,
[x, y, z].
If it had been a
bigger sphere, then you
would have to divide
this by the radius
to scale it appropriately.
All right.
So we've got curl
F. We've got n.
So the integral that we
want is this double integral
over the surface
of curl F dot n.
So that's x plus 2y plus z,
with respect to surface area.
OK.
Well, now we've just
got a surface integral.
It's over a hemisphere.
Not a terrible thing
to parametrize.
So that's what we should do.
We should go in, we
should parametrize it,
and then we should just compute
it like a surface integral,
like we know how to do.
So before we start
though, I want
to make one little observation.
Well, maybe two
little observations.
We can simplify this.
All right?
x.
We're integrating x over
the surface of a hemisphere
centered at the origin.
This hemisphere is
really symmetric.
And on the back
side-- the part where
x is negative-- we're getting
negative contributions from x.
And on the front
side-- where x is
positive-- we're getting
positive contributions from x.
And because this sphere
is totally symmetric,
those just cancel
out completely.
So when we integrate x
over the whole hemisphere,
it just kills itself.
I mean, the negative parts
kill the positive parts.
We just get 0.
Similarly, this hemisphere is
symmetric between its left side
and its right side, and so
the parts where y are negative
cancel out exactly the
parts where y are positive.
So as a simplifying
step, we can realize,
right at the beginning,
that this is actually
just the integral over S of z
with respect to surface area.
Now, if you didn't
realize that, that's OK.
What you would have
done is you would
have done the parametrization
that we're about to do.
And in doing that
parametrization,
you would have found that you
were integrating something like
cosine theta between 0 and
2*pi, or something like this.
And that would have given you 0.
So you would have
found this symmetry,
even if you didn't
realize it right now,
you would have found it in
the process of computing
this integral, but it's
a little bit easier on us
if we can recognize
that symmetry first.
Now, notice that z doesn't
cancel, because this is just
the top hemisphere, so it
doesn't have a bottom half
to cancel out with.
Right?
So the z part we can't
use this easy analysis on.
If we integrated this z
over the whole sphere--
if we had the other half
of the sphere-- well,
then that would also give us 0.
But we only have the
top half of the sphere.
So it's going to give us
something positive, because z
is always positive up there.
OK, so let's actually set
about parametrizing it.
We want to parametrize
the unit sphere.
Well, OK.
So we have our standard
parametrization
that comes from
spherical coordinates.
So rho is just 1.
Right?
So x is equal to, it's
going to be cosine--
You know what?
I always get a little
confused, so I'm just
going to check, carefully, that
I'm doing this perfectly right.
x is going to be
cosine theta sine phi.
Good.
y is going to be
sine theta sine phi.
And z is going to be cosine phi.
So that's our parametrization.
But we need bounds, of
course, on theta and phi
in order to properly describe
just this hemisphere.
So let's think.
So for phi, we
want the hemisphere
that goes from the z-axis
down to the xy-plane.
So that means we
want 0 to be less
than or equal to phi to be less
than or equal to pi over 2.
Right?
That will give us
just that top half.
And we want the whole thing.
We want to go all
the way around.
So we want 0 less than or equal
to theta less than or equal
to 2*pi.
OK, so this is what
x, y, and z are.
These are the bounds for our
parameters phi and theta.
Now, the only
other thing we need
is we need to know what dS is.
So in spherical
coordinates, we know
that dS-- I'll put it
right above here-- so dS
is equal to sine
phi d phi d theta.
Let me again just double-check
that, that I'm not
doing anything silly.
So dS is equal to sine
phi d phi d theta.
So we've got our
parametrization.
We've got our bounds
on our parameters.
We know what dS is.
And we have the integral
that we want to compute.
So now we just have to
substitute everything
in and actually compute it
as an iterated integral.
Great.
So let's do that.
So, this integral
that we want, I'm
going to write a big
equal sign that's
going to carry me
all the way up here.
That's an equal sign.
All right.
So our integral, the
integral over S of z
with respect to surface area.
So z becomes cosine phi.
So we've got our double integral
becomes an iterated integral.
z becomes cosine phi.
dS becomes sine
phi d phi d theta.
And our bounds.
So let's see: phi we said is
going from 0 to pi over 2.
Zero, pi over 2.
And theta is going
from 0 to 2*pi.
OK.
So now we just have a
nice, straightforward
iterated integral
here to compute.
So let's do the inner one first.
So we're computing--
the inner integral
is the integral from 0 to pi
over 2, of cosine phi sine phi
d phi.
And OK.
So there are a bunch
of different ways
you could do this.
If you wanted to get fancy, you
could do a double-angle formula
here, but that's really
more fancy than you need.
Because this is like sine
phi times d sine phi, right?
So this is equal
to-- another way
of saying that is you can
make the substitution u equals
sine phi.
Anyhow, this is all Calc I
stuff that hopefully you're
pretty familiar with.
So OK.
So this is equal
to-- in the end,
we get sine squared phi over
2, between 0 and pi over 2.
OK.
So we plug this in.
So sine squared pi
over 2, that's 1/2,
minus-- sine squared
0 over 2 is 0 over 2.
So it's just 1/2.
So the inner integral is 1/2.
So let's see about
the outer one.
The outer integral is just the
integral from 0 to 2*pi d theta
of whatever the
inner integral was.
Well, the inner
integral was 1/2.
So the integral from 0
to 2*pi of 1/2 is pi.
Straightforward.
Good.
So OK.
So that's what the
surface integral gives us.
So let's go back
here and compare.
So way back at the beginning
of this recitation,
we did the line
integral for this circle
that's the boundary of this
hemisphere, and we got pi.
And just now what we did is
we had the surface integral--
the associated surface integral
that we get from Stokes'
Theorem, this curl F dot n dS.
So we computed F
and curl F and n.
And then we'd noticed a
little nice symmetry here.
Although if you
didn't notice it,
you should have had no trouble
computing the extra terms
in the integral that you
actually ended up with it.
It would've been another
couple of trig terms
there after you made
the substitution.
So we parametrized
our surface nicely.
Because it's a sphere,
it's easy to do.
And then we computed
the double integral
and we also came out with pi.
And we better have
also come out with pi,
because Stokes' Theorem
tells us that the line
integral and the
surface integral
have to give us the same value.
So that's great.
So that's exactly what we
were hoping would happen.
And now we've sort of
convinced ourselves, hopefully,
that through an
example now, we have
a feel for what sorts of things
Stokes' Theorem can do for us.
I'll end there.
