Uploaded by UCBerkeley on 01.12.2009

Transcript:

OK.

How are you guys doing?

Did you have a good Thanksgiving break?

Yes?

Are you ready for the last week of classes?

Yeah, OK.

It's hard to believe it is the last week, but actually next

week I will give a bonus lecture, I will give a -- I'll

give a review lecture on Tuesday next week.

And also I remind you that there's a lot of materials for

the final exam, which are available on bSpace, and also

on the class home page, which I hope by now you

know how to find.

All right.

And now we go back to the study of Stokes' theorem.

So we've been slowly laying the foundations, the necessary

foundations for Stokes' theorem and finally, it's time to sort

of reap the benefits of all of that, and finally today we'll

be able to state the result and to see what it means, and what

the applications of this result are.

But first of all, I would like to revisit the material which

we talked about last week, OK.

A week ago we talked about -- we talked about surface

integrals, integrals over the surfaces, OK.

And the point was that actually we studied two

types of integrals.

Two types of integrals.

Over general surfaces in the 3-dimensional space, as a good

working example of such a surface, we'll look at

the upper hemisphere.

It's a surface with a boundary, which is a circle.

So, you can actually, you can think that there is a -- it's

actually a arc of a sphere which is centered at the

origin, so you could also imagine that there is a

coordinate system here, xyz, and so the surface, what I'm

talking about is the this upper hemisphere, the surface of the

upper part of the surface of the sphere, OK, which has a

boundary which is a circle.

So what we're trying to understand is how to integrate

over such surfaces, surfaces of this type, and relate integrals

over such surfaces to integrals over their boundaries.

That is the point of Stokes' theorem.

So let's denote this by m, and m stands for membrane, because

I explained last time that integrals over surfaces of the

second type can be viewed as fluxes over fluid

through membrane.

But here I will recall that there are two types of such

integrals that we have looked at.

The first type -- the first type you pick a function and

then you integrate over m this function with a measure

which we denote by ds.

We talked about the interpretation of this -- this

has to do with the surface area, and more generally

with the mass.

So this represents the mass of the surface, think of the

surface of aluminum foil, say, and f is the mass

density function.

And when you integrate the density function, you

get the total mass.

It can also be a charge density function, then you would be

integrating total charge.

So that's the first type.

The important thing here is you integrate a function.

And the second type is when you start with a vector field.

So it's not a function, it's much more than a function

-- actually, we represent it by triple functions.

It has components P, Q, and R, where each of these three is a

function of its own, in its own right.

In this case, we also have double integral, which

we discussed last time.

And that's what's called a flux of a vector field, or a surface

integral of a vector field.

Which we denote like this -- e dot ds.

So this is called, sometimes called the flux of e

through this membrane.

And it represents the rate of flow of a fluid through a

membrane m if f is a velocity vector field for this fluid.

I explained last time.

And now, of course, even if you were not here, you can actually

watch the lecture on multiple places, which is good -- I

think it's a good resource.

OK, so it makes it easier for me to just say

I did it last time.

If it was not recorded, that would be a little unfair, but

now I know that if you don't remember, you can always go

back and watch it again.

Now, how to compute this.

So the key to computing this is to, as well as the other

one, is to parameterize the surface, right.

So parameterize it, which means that we write x as some uv, and

y as some y of uv, and z as some function, z of uv where u

and v are in the -- belong to a certain domain now

on the plane.

Parameterize.

When I say parameterize it, I mean parameterize m.

And so when you parameterize it, you actually get a

very explicit formula for this integral as just a

double integral over d.

So, we get an integral, which is now in the realm on the

previous chapter of this course, something which

we studied before.

OK?

Now the key to this is the following, that the way we

introduce this integral -- we actually wrote it as an

integral of the first type, wrote it as an integer of the

first type over m where what we do is we take the dot product

between e and n where n is the unit's normal vector

field for the surface.

When we do that, when we take the dot product, we convert our

vector field into a function.

This becomes the function and that function we can now

integrate in the sense of the integral of the first type.

So this is how we introduced it.

Today I will explain a different way to think about

this where you don't have to use the concept of the first --

our integral of the first type to understand the integral

of the second type.

But this actually, this brings out the question as to what

this m is, OK, this normal vector.

So in my picture -- let me draw this picture again.

In my picture, this vector field, m would look like this

-- at each point of the surface we would have a vector which

kind of sticks out, and which is normal -- normal means

perpendicular, perpendicular to the surface.

If you have a surface and you have a point on the surface,

you can talk about the tangent plane -- right, we talked

about this, tangent planes.

We even know equations for tangent planes, how to find

equations for tangent plane.

So we also have the notion of a normal line.

A normal line is perpendicular to the tangent plane, right.

So at this point you have a tangent plane -- I'm not going

to draw this, just not to mess up the picture, I don't

want to make it too heavy.

But there is a tangent plane here, right, but it is

also a normal line.

So it's perpendicular to the tangent plane.

When something's perpendicular to the tangent plane at the

point, we might as well say it's perpendicular

to the surface.

What else can it mean but to be perpendicular to the surface.

It just means being perpendicular to

the tangent plane.

On this line there are many vectors.

If you have one vector on this line, you can pick an a

multiple of this vector.

So what we'd like to do is we'd like to specify a normal vector

at each point, and we want to specify it with as little

ambiguity as possible.

So the way to get rid of the ambiguity, get rid of the

redundancy is to normalize the vector, to make

it a unit vector.

If you have any vector, it has certain magnitude -- unless,

you know, if it's not 0, you can't divide by the magnitude

and you get a vector of length 1, or magnitude 1.

Such a vector's called unit vector.

So, out of all the vectors along this normal line, we can

say let's take a unit vector.

The problem is that when you do that you still have a

small ambiguity, it is a much more ambiguity.

Instead of instantly many vectors, you end up with two --

there are two possibilities.

There is this one, but there is also one which goes in

the opposite direction.

So it's kind of a mirror image of this one.

And just by saying it's a unit vector, we can't

kill this redundancy.

Both of this are bona fide unit normal vectors for the surface.

If you think about this upper hemisphere, one of them sticks

out, goes outside, and the other one goes inside.

So that is two choices at each point -- at each point

there are two choices.

And what we'd like to do is we'd like to make a choice at

each point along this entire surface so that this choice

is continuous, smooth.

In other words, we don't want to be in a situation where we

choose a unit vector at this point to go outside, outward,

and then at the nearby point to make it go inside,

that wouldn't be smooth.

So when we talk about normal vector fields, we mean that

we want it to vary in a smooth way.

So that's what this m Is, m this formula.

This m is the unit normal vector field.

And then I will emphasize very smoothly along m.

That vector field is called orientation.

I would like to contrast this to the notion of

orientation of a curve.

In the case of a curve, if you have a curve going from point a

to point b, we talked each orientation as a direction

along the curve.

And if the curve has boundary points, specifying orientation

is the same as specifying which one is the initial point and

which one is the end point.

So here, this would be the initial point, this

would be the end point.

If the curve does not have a boundary, for example, a

circle, we could say it's counterclockwise or

it's clockwise.

But let's look at this one which have end points,

it has a boundary.

In this case, by analogy to this, we observe that actually

in this case there are also two possible orientations -- we

can go this way or we can go th that way.

So there's a lot of similarity with the notion of orientation

for curves and surfaces.

The main difference is that orientation for a curve

goes along the curve -- you go with the flow.

And here, you also go with the flow, but the point is that the

flow goes across the surface usually.

You can't flow along the surface.

Suppose you can maybe, but it makes much more sense to

flow across the surface.

As I explained you could have a pipe, you know, and this would

be a membrane, which kind of covers this pipe, and you would

be measuring the amount of fluid which passes through it.

So it's much more natural to talk about orientation

across rather than along in the case of surfaces.

Unlike curves where we should be going along

means a lot more sense.

So this is one important difference between orientation

for curves and surfaces.

There is a similarity, right -- there's a similarity that's in

both cases, two choices, so here there are two choices.

Here there are two choices.

And in this case, there are also two choices.

One indicated in white, and one indicated in red.

And if you think about it once you choose it at one point,

then basically there is no choice at nearby points,

because you have to very smoothly, so you know nearby

would be like this, nearby would be like this, like this,

like this, because you don't allow jumps.

If you don't allows jump, once you know it's at one point,

once you specified at one point of your surface, you

know it everywhere.

But then the question is can we actually define it everywhere

in a consistent way.

And this is, here we kind of get surprised because in the

case of a curve there are always orientations.

For any curve you can always choose two orientations, one

goes this way, one goes that way.

But in the case of surfaces, it actually exists surfaces which

have none, which don't have any orientation.

And this Is actually kind of cool.

You must have heard about Meryl Streep, right.

Have your heard about Meryl Streep?

Yes, OK.

So here is a little demonstration for you.

So, we need a close-up.

Sometimes I'm delayed so that's why I'm saying it and I'm

waiting for a few seconds.

So here is a surface, I mean this is a surface, and I

colored it in two ways -- on one side it's white, on the

other side it's pink, OK.

So now what I'm going to do is I'm going to make a surface, I

could glue a surface out of it in different ways.

So for example, I can glue it in this way.

If I glue it in this way it's like a cylinder, it's like a

little cylinder -- not so, I mean kind of very low height.

And in this case if you look you could see that actually it

clearly has two sides, there's an inner side and there's

an outer side -- one is pink and one is white.

So when you talk about orientation you could also

think in terms of choosing the side of the surface.

I'm thinking about vectors sticking out, but once you have

a vector sticking out, you have the notion of what's

outside and what's inside.

So let's say that if it sticks out this way then that's the

outside of surface, and if it sticks out this way this would

be the outside and that would be the inside.

So it's like, you know, if you put your teeshirt on inside

out, this kind of stuff.

So it's a clear notion, inside or outside of a surface.

So that's what we're talking about orientation is saying

which side we consider as inside and which one you

consider outside, because, of course, after all it's not

clear, you can choose -- you can make your choice, and

sometimes you can have a teeshirt which actually looks

a little better if you wear it inside out.

It kind of looks cool.

But imagine you have a shirt, which only has one -- a

teeshirt which only has one side.

So that would be difficult to wear.

Now, but actually you can create such a, not necessarily

a teeshirt, but you can create such a surface very easily by

taking this thing strip and gluing the ends kind of by

twisting by 180 degrees.

If you do that you say OK, so let's say that this is the

-- this is the outside.

But then you trace it, if you trace it, the outside, you

trace it, you trace it, and then suddenly it becomes

white, and when you make the full circle you come back

on the opposite side.

So if this was outside, outside it should be the notion that if

we follow it -- you know, on my shirt, if I follow it

everywhere, it's going to be on the same side always.

I will never be able to jump on the other side, one

would hope anyway.

And that's the way it's supposed to be, right.

But here you go, you make a full circle and you come

back on the opposite side.

So that means this is outside but also this is outside.

So that means -- but then which one is it?

So it means at this point it's not defined.

There is no global notion, there's no global notion

of inside and outside.

You see what I mean?

So in this case, we will say that the surface is not

orientable, and this is really remarkable phenomenon.

Again, in the case of curves, all curves are orientable, it's

just that there is a redundancy -- there are two possible, or

ambiguity, there are two possible orientations, you

have to make a choice.

For surfaces actually we don't even have that luxury.

There are many surfaces which are not orientable for which

you could not consistently say which is this inside and which

is outside, and that's example.

Any questions?

So in this case, if the surface is like this,

there is no orientation.

There is no unit vector that you can make which varies

smoothly and gives it a normal vector at each point, because

if I were to give such a vector, let's say again, here,

the vector will be sticking out this way, but if I varied

smoothly, I would be forced to come out here on this side.

So that means it becomes ambiguous, and at this point

there's no -- it has to be simultaneously this and that,

and that's not allowed.

So it does not have a consistent orientation.

So, in the case of such a surface, this integral is

actually not well defined, you can not set up an integral of

vector field over a surface which is not orientable,

which does not have this type of orientation.

In other words, the consistent notion of inside and outside.

So we have to be aware of it.

Now, the good news is that most surfaces are orientable,

most surfaces that are of interest to us.

For example, if we are thinking about really a fluid flowing in

a pipe and you're thinking of something which is like a cover

at the end of the pipe, it is going to be orientable.

So, we're not going to be too handicapped by the fact that

for non-orientable surfaces this is not well-defined.

All the surfaces we will consider in this class for sure

will be orientable, but we have to be aware of the fact that

actually we should not take this for granted.

That actually a surface that may not have a particular, even

one orientation -- of course, if it has one, it will

have to have two, right.

The point is that if n is one of them, the negative n will

also such an object, also such an orientation.

So it can't be that you have just one orientation -- you

either have none or you have exactly two, always.

And in all of the cases which we will consider here, our

surfaces will have two orientations.

And this will give us a little bit of a headache, because when

I write now a Stokes' formula, I will have to be very precise

in specifying the orientations on the left and right

hand side, so as to make the equation correct.

So, let's go back to this.

This integral.

So the integral is actually -- let's suppose that

it is orientable.

So this is important.

It's not a given that the surface is but let us

assume from now on that m is orientable.

That is to say m has an orientation vector field, m.

And then it has necessarily two of them -- it has two

orientations, n and negative n.

OK, let's assume that.

And so pick one, we'll pick one orientation, which

is what I call n.

In this case we have this integral, we have

this double integral.

Like of this, so you'll have e dot ds, which can be defined by

the formula here, I just rewrite it, e dot n, ds.

Now, there are two choices, so if you take the second choice,

the second choice means instead of m you take negative n.

So, if we choose negative m, we just get minus

of what we had before.

If we choose the other orientation, this will just

result in overall size.

So in that sense you can say well, it's not

such a huge ambiguity.

You don't get sort of a totally different answer, you get

something up to sign.

But it wouldn't be serious if we, in this course we said

all our results were defined up to sign.

We have to be precise, we have to be able to say is it 10 or

negative 10, because sometimes your life depends on it to know

it's positive or negative.

So that's why we take this seriously and we say the

definition of such an integral has to include orientation.

So if you do a homework problem, it has to say the

integral vector field over the surface with respect

to which orientation.

And then there is standard terminology.

Standard terminology is you can say orientation is upward.

Upward means that the vertical component of

vector field is positive.

So the white orientation on that picture is upward.

Or it could be downward, that would be the red one.

And oftentimes if we have a closed surface, for example,

not the entire sphere, but say so it's not the upper

hemisphere but the entire sphere, you could say outward

or which also has a clear meaning.

OK, so let's pick one.

How do we actually compute this?

We compute it by parameterizing m by choosing two existing

parameters u and v, and then when we parameterize it, so

this integral becomes -- we can write it very

concretely as follows.

It's e dot ru cross rv over d, where d is the parameterizing

domain on the uv plane.

On the plane of these coordinates u and v.

What is r?

So, r is a vector which we get by combining the three

functions which parameterize our curve -- x of uv i, plus

y of uv, j plus z of uv k.

So these are the three functions.

You put them together into vector field.

r sub u simply means that you take derivatives with respect

to u -- maybe you write dx du i plus, and so on.

And likewise r sub v means that you take dx dv

i plus, and so on.

So once you have your parameterization, once you have

your parameterization, you have this r, and you have r sub u by

taking simply partial derivatives of these three

functions with respect to u.

And you have r sub v.

So then you can just take the cross product. and when you

take the cross product you get another vector field, which you

can then dot with your original vector field.

This will give you a function which you then integrate over

the parameterizing domain.

And, of course, the point here is that both the original

integral and this integral are integrals over surfaces, but

there's a huge difference.

This surface is actually on the plane, it's flat.

So, by this formula, we express a complicate integral over a

very complicate surface, which could be very curved like a

sphere or part of a sphere as an integral over domain on

the plane, which is flat.

So that's the definition which we talked about last time.

And that there is some special cases where, for example, your

surface is a graph of a function, we have some -- the

formula simplifies and we have various formulas which kind of

give us a shortcut to the answer.

So finally, we are ready to state the Stokes' theorem by

using this notion of a surface integral.

So let me draw this one more time.

So you have our surface m and we have the boundary, the

circle boundary, which I'll call b of m.

So, on the left hand side of the Stokes' formula, we will

have a line integral over vector field over

this boundary.

So we'll have a vector field, f, which will be defined in a

three-dimensional space, in particular, will be defined

on this entire surface.

And on the right hand side, we'll simply take the line

integral of this vector field over the boundary.

So this is something we learned before, so I don't have

to say much about it.

This is just has become by now the usual thing, the line

integral of vector field, which is again, to calculate it we

need to parameterize the curve of and then there's a very

simple formula which allows you to take this integral.

And now on the left hand side, we're going to have an integral

which is a double integral and it's going to be an

integral over m.

And then the integral, it will be the integral of another

vector field, and that vector field will be nothing

but the curl of f.

So you have your vector field and it appears on

the right hand side.

But on the left hand side you have another vector field.

This is something which I was denoting by E in all my

previous formulas, so we have double integrals of a vector

field ef over a surface.

But now take as e the curl of your original vector field

which was given to you from the beginning, f. f is

given, take its curl.

That's another vector field.

And take the integral of the curl.

And so the markable statement is that you always have this

equality, you have this equality of double integral

and the single integral over the surface and over the

boundary of the surface.

Now, what I have said is still incomplete, and the reason it's

incomplete is that both of these integrals depends on

the choice of orientation.

Did you have a question about this?

[INAUDIBLE]

The curl f would be what I call here e.

[INAUDIBLE]

So, I purposefully called this vector field e, and I called

that vector field f because on the left hand side when we do

the double integral, we're actually integrating

not f but curl f.

So that's why I'm using a different notation here so

as not to confuse you.

We are given the vector field, f, but we are sticking that

vector field on the right hand side into a single integral,

into a line integral, right?

Whereas on the left hand side we do curl, and curl is

another vector field which we know how to calculate.

And then we'll actually see what the meaning of that

curl is we'll get a better explanation of what the meaning

of the vector field is.

OK, so that's the formula, but as I said, on both sides we

have two integrals which, as I explained, requires the

choice of orientation.

So, in other words, depending on how I choose orientation on

m, I want either get -- I will get plus minus some answer.

I will get some answer or it's negative, and if I want to pin

it down, I have to say which orientation I should take.

Both integrals are well-defined.

There is no canonical choice of orientation, right?

There's no canonical choice of orientation.

So there are two possibilities, and the two possible answers

they differ by sign. but I have to say which one I should take

to make this well-defined.

Likewise here, I also have two different answers.

In the case at hand, I have this circle.

I can choose orientation which goes counterclockwise or

clockwise, if you look from above.

So again, there are two different answers

which differ by sign.

If I'm saying that these two things are equal to each other,

I better be precise and say how the orientations should be

defined so as to make this into an equality, because if I make

the wrong choices, instead of getting this formula, I'll get

this equal to negative of that.

In other words, if I wanted to be careless about orientation,

I would not be able to state this formula at all.

The best I could do I would be able to say this is equal to

plus minus, and if I did that I should be fired.

So, we don't want it to happen -- well, maybe

you do but I don't.

So that's why I'm going to explain how to choose

orientation so that we don't have this ambiguity.

And of course, the point is, you probably guess already

because this is very similar to the way, to the problem which

already arose when we talked about the Green's theorem,

which is actually kind of a little brother of the Stokes'

theorem we'll talk about in a minute.

In Green's theorem the results of this issue of orientation,

and the way we handled that was by saying -- was the following.

So, in Green's theorem, we also have a two-dimensional

region, but it was flat.

This is flat, this is not flat.

Well, I have drawn it on a flat blackboard, but 3D, you know.

I was just reading about this new movie, Avatar, it's going

to be in 3D, so I was thinking maybe some years from now

lectures like calculus will be given also in 3D, so you will

all be wearing these cool classes and you will actually

see the -- i'll will be able to draw 3-dimensional surfaces.

I should talk to Cameron about this.

I'm sure he is a huge fan of multi-variable calculus.

Or if he isn't he should be because he uses a lot of

special effects, and of course, how can you do special effects

if you don't use calculus.

Anyway, so in Green's theorem you have a region like this.

And the way we handle this is we said once you have this

region, the orientation you define on the boundary is such

that when you walk on the boundary, the region

is to your left.

If you walk on the boundary the region is to your left.

So it would be like this, it would be counterclockwise

in this picture.

So, I would like to say the same.

I would like to say that if I walk on the boundary, if

somebody walks on this boundary then the region should be along

orientation, with respect to the orientation.

The relation should be such that as somebody walks along

this boundary, the region itself, the surface

is to the left.

But the question is how do you expect that person

to walk on this?

There are two possibilities.

It could be that they walk like this or they could be walking

from the other side of the blackboard.

A priori there is no choice.

I mean there are two choices of apriori, there is no canotical

choice, there is no given choice.

In the case of a plane, we do have an obvious orientation, we

do have a special orientation, and the reason is the

following, because there is an orientation which kind of

sticks out towards us.

An other way to say is that there is an orientation

which we call k.

So we draw this coordinate system, which is i, j and then

there is a k which is obtained by the Corkscrew rule,

the right hand rule, whatever you call it.

Remember this picture?

This was one of my fondest memories of this class, the

discovery of this visual aide.

This is how you show the normal.

So the point is that you can't just say it seems like I

can actually move this.

You can already appreciate the problem in giving

it orientation.

The orientation could be like this or it could be like this.

But the point is when we actually draw something on a

2-dimensional plane, we look at it from a certain direction.

We have to be looking at it from a certain direction.

We either are on this side and we're looking on this side, or

we are there -- and let me tell you, there's not a whole lot of

space behind that board so we should be on this side.

Once we are looking at it from this side, we can give it

any coordinate system, any orientable coordinate system.

In fact, this is not even canonical either, but I

can rotate it and so on.

But whichever coordinate system, xy, I choose, the cross

product of the i and j will be looking in this direction,

will be sticking this way.

So that's why when we are having a region on the

plane, we actually have a canonical orientation.

That's what we call vector k, k being the cross product of i

and j in this order -- first x -- this is a unit vector

in the x direction.

This is a unit vector in the y direction.

We take the cross product in this order, and so that's what

gives us the z-axis, or the unit vector along the z-axis.

So we're kind of lucky in this case, we don't we

have to make a choice.

The choice is made by the fact that we're actually absorbing

it in a certain way.

We are looking at this blackboard in a certain way.

But if we have now a surface in the 3-dimensional space, for

example, I could just put this blackboard somewhere in the

middle of this class, and some of you will be on one side and

some of you will be on the other side.

So some of you will be looking from one direction, some of

them will be looking from the opposite direction.

So some of you will be one orientation, for some of you,

it would be natural to take the other orientation.

You see, so in general there is a choice of orientation.

But once you make that choice, if you make it like this, you

can't say if a person walks so that their head is -- points in

the right direction, up, then the domain is on the left.

Then it actually makes sense.

So, somebody could be walking like this, their head would be

this way, right, and so they will have the region on their

left, so that would be we'd get the old rule.

If, on the other hand, I will just say let's choose the other

orientation, which goes in that direction, I would still have

the notion of somebody walking along the boundary, but that

person would be on this side, right, and then it would still

make sense to say that the region is to their left, which

would then mean the opposite direction.

Well, you'd have to really trust me that this

is how it works.

You can just imagine it, if it were made of

glass you will see it.

So, the rule in general is like this.

If the person is walking along this pointing in the direction

of the orientation, if this is the orientation and you are not

upsidedown, you have to be oriented in the right way, your

head should be above your feet basically, with respect

to this orientation.

I'll try to not to make too many jokes about this, but I'm

sure you can do them yourself.

If you walk this way, the surface should be to your left,

so this is how it works.

And if I chose a different orientation, which would be

inward, then I would have to draw this little fellow

pointing this direction and then it would have to walk

so that the region is to their left.

So that's the rule.

Is that clear?

So, it's pretty clear.

So that's the rule I choose, and if I choose that rule,

which it will give me a consistent formula, it will

give me a choice -- whatever choice I take here, once they

make the choice orientation M, I get the orientation on

the boundary as well.

And it is for these two choices that these two integrals

will be equal to each other.

So that's a statement of Stokes' theorem.

So, the next thing I would like to do is I would like to kind

of try to demystify this formula, because at the moment

it looks pretty bizarre in a way because you have this

strange -- you have two different types of integrals.

So one of them is of the vector field, the other

one is this curl, OK?

So would like what I would like to do is I'd like to explain

how this formula actually fits in this general principle,

which I talked about earlier of which we have now seen

several examples.

So I would like to convince you that this formula is actually

special case of this general principle, just like the

fundamental theorem for line integrals and the

Green's theorem.

So that's the next thing we are going to do.

So, in order to explain this, I would like to actually give a

slightly more besides formula for a surface integral, so this

is going to be a slightly long and somewhat boring

calculation, tedious calculation, but bear with me

because at the end of the calculation I hope that I will

be able to demystify this, demystify this formula, and you

will see that really it is part of a pattern, which is much

more, how should I say, much more convincing.

So the first thing I'd like to do is I want to go back to

general surface integral.

So here, E doesn't have to be the curl.

It doesn't have to be the curl.

It could be some vector field.

And I would like to write a slightly more explicit formula

for it, assuming that I have a parameterization.

So, let me recall what I wrote on the top, in the

top right corner there.

I wrote that this is equal to the integral over z, the

parameterizing domain, of e dot ru plus rv.

Where ru and rv you have -- well, the beginnings of ru and

rv you have on the blackboard.

So now I want to actually calculate rv plus rv for you.

So I'm going to calculate what ru plus rv is.

So this is just they use a cross product, this is just

they use a cross product where I put dx, du, dy, du and dz du

in the first row, and I put dx dv, dy dv, and dz dv

in the second row.

So now I just open, I just write down what it means.

So let me just start with I, so this is going to be dy du, dz

dv minus dy dz, dz du -- this is times i.

Maybe here plus, so now j, so I have to look at this, and I

have to remember this sign.

So it's going to be dz du -- it's going to

take a little time.

But you will like it when it's over, trust me,

so it's worthwhile to do this calculation.

dz dz, dx du, j.

You have to do it once to believe it and then after this

a lot of things become simpler.

Plus finally, I have to do the k, and the k is dx du, dy

dz, minus dx dz, dy du.

Check this.

Are you with me on this?

This is good?

If you agree then, you know, you will not have a chance

to disagree with me later.

It's your last chance.

So, if we agree on this, have you ever seen

this formula before?

You're under oath, so have you ever seen this formula before?

What is it called, this kind of expression.

It starts with a J..

It's Jacobian.

It is a Jacobian for the coordinates y and z with

respect to u and v.

This is a Jacobian for the coordinates z and x with

respect to u and v, and this is a Jacobian for x and y

with respect to u and v.

Everybody agrees with that?

So, I actually can rewrite this in a more compact way by using

the notation which we introduced when we

talked about Jacobians.

OK?

So, the notation for the Jacobians is d -- you put the

dell like this, and you put y and z over dell of xy -- uv.

This is a notation for the Jacobian, which we have used

extensively about a month ago.

So this is times i.

Plus we have the Jacobian for zx over uvj plus the Jacobian

over xy uv, and then you put k.

So, this is actually the explicit formula for this cross

product, and this cross product answers in the definition.

So now let me actually take the dot product so that I

get the entire expression.

For this I have to also say what is e.

So let's say that e is ai plus bj plus ck, and what do I get?

I get e dot ru plus rv -- which is my integrant in the surface

integral -- is a times the Jacobian plus b, OK?

So that's the formula.

Here is actually explicit formula for what you should

integrate when you take double integral of vector field, an

explicit formula, which is written entirely in terms of

the components of your original vector field, and the Jacobian

of the coordinate change from iv to all possible pairs of

coordinates amongst xyz.

Now, how can you remember this without memorizing the formula.

It's very easy.

So first of all you have abc, right, these are the components

of the original vector field.

And then you also have xy and z, the z coordinates.

So you have to remember the structure of the formula.

There are three terms, each term responds to one of the

three components of the vector field, and then you have to

remember which Jacobian to put -- you see there are three

different Jacobians.

but there is also subtle point, which is in the Jacobian you

also have to remember in which order you put the coordinates.

You may remember from doing double integrals by using

general coordinate changes, that if you switch the

coordinates, the Jacobian gets the minus side, right -- the

Jacobian gets the minus side.

So here it's important that I put yz over uv

and not zy over uz.

So first thing you note is that if you take a, the variables

which get involved -- so a responds to x.

So to get the variables involved, you have

to cross it out.

So you have ayz here, then you have b with xz, and c with xy.

So that the two coordinates which you take which appear

together with the component of the vector field are the

remaining two components.

So a is the x component.

So the coordinates which are relevant to this term of the

formula are y and z, and then you have to ask yourself in

which order, and the order should be what mathmaticians

call a cyclic order, which means this is first, this is

second, this is third, this is fourth and so on, so it

kind of goes in a circle.

So, for example, y and z have to appear in this order, yz.

But z and x should appear -- you're at z, what's the next

one, the next one is the first one -- it's a cyclic order.

So it's like zx.

And then xy is clear.

So xy, yz, zx.

And then you put the remaining component of the vector field.

So that's what the meaning of this formula is.

So that's the first thing we should agree on, the

double integral can be written in this form.

But now let's go back to Stokes' formula.

In Stokes' formula there is an additional complication, as if

this formala was not complicated enough.

We actually have to take the surface integral not of the

vector field itself, but with curl.

So gee, we have to do one more cross product.

But actually, the funny thing is that when you do that cross

product and you combine it with this stuff, which you can

imagine is going to look horrible, right, actually it

will look better in some sense, or you will get something

which will have much more conceptual explanation.

So let's do that, let's do that.

I promised that I would explain it to you, so let

me explain this to you.

So, we're halfway there.

What we need now is we need to take the -- apply this formula

to the left side of Stokes' theorem where we are just doing

it to the curl of the vector field and not to vector

field as before.

So, if E is curl of F, what are, what does it look like?

So we have to again do this -- i, j, k, then we

put d dx, d dy, d dz.

And then we put P, Q, R.

So what do we get?

We get dR dy minus dQ dz times I plus dP dz minus dR dx

plus dR dx minus dP dz.

But please check that I did it correctly.

Because if I mess up one term then it will

not come out right.

[INAUDIBLE]

OK, good.

That's right.

And in fact for good reason because this is expression

of Green's theorem.

So we know that Green's theorem actually is a special case of

this when only this term shows up.

Dz dx and dz dy exactly, good.

I think now it's correct.

So now I want to put this in here.

In other words, this will be my a and this will my b,

and this will be my c.

So you would think this would look awful, and in some sense

it will, but then I will explain another way to get that

expression in a much more, in kind of a much more

conceptual way.

And this will, at least for me, and I hope for you, this will

demystify this expression and this result.

OK.

So let me put this into this formula.

And I really want to do it in real time instead of showing it

to you already written before, because if I show you you'll be

like OK, whatever, you know, this isn't some derivative, so

he says it works and it works, but I want you to really

follow me and to really see how it works.

So, dr dy -- so please keep track of what I'm writing,

because as you see, I can make mistakes.

But now I want to do one more thing, actually, to simplify,

to kind of accelerate this lesson.

At the end of the day I'm going to integrate this, right, I'm

going to integrate this -- I'm not just taking this

expression, I'm actually integrating this, and I'm

integrating it with respect to du dv.

So let me actually then multiply this by du dv.

But if I do that, then this becomes A dy dz, right?

Because remember -- let me write it here.

When we did double integrals, this was the formula

which we used.

If we integrate -- the way we did it was for xy,

we did it for dx dy.

But the same would apply if I do it for xz or yz.

In other words, when we integrate a function in x and

y, we said it can be written as an integral over u and v, but

you have to insert this factor.

I'm sorry, this doesn't look right.

What I mean is, of course -- so the way we wrote it is we said

that when you have a double integral function, f, this is

equal to -- what I'm doing is just stripping it of the

integral and move the function.

I'm just saying that this is really the identity which

we use and we proved it.

So what I'm going to do is whenever I have this ratio

because I now multiply overall by du dv, I'm just going

to replace this by dy dz.

I will replace this by dzds, and I'll replace this by dfdy.

Can I do that?

You agree that this is OK?

Legitimate?

Yeah?

OK.

So was the purpose -- I'm just trying to remind you that this

was the purpose of introducing these Jacobians in first place.

Because we wanted to replace things like dx dy by du dv.

And we knew that this was the price to pay.

We had to insert this factor.

So might as well replace this expression by this expression.

And likewise for the remaining two summits.

So, this is a formula I'm going to use.

And I'm going to now take this formula and say that a is, a,

b, c, are those components, which I got from the curl.

So now here's what it's going to look like.

So, b, now, dP dz, minus dR dx.

dQ dx minus dP dy.

OK?

Check that I didn't shortchange you.

This is the formula, right.

I just take each of the three components -- so the easy way

to check is that it should be yz, yz, this should be zx,

zx, this should be xy, xy.

And then it should be dR, QP, RQ.

So this is what I get.

So this is what this expression stands for.

When I integrate this, I'm integrating this expression.

This is the left hand side.

This is the left hand side more precisely if I put now the

double integral over m, this is the left hand side of

Stokes', of Stokes' theorem.

And now I want to write the right hand side, and the right

hand side of course we know, we've known for a while.

The right hand side I recall the dR is dx i

plus dy j plus dz k.

And so f dR -- and f is PQ -- I should have said it, but f

is PQr, but this is what I was using.

So, f is Pi plus Qj plus Rk.

So, f dR, is, of course, then just P dx plus Q dy plus R dz.

So the right hand side is the integral over the boundary of

M, of P dx plus Q dy plus R dz.

This is actually a familiar expression for line

integrals, right?

For line integrals, we've known this all along, that

line integrals could be written in this way.

If we did it often in two variables then it would

be just P dx plus Q dy.

For three variables, it's P dx plus Q dy plus R dz.

So what we are saying now or what Stokes and others have

told us is that this integral over the boundary is equal to

that integral over the surface.

So how can relate these two expressions?

So there is actually a very simple way, which I already

explained in the context of Green's theorem, and this

really holds the notion of differential.

So remember the notion of, remember the notion of

differential of a function.

The notion of differential as a function was like this.

If f is a function, then the differential df is df dx dx,

plus df dy dy, plus df dz dz, right?

Remember this?

This is the differential.

And now I would like to apply the same operation, d,

to this expression.

So now let's apply -- This is the right hand side of Stokes.

So now what I want to do is, I want to apply d to

P dx plus Q dy plus R dz.

What am I going to get?

The rule is very simple.

Just apply it to the function which stands in front the same

way you would apply it here when you can calculate the

differential, and just the multiply by whatever remains,

for example, dx here.

So let's apply it to the first one.

What are we going to get?

We're going to get -- let's apply it to this.

So let's think it means that I take the differential of

P -- let me do it slowly.

So, dP dx, then plus dQ dy plus dR dz.

So what is dP?

dP is dP dx, dx, plus dP dy, dy, plus dP dz, dz.

And now I have to multiply this by dx.

Let me just put it here.

So I'll have dx dx, dy dx, dz dx.

This is what the first term will give me.

I just take the differential dP and I multiply by dx, so

there is always this dx.

But otherwise it's exactly the terms that I get

in the differential.

Is that clear?

Any questions?

That's the expression.

Let me do it a second time. dQ -- so it's going to be dP dx

dx, plus -- sorry, dQ, plus dQ dy dy.

Plus dQ dz dz.

Let me open the brackets.

When I open the brackets I have to put dz here, dz here -- and

here I have two dz's actually -- dy, right, sorry, I'm

already jumping to the last line, dy.

So now the last line. dR dx dx, plus dR dy -- let me leave some

space. dR dy dy, plus dR dz dz.

Let me open the bracket.

dz, dz, dz.

OK, now we're almost there.

So first of all, if you have that same variable twice,

you have to get rid of it.

I already explained once, because if you have two

coordinates which are not independent from each other,

you can not think of them as two new coordinates.

In other words, the area of the parallelogram

which the span is 0.

So this has to go, this has to go, and this has to go.

So I have nine terms of which only six terms will survive,

and I claim that these six terms will be equal to this,

to the six terms here.

So let's check that.

Let's start with xy.

dx dy.

Where do I have dx dy?

I have dx dy here, and I have dQ dx, and I have dy dx.

But remember as we discussed before, if you have dy dx,

it's the same dx dy, but you have to put a minus sign.

So you get minus dP dy.

dQ dx minus dP dy.

Voila.

I hope the rest works as well.

I'll return to this issue of switching and getting the sign

in a minute, but I just want to convince you that we

get the right formula.

OK, next is the dy dz.

So here is dz dy, here is dy dz. dR dy.

dQ dz, but we have to swtich them here so we get the minus

sign, which is correct.

And finally, dz dx, so this is dz dx and this is dz dx. dz dx.

You get dP dz.

And here you get them in the wrong order, so you have

to put a minus sign.

Minus dR dx -- it works.

So, this very complicated formula, this very complicated

formula, which I got by this very long and tedious

calculation is nothing but z of P dx plus Q dy plus R dz.

This is what it is.

And now I can rewrite Stokes' formula in a much nicer way,

in a much nicer way by using this operation.

Namely, the left hand side is just P dx plus Q dy plus R dz

over the boundary of M, and the left hand side is the integral

of d of this P dx plus Q dy plus R dz over M, where the

differential, d, is understood in the way I have explained.

Right, so this is what it boils down to.

When certainly now it looks much more beautiful, there's no

curls, there are no adhock definitions of integrals

and stuff like this.

You've seen a clear pattern.

Let me call this omega.

And this is the integral of omega over b of M, and this

is the double integral of d omega over M.

Where d now acquires very concrete meaning

as this operation.

So this is the general principle which I have been

invoking throughout this lecture, and I told you that I

will explain that all of the formulas that we are studying

in this course, in this part of this course, can be thought of

as just special cases of this general formula.

And now I can make it much more precise in this particular

case, and you could see that this is how it works.

Any questions about this?

Yes?

Do we have to worry about simply connected regions?

Do we have to worry about simply connected regions.

At this moment, no.

So the simply connectedness shows up when we try to

determine whether a given vector field is conservative,

right, whether on the plane -- given a vector field you try to

see whether it is conservative and then you want to find the

potential function, so that moment we have to

worry about it.

But this issue of simply connectedness is sort of

one dimension lower.

It has to do rather with some [UNINTELLIGIBLE WORDS]

theorem for line integrals than Stokes' theorem, because in

that issue there was no, it was about finding line interval.

So here, no, the short answer's no, we don't

have to worry about that.

[INAUDIBLE]

Any orientable -- smooth orientable surface

was a boundary.

[INAUDIBLE]

That's right.

This is a very powerful equation.

Now, this derivation that I have given you, you're

not responsible for it.

In other words, it's not going to be on the exam and so on.

I just wanted to explain for you so you can understand a

more conceptual way of viewing and appreciating this formula.

And I think that really clarifies a great deal.

Now, going back to what you have to know and what you are

responsible for, I want to give you an example -- I want to

give you an example of the application of the Stokes'

formula, which is very similar to what you will have on your

homework for this lecture.

So example is actually from the book at 16.8.8.

So you are given the vector field f, which is e to the

negative xi plus e to the xj plus e to the zk.

You're asked to evaluate line integral of this vector field

over a triangle, which is given by -- which is a part of that

plane 2x plus y plus 2z equals 2, which is contained in the

first octant, which is like for positive x, y, and z.

So what it looks like it's like this, here you have 1, here

you have 2, and here you have 1, and it's this triangle.

This is a triangle which is -- this is triangle intersection

of this plane, this plane, with three coordinate planes.

You see because for example you can put -- if you put y and

z equal 0 you get x equal 1.

If you put x and z equal 0, you get y equal 2

and then z equal 1.

So this is xyz.

So you're supposed to evaluate this integral, which is fine.

It's not too complicated.

What you need to do is you break it in half -- you have

to say which orientation.

So I have to be careful, so the orientation is

counterclockwise, like this.

So you can evaluate by breaking into three pieces by

parameterizing each of these line segments and then just

doing this line integral, but it's going to take a while,

because you have to do three different integrals, you have

to parameterize, so there's some work involved.

So what Stokes' theorem does, it gives you

a shortcut to this.

It allows you to evaluate this integral in a much more direct

way, because you realize that this curve is actually the

boundary of this surface -- this will be our M.

And so by using Stokes' theorem you can actually rewrite this

as an integral over M of curl of f dot ds.

So, I'm not going to use any of this fancy machinery, I'm just

going to use the formula the way it's written in the

book -- curl f dot ds.

So, instead of using this integral, we might as well do

this integral, and in fact it is a very good idea because

curl f is not exactly 0, so that would be like maybe the

first example for you would be to write this in such a way

that the curl is 0, and then you actually say OK,

so you don't have to calculate anything.

So in this particular case it's almost 0, there

is one component.

If I calculate it correctly it is e to the x times k.

Right, because this one you take derivative with respect to

y and z, here you take respect to x and z, so this is what

gives you that, and here you take, there is an x and

y, so they disappear.

So it's almost 0.

If it were 0, you just get 0.

So that's already -- no matter how complicated this curve is,

if the curl of the vector field is 0, you will actually get 0.

So that's a wonderful application of this result,

where you can calculate the integral very quickly without

doing any calculation.

Here it's almost 0, so what we need to do is we need to

evaluate the surface integral of this vector field.

OK?

So for this we have to choose -- we have to use the formula.

And the formula is that in this case you can say that this

plane is a graph of a function, you can write it as z equals

1 minus x minus 1/2 y.

Let's call this g of xy, and then we know that if you have

a vector field E, which is A, B, C.

Or maybe let's go with P, Q, R, because I think this is

the way I wrote it before.

Then the surface integral of E dot ds in such a case is equal

to double integral over the base, over the domain in x and

y, where here you put P minus P dg dx, minus Q dg dy plus R.

Right?

Remember this formula?

So in this case you only have, P and Q are 0, so you only have

R, which is equal to E to the x.

And so you end up with a double integral of E to the x, dx dy

over the triangle on the plane, which is the projection of this

triangle onto the xy plane.

So it has x-coordinate 1 at this endpoint, and

y-coordinate 2 at this end.

And this is your D.

So then this is very easy to calculate.

So I'm actually out of time so I stop here, and we'll

continue on Thursday.

How are you guys doing?

Did you have a good Thanksgiving break?

Yes?

Are you ready for the last week of classes?

Yeah, OK.

It's hard to believe it is the last week, but actually next

week I will give a bonus lecture, I will give a -- I'll

give a review lecture on Tuesday next week.

And also I remind you that there's a lot of materials for

the final exam, which are available on bSpace, and also

on the class home page, which I hope by now you

know how to find.

All right.

And now we go back to the study of Stokes' theorem.

So we've been slowly laying the foundations, the necessary

foundations for Stokes' theorem and finally, it's time to sort

of reap the benefits of all of that, and finally today we'll

be able to state the result and to see what it means, and what

the applications of this result are.

But first of all, I would like to revisit the material which

we talked about last week, OK.

A week ago we talked about -- we talked about surface

integrals, integrals over the surfaces, OK.

And the point was that actually we studied two

types of integrals.

Two types of integrals.

Over general surfaces in the 3-dimensional space, as a good

working example of such a surface, we'll look at

the upper hemisphere.

It's a surface with a boundary, which is a circle.

So, you can actually, you can think that there is a -- it's

actually a arc of a sphere which is centered at the

origin, so you could also imagine that there is a

coordinate system here, xyz, and so the surface, what I'm

talking about is the this upper hemisphere, the surface of the

upper part of the surface of the sphere, OK, which has a

boundary which is a circle.

So what we're trying to understand is how to integrate

over such surfaces, surfaces of this type, and relate integrals

over such surfaces to integrals over their boundaries.

That is the point of Stokes' theorem.

So let's denote this by m, and m stands for membrane, because

I explained last time that integrals over surfaces of the

second type can be viewed as fluxes over fluid

through membrane.

But here I will recall that there are two types of such

integrals that we have looked at.

The first type -- the first type you pick a function and

then you integrate over m this function with a measure

which we denote by ds.

We talked about the interpretation of this -- this

has to do with the surface area, and more generally

with the mass.

So this represents the mass of the surface, think of the

surface of aluminum foil, say, and f is the mass

density function.

And when you integrate the density function, you

get the total mass.

It can also be a charge density function, then you would be

integrating total charge.

So that's the first type.

The important thing here is you integrate a function.

And the second type is when you start with a vector field.

So it's not a function, it's much more than a function

-- actually, we represent it by triple functions.

It has components P, Q, and R, where each of these three is a

function of its own, in its own right.

In this case, we also have double integral, which

we discussed last time.

And that's what's called a flux of a vector field, or a surface

integral of a vector field.

Which we denote like this -- e dot ds.

So this is called, sometimes called the flux of e

through this membrane.

And it represents the rate of flow of a fluid through a

membrane m if f is a velocity vector field for this fluid.

I explained last time.

And now, of course, even if you were not here, you can actually

watch the lecture on multiple places, which is good -- I

think it's a good resource.

OK, so it makes it easier for me to just say

I did it last time.

If it was not recorded, that would be a little unfair, but

now I know that if you don't remember, you can always go

back and watch it again.

Now, how to compute this.

So the key to computing this is to, as well as the other

one, is to parameterize the surface, right.

So parameterize it, which means that we write x as some uv, and

y as some y of uv, and z as some function, z of uv where u

and v are in the -- belong to a certain domain now

on the plane.

Parameterize.

When I say parameterize it, I mean parameterize m.

And so when you parameterize it, you actually get a

very explicit formula for this integral as just a

double integral over d.

So, we get an integral, which is now in the realm on the

previous chapter of this course, something which

we studied before.

OK?

Now the key to this is the following, that the way we

introduce this integral -- we actually wrote it as an

integral of the first type, wrote it as an integer of the

first type over m where what we do is we take the dot product

between e and n where n is the unit's normal vector

field for the surface.

When we do that, when we take the dot product, we convert our

vector field into a function.

This becomes the function and that function we can now

integrate in the sense of the integral of the first type.

So this is how we introduced it.

Today I will explain a different way to think about

this where you don't have to use the concept of the first --

our integral of the first type to understand the integral

of the second type.

But this actually, this brings out the question as to what

this m is, OK, this normal vector.

So in my picture -- let me draw this picture again.

In my picture, this vector field, m would look like this

-- at each point of the surface we would have a vector which

kind of sticks out, and which is normal -- normal means

perpendicular, perpendicular to the surface.

If you have a surface and you have a point on the surface,

you can talk about the tangent plane -- right, we talked

about this, tangent planes.

We even know equations for tangent planes, how to find

equations for tangent plane.

So we also have the notion of a normal line.

A normal line is perpendicular to the tangent plane, right.

So at this point you have a tangent plane -- I'm not going

to draw this, just not to mess up the picture, I don't

want to make it too heavy.

But there is a tangent plane here, right, but it is

also a normal line.

So it's perpendicular to the tangent plane.

When something's perpendicular to the tangent plane at the

point, we might as well say it's perpendicular

to the surface.

What else can it mean but to be perpendicular to the surface.

It just means being perpendicular to

the tangent plane.

On this line there are many vectors.

If you have one vector on this line, you can pick an a

multiple of this vector.

So what we'd like to do is we'd like to specify a normal vector

at each point, and we want to specify it with as little

ambiguity as possible.

So the way to get rid of the ambiguity, get rid of the

redundancy is to normalize the vector, to make

it a unit vector.

If you have any vector, it has certain magnitude -- unless,

you know, if it's not 0, you can't divide by the magnitude

and you get a vector of length 1, or magnitude 1.

Such a vector's called unit vector.

So, out of all the vectors along this normal line, we can

say let's take a unit vector.

The problem is that when you do that you still have a

small ambiguity, it is a much more ambiguity.

Instead of instantly many vectors, you end up with two --

there are two possibilities.

There is this one, but there is also one which goes in

the opposite direction.

So it's kind of a mirror image of this one.

And just by saying it's a unit vector, we can't

kill this redundancy.

Both of this are bona fide unit normal vectors for the surface.

If you think about this upper hemisphere, one of them sticks

out, goes outside, and the other one goes inside.

So that is two choices at each point -- at each point

there are two choices.

And what we'd like to do is we'd like to make a choice at

each point along this entire surface so that this choice

is continuous, smooth.

In other words, we don't want to be in a situation where we

choose a unit vector at this point to go outside, outward,

and then at the nearby point to make it go inside,

that wouldn't be smooth.

So when we talk about normal vector fields, we mean that

we want it to vary in a smooth way.

So that's what this m Is, m this formula.

This m is the unit normal vector field.

And then I will emphasize very smoothly along m.

That vector field is called orientation.

I would like to contrast this to the notion of

orientation of a curve.

In the case of a curve, if you have a curve going from point a

to point b, we talked each orientation as a direction

along the curve.

And if the curve has boundary points, specifying orientation

is the same as specifying which one is the initial point and

which one is the end point.

So here, this would be the initial point, this

would be the end point.

If the curve does not have a boundary, for example, a

circle, we could say it's counterclockwise or

it's clockwise.

But let's look at this one which have end points,

it has a boundary.

In this case, by analogy to this, we observe that actually

in this case there are also two possible orientations -- we

can go this way or we can go th that way.

So there's a lot of similarity with the notion of orientation

for curves and surfaces.

The main difference is that orientation for a curve

goes along the curve -- you go with the flow.

And here, you also go with the flow, but the point is that the

flow goes across the surface usually.

You can't flow along the surface.

Suppose you can maybe, but it makes much more sense to

flow across the surface.

As I explained you could have a pipe, you know, and this would

be a membrane, which kind of covers this pipe, and you would

be measuring the amount of fluid which passes through it.

So it's much more natural to talk about orientation

across rather than along in the case of surfaces.

Unlike curves where we should be going along

means a lot more sense.

So this is one important difference between orientation

for curves and surfaces.

There is a similarity, right -- there's a similarity that's in

both cases, two choices, so here there are two choices.

Here there are two choices.

And in this case, there are also two choices.

One indicated in white, and one indicated in red.

And if you think about it once you choose it at one point,

then basically there is no choice at nearby points,

because you have to very smoothly, so you know nearby

would be like this, nearby would be like this, like this,

like this, because you don't allow jumps.

If you don't allows jump, once you know it's at one point,

once you specified at one point of your surface, you

know it everywhere.

But then the question is can we actually define it everywhere

in a consistent way.

And this is, here we kind of get surprised because in the

case of a curve there are always orientations.

For any curve you can always choose two orientations, one

goes this way, one goes that way.

But in the case of surfaces, it actually exists surfaces which

have none, which don't have any orientation.

And this Is actually kind of cool.

You must have heard about Meryl Streep, right.

Have your heard about Meryl Streep?

Yes, OK.

So here is a little demonstration for you.

So, we need a close-up.

Sometimes I'm delayed so that's why I'm saying it and I'm

waiting for a few seconds.

So here is a surface, I mean this is a surface, and I

colored it in two ways -- on one side it's white, on the

other side it's pink, OK.

So now what I'm going to do is I'm going to make a surface, I

could glue a surface out of it in different ways.

So for example, I can glue it in this way.

If I glue it in this way it's like a cylinder, it's like a

little cylinder -- not so, I mean kind of very low height.

And in this case if you look you could see that actually it

clearly has two sides, there's an inner side and there's

an outer side -- one is pink and one is white.

So when you talk about orientation you could also

think in terms of choosing the side of the surface.

I'm thinking about vectors sticking out, but once you have

a vector sticking out, you have the notion of what's

outside and what's inside.

So let's say that if it sticks out this way then that's the

outside of surface, and if it sticks out this way this would

be the outside and that would be the inside.

So it's like, you know, if you put your teeshirt on inside

out, this kind of stuff.

So it's a clear notion, inside or outside of a surface.

So that's what we're talking about orientation is saying

which side we consider as inside and which one you

consider outside, because, of course, after all it's not

clear, you can choose -- you can make your choice, and

sometimes you can have a teeshirt which actually looks

a little better if you wear it inside out.

It kind of looks cool.

But imagine you have a shirt, which only has one -- a

teeshirt which only has one side.

So that would be difficult to wear.

Now, but actually you can create such a, not necessarily

a teeshirt, but you can create such a surface very easily by

taking this thing strip and gluing the ends kind of by

twisting by 180 degrees.

If you do that you say OK, so let's say that this is the

-- this is the outside.

But then you trace it, if you trace it, the outside, you

trace it, you trace it, and then suddenly it becomes

white, and when you make the full circle you come back

on the opposite side.

So if this was outside, outside it should be the notion that if

we follow it -- you know, on my shirt, if I follow it

everywhere, it's going to be on the same side always.

I will never be able to jump on the other side, one

would hope anyway.

And that's the way it's supposed to be, right.

But here you go, you make a full circle and you come

back on the opposite side.

So that means this is outside but also this is outside.

So that means -- but then which one is it?

So it means at this point it's not defined.

There is no global notion, there's no global notion

of inside and outside.

You see what I mean?

So in this case, we will say that the surface is not

orientable, and this is really remarkable phenomenon.

Again, in the case of curves, all curves are orientable, it's

just that there is a redundancy -- there are two possible, or

ambiguity, there are two possible orientations, you

have to make a choice.

For surfaces actually we don't even have that luxury.

There are many surfaces which are not orientable for which

you could not consistently say which is this inside and which

is outside, and that's example.

Any questions?

So in this case, if the surface is like this,

there is no orientation.

There is no unit vector that you can make which varies

smoothly and gives it a normal vector at each point, because

if I were to give such a vector, let's say again, here,

the vector will be sticking out this way, but if I varied

smoothly, I would be forced to come out here on this side.

So that means it becomes ambiguous, and at this point

there's no -- it has to be simultaneously this and that,

and that's not allowed.

So it does not have a consistent orientation.

So, in the case of such a surface, this integral is

actually not well defined, you can not set up an integral of

vector field over a surface which is not orientable,

which does not have this type of orientation.

In other words, the consistent notion of inside and outside.

So we have to be aware of it.

Now, the good news is that most surfaces are orientable,

most surfaces that are of interest to us.

For example, if we are thinking about really a fluid flowing in

a pipe and you're thinking of something which is like a cover

at the end of the pipe, it is going to be orientable.

So, we're not going to be too handicapped by the fact that

for non-orientable surfaces this is not well-defined.

All the surfaces we will consider in this class for sure

will be orientable, but we have to be aware of the fact that

actually we should not take this for granted.

That actually a surface that may not have a particular, even

one orientation -- of course, if it has one, it will

have to have two, right.

The point is that if n is one of them, the negative n will

also such an object, also such an orientation.

So it can't be that you have just one orientation -- you

either have none or you have exactly two, always.

And in all of the cases which we will consider here, our

surfaces will have two orientations.

And this will give us a little bit of a headache, because when

I write now a Stokes' formula, I will have to be very precise

in specifying the orientations on the left and right

hand side, so as to make the equation correct.

So, let's go back to this.

This integral.

So the integral is actually -- let's suppose that

it is orientable.

So this is important.

It's not a given that the surface is but let us

assume from now on that m is orientable.

That is to say m has an orientation vector field, m.

And then it has necessarily two of them -- it has two

orientations, n and negative n.

OK, let's assume that.

And so pick one, we'll pick one orientation, which

is what I call n.

In this case we have this integral, we have

this double integral.

Like of this, so you'll have e dot ds, which can be defined by

the formula here, I just rewrite it, e dot n, ds.

Now, there are two choices, so if you take the second choice,

the second choice means instead of m you take negative n.

So, if we choose negative m, we just get minus

of what we had before.

If we choose the other orientation, this will just

result in overall size.

So in that sense you can say well, it's not

such a huge ambiguity.

You don't get sort of a totally different answer, you get

something up to sign.

But it wouldn't be serious if we, in this course we said

all our results were defined up to sign.

We have to be precise, we have to be able to say is it 10 or

negative 10, because sometimes your life depends on it to know

it's positive or negative.

So that's why we take this seriously and we say the

definition of such an integral has to include orientation.

So if you do a homework problem, it has to say the

integral vector field over the surface with respect

to which orientation.

And then there is standard terminology.

Standard terminology is you can say orientation is upward.

Upward means that the vertical component of

vector field is positive.

So the white orientation on that picture is upward.

Or it could be downward, that would be the red one.

And oftentimes if we have a closed surface, for example,

not the entire sphere, but say so it's not the upper

hemisphere but the entire sphere, you could say outward

or which also has a clear meaning.

OK, so let's pick one.

How do we actually compute this?

We compute it by parameterizing m by choosing two existing

parameters u and v, and then when we parameterize it, so

this integral becomes -- we can write it very

concretely as follows.

It's e dot ru cross rv over d, where d is the parameterizing

domain on the uv plane.

On the plane of these coordinates u and v.

What is r?

So, r is a vector which we get by combining the three

functions which parameterize our curve -- x of uv i, plus

y of uv, j plus z of uv k.

So these are the three functions.

You put them together into vector field.

r sub u simply means that you take derivatives with respect

to u -- maybe you write dx du i plus, and so on.

And likewise r sub v means that you take dx dv

i plus, and so on.

So once you have your parameterization, once you have

your parameterization, you have this r, and you have r sub u by

taking simply partial derivatives of these three

functions with respect to u.

And you have r sub v.

So then you can just take the cross product. and when you

take the cross product you get another vector field, which you

can then dot with your original vector field.

This will give you a function which you then integrate over

the parameterizing domain.

And, of course, the point here is that both the original

integral and this integral are integrals over surfaces, but

there's a huge difference.

This surface is actually on the plane, it's flat.

So, by this formula, we express a complicate integral over a

very complicate surface, which could be very curved like a

sphere or part of a sphere as an integral over domain on

the plane, which is flat.

So that's the definition which we talked about last time.

And that there is some special cases where, for example, your

surface is a graph of a function, we have some -- the

formula simplifies and we have various formulas which kind of

give us a shortcut to the answer.

So finally, we are ready to state the Stokes' theorem by

using this notion of a surface integral.

So let me draw this one more time.

So you have our surface m and we have the boundary, the

circle boundary, which I'll call b of m.

So, on the left hand side of the Stokes' formula, we will

have a line integral over vector field over

this boundary.

So we'll have a vector field, f, which will be defined in a

three-dimensional space, in particular, will be defined

on this entire surface.

And on the right hand side, we'll simply take the line

integral of this vector field over the boundary.

So this is something we learned before, so I don't have

to say much about it.

This is just has become by now the usual thing, the line

integral of vector field, which is again, to calculate it we

need to parameterize the curve of and then there's a very

simple formula which allows you to take this integral.

And now on the left hand side, we're going to have an integral

which is a double integral and it's going to be an

integral over m.

And then the integral, it will be the integral of another

vector field, and that vector field will be nothing

but the curl of f.

So you have your vector field and it appears on

the right hand side.

But on the left hand side you have another vector field.

This is something which I was denoting by E in all my

previous formulas, so we have double integrals of a vector

field ef over a surface.

But now take as e the curl of your original vector field

which was given to you from the beginning, f. f is

given, take its curl.

That's another vector field.

And take the integral of the curl.

And so the markable statement is that you always have this

equality, you have this equality of double integral

and the single integral over the surface and over the

boundary of the surface.

Now, what I have said is still incomplete, and the reason it's

incomplete is that both of these integrals depends on

the choice of orientation.

Did you have a question about this?

[INAUDIBLE]

The curl f would be what I call here e.

[INAUDIBLE]

So, I purposefully called this vector field e, and I called

that vector field f because on the left hand side when we do

the double integral, we're actually integrating

not f but curl f.

So that's why I'm using a different notation here so

as not to confuse you.

We are given the vector field, f, but we are sticking that

vector field on the right hand side into a single integral,

into a line integral, right?

Whereas on the left hand side we do curl, and curl is

another vector field which we know how to calculate.

And then we'll actually see what the meaning of that

curl is we'll get a better explanation of what the meaning

of the vector field is.

OK, so that's the formula, but as I said, on both sides we

have two integrals which, as I explained, requires the

choice of orientation.

So, in other words, depending on how I choose orientation on

m, I want either get -- I will get plus minus some answer.

I will get some answer or it's negative, and if I want to pin

it down, I have to say which orientation I should take.

Both integrals are well-defined.

There is no canonical choice of orientation, right?

There's no canonical choice of orientation.

So there are two possibilities, and the two possible answers

they differ by sign. but I have to say which one I should take

to make this well-defined.

Likewise here, I also have two different answers.

In the case at hand, I have this circle.

I can choose orientation which goes counterclockwise or

clockwise, if you look from above.

So again, there are two different answers

which differ by sign.

If I'm saying that these two things are equal to each other,

I better be precise and say how the orientations should be

defined so as to make this into an equality, because if I make

the wrong choices, instead of getting this formula, I'll get

this equal to negative of that.

In other words, if I wanted to be careless about orientation,

I would not be able to state this formula at all.

The best I could do I would be able to say this is equal to

plus minus, and if I did that I should be fired.

So, we don't want it to happen -- well, maybe

you do but I don't.

So that's why I'm going to explain how to choose

orientation so that we don't have this ambiguity.

And of course, the point is, you probably guess already

because this is very similar to the way, to the problem which

already arose when we talked about the Green's theorem,

which is actually kind of a little brother of the Stokes'

theorem we'll talk about in a minute.

In Green's theorem the results of this issue of orientation,

and the way we handled that was by saying -- was the following.

So, in Green's theorem, we also have a two-dimensional

region, but it was flat.

This is flat, this is not flat.

Well, I have drawn it on a flat blackboard, but 3D, you know.

I was just reading about this new movie, Avatar, it's going

to be in 3D, so I was thinking maybe some years from now

lectures like calculus will be given also in 3D, so you will

all be wearing these cool classes and you will actually

see the -- i'll will be able to draw 3-dimensional surfaces.

I should talk to Cameron about this.

I'm sure he is a huge fan of multi-variable calculus.

Or if he isn't he should be because he uses a lot of

special effects, and of course, how can you do special effects

if you don't use calculus.

Anyway, so in Green's theorem you have a region like this.

And the way we handle this is we said once you have this

region, the orientation you define on the boundary is such

that when you walk on the boundary, the region

is to your left.

If you walk on the boundary the region is to your left.

So it would be like this, it would be counterclockwise

in this picture.

So, I would like to say the same.

I would like to say that if I walk on the boundary, if

somebody walks on this boundary then the region should be along

orientation, with respect to the orientation.

The relation should be such that as somebody walks along

this boundary, the region itself, the surface

is to the left.

But the question is how do you expect that person

to walk on this?

There are two possibilities.

It could be that they walk like this or they could be walking

from the other side of the blackboard.

A priori there is no choice.

I mean there are two choices of apriori, there is no canotical

choice, there is no given choice.

In the case of a plane, we do have an obvious orientation, we

do have a special orientation, and the reason is the

following, because there is an orientation which kind of

sticks out towards us.

An other way to say is that there is an orientation

which we call k.

So we draw this coordinate system, which is i, j and then

there is a k which is obtained by the Corkscrew rule,

the right hand rule, whatever you call it.

Remember this picture?

This was one of my fondest memories of this class, the

discovery of this visual aide.

This is how you show the normal.

So the point is that you can't just say it seems like I

can actually move this.

You can already appreciate the problem in giving

it orientation.

The orientation could be like this or it could be like this.

But the point is when we actually draw something on a

2-dimensional plane, we look at it from a certain direction.

We have to be looking at it from a certain direction.

We either are on this side and we're looking on this side, or

we are there -- and let me tell you, there's not a whole lot of

space behind that board so we should be on this side.

Once we are looking at it from this side, we can give it

any coordinate system, any orientable coordinate system.

In fact, this is not even canonical either, but I

can rotate it and so on.

But whichever coordinate system, xy, I choose, the cross

product of the i and j will be looking in this direction,

will be sticking this way.

So that's why when we are having a region on the

plane, we actually have a canonical orientation.

That's what we call vector k, k being the cross product of i

and j in this order -- first x -- this is a unit vector

in the x direction.

This is a unit vector in the y direction.

We take the cross product in this order, and so that's what

gives us the z-axis, or the unit vector along the z-axis.

So we're kind of lucky in this case, we don't we

have to make a choice.

The choice is made by the fact that we're actually absorbing

it in a certain way.

We are looking at this blackboard in a certain way.

But if we have now a surface in the 3-dimensional space, for

example, I could just put this blackboard somewhere in the

middle of this class, and some of you will be on one side and

some of you will be on the other side.

So some of you will be looking from one direction, some of

them will be looking from the opposite direction.

So some of you will be one orientation, for some of you,

it would be natural to take the other orientation.

You see, so in general there is a choice of orientation.

But once you make that choice, if you make it like this, you

can't say if a person walks so that their head is -- points in

the right direction, up, then the domain is on the left.

Then it actually makes sense.

So, somebody could be walking like this, their head would be

this way, right, and so they will have the region on their

left, so that would be we'd get the old rule.

If, on the other hand, I will just say let's choose the other

orientation, which goes in that direction, I would still have

the notion of somebody walking along the boundary, but that

person would be on this side, right, and then it would still

make sense to say that the region is to their left, which

would then mean the opposite direction.

Well, you'd have to really trust me that this

is how it works.

You can just imagine it, if it were made of

glass you will see it.

So, the rule in general is like this.

If the person is walking along this pointing in the direction

of the orientation, if this is the orientation and you are not

upsidedown, you have to be oriented in the right way, your

head should be above your feet basically, with respect

to this orientation.

I'll try to not to make too many jokes about this, but I'm

sure you can do them yourself.

If you walk this way, the surface should be to your left,

so this is how it works.

And if I chose a different orientation, which would be

inward, then I would have to draw this little fellow

pointing this direction and then it would have to walk

so that the region is to their left.

So that's the rule.

Is that clear?

So, it's pretty clear.

So that's the rule I choose, and if I choose that rule,

which it will give me a consistent formula, it will

give me a choice -- whatever choice I take here, once they

make the choice orientation M, I get the orientation on

the boundary as well.

And it is for these two choices that these two integrals

will be equal to each other.

So that's a statement of Stokes' theorem.

So, the next thing I would like to do is I would like to kind

of try to demystify this formula, because at the moment

it looks pretty bizarre in a way because you have this

strange -- you have two different types of integrals.

So one of them is of the vector field, the other

one is this curl, OK?

So would like what I would like to do is I'd like to explain

how this formula actually fits in this general principle,

which I talked about earlier of which we have now seen

several examples.

So I would like to convince you that this formula is actually

special case of this general principle, just like the

fundamental theorem for line integrals and the

Green's theorem.

So that's the next thing we are going to do.

So, in order to explain this, I would like to actually give a

slightly more besides formula for a surface integral, so this

is going to be a slightly long and somewhat boring

calculation, tedious calculation, but bear with me

because at the end of the calculation I hope that I will

be able to demystify this, demystify this formula, and you

will see that really it is part of a pattern, which is much

more, how should I say, much more convincing.

So the first thing I'd like to do is I want to go back to

general surface integral.

So here, E doesn't have to be the curl.

It doesn't have to be the curl.

It could be some vector field.

And I would like to write a slightly more explicit formula

for it, assuming that I have a parameterization.

So, let me recall what I wrote on the top, in the

top right corner there.

I wrote that this is equal to the integral over z, the

parameterizing domain, of e dot ru plus rv.

Where ru and rv you have -- well, the beginnings of ru and

rv you have on the blackboard.

So now I want to actually calculate rv plus rv for you.

So I'm going to calculate what ru plus rv is.

So this is just they use a cross product, this is just

they use a cross product where I put dx, du, dy, du and dz du

in the first row, and I put dx dv, dy dv, and dz dv

in the second row.

So now I just open, I just write down what it means.

So let me just start with I, so this is going to be dy du, dz

dv minus dy dz, dz du -- this is times i.

Maybe here plus, so now j, so I have to look at this, and I

have to remember this sign.

So it's going to be dz du -- it's going to

take a little time.

But you will like it when it's over, trust me,

so it's worthwhile to do this calculation.

dz dz, dx du, j.

You have to do it once to believe it and then after this

a lot of things become simpler.

Plus finally, I have to do the k, and the k is dx du, dy

dz, minus dx dz, dy du.

Check this.

Are you with me on this?

This is good?

If you agree then, you know, you will not have a chance

to disagree with me later.

It's your last chance.

So, if we agree on this, have you ever seen

this formula before?

You're under oath, so have you ever seen this formula before?

What is it called, this kind of expression.

It starts with a J..

It's Jacobian.

It is a Jacobian for the coordinates y and z with

respect to u and v.

This is a Jacobian for the coordinates z and x with

respect to u and v, and this is a Jacobian for x and y

with respect to u and v.

Everybody agrees with that?

So, I actually can rewrite this in a more compact way by using

the notation which we introduced when we

talked about Jacobians.

OK?

So, the notation for the Jacobians is d -- you put the

dell like this, and you put y and z over dell of xy -- uv.

This is a notation for the Jacobian, which we have used

extensively about a month ago.

So this is times i.

Plus we have the Jacobian for zx over uvj plus the Jacobian

over xy uv, and then you put k.

So, this is actually the explicit formula for this cross

product, and this cross product answers in the definition.

So now let me actually take the dot product so that I

get the entire expression.

For this I have to also say what is e.

So let's say that e is ai plus bj plus ck, and what do I get?

I get e dot ru plus rv -- which is my integrant in the surface

integral -- is a times the Jacobian plus b, OK?

So that's the formula.

Here is actually explicit formula for what you should

integrate when you take double integral of vector field, an

explicit formula, which is written entirely in terms of

the components of your original vector field, and the Jacobian

of the coordinate change from iv to all possible pairs of

coordinates amongst xyz.

Now, how can you remember this without memorizing the formula.

It's very easy.

So first of all you have abc, right, these are the components

of the original vector field.

And then you also have xy and z, the z coordinates.

So you have to remember the structure of the formula.

There are three terms, each term responds to one of the

three components of the vector field, and then you have to

remember which Jacobian to put -- you see there are three

different Jacobians.

but there is also subtle point, which is in the Jacobian you

also have to remember in which order you put the coordinates.

You may remember from doing double integrals by using

general coordinate changes, that if you switch the

coordinates, the Jacobian gets the minus side, right -- the

Jacobian gets the minus side.

So here it's important that I put yz over uv

and not zy over uz.

So first thing you note is that if you take a, the variables

which get involved -- so a responds to x.

So to get the variables involved, you have

to cross it out.

So you have ayz here, then you have b with xz, and c with xy.

So that the two coordinates which you take which appear

together with the component of the vector field are the

remaining two components.

So a is the x component.

So the coordinates which are relevant to this term of the

formula are y and z, and then you have to ask yourself in

which order, and the order should be what mathmaticians

call a cyclic order, which means this is first, this is

second, this is third, this is fourth and so on, so it

kind of goes in a circle.

So, for example, y and z have to appear in this order, yz.

But z and x should appear -- you're at z, what's the next

one, the next one is the first one -- it's a cyclic order.

So it's like zx.

And then xy is clear.

So xy, yz, zx.

And then you put the remaining component of the vector field.

So that's what the meaning of this formula is.

So that's the first thing we should agree on, the

double integral can be written in this form.

But now let's go back to Stokes' formula.

In Stokes' formula there is an additional complication, as if

this formala was not complicated enough.

We actually have to take the surface integral not of the

vector field itself, but with curl.

So gee, we have to do one more cross product.

But actually, the funny thing is that when you do that cross

product and you combine it with this stuff, which you can

imagine is going to look horrible, right, actually it

will look better in some sense, or you will get something

which will have much more conceptual explanation.

So let's do that, let's do that.

I promised that I would explain it to you, so let

me explain this to you.

So, we're halfway there.

What we need now is we need to take the -- apply this formula

to the left side of Stokes' theorem where we are just doing

it to the curl of the vector field and not to vector

field as before.

So, if E is curl of F, what are, what does it look like?

So we have to again do this -- i, j, k, then we

put d dx, d dy, d dz.

And then we put P, Q, R.

So what do we get?

We get dR dy minus dQ dz times I plus dP dz minus dR dx

plus dR dx minus dP dz.

But please check that I did it correctly.

Because if I mess up one term then it will

not come out right.

[INAUDIBLE]

OK, good.

That's right.

And in fact for good reason because this is expression

of Green's theorem.

So we know that Green's theorem actually is a special case of

this when only this term shows up.

Dz dx and dz dy exactly, good.

I think now it's correct.

So now I want to put this in here.

In other words, this will be my a and this will my b,

and this will be my c.

So you would think this would look awful, and in some sense

it will, but then I will explain another way to get that

expression in a much more, in kind of a much more

conceptual way.

And this will, at least for me, and I hope for you, this will

demystify this expression and this result.

OK.

So let me put this into this formula.

And I really want to do it in real time instead of showing it

to you already written before, because if I show you you'll be

like OK, whatever, you know, this isn't some derivative, so

he says it works and it works, but I want you to really

follow me and to really see how it works.

So, dr dy -- so please keep track of what I'm writing,

because as you see, I can make mistakes.

But now I want to do one more thing, actually, to simplify,

to kind of accelerate this lesson.

At the end of the day I'm going to integrate this, right, I'm

going to integrate this -- I'm not just taking this

expression, I'm actually integrating this, and I'm

integrating it with respect to du dv.

So let me actually then multiply this by du dv.

But if I do that, then this becomes A dy dz, right?

Because remember -- let me write it here.

When we did double integrals, this was the formula

which we used.

If we integrate -- the way we did it was for xy,

we did it for dx dy.

But the same would apply if I do it for xz or yz.

In other words, when we integrate a function in x and

y, we said it can be written as an integral over u and v, but

you have to insert this factor.

I'm sorry, this doesn't look right.

What I mean is, of course -- so the way we wrote it is we said

that when you have a double integral function, f, this is

equal to -- what I'm doing is just stripping it of the

integral and move the function.

I'm just saying that this is really the identity which

we use and we proved it.

So what I'm going to do is whenever I have this ratio

because I now multiply overall by du dv, I'm just going

to replace this by dy dz.

I will replace this by dzds, and I'll replace this by dfdy.

Can I do that?

You agree that this is OK?

Legitimate?

Yeah?

OK.

So was the purpose -- I'm just trying to remind you that this

was the purpose of introducing these Jacobians in first place.

Because we wanted to replace things like dx dy by du dv.

And we knew that this was the price to pay.

We had to insert this factor.

So might as well replace this expression by this expression.

And likewise for the remaining two summits.

So, this is a formula I'm going to use.

And I'm going to now take this formula and say that a is, a,

b, c, are those components, which I got from the curl.

So now here's what it's going to look like.

So, b, now, dP dz, minus dR dx.

dQ dx minus dP dy.

OK?

Check that I didn't shortchange you.

This is the formula, right.

I just take each of the three components -- so the easy way

to check is that it should be yz, yz, this should be zx,

zx, this should be xy, xy.

And then it should be dR, QP, RQ.

So this is what I get.

So this is what this expression stands for.

When I integrate this, I'm integrating this expression.

This is the left hand side.

This is the left hand side more precisely if I put now the

double integral over m, this is the left hand side of

Stokes', of Stokes' theorem.

And now I want to write the right hand side, and the right

hand side of course we know, we've known for a while.

The right hand side I recall the dR is dx i

plus dy j plus dz k.

And so f dR -- and f is PQ -- I should have said it, but f

is PQr, but this is what I was using.

So, f is Pi plus Qj plus Rk.

So, f dR, is, of course, then just P dx plus Q dy plus R dz.

So the right hand side is the integral over the boundary of

M, of P dx plus Q dy plus R dz.

This is actually a familiar expression for line

integrals, right?

For line integrals, we've known this all along, that

line integrals could be written in this way.

If we did it often in two variables then it would

be just P dx plus Q dy.

For three variables, it's P dx plus Q dy plus R dz.

So what we are saying now or what Stokes and others have

told us is that this integral over the boundary is equal to

that integral over the surface.

So how can relate these two expressions?

So there is actually a very simple way, which I already

explained in the context of Green's theorem, and this

really holds the notion of differential.

So remember the notion of, remember the notion of

differential of a function.

The notion of differential as a function was like this.

If f is a function, then the differential df is df dx dx,

plus df dy dy, plus df dz dz, right?

Remember this?

This is the differential.

And now I would like to apply the same operation, d,

to this expression.

So now let's apply -- This is the right hand side of Stokes.

So now what I want to do is, I want to apply d to

P dx plus Q dy plus R dz.

What am I going to get?

The rule is very simple.

Just apply it to the function which stands in front the same

way you would apply it here when you can calculate the

differential, and just the multiply by whatever remains,

for example, dx here.

So let's apply it to the first one.

What are we going to get?

We're going to get -- let's apply it to this.

So let's think it means that I take the differential of

P -- let me do it slowly.

So, dP dx, then plus dQ dy plus dR dz.

So what is dP?

dP is dP dx, dx, plus dP dy, dy, plus dP dz, dz.

And now I have to multiply this by dx.

Let me just put it here.

So I'll have dx dx, dy dx, dz dx.

This is what the first term will give me.

I just take the differential dP and I multiply by dx, so

there is always this dx.

But otherwise it's exactly the terms that I get

in the differential.

Is that clear?

Any questions?

That's the expression.

Let me do it a second time. dQ -- so it's going to be dP dx

dx, plus -- sorry, dQ, plus dQ dy dy.

Plus dQ dz dz.

Let me open the brackets.

When I open the brackets I have to put dz here, dz here -- and

here I have two dz's actually -- dy, right, sorry, I'm

already jumping to the last line, dy.

So now the last line. dR dx dx, plus dR dy -- let me leave some

space. dR dy dy, plus dR dz dz.

Let me open the bracket.

dz, dz, dz.

OK, now we're almost there.

So first of all, if you have that same variable twice,

you have to get rid of it.

I already explained once, because if you have two

coordinates which are not independent from each other,

you can not think of them as two new coordinates.

In other words, the area of the parallelogram

which the span is 0.

So this has to go, this has to go, and this has to go.

So I have nine terms of which only six terms will survive,

and I claim that these six terms will be equal to this,

to the six terms here.

So let's check that.

Let's start with xy.

dx dy.

Where do I have dx dy?

I have dx dy here, and I have dQ dx, and I have dy dx.

But remember as we discussed before, if you have dy dx,

it's the same dx dy, but you have to put a minus sign.

So you get minus dP dy.

dQ dx minus dP dy.

Voila.

I hope the rest works as well.

I'll return to this issue of switching and getting the sign

in a minute, but I just want to convince you that we

get the right formula.

OK, next is the dy dz.

So here is dz dy, here is dy dz. dR dy.

dQ dz, but we have to swtich them here so we get the minus

sign, which is correct.

And finally, dz dx, so this is dz dx and this is dz dx. dz dx.

You get dP dz.

And here you get them in the wrong order, so you have

to put a minus sign.

Minus dR dx -- it works.

So, this very complicated formula, this very complicated

formula, which I got by this very long and tedious

calculation is nothing but z of P dx plus Q dy plus R dz.

This is what it is.

And now I can rewrite Stokes' formula in a much nicer way,

in a much nicer way by using this operation.

Namely, the left hand side is just P dx plus Q dy plus R dz

over the boundary of M, and the left hand side is the integral

of d of this P dx plus Q dy plus R dz over M, where the

differential, d, is understood in the way I have explained.

Right, so this is what it boils down to.

When certainly now it looks much more beautiful, there's no

curls, there are no adhock definitions of integrals

and stuff like this.

You've seen a clear pattern.

Let me call this omega.

And this is the integral of omega over b of M, and this

is the double integral of d omega over M.

Where d now acquires very concrete meaning

as this operation.

So this is the general principle which I have been

invoking throughout this lecture, and I told you that I

will explain that all of the formulas that we are studying

in this course, in this part of this course, can be thought of

as just special cases of this general formula.

And now I can make it much more precise in this particular

case, and you could see that this is how it works.

Any questions about this?

Yes?

Do we have to worry about simply connected regions?

Do we have to worry about simply connected regions.

At this moment, no.

So the simply connectedness shows up when we try to

determine whether a given vector field is conservative,

right, whether on the plane -- given a vector field you try to

see whether it is conservative and then you want to find the

potential function, so that moment we have to

worry about it.

But this issue of simply connectedness is sort of

one dimension lower.

It has to do rather with some [UNINTELLIGIBLE WORDS]

theorem for line integrals than Stokes' theorem, because in

that issue there was no, it was about finding line interval.

So here, no, the short answer's no, we don't

have to worry about that.

[INAUDIBLE]

Any orientable -- smooth orientable surface

was a boundary.

[INAUDIBLE]

That's right.

This is a very powerful equation.

Now, this derivation that I have given you, you're

not responsible for it.

In other words, it's not going to be on the exam and so on.

I just wanted to explain for you so you can understand a

more conceptual way of viewing and appreciating this formula.

And I think that really clarifies a great deal.

Now, going back to what you have to know and what you are

responsible for, I want to give you an example -- I want to

give you an example of the application of the Stokes'

formula, which is very similar to what you will have on your

homework for this lecture.

So example is actually from the book at 16.8.8.

So you are given the vector field f, which is e to the

negative xi plus e to the xj plus e to the zk.

You're asked to evaluate line integral of this vector field

over a triangle, which is given by -- which is a part of that

plane 2x plus y plus 2z equals 2, which is contained in the

first octant, which is like for positive x, y, and z.

So what it looks like it's like this, here you have 1, here

you have 2, and here you have 1, and it's this triangle.

This is a triangle which is -- this is triangle intersection

of this plane, this plane, with three coordinate planes.

You see because for example you can put -- if you put y and

z equal 0 you get x equal 1.

If you put x and z equal 0, you get y equal 2

and then z equal 1.

So this is xyz.

So you're supposed to evaluate this integral, which is fine.

It's not too complicated.

What you need to do is you break it in half -- you have

to say which orientation.

So I have to be careful, so the orientation is

counterclockwise, like this.

So you can evaluate by breaking into three pieces by

parameterizing each of these line segments and then just

doing this line integral, but it's going to take a while,

because you have to do three different integrals, you have

to parameterize, so there's some work involved.

So what Stokes' theorem does, it gives you

a shortcut to this.

It allows you to evaluate this integral in a much more direct

way, because you realize that this curve is actually the

boundary of this surface -- this will be our M.

And so by using Stokes' theorem you can actually rewrite this

as an integral over M of curl of f dot ds.

So, I'm not going to use any of this fancy machinery, I'm just

going to use the formula the way it's written in the

book -- curl f dot ds.

So, instead of using this integral, we might as well do

this integral, and in fact it is a very good idea because

curl f is not exactly 0, so that would be like maybe the

first example for you would be to write this in such a way

that the curl is 0, and then you actually say OK,

so you don't have to calculate anything.

So in this particular case it's almost 0, there

is one component.

If I calculate it correctly it is e to the x times k.

Right, because this one you take derivative with respect to

y and z, here you take respect to x and z, so this is what

gives you that, and here you take, there is an x and

y, so they disappear.

So it's almost 0.

If it were 0, you just get 0.

So that's already -- no matter how complicated this curve is,

if the curl of the vector field is 0, you will actually get 0.

So that's a wonderful application of this result,

where you can calculate the integral very quickly without

doing any calculation.

Here it's almost 0, so what we need to do is we need to

evaluate the surface integral of this vector field.

OK?

So for this we have to choose -- we have to use the formula.

And the formula is that in this case you can say that this

plane is a graph of a function, you can write it as z equals

1 minus x minus 1/2 y.

Let's call this g of xy, and then we know that if you have

a vector field E, which is A, B, C.

Or maybe let's go with P, Q, R, because I think this is

the way I wrote it before.

Then the surface integral of E dot ds in such a case is equal

to double integral over the base, over the domain in x and

y, where here you put P minus P dg dx, minus Q dg dy plus R.

Right?

Remember this formula?

So in this case you only have, P and Q are 0, so you only have

R, which is equal to E to the x.

And so you end up with a double integral of E to the x, dx dy

over the triangle on the plane, which is the projection of this

triangle onto the xy plane.

So it has x-coordinate 1 at this endpoint, and

y-coordinate 2 at this end.

And this is your D.

So then this is very easy to calculate.

So I'm actually out of time so I stop here, and we'll

continue on Thursday.