Mathematics - Multivariable Calculus - Lecture 20

Uploaded by UCBerkeley on 17.11.2009

The general principle looks like this.

We have two integrals, one on the left-hand side and one
on the right-hand side.
On the right-hand side, will have an integral of some
quantity which is to be decided and which depends
on the circumstances.
Which dimension we are, what we'd like to
integrate, and so on.
On the lefthand side we integrate some sort of
derivative of that object.
Which I just schematically denote by d omega.

Now the next question is what do we integrate over?
And we're integrating on the left over some domain, D.
And on the right, we integrate over the
boundary of this domain.
This a boundary of D.
So this formula presents the kind of trade-off between an
algebraic operation-- namely taking the derivative going
from right to left, and the geometric operation which is
taking the boundary of your region.
So now would like to see how this general principle plays
out in different contexts.
And what we discussed last week was, in some sense,
the simplest possible context of this.
In this context, we took D to be one-dimensional.

So, its boundary b of D is zero-dimensional and
consists of two points.
So here is a picture.

Here is D.
It's a curve in this case.
And being a curve we usually denote it by c.
So c for curve.
And the boundary has two points.
This --and also there's an important aspect of this
picture which is that we have to choose orientation.
We have to say which is the initial point and which
is the end point.
We travel from here to here or from there to this point.
So let's choose orientation like this.
We travel from here to here.
Let's call this point a and this point b.
So, the formula which we learned last time
looks as follows.
We have --on the left we have-- that line integral
of vector field f.
So, f is a vector field.

Actually, let me rephrase that.
So, let me start from the right-hand side.
Because the point is that the algebraic object, omega,
actually is found on the right-hand side not on
the left-hand side.
So, we should really start on the right-hand side.
And, on the right-hand side, I have the values --first of all
a function f on the plane.
And what I have on the right-hand side is just that
I evaluate my function at the end points.
The point b.
And I also evaluated at the point a.
And I take the difference.
So this one comes with coefficient plus 1, and this
one with coefficient minus 1.
And that is a testimony to the that fact that this is the
end point and this is the initial point.
Which is due to the choice of orientation that we make.
This is the right-hand side.
And I can really think of this as kind of integration of the
function f over the boundary.
Over the boundary.
Even though it's a zero-dimensional boundary.
So integration doesn't really make sense.
What I'm talking about is really evaluating the function
at these two end-points.
But it does fit the general formula.
Because I have a function.
That's my omega in this case.
And I have the boundary which two points b and a.
Integration over the boundary is simply presented by the
difference of the values of f at b and a.
What about the left-hand side?
On the left-hand side, I'm supposed to take some
sort of derivitive of omega --so d omega.
I have to integrate it over my curve c.
So my domain now is curve c.
And what I'm going to integrate is the --I'm going to take the
line integral of the gradient vector field of my function.
So this is this derivitive --this is the real incarnation
of the derivative of my function in this
particular context.
In different context this will mean different things.
In this particular context, the derivative simply means taking
the gradient vector field.
And then I simply integrate that gradient vector
field over the curve.
So what I get is the fundamental theory for
line integrals which we talked about last time.
And my point now was to emphasize how it fits
this general principle.
So you can think of this as the first application of this
general principle which is explained on this board.
All right.
So what we're going to do now is we're going to
make the next step.
But you have a question.

The omega.

Well, I'm just saying that --if you don't like an upper arrow,
I will make it a sideways arrow like this.
Is that better?
Don't read too much into the arrow.
It's just to say what the omega is in this context.

Omega in this case is a function f which
is what I wrote.
That's omega in this case.
Any other questions?
By the way, for our worldwide viewing audience --and we have
a pretty wide worldwide audience as it turns out.
The lectures --all the previous lectures starting from Lecture
One are now being uploaded on You Tube.
So everybody can now learn what happened previously on Math 53.
You know, the struggle, the intrigue, the romance, the
broken hearts. [LAUGHTER]
It's all there for our viewing audience.

Seasons One and Two --the unforgettable Seasons
One and Two. [LAUGHTER]
OK let's go back to Season Three though.
I have to say, I saw lecture one today.
And it sort of made me a little nostalgic about it.
But I remember the good old days when I first came to this
classroom and saw all of you.
It was a priceless moment.
So what we'd like to do now is we want to make the next step.
We would like to generalize it when the domain is
two-dimensional instead of being one-dimensional.
So now we generalize.

Domain D is two-dimensional.
And so what I'd like to do is actually kind of make a chart
and make a comparison between the one-dimensional and the
two-dimensional cases.
So this will be this will be the one-dimensional
case, and this will be two-dimensional case.

So in the one-dimensional case, first of all, what is D?

D is the curve which I denoted now by c because it's a curve.
So let me draw another picture.

So, what should be the analog of this when we generalize
it to two-dimensional case?
It should be a two-dimensional region, obviously.
This is one-dimensional.

The analog of this in two dimensions would have to be
just a two-dimensional region.
So what is it going to look like?
Well it's going to be some region on the plane.
Like this.
When I draw this curve, I really mean in the
interior of this.
This will be D.
And the curve will actually show up now as the boundary.
See this is a very important point.
But now the domain is two-dimensional but the
boundary's one-dimensional.
So now the boundary is an object which kind of
looks like the original domain in the example.
So that's the domain.
So, this two-dimensional region.
This is the region --two-dimensional region.
I will just keep denoting it by D.
We will not change notation.
OK what about the boundary?

The boundary here, as we discussed, just consists
of two points.
Two points.

So, I'll just repeat.
I'll just draw little bit.
That's two points, right?
What about here?
Here the boundary is, as I said, is this yellow curve.
So, let me draw it one more time, approximately.
See now I'm not shading the interior of this picture.
I'm really focusing on the curve.
The curve is one-dimensional, and the curve shows up
as the boundary of this two-dimensional region.
This is a curve b of d.
Next, what is omega And again, I would like to think in
terms of this formula.
So, I would like to define all of this data and then see
what kind of equality I get.
So, what is omega.
Here omega was f --was the function f.
The derivitive of omega was just the gradient of f
--the vector field which is a gradient of f.

And so what did I do?
I integrated f over the boundary.
So I evaluated my function of the point bma.
But on the left I was taking the line integral over
the curve itself.
That is a line over the curve itself.
Over the gradient of the function f.
So now the question is, what should I do --what should I
take-- in this two-dimensional generalization now?
So I have to take something as omega I have to
take something which I compare with this curve.
Because omega is supposed to be integrated on
the right-hand side.
Omega is supposed to be integrated in the right-hand
side of the formula.
And, on the right-hand side, I integrate over the boundary.
Over the boundary.
Not over the region itself --which is two-dimensional--
but over the boundary.
And that's this curve.
So I have to think of the most general integral that
I can do over a curve.
And we've learned such integrals.
Those are line intervals.
So, I should be integrating a vector field.
That's a natural thing to do because, in this analysis so
far, we're integrating --over curves-- we're integrating
vector fields.
Functions we also integrate, but in this case it
seems more reasonable to integrate a vector field.
So what I'm going to take is a vector field f.
And this vector field is going to have two
components. p and q.
And that's going to be a vector Field.

So I will integrate it on the right-hand side
over this boundary.
But I also have to say, what is the derivitive of this?
What do I mean by the derivative of this
vector field?
That should be some expression involving the derivitives of p
and q which I would integrate over two-dimensional
region itself.
And that would give me the left-hand side.
So I have to decide what this is.
I have to find out what this is and, then I
can build the formula.
And it's not obvious from the beginning.
It's not obvious this what that derivitive should be.
Well, likewise, it was not so obvious that this
should be the gradient.
So let me just give you an answer, and then we'll see
what is the meaning of this.
The answer for this derivitive is given by the following
formula which at first might look kind of strange to you.
But, the will give you kind of a explanation how
this would be derived.
And the expression is the following.
It's actually going to be a function.
Whereas as omega here is a vector field, the derivitive
here is actually going to be a function.
So it's kind of opposite.
The operation of differentiation, in this
context, does something opposite to what it did
in one-dimensional case.
In the one-dimensional case, the original object was a
function, but its derivitive turned out to be a
gradient vector field.
And now we start with the vector field, but we
end up with a function.
That function is a combination of partial derivitives
of P and Q.
Namely, dQ dx minus dP dy.

So, in some sense it is like a derivitive.
It's not a derivative in any obvious sense because we
haven't talked really about derivitives of vector fields.
All we know that it should be some combination of derivatives
of the two components of vector field.
And there it is.
It is a combination of partial of the two components.
Namely, I take the second component q differentiate
this back to x.
I take the first component p differentiate this back to y,
and I take the difference.
And, of course in the discussions it is assumed that
all my functions --my vector fields are differentialable,
and in fact have continuous partial derivitives.
So this actually makes sense.
OK, so let's accept this.
Let's accept this formula for now.
I'll get back in a minute to of the meaning of this
form of how to derive it.
But let's accept it for now.
And so let's now see if our brand-new example of this
general formula, in the case, indicates when the domain
is two-dimensional.
Let's see what it's going to look like.
Well, again, on the right-hand side, we know what it is
on the right-hand side.
On the right-hand side we integrate over this curve.

Let me draw another replica of this curve.

That's the boundary of my two-dimensional
national domain.
And I'm taking the line-integral of
this vector field.
I could write it as f dot dr, if you want.
But because I have already represented F as a vector field
with two components P and Q, I might as well just write
it in terms of P and Q.
So I will write it as P dx plus Q dy.
And this is certainly just this expression.
This is the same expression.
So that's what I integrate on the right-hand side.
What about the left-hand side?
On the left-hand side now, I should integrate over my domain
D, and actually someone pointed out last time --at the last
lecture-- that, in this formula, I didn't keep track
of how many times I put the integration sign.
It was an approximate formula.
I just wanted to use a sign of integration --just one sign
regardless of how many times I have integrate.
But now I'm writing a precise formula.
This was just a guiding principle.
Now I'm writing a precise formula.
So I would like to keep track of the number integrals which
is the dimension of the domain over which I'm integrating.
So now my domain is two-dimensional.
And therefore, it should really be a double integral.
It should really be a double integral.
It should be a double integral of the derivitive of omega.
But the derivitive of omega have given by this formula.
So what I get is dQ dx minus dP dy dA.

And that's the formula I get.
So now, if you've never seen this before, it
looks kind of surprising.
So, it's a formula which equates the line-integral
of a vector field to a certain double integral of
function over the domain.
Here over the domain.
Here over the boundary of this domain.
Perhaps let me emphasize this by drawing again the domain.

This is the domain and this is its boundary.
So that is actually the theorem.
Which is the analog of the Fundamental Theorem for
Line-integrals in the Two-dimensional Case.
And it is called Green's Theorem.


That's right.
Very good.
So they're still quite a few loose ends here which I'm
going to now unravel.
And the first question is a very legitimate one.
Which is, what orientation do I take on the right-hand side?
This is a very good question because, if I just write this
formula, it actually does not make sense.
Because a line integral of a vector field is only
well-defined when I've prescribed the orientation
of the curve over which I integrate.
And if I change the --well there are two possibility

In this particular case, I'm dealing with the closed curve.
A curve which does not have a boundary as you see.
The way I will be referring to the orientation is, I'll be
saying it's clockwise or counter clockwise.
So there are two possible orientations.
In other words, I can go like this and go around
or I can go like this.
There's only two possible choices.
So you can say, well what's the big deal?
It doesn't matter which choice I take.
There are not so many choices.
Well, it does matter because I get two different
answers this way.
Granted, the two answers only differ by a sign.
But, of course, sign is important because if I want
to say that this is actually equal to this, I have to say
under what assumptions.
In particular, under what choice of orientation.
And so the point here is that the orientation has to be
chosen counter clockwise.
For this formula to be correct, I should take
it counter clockwise.
This is not to say that the integral with opposite
orientation is not well defined.
It is also well defined.
It's just not going to be equal to this.
It will be equal to minus this.
So that's a very important point.
The second point that I would like to make is this
distinction between closed curves and open curves --or
curves with boundary.
You see, there's a difference --this is a curve.
And there is also a curve in our previous example.

These are both curves.
And the word curve here refers to the fact that they are
one-dimensional objects.
But they are very different.
This one has boundary.
Has a non-empty boundary.
It has a boundary which consists of two points.
But this one does not have a boundary --or more
appropriately it has an empty boundary.
Such curves are called closed curves.
And it is interesting that it is exactly such a curve which
shows up as a boundary of a two-dimensional region.
The curve like this is called a closed curve.
It has empty boundary.

And you should contrast that to a curve with
a non-empty boundary.
This is not closed curve.
This is a curve --well, terminology here is slightly
misleading because a mathmetician would also say
that this is closed because it does include the
boundary points.
But when we say closed in this class, it will
mean a curve like this.
It closes on itself.
This one the, case strictly speaking, is also closed
because I cannot say it's an open curve because I include
the boundary points.
So let just say it's a curve boundary points.
So, curve with two boundary points.
And of course that immediately begs the question, is it
possible to have such a curve -- not a curve like this, but
such a curve -- as a boundary of something two-dimensional?
What do you think?

Well, any other answers?
Not that there are too many answers here, right?
If you think about it, you'll see that it looks like I'm
giving you an example of this curve being a boundary
of the shaded region.
That's not true.
What about this part?
What about this part?
Now, granted, if I included this and this, this would be
the boundary of something, but then it would have to go --it
would have to go to infinity Or it would have to
close somewhere.
So the regions which we will consider in this in this case
--the regions like this-- the regions to which we will
apply Green's Theorem.
These regions are going to be bounded.
They don't go to infinity.

That's right.
Let's go back to this.
The question is, is counter clockwise defined as positive?
So. the terminology, you know we are free to make any
terminology we like.
We can call it positive, negative, whatever we want.
Some people like counter clockwise, some people
like clockwise.
The standard terminology for this is counter clockwise
orientation --counter clockwise.

And this terminology is that it's also called positive.
But it's a matter of choice.
I am free to take line-intervals over
any curve I what.
I could take it over a curve oriented this
way, oriented that way.
My interest here is to choose orientation for which
this theorem is correct.
And I know that if for one of them it is correct, it cannot
be correct for the other one.
So, I have to make a choice.
And the way --actually the better way to think
about it is as follows.
Which will actually --will lead to slight
generalization on this.
For which this rule that I will explain now will make more
sense --or will be more easy to remember.
Right now it looks kind of ad hoc.
Why do I choose it counter clockwise?
So the actually rule is the following.
If you imagine somebody walking --this is just a sketch--
somebody's walking on the boundary of this domain.
So, it's like more realistically I would have to
draw this domain on this stage and then I would be walking on
the bounty boundary.
The rule is that the orientation of my path
should be such that the domain is to my left.
You have to chose some way, right?
So, this is well-defined.
If you walk this way it's to your left, if you walk this
way it's to your right.
It has to be to your left.
Orientation to the left.
Domain to the left.

It does sound a little ad hoc.
But actually it all makes sense in the end.
So, believe me.
It's a good rule.
Anyway, it is a rule for which everything works.
If you chose a different one, you can do that but please put
a sign --change the sign-- on either side --on one
side, not on both sides.
So let, me go back to the issue of boundaries.

First of all, the domains that we will consider --domains like
this-- will have to be bounded.
Bounded is something else --bounded is something which
that doesn't go away to infinity --off to infinity.
See what I mean?
So, the entire plane is not bounded.
And this is bounded.
This is something we talked about when we talked
about double integrals.
When you do double integrals we usually do them
over bounded domains.
If you have a bounded domain, its boundary would have
to be a closed curve.

So this is important.
This is a very important observation --maybe its
not so obvious at first-- but it is correct.
There's one other subtle point which, I guess, I should
mention which is that it looks like this curve you can say its
not really a curve, its a combination of four
different curves.
This one looks kind of smooth --although I kind of messed
it up so its not so smooth.
I was trying to draw a smooth curve.
This certaintly is not smooth.
It has four corners at least.
But, in fact, for all intents and purposes,
these are both curves.
In other words, I will think of this entire thing as one curve.
This curve will have four different segments
which are smooth.
And such curves we will call them piecewise smooth.
In other words, they're built from pieces which are smooth.
The curve itself --the entire curve is not smooth, but it's
the union of pieces which are smooth.
In this case four.
So when I talk about the boundary of this region, I can
say that it's a combination of the four pieces.
But I will usually just say it is this curve.
I will ignore the fact that there are some corners in
there, because after all I can draw it in one
stroke like this.
So it's not really --it doesn't add much value to
say how many corners it has.
It's not going to matter for us in the end.
When we calculate integrals over a curve like this, of
course, we will do it as a sum of integrals over each of
the smoke smooth parts.
But that's sort of calculational matter.
It's a technical issue.
It's not a fundamental theoretical issue.
The critical issue is that the boundary of this domain
is the closed curve.
A boundary of a domain like this on the plane can not
consist of a single curve like this which has a
boundary of its own.
In other words, the boundary of the boundary is empty.
If you know what I mean.
Boundary --if you take boundary of something-- and then you
take a boundary again. --This is empty.
I'll use the standard notation for empty.

A boundary can be non-empty.
This curve has a boundary of its own.
But then itself can not appear as boundary of a
two-dimensional domain.
If a curve appears as a boundary of a two-dimensional
domain it has to be closed.
It's own boundary should be empty.
See what I mean?
Any questions about this?

And the question is, how does this apply to Green's Theorem?
Here's how it applies to Green's Theorem.

In the case of Green's Theorem, the curve I have is a
boundary of something.
So this principle, which I explained, the boundary of a
boundary is empty means that this one will have
an empty boundary.
Will have no boundary.
So that means that, in Green's Theorem, I will --on the
right-hand side-- will always integrate the closed curves.

No, we could apply Green's Theorem to this one as well.
Let's look at this example.
Let's suppose that my domain here is a square.
This is my domain.
I could apply Green's Theorem to this domain, but then on
the right-hand side --on the right-hand side I will have
this square as my boundary.

This boundary is slightly different in this than the
boundary in the previous example in the sense that, its
a curve which has corners.
But it is a curve.
And it is a closed curve.
This curve does not have any end points.
This is a closed curve.
Closed curve.
Empty boundary.

These points --it has the special points.
I grant you that.
It has some special points which are different from
the rest because these are the singular.
They are not smooth.
But these are not boundary points.
This is not a boundary.
It's in the middle.
If I had something like this, those would be boundary points.
But this is still not a boundary point.
If I have something like this, these are the boundary points.
Report You have a question?
It's empty.

No this is not the definition of a closed curve.
This is a general principle which is if I start with a
two-dimensional object and take its boundary, and then take the
boundary of that boundaryl, this will be empty.
This is illustrated here.
I start with a two-dimensional domain, I take it's boundary.
I get a curve which already has no boundary.

I know it's a little confusing so, I'll just let this
sink in --as they say.
This is an important point.
This is an important distinction.
So from this point of view, Green's Theorem can be --if you
have a line-integral, and you would like to use Green's
Theorem to understand that integral, you have to make sure
that you're integrating over a closed curve.
In Green's Theorem, on the right-hand side, you will
never have a curve like this.

You will never have a curve which has boundary because
the curve in Green's Theorem is not independent object.
It is a boundary of something on the left-hand side and
because it is a boundary of something on the left-hand
side, it has no you boundary of its own.

What's wrong with integrating over an infinite
two-dimensional domain?
What's wrong with is that the integral might diverge.
It might actually be infinite.
It may not be well defined.
If we integrate over a bounded domain like this, which sort of
doesn't go off to infinity, we are assured that, if our
function is defined everywhere on this domain, we are assured
to get the final answer.
Whereas if we integrate over something infinite, we
are not assured of it.
For instance, take function one.
If you function one, then the integral over any domain is
going to be just the area of that domain.
If the domain is bounded, the area is finite.
If it is all bounded, it's infinite.
It's not well defined.
But it could be well defined.
That's right.
It could be well find defined.
So oftentimes we are forced to consider integrals
over infinite domains.
And the way we do it is quite obvious.
We approximate the infinite domain by finite --bigger
and bigger finite domains.
And then we integrate over those finite domains.
And we take the limit.
And you're familiar with the notion of limits, so you see
roughly how it would have to be defined.
But we are not assured, in general, that the answer --that
limit would exist-- and so the integral would be well defined.
Here we don't want to deal with these issues.
So from the beginning we restrict ourselves to domains
like this which are finite.
Any other questions?

Let's talk about this.

So the question is suppose you have two curves which connect.
How do we calculate the boundary of the sum
of these two curves?
So, for example, let's suppose we have a curve like this.
So what is the boundary of this curve.
Just from what I've said, from a induitive what boundary
means, you can see that its to be this, depending on
what the orientation is.
Let me put orientation like this.
So the boundary is this minus this.
So that's the boundary of this total curve.
Let's call this point --I want to call it a,b,c but I don't
want to do that because I have c for this.
Let me call them alpha, beta, and gamma.
I think as long as it's not epsilon and delta
you're fine with it.
The boundary of c is gamma minus alpha.
But, on the other hand, you can think of this curve c as the
sum of this curve and this curve.
Let me just draw it.
I mean it's not shifted.
It's just two difference sheets of paper.
I can think of this curve as the sum of this
one and this one.
So let me see, well what will I get if I treat it like this?
It will be c1 and c2.
The boundary of c1 would be beta minus alpha.
And the boundary of c2 would gamma minus beta.

So, if I take the sum, the boundary of c1 c2 is going
to be beta minus alpha plus gamma minus beta.
So you see the betas cancel out.
And they get the same result.
So it's all consistent.
The boundary of this curve is gamma minus alpha no
matter how you look at it.
You look at it as just a boundary of this curve.
We look at it as a boundary of the sum of these two curves.
Because, the point is, that you have to keep track of signs.
If you keep track of the signs --actually this comes with
minus, this comes with plus, this comes with minus,
this comes with plus.
And so these two guys annihilate each other.
They kill each other.
Does this answer your question?
All right.
So let's move on, and let's talk more about
the Green's Theorem.

Now that we have figured out --more or less-- what the
meaning of the boundary is and what kind of curves we can get
on the boundary in a two-dimensional region, let's
talk about this very strange-looking expression
that we integrated on the left-hand side.
So on the left side, again, we're integrating this
dQ dx minus dP dy.
And I want to write it as dx dy.
This is what we call dA when we talked about doubled integrals.
I'm just re-writing the formula.
But just a slightly different way.
I want to focus now on the integrands, on the
algebraic objects.
We have discussed the geometric objects, now let's discuss
the algebraic objects.
So, in what sense is this a derivative Why should we think
that this is a derivative Let me give you kind of an
intuitive explanation of this.
But let's do it again by analogy --by analogy with
the one-dimensional case.
In the one-dimensional case, we were integrating a function on
one side, and we're integrating the gradient on the other side.
So, in the one-dimensional case --this is a
two-dimensional case.
And the one-dimensional case again.
I just want to put them next to each other.
Nabla f dr equals --well let's write it like this.
f of B minus f of A.
But what is nabla f dot dr?
We can write it in coordinates.
It is dP dx dx, plus dQ dy dy.

That's what we call nabla f dot dr.

So you see this is a derivative --this is what
you call a derivitive.
And, in fact, if you look at this formula, what do you see?
What do you recognize?
I should have sign change?
No, I shouldn't.
Here I have the sign.
But this is two different formulas, right?

Why do you think that I should change the sign?
Nabla f is a vector field which has two components,
dP dx i plus dQ dy j.
When I take line integral of this vector field, I have to
take the first component, multiply by dx.
The second component multiply by dy.
That's what I integrate.
My question is, have you ever in an expression
like this before?
In Season Two it made its appearance.
You know the answer.
Somebody say it.
Thank you.
So this is a differential.
So, in fact, this is df.
This is df, the differential of df.

So you can think that you are doing the line-integral of the
gradient of the function, but actually a better way to think
about it is the integral of the differential.
The differential is an example of what mathmeticians
call differential forms.
You see that every time we differentiate, we have to have
some d's, dx or dy, here.
Only one, because we're doing a single integral.
Or dx dy, both of them, because we are doing double integral.
Sometimes we have dx dy dz.

All those objects are mathematically called
differential forms.
And actually a slightly more consistent theory of
integration can be developed in which the objects
that we integrate are differential forms.
We take a slightly different point of view because we want
to think about these integrals more in terms of applications
like work done by force, and other things like a flexible
vector filed that we will talk about later.
And this is the reason why we're thinking about
integrating vector fields rather than differential forms.
But, algebraically, it's easier to think of this as integrating
the differential form.
Which is just the differential of the function.
So, in the case --in the one-dimensional case-- in the
case of Fundamental Theorem for Line-intervals, what we're
doing passing from the right-hand side --where we have
a function-- right-hand side we have a function.
On the left-hand side we have what I call the derivitive.
This derivitive, actually, is just the differential of that.
You could think of integrating just the differential
of this function.
Shouldn't be a derivitive of what?

I'm sorry.
That's right.
I made a mistake.
It's df.
That's right. sorry.
That's why you did not answer my question.
It was my fault.
I got carried away with all this df dP and df myself.
Of course, there's no there is no P and Q in this formula.
P and Q shows up here, because we have P and Q here.
Here what we're given is just a function f.
And what I should have written is just a dirivitive
of f in both cases.
Now it's much better.
Do you see now what I mean?
Is that better?

So what did we do to get this formula from
f? what did you do?
We took the function f, and we took two possible
derivitives of f.
Respect to x and respect to y.
So we took f and we replaced it by df dx plus df dy, dy.

So what is this operation?
I differentiate, I take partial derivitive.

And, at the same time, multiply by dx.
x partial derivitive and multiply by dx plus y partial
derivitive and multiply by dy That's what it is.
So there are two possible partial derivitives.
I take partial derivitives, multiply by dx.
Take second partial derivitives multiply by dy.

Let's take this as the guiding principle.
Now I would like to apply it here.
So now my starting point will be pdx plus qdy.

And I want to see what I can get from this by applying
that same procedure.

So let me start with the first term, pdx.
When I start with pdx, I apply the same rule.
So I take the derivitive of P with respect to x,
then multiply by dx.
And I keep the old dx, which I had before.
So I just treat it like this.
I've applied my differential to P, but I keep this
extra factor as well.
So I get this term, plus dP dy.
And then I have dy dx.

Have you ever seen dx dx?
And for a very good reason.
We see dx dy, because this represents the area of an
elementary -- of a parallelogram with
sides delta x delta y.
The area is delta x delta y.
And this is what eventually gives us this term, dx
dy, which we call dA in the double integral.
But delta x delta x makes no sense.
Delta x delta x does not present any area whatsoever.
To get an area, you have to have increments in two
transversal coordinates.
Like x and y.
Or maybe r and theta, if you switch to polar
coordinates if you like.
Or dx dy dz, if you're doing triple integrals.
But delta x delta x makes no sense.
And neither does this.
So whenever you have dx dx, this just is prohibited.
So you have to cross it out.
Or more properly, you should say it is equal to 0.
It sort of represents the area of this segment.
Which is 0.
To have an area, you have to have something two-dimensional.
Now this, on the other hand, is fine.
It's dy dx.
The problem is that we switched the order in which we
normally take the area.
Which is usually dx dy.
So the rule, in fact, is that this should be replaced
by minus dP dy dx dy.
This is an issue of orientation.
Remember we talked about different types of orientation.
And we know that when we talked about general changes of
coordinates on the double integrals, we saw that if you
switch two coordinates, the Jacobian you get under the
integral gets a minus sign.
We kind of skirted that issue by saying we'll just take the
absolute value of the Jacobian.
But, at this point, we can't avoid this issue.
And the solution to this issue is to say that dy dx is
equal to negative dx dy.
So when we apply this procedure to pdx, this is what we get.
But I also have a second term.
Q dy.
So let me calculate what I get for Q dy.

For Q dy.
I get dQ dx, dx dy.
This is good.
First of all, there are two different variables.
And second of all, they are already in the right order.
I do nothing.
And I also have dQ dy dy dy.

Again, I got dy dy.
It makes no sense.
Cross it out.

You see what I'm doing?
I'm just taking Q.
And Q gives me this plus this.
But I also have the second factor dy, which I keep.
And it is the second factor which tells me that the second
term should disappear.
So let me collect all the terms that I got.
If I start with P dx plus P dy, I end up with minus
dP dy, dQ dx, dx dy.

And that's exactly the formula that I get on the left.
Well I write it as dQ dx minus dP dy.
I just wish switched the word.

So, the rule is, each of the functions you have, you should
take partial with respect to x and multiply by dx.
Plus partial with respect to y multiply by dy.

So, here's how it works for this term.
I take partial with respect to x and multiply by dx.
But don't forget I had dx from the beginning also.
So I just keep it.
Plus, partial with respect to y multiply by dy.
And keep the dx.
That's what I have.
But, in addition, I have this term.
For this term, I get this answer.
So I get, a priori, four terms.
But, out of these four terms, only two are non-zero.
This one and this one.
I put them together and I get this answer.
And that's exactly the answer which Green's Theorem tells
me I should put here.
So, this is the explanation of this expression.
This expression, at first glance, looks unnatural.
But now I have shown that you can obtain it in exactly the
same way as the expression for the differential of a function.
Which certainly is quite natural.
We know that the differential function is actually
very meaningful.
That's right.
So, this is a subtle point.
Question is why do I put the minus sign?
This is a subtle point.
When we did double integrals, we actually kind of swept
this under the rug.
Because we often times would write dx dy.
And often times We also write dy dx, depending on which order
of integration we choose.
Because, what we did, we represented the
double integrals as integrated intervals.
We would first integrate over x and then over y.
Or, first over y and then over x.
Depending on that, we would just write dx dy or dy dx.
But, in fact, the correct meaning of this expression
should involve also the choice of orientation.
And in the orientation --if you do that-- dA, the measure of
integration for double integrals should
really be dx dy.
And dy dx should be minus that.
The reason why we didn't do it is because essentially we were
actually always integrating with respect for
the same measure.
Just the fact that we do the integral in one order or the
other order, does not introduce the sign.
If we do the integration over surfaces, keeping track of the
orientation, then this would actually become much
more essential.
Our next topic will be more general surface integrals,
where we will talk about the orientation.
In a moment this will become more clear.
Now it looks a little strange, why we didn't care about it
before and now suddenly I care about it.
Well because we didn't have to explain it before,
but now we have to.
So that's roughly the answer.
The question is about line integrals.
In the case of a line integral, we were supposed to take this
vector field df dx i plus df dy j.
And we are supposed to take the dot of this with dr.
When you take the dot with dr, you get this expression.
I have explained now this expression as one obtained
from f by this rule, by this procedure.
The reason why I explain this, is I wanted to show there is
some uniformity between these two formulas.
Without this explanation, these two formulas look
completely disjoined.
In this case, it looks very arbitrary.
Very ad hoc.
So it looks like you have to simply memorize
this expression.
There's no other way to see why is it dQ dx?
Why is it not dQ dy minus dP dx, for example?
That is the natural question which arises here.
What I tried to explain is that, actually, this is not an
arbitrary expression which can be obtained by some
universal procedure.
That procedure can also be used to explain the
formula from last week.
This is something which is not in the book.
You don't have to memorize it.
I think it's a good idea to memorize it --not memorize
it-- to understand it because then you don't have to
memorize this formula.
If you understand this procedure, this just fall
as I have just explained.
But you don't have to do it.
I wanted to give you an intuitive explanation of
what this formula is about.
The connection between the two formulas.
But, if you like, you may just memorize these two
formulas separately.
And not necessarily think of them as two special cases,
this is something general
You can think of this as integral of a vector field.
That's right.
You're asking what is the integral of a function
f over a curve?
So that's a very good question too.
So remember when we talked about line integrals there were
two types of line integrals.
There were line integrals of vector fields and line
integrals of functions.
And then line integrals of functions was kind of like
one of those cousins nobody wants to talk about.
It's like it's there somewhere but suddenly we don't
talk about them anymore.
The reason is, this it is very useful --those integrals
are very useful.
But those integrals aren't as easy to fit into this formula.
But since you've asked me, those were not exactly
integrals of functions.
We kind of thought of them as integrals of functions.
We never wrote integral of f.

WeThis never wrote integral of f.
This doesn't make sense unless you integrate over points.
For example, here, you can think of this as an integral
over two points of a function.
Which would be just evaluating one point minus evaluating
the other point.
We never write this as an integral over a curve.
Or over a two-dimensional domain for that matter.
Over a curve, we can only integrate f ds.
You have to put some measure here.
You have to put ds or dx or dy, something like this.
ds is actually also a differential form.
It's a kind of object like dx and dy.
And I spent some time about a week ago talking about this.
You have to remember this.
And if you treat this not as an integral of a function, but an
integral of this expression, then it actually fits into
this general theory.


Where did it come from?

For example, one would go from a function to differential.
It's just the way it is.
That's a formula which gives you --the formula for the
differential involves partial derivatives with respect to
x dx plua the same for y.
The formula was explained before because this was the
formula for the linear approximation of the function.
Now we're developing a formula which allows us to relate
integrals over domains and their boundaries.
And it turns out that within that formula, there is a
certain procedure that you have to make.
Starting with an object on the right-hand side, to get the
object on the left-hand side.
And we have now seen two example of this in action.
This is the first example which was last week.
You actually have a function, and on the left you have
to take a differential.
And now as a new example.
Here we start with a vector field with components p and q.
And then that's this expression.
But if I had just written this expression without
any explanation, you would way what does it mean?
How did you get this formula?
I have tried to give you an explanation of that formula in
a way which will also makes it parallel and analogous
to this formula.
Let's move on now and do a few examples of this.

As you can imagine, this actually has some very
important application.
Because you can now trade an integral on one side of this
formula for an integral on the other side.
And oftentimes one of these integrals is very complicated
and maybe not even possible to evaluate easily.
And the other one may be very easy to evaluate.

That's what we are going to do.
We're going to apply this formula.

So here's an example.
Suppose your curve c is a boundary over a
parallelogram like this.

This is your curve c, and it will be positively oriented
which means that you will go counter clockwise.

This is your curve c.
And suppose you're asked to evaluate the line integral over
c of the following expression. xe to the minus 2x.

plus x cubed plus 2x squared y squared dy.

This is exactly the kind of expression that I'm talking
about in this formula --in Green's Theorem.
General expression is pdx plus qdy.

But, as I already explained, in order to apply Green's Theorem,
you have to make sure that the curve over which you
integrate here is closed.
Has no boundary.
So that's let's see.
Can we actually have a chance to apply Green's Theorem
to this line integral?
For this we have to make sure that the boundary
of this curve is empty.
Is it empty?
Yes, it is empty.
It is a closed curve.
So we are in good shape.
We can actually replace it by the left-hand side
of Green's Theorem.
Why would you want to do this?
Well someone might say, I'll just calculate this like this.
And to this I would say, good luck.
Because first of all, you have to break it into four parts.
You have to integrate over each of the segments.
Each of the segments you'll have to parameterize.
Once you parameterize it, you'll have to substitute
parameterization here.
And then go and find the derivative from
those functions.
Maybe, in this case, it's not so bad, but I could have
written something much more complicated for which you
wouldn't be able to find a derivative.
Which I will on the exam.
And that would be the first indication for you to know
that you shouldn't do it in a straightforward way.
You should not do it directly.
It's just way too complicated.
So you could think of other ways to evaluate this.
And here it is beneficial to you as Green's Theorem.
What does Green's Theorem tell me?
It tells me that this is equal to the double integral over the
interior of this curve which is the parallelogram itself.
This is d.
Of what?
I have to take --let me write it here-- I have to
take dQ dx minus dP dy.

And what are the p and q?
this is p, and this is q.
So, let's take the derivative of q with respect to x.

I get 3x squared plus 4xy squared.

And now I have to take also dP dy.

And, lo and behold, p actually does not depend on y at all.
It only depends on x.
So we kind of lucked out here.
This complicated looking function will actually
disappear when we go to the other side of Green's Theorem.
This actually will be 0.
This is 0.
So I don't have to add anything to this.
In general, I have to take dP dy.
But, dP dy is 0.

Everybody everybody with that? dP dy is 0.
It does not depend on y.
Its partial derivative with respect to y is 0.
So this is already simplification.
So, you can imagine here is the quickest exercise you could
give to somebody which looks complicated, but actually is
the easiest to calculate.
That would be, if I actually wrote here something
which only depends on y.
Let's suppose I gave you a formula here I wrote a
very complicated function but only depended on y.
I would write some square root of 1 plus e to the
sine y, or something.
And then you look at it --it looks horrible.
But then you apply Green's Theorem and you see that
dQ dx is actually 0.
No matter how complicated this function could be in y, if it
doesn't depend x it would be 0.
And this would be 0, this would be 0 and the answer would be 0.
So, you don't have to calculate anything.
You already see the power of this method.
That sometimes you get the answer right away even though
the original integral looks very complicated.
This would be sort of a gimmick.
If I gave you this, it would be sort of a gimmick.
And you would be saying he's playing with us.
This is sort of a more serious example, right?
Which is kind of intermediate level.
Where one of the functions disappears, and the other
one does not disappear.
So you actually to have to calculate something meaningful.
You have to calculate this.
Of course, dx dy.
Or da if you will.
So now we have to remember what we learned about
double integrals.
You have a question.

So the question is do I have to put dx dy?
There are two issues here with dx dy.
So in this discussion of this procedure, you
have to keep of this.
And that is the discussion which goes into the issue of
the meaning of the formula.
That's the kind of fundamental stuff about deriving
these formulas.
This is one aspect.
The other aspect of this is, when we do double integrals,
we treat them as integrated intervals.
Where we first integrate over one of the two variables, and
then we integrate over the second variable.
Both of them are okay.
They don't introduce any sign.
The switching that I'm talking about is a switching in the
formula when you pass from here to here.
But once you're going to double integral you
can do it either way.
Don't worry about it.
The sign only happens in this discussion to explain why
there's a minus here and there's no minus here.
But when you're actually doing the double integral over an
expression like this, the rules would which we have used to
evaluate these integrals.
And those rules involve integrating over x and
then over y or over why and then over x.
Both of these rules will give you the same answer.
Let's actually evaluate.
So you have -- So this is going to be --now we do it
as an integrated interval.
And as I said, you can choose whichever you like for
the first variable, and whichever you like in
the second variable.
So let me see.
Let me say here do y from 0 to 2.
Here x will be from 0 to 5.
Here we have 3x squared plus 4x y squared.
Here I have dx, and then I'll have dy.

So here you will have x cubed plus 2 x squared
y squared from 5 to 0.

Which is --what's 5 cubed?
125 plus 50 y squared.

Tell me if I make a mistake.
And then you do the outer integral.
I'm doing this because I'm trying to remind you because
it's something we haven't done since the exam.
So I hope you have forgotten.
We certainly have to use this knowledge in this discussion.
125 plus 50 y squared dy.
And that's going to be 125y plus 50 over 3 y cubed
between 2 and 0.
So you have 250 plus 50/3 times 8 This is how you compute.
That's right.
So this is a very easy way to calculate what would have been
a nightmare calculation.
If you had approached this in the most direct way.
This is the answer for the integral of this vector
field over this curve.
Over this curve.
We're lucky here.
I want to emphasize it one more time.
This is something which we could not apply to the
general line integral.
We can only apply it to a line integral over a curve like
this which has no boundaries.
My question was to calculate over this segment, or even over
this, or even over this, something which has a boundary.
I would not be able to this.
Of course, what I could have done, if the integral was like
this, this, this, I would say, okay, I'm missing this part.
But it this part were there, it would be there.
So let me say that this then is equal to this
part minus this integral.
So I would be able to trade this integral over three
segments for a double integral plus or minus the integral
over one segment.
So in other words, not everything is lost even if
your curve is not bounded --it is not closed.
You could trade a complicated curve for an easy curve.
But here we're actually lucky because the curve is close, so
we would just trade the single integral --line integral for a
double integral is much simpler to calculate.

what look like a square?

Does this curve look like a square?
Does it have to look like a square?
First of all, it looks like a paralellogram not a square.
At least meant it to have unequal sides.
To answer your question, let's do another example where the
curve is not a square or a parallelogram for that matter.
Let's do another example.

Suppose that you have to evaluate the line integral
of something which has two components.
I would like to draw a picture like this.

I would like to say that here, again, I could
use Green's Theorem.

This is equal to pdx plus qdy.

Where the region --my two-dimensional region-- will
be the [? analise ?], which is confined between
these two circles.
The first circle, let's say, would have radius 1, and the
second circle would have radius 2.
In all the previous discussion, I assumed that my it region, so
to speak, didn't have holes.
In other words, the boundary consists of
only a single curve.
And now this region has a hole.
So the boundary, now, consists of two circles.
There's an outer circle and there's an inner circle.
But, it turns out, that Green's Theorem is applicable to
such regions as well.
But you have to be very careful with the way you put
orientation on the boundary.
That orientation on the exterior circle, we have
already figured is that counter-clockwise.
That's going to stay the same in all cases.
But now the question is which orientation you should
put on the inner circle.
Because this is exactly a situation where the counter
clockwise, clockwise rule doesn't apply.
Counter clockwise rule only applies to the
exterior boundary.
But for the inner boundaries, it's going to be the opposite.
It's going to be clockwise.
The reason is that the universal rule in both
You have to think of a person walking here.
And the domain has to be always on the left.
So that means that you have to walk counter clockwise outside.
If you go counter clockwise on the inner circle, on you left
will not be the domain but will be something else.
So you have to walk like this here.
Like this here to have the domain on the left.
Which brings us to this second example in which --in which,
let's say let's say you have some function.
Where did I put it?
Oh yes, here it is.
Brings me to the second example.

Let's say you have the following concrete functions.
This c now represents the union of both of these
boundary curves.
Taken with appropriate orientation.
And so let's say you are evaluating the following thing.
You have cosine x dx plus x squared sine y dy.

So, this P and this is Q.
So we use Green's Theorem.
And Green's Theorem tells us that this is going to be an
integral over this [? analise. ?]
Again dQ dx minus dP dy.
Which is 2x sin y minus dP dy.
So again, I chose the function to be such that actually this
disappears, to make it easier.

And now, I need to evaluate this double integral.
Of course, here you have to remember all the tricks which
you've learned in dealing with double integrals.
When you do double integrals, you have this entire tool
box available to you.
You don't have to do it in a straightforward
way, first x, then y.
You could, for example, use polar coordinates.
And surely this region calls for polar coordinates.
You need to switch to polar coordinates, and the
integral that you will get will be as follows.
You will have 0 to 2 pi over theta.
And then from 1 to 2.
I recall that this has radius 1, this has radius 2.
So, theta you integrate from 0 to 2 pi.
But r you integrate from r to 2.
You substitute the polar coordinates.
So you get 2r cosine theta.

And then you have sine of r sine theta.
Then you have r dr d theta.
Q dx.
Maybe I made it a little to complicated.
Let me just get ride of this. [LAUGHTER]
Just for the sake of practice, though.
I think it's good enough.

But don't forget to put an extra r in da.
So then you have this extra r. dr, d theta.
And then, of course, you can already do very easily.
So, we'll continue on Thursday.