Uploaded by UCBerkeley on 10.11.2009

Transcript:

Have you recovered after the midterm?

You actually did pretty well.

I was happy about it.

Good job, keep on doing it.

You know these lectures are on web cast dot berkeley dot edu.

They're also on iTunes and in the iTunes version I

sing all the formulas.

And now our lectures are on YouTube.

So I'd like to welcome our new additions, new

addition to our audience.

Our YouTube audience.

I think it's great because actually I tried to search for

math courses on YouTube and I was surprised that there are

actually very few of those.

A few people want to learn about multivariable calculus,

you've come to the right place.

And even if you don't want to learn about multivariable

calculus, you've come to the right place.

You know, unfortunately there's only one lecture available for

now and it's like when you start watching a television

series and it's in its third season and you don't know what

happened in the first season or the second season.

It's exactly the same here because I would say first

season is from the beginning to the first midterm and second

seasons is from first midterm to the second midterm and now

we start our third season.

I personally think the first season with the best one but

I think the third one will be the best.

Of course all the lectures are available on bypassed webcast

at dot berkely dot edu and I was told that they will all be

available on YouTube perhaps this week there were some

problems without uploading.

So if you want to know what happened previously on

math 323, stay tuned.

Having said that, what is this third season, so to

speak, what is it about?

I personally think this is the best one, it's the most

interesting one, because that's where we will actually be

able to apply all the stuff we've learned up to now.

This has been a sort of a recurring theme in this course,

I've said it many times, that calculus you can basically

break into two parts.

Differential calculus and integration calculus.

And in some sense integration and differentiation are

operations which are inverse to other.

In some sense you can say very roughly that integration is

like differentiation inverse.

This is the thesis.

I will have to follow up on this and I will have to explain

what exactly this means in the context of multivariable

calculus.

We all know what that means in the context of single

variable calculus.

In the one variable calculus we have the beautiful formula

which relates integration and differentiation which is called

the fundamental theorem of calculus or single

variable calculus or the Newton-Leibniz formula.

And this formula looks as follows: on the left hand

side you have an integral.

And to make this formula to support this thesis, I would

like to write it in this way.

Integral of a function which itself is a derivative

of another function.

So you put a function here which is derivative of some

function f so you put f prime.

And you integrate this function from a to b.

And on the right hand side we simply have the values of this

function at the end points.

So we have f of b minues f of a.

So I would like to think about this formula in

the following way.

On the left we integrate our function over an

interval from a to b.

So, on the left I have an integral over this domain

which is just an interval.

Well in a single -- and by the way no walking in front of

the front row it looks really bad on YouTube.

On the left we have an integral over this domain and because we

are in single variable calculus now there are not so many

domains to go around.

The only domains you can think of are intervals or perhaps

unions of intervals.

So that's what we have on the left side.

And on the right hand side we have these two points which I

would like to think of as the boundary of this interval.

So this is an interval and this is the boundary

of this interval.

Two points, let me put them as dots.

Something interesting happens here. f of b enters with

a plus sign and f of a enters with a minus sign.

So actually this point a comes with a plus sign of this point

comes with a minus sign.

And we can actually read this off from the orientation

of this interval.

This interval has an orientation.

We integrate from a to b, so there is an orientation.

So the arrow points from a to b, therefore the end point

comes with a plus and the initial point a

comes with a minus.

So you see this already indicates that orientation

is going to play an important role.

But for now the essential point is the following: on the left

you have a domain interval, on the right you have the

boundary of this domain.

That's about the region of integration.

In some sense you can think of the right hand side also as

integration, it's just that you are integrating over zero

dimension domains which means simply evaluating the function.

What do negative and positive mean here?

For now it's just notation that I've put.

I assigned plus to this point which is point b and I

assigned minus to this point which is point a.

The practical meaning of this is the following: that when I

evaluate the integral I evaluate at this point with the

sign plus and at this point with the sign minus, which

is what is really here.

There is some rule here that some part of the boundary comes

with coefficient plus 1 and some part of the boundary comes

was coefficient minus 1.

But integral, is in a way, a sort of the

union of two things.

The integrand, the object you integrate, and the domain

of integration, the domain over which you integrate.

You cannot make an integral with just one of the two.

You need both.

The integrand and the domain of integration.

In this formula there is a trade off between something

that happens to the integrand and something that

happens to the domain.

We see now what happens to the domain, when the go from left

hand side to the right hand side we pick the

boundary of the domain.

On the other hand, let's look at the integrand.

On the right, the integrand is f and on the left the

integrand is f prime.

What we do when we go from right to left

is we differentiate.

Here we take the boundary.

And this is the basic structure of the formula.

So what we would like to do is to find analogues of this

formula in multivariable calculus.

That's what they're going to do in the remaining

part of the semester.

In other words, what we would like to do is to prove formulas

of the following type: on the left we'll have an integral of

some quantity and on the right we'll have an integral

of some quantity.

On the right it will be integral of some constant,

let's say omega, I will explain what kind of

object this will be.

In some cases it will be a function.

We will also have more general options integrated here.

On the left will have some sort of derivative of this object.

This is just a schematic notation for now.

Don't read too much into it.

Something like taking the derivative, for example

here just taking f prime.

So that's as far as the integrand goes.

And as far as the integration domain goes, here we will have

some domain D and here we will have the boundary of D.

You had a question.

So here I'm talking about the domain of integration.

We have a formula.

A formula has a left hand side and a right had side.

And each of them is an integral of something the

function over domain.

So i'd like to trace here on this blackboard what happens to

the domain and what happens to the boundary as we go from left

to right or from right to left.

And I observe that for the domain on the left we start

with an interval but on the the right we take the boundary

of this interval.

It's like going from the from the left hand side to the

right hand side for domains corresponds to taking

the boundary.

But as far as the integrand is concerned, on the right I

have some function f and on the left I have f prime.

So going from right to left I differentiate.

So this formula could be thought of as a trade-off

between taking the boundary and differentiation.

Taking the boundary at the level of domain, taking the

derivative or differentiating the integrand.

So this is the basic structure -- this is how I would like to

think about the fundamental theorem of calculus -- of

one variable calculus -- which is written here.

I would to first of all think of taking an integral as

combining or pairing two different types of objects.

One is algebraic, say a function, the other one

geometric, a domain.

That's the first thing and if I think about integration in this

way, then I clearly see what is the structure of this formula.

At the level of domains it's going to the boundary, at the

level of functions it's taking the derivative.

So this suggests that perhaps if I am on not on the line but

on the plane or in three dimensional space, if I'm doing

multivariable calculus, now I should also have a, formula

like this where on the right I would have some object

integrated and on the left I would have some

object integrated.

But the correspondents would be like in the fundamental theorem

of one variable calculus.

That is to say at the level of domains we go from the domain

to its boundary, and at the level of integrands we go from

omega, whatever it is, this we will have to discuss in more

detail, to its derivative of some sorts.

That will be the basic structure of the formulas

that we will derive.

And they will be the analogues of the fundamental theorem of

calculus and the will allow us to learn a lot of things

about integrals.

And we will also talk about various real-life applications

of these integrals.

What do these integrals mean?

And various applications of those integrals.

And such formulas will allow us to express integrals of one

kind in terms of integrals of the other kind.

That's basically the outline of what we're going to do next.

So now the question is to discuss what kind of objects

we should be integrating and over what?

The integrands and the domains of integration.

What are the most general objects which we should

introduce which we could then use to obtain a formula like

this and what is the meaning of the corresponding integrals.

These are the question that we will have to ask now.

So let's remember first of all what kind of integrals we

have learned up to now in multivariable calculus.

We have learned, in the second season so to speak, in the last

month or so, we've learned how to integrate functions --

integrals, double integrals of functions and triple

integrals of functions.

Over some regions in the two dimensional space on plane and

over solids, three dimensional regions in the three

dimensional space.

But in those cases we have integrated functions.

It turns out that this type with of integrals are not

sufficient to do justice to the kind of formulas

which I promised you.

We will have to generalize these integrals first.

And there are two types of limitations that we have now.

Two types of things that will have to relax and consider

more general integrals.

We will generalize these integrals in two ways.

First of all we will learn how to integrate more general

objects and functions.

We will integrate vector fields.

We will learn how to integrate vector field.s.

We will talk about what they are, vector fields.

And what is the meaning of the corresponding integral.

And that'll be the first kind of generalization

that we will make.

The second generalization is that we will consider

more general domains.

They will be roughly, the difference is that

they will be curved.

We will consider most general curved domains, whereas the

domains we have looked at up to now are flat domains.

Let me explain what I mean by this.

Flat versus curved.

You can just illustrate this.

The kind of regions that we consider here, the two

dimensional regions are -- it could be some region

on the plane like this.

And someone might say well this is curved.

It kind of looks curved.

That would be misleading because what is curved is not

the domain itself, which is the interior of this

curve, but the boundary.

In other words, in this subject, one of the most

important things is to distinguish domains

and their boundaries.

That's why I would like to emphasize it and I would

like to trace this boundary in a different color.

Yello.

This is a boundary and this is a domain.

These are two different things.

This is a domain.

For example the domain is two dimensional, it's on the plane.

But the boundary is one-dimensional.

And this is always the case.

The boundary of an n-dimensional domain is going

to n minus one-dimensional, you lose one dimension when you

pass through the boundary.

Same things happened here.

Here the domain was an interval.

It was one dimensional and the boundary was zero dimensional,

it consisted of two points.

Maybe I should make them yellow also to keep the analogy.

So the boundary here is curved.

It is curved, I wouldn't disagree with that.

But the domain itself is flat.

It's flat in the sense that if you look at any point here,

and you look at the small neighborhood of this point.

If you look at a small neighborhood of this point you

don't know what the boundary looks like, so you don't have

any illusions about the domain.

You just look at the very small part of this domain here.

And you see that it's just the same as a small part of the

plane around this point.

And the plane is flat. r2, the blackboard is flat.

So locally this domain everywhere looks like

the blackboard itself, so it is flat.

You should contrast that with the boundary itself.

The boundary is not flat.

A flat one-dimensional domain would be an integral

or a line segment.

So this will be flat.

For example this is flat, this is also flat.

So you could have, of course, the domain which is flat and

its boundary is also flat.

It migh have some corners.

So you see what I mean for the one-dimensional boundaries,

here it's curved, here it's flat.

Then you could ask, well give me an example of a two

dimensional curved doman and that's really easy,

think of a sphere.

A sphere is curved.

Because a sphere cannot be, it's two dimensional, it's the

same dimension as the plane, same dimension as the

blackboard, but you can't put the sphere inside

the blackboard.

I mean you can project that but you will have to make some

violent transformation to the object.

You will change its geometry.

So that's the point.

A curved two dimensional object cannot be fit in

a two-dimensional plane.

Anything that is two dimensional and that fits

in the two dimensional plane is flat.

There are a lot of two dimensional objects like a

sphere or a [UNINTELLIGIBLE]

and many other things, if you can imagine, like the surface

of the ocean, those are also two dimensional.

But there are not like the plane, so they are curved.

So far we have not really talked about integrating

over such domains and we'll need to do that here.

What's the boundary of a sphere?

That's a very good question.

So before answering this question let's talk about a

simpler question right but analogous question, in

dimension one lower.

There will be many analogies between objects of different

dimensions but always it's easier to look at the

situation where dimension is lower than higher.

So oftentimes is better to start with lower dimensions the

proceed to higher domensions.

So a good question here would be what about boundaries of

one dimensional domains.

So if you just have an interval like this, we

know what the boundary is.

It's very similar to what we discussed in one

variable calculus.

So if your domain is like this then the boundary

consists of two points.

But here's an example of a domain which does not have a

boundary or a more proper was to say it has an

empty boundary.

A circly.

That's a question that's similar to your question.

You asked me what is the boundary of a sphere.

So a simpler question would be what is the

boundary of a circle.

Boundary of a circle is empty.

So you see the difference between this and this.

In some sense you can think of you can obtain this by

combining these two end points and closing these

two end points.

It actually all make sense because the boundary, as we

discussed, in the boundary these points will enter

with different signs.

This one, if we orient it like this.

This one with sign plus and this one with sign minus.

When we close them they will sort of cancel each other.

That's consistent with our natural observation that a

circle has no boundary.

Likewise a sphere has no boundary or has empty boundary.

But if you take the upper hemisphere -- the upper

hemisphere has a boundary which is a circle.

Upper hemisphere is like a dome and the dome has a boundary

which is a circle.

There are many things that you need to be able to visualize

here and to understand and to see that there are certain

objects which do have boundaries.

There are objects which don't have boundaries.

You should always keep track of where you are.

If you are talking about the domain, the boundary inside

three dimensional space, two dimensional space.

There are many different options.

This I have now explained.

The second aspect is that we will integrate over more

general domains which in practical terms means that we

talk about single integrals.

Up to now we would only consider integrals like this.

In other words, we just integrate over an interval.

But now we will learn how to integrate over curved

domains like this.

Curved one-dimensional, so that means this yellow.

And it could be closed like this.

By the way, this has no boundary either, just

like the circle.

Neither of them has a boundary.

This one has a boundary.

And of course you could have a curved curve, a general

curve which has a boundary.

This minus this as a boundary.

So there's a whole variety.

There are closed curves without boundary there are curves

with the boundary.

The curves could be flat, like this.

Or the could be curved.

We'll have to learn how to integrate over all of them.

And that's just a single integral.

For one dimensional integrals.

For two dimensional integrals we will have to learn how to

integrate over general surfaces in r3.

For example over a sphere, or hemisphere, or over

[UNINTELLIGIBLE].

In case you get worried, that's it.

Because when we comes to three dimension, to have and a curved

three dimensional domain we have to go to four

dimension space.

And that's subject for the next course.

Season four.

We don't do it here.

We stop at three dimension.

We only consider in this class objects which can be embedded,

immersed in a two dimensional or three dimensional space.

And therefore as far as the three dimensional domains

are concerned, they're all going to be flat.

For example, the interior of the sphere or a ball is

three-dimensional and unlike this sphere, which is two

dimension and curved, the ball is three dimensional and flat

because it just sits in a three dimensional space

which is flat.

To have something curved which three dimensional you'd have to

introduce one more dimension, you'd have to think of -- but

it will be very difficult to visualize, so we're

not to do that.

So now we need to talk about the integrand, we need to talk

about the objects that we will integrate over those domains.

Question.

That's right.

I would take this, so the question is what would be the

analogue of pass through the boundary of this picture.

So on this picture on the left-hand side we would have

that a double integral over this domain and on the right

hand side of the formula we would have a single integral

over the boundary of this domain.

No inside will be on the left.

When I say domain I mean the interior.

That's the shaded area here, that's the domain.

It's two dimensional.

An example of a kind of formula which we will prove

will look as follows.

On the left we'll have a double integral over this, let's call

this r of something, which I'm not going to specify now, and

on the right-hand side we'll have the integral over this,

over the boundary of r of something else.

Any other questions?

Right, what did I mean when I said that the upper

hemisphere has a boundary which is a circle.

I don't have good visual aids, unfortunately, so I have to

just use my hands like this.

Think of an upper hemisphere.

Let me try to draw this.

It's going to look something like this.

That's the dome.

But it does have a boundary, which is this.

The circle at the bottom is its boundary. [UNINTELLIGIBLE]

The boundary means the following: the difference

between a point here and the point in the interior is the

following: if you have a point in the interior can go in

any direction you like.

There are two degrees of freedom, two independent

directions in which you can go.

By the way the same here if you take a point in the interior

you can go in any direction, but if you are on the boundary

you can also go in all directions but when I say in

all directions I mean to go in some directions and stay

within the domain.

If you want to stay within the domain, you can go inside, you

can go along the boundary, but you're prohibited

from going outside.

So some directions are prohibited.

That's what it means to be on the boundary.

Likewise if you look at the circle at the base of

this dome you see that you cannot go down.

This is prohibited.

You can only go up on the dome or you can go

along on the boundary.

That's the difference between these points.

But even if you have a point very close to the boundary you

can go down a little bit.

And other questions? [UNINTELLIGIBLE]

Not every, because see I have not put the integrand yet.

We'll have to learn that.

See the trick is that at the level of domains you take the

boundary but that the level of the integrand you take

the derivative this way.

So, in other words, what you will have here, schematically I

write it as a derivative of something, which is on

the right hand side.

So it's not the most general integral you see here.

It's the integral of something which is derivative of

something from the right.

So this is not a formula which allows you to express any

double integral as a single integral, but only allows you

to express a double integral of something which

is a derivative.

As a single integral of the quantity of which

it is a derivative.

But we'll talk about this more later so I don't wnt to spend

too much time making it precise at the moment.

What we need to do now is describe the kind of objects

that we'll have to integrate to really do justice to

this kind of formula.

And the point is that it's not enough to just integrate

functions. we have to include more general objects, which

are called vector fields.

And everybody knows what this concept is, maybe not the term.

It's actually a very intuitive concept which

I'm going to explain.

The way to think about is as follows: first of all

let's remember what is a vector to begin with.

What is a vector?

Everything I say can be applied in r2 and r3, on the plane and

in space, but it'll be easier to stay in r2, because there's

fewer components and so on.

Let me say it first in r2 and then we'll get

generalize into r3.

In r2, on the plane.

What is a vector in r2?

A vector in r2, as we know -- well first of all we can draw

it like this, it's an object which has not only magnitude

but also direction.

This is a vector.

Algebraically we are presented by a pair of numbers,

we write it as P,Q.

It's not just a single number, it can be represented but

a pair of numbers once you choose a coordinate system.

P,Q a and we also use a different notation Pi plus Qj,

you remember we've learned, i and j was our notation for the

unit vectors on the x-axis and along the y-axis.

That's a single vector, that's just one vector.

But suppose you would like to attach a vector to

each point on the plane.

Suppose we have attached the vector, a quantity like this,

but to point on the plane.

So what is this going to look like?

Say this point will have this vector and at this point we'll

have that vector and then this have this vector and this

point will have this vector.

So a priori they can all be different.

There should be no connection.

The simplest situation would be just that assigning the same

one to all points, but that's kind of boring so you can

assign them in a very complicated way.

This actually occurs in practice, in reality.

In many different ways.

The simplest way to think about it is wind maps.

It's funny because in the book, the picture which illustrates

vector fields is the picture of the wind map of San

Francisco Bay Area.

So I guess for good reason because our winds change and

have an interesting pattern.

Some years ago, I used to do a lot of windsurfing, and I

remember there was this web site where you could go and you

get a beautiful picture of the winds, real-time, all across

the San Francisco Bay.

If you want to wind surf.

Let's say that this is a bay, roughly.

This is San Francisco, this is the Golden Gate Bridge

This is San Francisco.

It's very schematic.

This is Berkeley.

I'm looking from the top.

It's important to say that, I'm looking from the air.

So there are many spots in the bay for wind surfing.

So let's say here, the Berkeley Marina, Emeryville, here you

have Treasure Island and so on.

If you look at this map you know right away that

the wind here is east point, eastward winds.

And then here will be north and here will be south.

That's the meaning -- you've all seen such maps.

You can imagine such maps.

Because the thing about wind it is that the wind does not just

have magnitude, the force of the wind, but also

has direction.

It can point in the northern direction, the east, the

west, south and so on.

Wind, at a given point, is a vector.

It's not a number, it's a vector.

But wind is everywhere, so therefore at each point you

have a vector, a wind vector.

So wind represents a vector field.

So this is a vector field.

So a vector field is an assignment of a vector to each

point on the plane, or maybe some given domain on the plane.

Not everywhere but I mean, of course -- say here I talk about

a particular domain which is on the map which is just San

Fransisco Bay area, I'm not talking about the entire United

States or the entire state of California.

So this is a vector field.

Is this clear?

Now let me contrast that to a function.

Up to now we've talked about functions we've been

integrating functions.

A function will be just -- a function means assigning

to each point not a vector but a number.

For example, think of barometric pressure.

At each point you have a barometric pressure and

barometric pressure is not a vector, it's a number.

So at each point you have a barometric pressure, at each

point you have a number, so you have a function.

What we normally call the function is assignment

over number of each point on the plane.

And vector field is almost like this, the only difference is

that we have assign to each point a number but a vector.

But in some sense, why not?

Okay so how would we represent such a vector field

algebraically?

Algebraically each of those guys, each of those vectors

would have two components but now these components

depend on the point.

So a point x,y on the plane would have a vector which

would have components P of x,y and Q of x,y.

So we will write P of x,y and Q of x,y and we will the square

brackets as we usually to.

Or we could also write it as P of x,y times i

plus Q of x,y times j.

Just like we did before.

PQ or Pi plus Qj.

The only difference is that now P and Q themselves

become functions.

So it's P of x,y Q of x,y.

And we will call this vector field f of x,y.

Oftentimes for practical purposes you could say that a

vector field is nothing but a pair of functions, so it's not

a single function, it's two functions, because a vector

has two components.

And now each component becomes a function of x,y.

But really reducing it to a pair of functions is not a good

idea because it kinds of strips the vector of its dramatic

meaning because a vector is much more than just

a pair of numbers.

First of all this pair of numbers can only be given once

we choose a coordinate system.

After all what are these numbers?

P and Q?

I have drawn them here.

Those numbers are the projections of the vector

onto the x and y-axis.

But if I choose a different coordinate system the vector

will stay the same, but the projections will be different.

So you have to remember this algebraic presentation really

depends on the choice of coordinates.

But the vector itself doesn't.

In other words, let's say here is a map of San

Francisco Bay Area.

San Francisco Bay Area doesn't know anything

about coordinates.

It exists without any coordinates.

Somebody will come and make the coordinate system like this,

but somebody else will come and make it like this.

Who knows?

Maybe for some purposes this will be more convenient.

The wind doesn't know about the coordinate system.

The wind will blow in this direction no matter what

coordinate system you prescribe.

So for wind is for real, it's a real dramatic object.

And the two numbers we assign, or two numbers which we used to

be represent this vector, is an abstraction, is a the way to

represent them algebraically.

This way of representing it algebraically depends on the

choice of coordinates, so that's why I'm saying that

reducing a vector to a pair of numbers, or reducing a vector

field to a pair of functions somehow doesn't do

justice to the object.

It sort of strips it of it dramatic meaning.

Nevertheless, in our calculations we will be working

with vector fields as though they were just represented by

these two function so for all practical purposes we just have

to integrate those functions or differentiate those

functions independently.

[UNINTELLIGIBLE]

Let's say at this point, this is point x,y, so

you have this vector.

So here's another point, which we'll have x,1

and y,1 coordinates.

At this point -- let me use a different chalk

to represent vectors.

So this vector goes like this, this is the vector

assigned to the point x,y.

The vector assigned to this point will be this vector,

so this vector will be P of x,1 y,1 comma Q of x,1 y,1.

A priori has nothing to do with this.

At each point you have a vector. [UNINTELLIGIBLE]

A function represents a bunch of numbers, so in other words,

this function is a number attached to each point x,y for

any x,y you have some number, and now you have these two

functions assembled into a vector.

So you have a vector each x,y.

The vector attached to x,y would be this vector, and that

vector attached to x1,y1 will be this vector, the vector

attached to x2, y2 will be this vector and so on.

It's represented by a formula.

Let's look at an example of a vector field.

The simplest example would be f of x,y is just i.

What does it mean?

That means that p in this case is 1, and q is zero.

From the point of view of this general formula, p times i plus

q times j, p is 1 and q is zero so I just write it as i.

What is the genetic of this formula?

It says that the vector attached to any point is

one and the same vector i.

So what does it look like?

Well at each point I just have vector i.

This is vector i.

A vector that is parallel to the x-axis and has magnitude 1.

So here I have this vector, and here I have this vector.

I cannot draw them all because I cannot account for all the

points on this blackboard, or even a small part of the

blackboard because I think there are too many of them.

So I'm just drawing a few of this.

This that's what's called a diagram, the vector field

diagram and it's not meant to represent the entire vector

field because inevitably no matter how may vectors I will

draw I will miss some points.

This is just meant to give you an idea as to what the

vector field looks like.

Of course, the point is that in this studies, in this course we

are going to consider the kind of vector fields that we will

consider will be nice, in the sense that the vector field in

general, the vectors will change from point to point, but

they will change it a continuous manner,

in a smooth manner.

So even if you know what the vectors look like at a certain

number of points, that should give you an idea as to what

vectors look like in between.

So this is the simples vector field it's called

a constant vector field.

Let's do something more complicated.

Question?

Yes.

If we say vector field on the plane.

That's right.

Very good question.

The question is do we have to assign a vector to every

point on the plane or not?

And it's the same kind of question as asking, suppose you

have a function does it mean that we have assigned a

number to each point.

And the answer is it depends.

By default if I just say this is a function on the line or

the plane, I'm saying that it is defined everywhere.

But we know that there are many interesting functions which

are not defined everywhere.

For example, even in a one dimensional context, say

functions f of x equals 1/x.

This is an important function and it's not defined at zero.

So this function assigns an number to every point x on the

line, except x equals zero.

Sometimes we only want to know the values of the function on

this certain domain, let's say on an interval from zero to

1 and we don't care what happens at other points.

Same goes with the vector fields.

If we say vector fields on the plane we are implicitly

saying that there's a vector attached to every point.

In many situations a formula that we write will not be

well defined at some points on the plane, right?

So then it means that the vector field is not defined or

it could be that we just just only care about the values

across a certain domain.

So if we only define it on that domain then we don't care

what happens outside.

Any other question?

So next example.

Let's say F of x,y equals x times i -- let me

use i -- xi plus yj.

So this is slightly more interesting.

This is certainly not a constant vector field because

now the function P of x,y is not a constant function

but it is function x.

And Q of x,y is a function y.

So what does it look like?

For example, let's say we take the point 1, zero.

If we take the point 1, zero.

I should put and arrow, when I talk about vectors I have to

put arrows on top that's the usual notation.

So let's say F at 1, zero.

This is the point 1, zero.

If I substitute 1, zero I get i, so here is

going to be vector i.

Just like in the previous example, but this is really

the only point where it is going to be a vector.

Let's take, for example, the zero, 1.

Zero, 1 we get j.

So this is a point zero, 1 and we get this vector and likewise

you could take negative 1, you could put plus minus 1 here and

this would be plus minus i, so this will go in this direction.

This will be negative i at this point.

And likewise here plus minus y will be plus minus j, so

this will be this vector.

So this already gives kind of tells you what the pattern is.

The pattern is that at each point, in fact, the vector

is going to go in the direction of the position

vector of that point.

For example, this is point 1, zero.

The position vector of this point is this vector.

And the value of our vector field is going to be

just that vector.

But it wouldn't be right to draw this vector starting at

this point, because we are drawing this picture with the

understanding that the vector that starts at this point is

the vector assigned at this point.

But, as we discussed many times, the same vector can

be presented in many different ways.

It can start with any point.

Here we start with the point to which it is attached.

This vector is attached to the point 1, zero

so we draw it here.

This vector is attached to the point zero, 1 as we did here.

So we do it here and more generally if we have a point

with coordinates x,y the vector attached to this point is going

to be xi plus wj, but that just happens to be, in this example,

it just happens to be the position vector of this point

which we normally draw like this.

So what should I draw then for this point?

I have to take this vector and transport it here so that

the initial point is here and not here.

I end up with -- transported here.

So that's the vector ij, this is vector xi plus ij, which is

prescribed to this point and it's the same as this vector.

So I can now kind of draw.

So you see they all move away from center, move

away from the origin.

At each point the direction is exactly away from the origin,

but the magnitude changes, the magnitude the is equal to the

distance from the origin, so if I go far away the magnitude

will be even higher.

If I am very close the magnitude is very small.

So that's the vector field that I get.

It's kind of an explosion.

It's not quite an explosion, if i have an explosion it actually

decays if you go farther away but here actually the

force increases.

The magnitude increases.

It's like the universe expanding.

That's right.

That's one of the scenarios.

We would like to think that it's expanding and it is

expanding now, but we don't know what will

happen eventually.

It might all collapse.

But not before the end of the semester, don't worry.

Unless some of those formulas are incorrect.

Let's suppose that you wanted the same kind of deal, you

wanted the vector field which also goes away from the origin

but you didn't want the magnitude to be higher and

higher, greater and greater.

Let's say you wanted the vector field where each vector

has magnitude 1.

This is very easy to do.

You could just convert each of those vectors into

a unit vector pointing in the same direction.

So third example would be, you would have to take x times i

plus y times j and divide by the magnitude.

And the magnitude is x squared plus y squared.

The square root of x squared plus y squared.

So now each of these vectors is a unit vector pointing in the

direction of the position vector like before.

So in other words the magnitude is not increasing but

stays fixed equal to 1.

Or if you want the magnitude vector field to decay you can

divide by a higher power of this.

So for instance you can also take F of x, y equals xi plus

yj but divide by -- actually you could divide by the square

of this, but it is a little better to the divide by the

cube and I will explain why. x squared plus y squared 3/2.

If we do that then the magnitude of this vector,

the length if you will, will be what?

When we take the magnitude we get square root of x squared

plus y squared in the numerator, we get --

actually, how do I want it?

I want to have twice the square of the length -- that's right.

No it was correct.

That's exactly what I want.

So the magnitude would then be 1 divided by x squared plus

y squared which is the square of the length.

Which is square of the x y.

So that corresponds to the vector field where the

magnitude is proportional or equal to the inverse square of

the length, of the distance.

And this is very typical for very important

forces of nature.

So this is going to be, the behavior of the vector field

is like the behavior of the force of gravity.

Or the electric force force, the electro-magnetic force.

The force of gravity is inverse proportion to the

square of the distance.

That's exactly what we get.

But because in the new numerator we have the position

vector itself, we divide by -- instead of dividing by the

second power we divide by the third power, so that when we

take the length there will be one power here, and

three powers here.

So on two powers on the left.

Because I would like to get the vector field, such that the

length of a vector is equal to 1, divided by the square

of the length of this.

Because this kind of behavior is characteristic of important

forces in nature, like the force of gravity.

If you have two masses, say sun and earth, two masses, two

objects, then the force of gravity is inversely

proportional to the square of the distance between them and

also proportional to the masses, or some coefficient

of proportionality.

Do we do what?

Do we cube every problem we do?

This is just one example.

I'm not saying that all vector fields look like this.

This is just one vector field.

Just trying to illustrate how you could produce new vector

fields by starting with some known vector fields.

So the next question is, so if that these are the vector

fields, next question is how do we have the integrate

these vector fields?

What I am going to explain it how to integrate vector

fields over curves.

That's right.

Why would we be interested in this.

That's the question.

This is a math class.

Of course, we wouldn't be doing that if weren't a good reason

for it and there is a good reason for it, which I am

going to explain now.

What I mean by this comment is that in math I'm not going to

say that we do useless stuff, on the contrary, we do

very useful stuff.

But sometimes it takes a while to really appreciate what

exactly this is good for.

You have to be patient.

But in this case actually you don't have to be patient

because I'm going to explain it right away.

So this is one of those cases.

So I would like to explain the real problem which will

actually lead us naturally to the notion of an integral of a

vector field over a curve.

And it's actually very easy to explain.

This has to do with calculating the work done by a force.

By the way, one important example of a vector field,

which I have kind of alluded to here, when I said the force of

gravity, I should say more forcefully, probably.

Which is that forces, in general, are represented by the

vector fields because a force also has a magnitude

and direction.

A force pushes you in a certain direction, not only, with

different levels of strength.

It also pushes you in a certain direction, pulls you in

a certain direction.

Like force of gravity.

Force of gravity pulls me down and not -- you know I'm not

flying out of this class, some of you may have wanted

but I'm staying here.

The only thing that happens is that this floor prevents me

from falling down, but if suddenly this floor would

disappear, I would go down.

I would go down, not up and not right, not left.

That means that the force of gravity, in this case, action

has a very specific direction, down.

Like the apple which fell on Newton's head.

So force has a certain direction.

Likewise electro-magnetic force, other forces, they are

represented by vector fields.

And if you have a force that oftentimes you may be

interested in the amount of work that a certain force, a

force vector field does in moving an object from

one point to another.

From point a to point b.

And the upshot is that this kind of quantity, the work done

by force is precisely represented by an integral over

a curve, where the curve would be the path along which the

force moved the object.

Let's say article, if it's an electromagnetic field.

Or if it's the force of gravity it would be any object.

But to explain why an integral, as always, we have to go back

to kind of the elementary situation where everything

is simplified.

Where the force is constant and the displacement is linear and

we look at the formula for the work and then we kind of

average it out over a general curve.

And that's how we get to the integral.

So the first question I'm going to ask is what is the amount of

work that is needed to move a particle from this point to

this point along a linear segment with some

displacement delta r.

If you have a constant vector field, if a force is constant,

forces everywhere are constant, like in my first

example where it was i.

What is the amount of work needed to move an article

from point a to point b.

And the constant force vector field, like this.

The value of the force is the same everywhere.

And actually it's very easy to illustrate this in

the following way.

I will do it not with a particle but with -- this

is what I'm talking about.

I'm moving this, unfortunately I have erased it, but I

hope you memorized it.

I'm moving this eraser from this point to this point

and I'm applying force.

Now the most efficient way to do it would be for the force to

be exactly parallel know the blackboard, but in reality I

don't do this, because I also want to -- I'm going to

push sort of against the blackboard.

So first of all I don't lose the eraser also

it erases better.

So in other words, oftentimes the displacement vector and

the force vector are not parallel to each other.

That's natural.

For example, I can go like -- there is a component of my

force which is moving it this way, but there is also

component which goes this way and keeps it close

to the blackboard.

I can do it like this, or I can do it like this.

A stupid thing to do would be to go like this.

Because then I wouldn't be able to move.

So in order to do this I have to have an non zero component

in this direction.

But the I can have some component in the

transverse direction.

The question is what is the work that has been done?

And the point is that what is essential for this process is

not that force itself, but only the horizontal component, the

component in the direction of displacement.

In other words, the work is equal to the displacement which

-- let's just write the word displacement times the

component of the force along the displacement vector.

In other words, what matters is how thoroughly I

erase the blackboard.

It doesn't matter how much I sweat doing this.

What I'm talking about is not how much work I have done,

but the question is what is the result?

And for the result it doesn't matter how much pressure I was

applying towards the board.

What matters is how much force I applied in the direction

of the displacement vector.

So that's why I get this formula.

But if so, it means that house algebraically I draw this is

the displacement vector, I have to draw it again -- and

so this is the force.

The force vector and this is displacement vector,

let's call it delta r.

And so the work done moving this particle is delta r times

f, but not f, but rather this is an angle theta, so this is

actually the length of this is f, a length of f times cosine

of the angle, times cosine of theta.

So I'm trying to say that this is just a dot

product between dr and F.

That's what I'm trying to say.

The work is a dot product of the displacement vector

and the force vector.

I this kind of simplified situation where displacement

is linear and the vector fields is constant.

If you want you can write it as F dr -- it doesn't matter.

Sorry, I should have written dr not delta r.

So that's that the idealized situation.

Now the general situation will be as follows: I will have

a more general curve, not necessarily a line segments so

it will be -- do it like this.

This will be some point a, and this b, and at each

point I will have some points like this.

So you see that the vector field changes.

It has different values at different points.

So let's suppose I would like to calculate the work done by

this force in moving this particle from this point to

this point along this curve.

So what should I do?

I should break it into small pieces, and this is a typical

thing, this should be almost reflex by now because this is

the kind of thing we always do when the talk about integrals.

We break it into small pieces and each small piece we

approximate by this idealized situation where thiss will be

-- displacement will be just linear.

And the vector field will be constant because if the vector

field varies in a continuous was, in a smooth way, more

precisely if it is differentiable.

If it's a good approximation on each, on a small subsegment

here, to think that the vector field doesn't changed so much.

So it's a good approximation to the replace it by the value at

any given point within this small integral.

The work will be equal to -- the total work be the sum of

the work on each of the small intervals, and this will be the

sum of F evaluated at some points along this, within

this interval dot delta ri.

so you know what will happen when this partition

becomes finer and finer.

The sum will converge to an integral.

This is very reminiscent of all kinds of integrals

that we've done up to now.

We've done double integrals, we've done integrals, we've

had triple integrals.

It's always the same idea.

If you have an expression like this, where you have a sum over

a small -- over this kind of partition then in the limit you

just replace this by an integral and this is valid if

the quantities involved satisfy some natural conditions

like being differentiable.

so the result of this is an integral of F dr -- I would

like to say from a to b, but it would be a little bit, it would

not be a good idea to write it like this.

When we were in one variable calculus, there was only one

way connect point a with point b, because we were going

along the single line that we are given.

Here we are on the plane, and there are many different ways

to connect these point and, of course, the answer will depend

on which way I'm going.

The longer the path, the longer the work, the larger the

work that will be done.

So instead I will write it like this, over C.

And I don't mean a, b, c.

I just want to denote the curve by letters C.

A and B denote the end points but C denotes

this entire curve.

It is important, however, to remember the orientation

of this curve.

We go from A to B and not from B to A.

If we were going from B to A, the calculation

would be different.

Let's talk about this formula.

So this was to convince you that an integral of a vector

field over a curve is important.

It gives you, the point is like the work done by a

force on a certain particle.

But now I have to explain in more detail what exactly

this integral stands for.

So I would like to explain it in more detail and also give

you a recipe on how to calculate it.

Let me write it down one more time.

F dr over c.

So for the dot product we have a nice formula in terms of the

components of the vector.

So let me remind you that f is a vector field

which looks like this.

It's P of x,y times i plus Q of x,y times j.

What is dr?

dr is the vector which is dx time i plus dy times j.

Let's take the dot product of these two.

The dot product is going to be P of xy dx plus Q of xy dy.

In retrospect the result looks very natural, because in single

variable calculus the only quantities that we were

allowed to integrate where integrals like this.

F dx, F of x dx.

First of all, there was a function which depended on only

one variable because it's single variable calculus, but

also there was only one dx.

We could only integrate against dx because there

were no other variables.

Now we are on the plane.

So there is dx and there is dy.

So we could integrate a quantity of the form P times

dx and we could integrate a quantity of the

form Q times dy.

And this is the most general expression which we could

actually integrate over a curve in two dimensional space.

Now we got this by thinking in terms of the work done by force

and by writing as a dot product of the force vector

field in dr.

But you see the end result is actually very nice and you we

could have actually written this end result from

the beginning.

As the most natural expression really that we could think of

integrating over a given curve.

Again because you have dx and dy and N of them could come

with some factor, some function, which will be some

function P and some function Q.

So now the only thing that remains is we need to explain

how to actually compute this and this is where the methods

which we developed at the very beginning of this course are

going to become very useful.

I'm talking about, of course, parametric curves.

To compute this integral, we need to parametrize our curve.

In other words, the curve was like this, but with

parametrizing, which means that we actually identify it with

a certain interval.

An interval on some auxiliary line with a coordinate

which we usually call t.

So we identified this curve on the plane with a straight

object, with a line segment on this auxiliary line.

Identify this point with this, and we identify this point with

this, and the point in the middle I identify

in a certain way.

So that there is a 1:1 correspondence between points

here and points here.

Algebraically we write it like this, we say x is f of t and y

is g of t along this curve, and we put some limits

on t, from A to B.

Here are the end points were actually points on the plane,

that's why I called them capital A and capital B because

actually capital A was actually a pair of numbers.

It was some, which actually now I can write as

f of a and g of a.

And this b is f of b and g og b.

Let's substitute this in this formula, so what do we get?

We will get integral now just over this integral, which is

great because this is already kind of an integral

which we did in single variable calculus.

Before it was an integral over this curve but now we write

this integral over this curve as an integral over the line

segment by using this parameterization.

We have to substitute, instead of x and y we have to

substitute f of t and g of t.

And then we have to write dx in terms of t.

But what is dx? dx is, of course, just f prime dt, just

by usual formula. dx is f prime of t dt, and dy

is g prime of t dt.

So I replace this dx by f prime of t dt.

And then I write the second one which is Q of f of t g

of t times g prime of t dt, then close the bracket.

Or, if you will, you could write it as a

sum of two integrals.

This is the first integral and this is the second one.

But you see I have, by using the parametric form for this

curve I have been able to the rewrite this integral which

involved x and y as an integral in one auxiliary

variable, namely t.

We have one minute to do an example of this.

So let's say compute the integral of z dx

plus x dy plus y dz.

On purpose I'm doing it in three dimensional space just to

show you that actually everything works out way in the

same way in the three dimensional space as in a

two dimensional space.

The difference is that now you have three variables, x, y, z

and likewise you have dx, dy, dz.

And so, let's say you have a curve in which x is t squared,

y is t cubed, z is t squared.

And t goes from zero to 1.

So the first the term -- so likewise put dz you will have

some h prime of t dt where h is the formula for

[UNINTELLIGIBLE].

So this will be an integral from zero to 1, z is equal to t

squared, dx is equal to 2t times dt, sorry it was z dx, x

is equal to t squared, dy is equal to 3t squared and finally

here I have t cubed and dz is 2t dt.

So you end up with a very simple integral

in one variable.

We'll stop here and we'll continue on Thursday.

You actually did pretty well.

I was happy about it.

Good job, keep on doing it.

You know these lectures are on web cast dot berkeley dot edu.

They're also on iTunes and in the iTunes version I

sing all the formulas.

And now our lectures are on YouTube.

So I'd like to welcome our new additions, new

addition to our audience.

Our YouTube audience.

I think it's great because actually I tried to search for

math courses on YouTube and I was surprised that there are

actually very few of those.

A few people want to learn about multivariable calculus,

you've come to the right place.

And even if you don't want to learn about multivariable

calculus, you've come to the right place.

You know, unfortunately there's only one lecture available for

now and it's like when you start watching a television

series and it's in its third season and you don't know what

happened in the first season or the second season.

It's exactly the same here because I would say first

season is from the beginning to the first midterm and second

seasons is from first midterm to the second midterm and now

we start our third season.

I personally think the first season with the best one but

I think the third one will be the best.

Of course all the lectures are available on bypassed webcast

at dot berkely dot edu and I was told that they will all be

available on YouTube perhaps this week there were some

problems without uploading.

So if you want to know what happened previously on

math 323, stay tuned.

Having said that, what is this third season, so to

speak, what is it about?

I personally think this is the best one, it's the most

interesting one, because that's where we will actually be

able to apply all the stuff we've learned up to now.

This has been a sort of a recurring theme in this course,

I've said it many times, that calculus you can basically

break into two parts.

Differential calculus and integration calculus.

And in some sense integration and differentiation are

operations which are inverse to other.

In some sense you can say very roughly that integration is

like differentiation inverse.

This is the thesis.

I will have to follow up on this and I will have to explain

what exactly this means in the context of multivariable

calculus.

We all know what that means in the context of single

variable calculus.

In the one variable calculus we have the beautiful formula

which relates integration and differentiation which is called

the fundamental theorem of calculus or single

variable calculus or the Newton-Leibniz formula.

And this formula looks as follows: on the left hand

side you have an integral.

And to make this formula to support this thesis, I would

like to write it in this way.

Integral of a function which itself is a derivative

of another function.

So you put a function here which is derivative of some

function f so you put f prime.

And you integrate this function from a to b.

And on the right hand side we simply have the values of this

function at the end points.

So we have f of b minues f of a.

So I would like to think about this formula in

the following way.

On the left we integrate our function over an

interval from a to b.

So, on the left I have an integral over this domain

which is just an interval.

Well in a single -- and by the way no walking in front of

the front row it looks really bad on YouTube.

On the left we have an integral over this domain and because we

are in single variable calculus now there are not so many

domains to go around.

The only domains you can think of are intervals or perhaps

unions of intervals.

So that's what we have on the left side.

And on the right hand side we have these two points which I

would like to think of as the boundary of this interval.

So this is an interval and this is the boundary

of this interval.

Two points, let me put them as dots.

Something interesting happens here. f of b enters with

a plus sign and f of a enters with a minus sign.

So actually this point a comes with a plus sign of this point

comes with a minus sign.

And we can actually read this off from the orientation

of this interval.

This interval has an orientation.

We integrate from a to b, so there is an orientation.

So the arrow points from a to b, therefore the end point

comes with a plus and the initial point a

comes with a minus.

So you see this already indicates that orientation

is going to play an important role.

But for now the essential point is the following: on the left

you have a domain interval, on the right you have the

boundary of this domain.

That's about the region of integration.

In some sense you can think of the right hand side also as

integration, it's just that you are integrating over zero

dimension domains which means simply evaluating the function.

What do negative and positive mean here?

For now it's just notation that I've put.

I assigned plus to this point which is point b and I

assigned minus to this point which is point a.

The practical meaning of this is the following: that when I

evaluate the integral I evaluate at this point with the

sign plus and at this point with the sign minus, which

is what is really here.

There is some rule here that some part of the boundary comes

with coefficient plus 1 and some part of the boundary comes

was coefficient minus 1.

But integral, is in a way, a sort of the

union of two things.

The integrand, the object you integrate, and the domain

of integration, the domain over which you integrate.

You cannot make an integral with just one of the two.

You need both.

The integrand and the domain of integration.

In this formula there is a trade off between something

that happens to the integrand and something that

happens to the domain.

We see now what happens to the domain, when the go from left

hand side to the right hand side we pick the

boundary of the domain.

On the other hand, let's look at the integrand.

On the right, the integrand is f and on the left the

integrand is f prime.

What we do when we go from right to left

is we differentiate.

Here we take the boundary.

And this is the basic structure of the formula.

So what we would like to do is to find analogues of this

formula in multivariable calculus.

That's what they're going to do in the remaining

part of the semester.

In other words, what we would like to do is to prove formulas

of the following type: on the left we'll have an integral of

some quantity and on the right we'll have an integral

of some quantity.

On the right it will be integral of some constant,

let's say omega, I will explain what kind of

object this will be.

In some cases it will be a function.

We will also have more general options integrated here.

On the left will have some sort of derivative of this object.

This is just a schematic notation for now.

Don't read too much into it.

Something like taking the derivative, for example

here just taking f prime.

So that's as far as the integrand goes.

And as far as the integration domain goes, here we will have

some domain D and here we will have the boundary of D.

You had a question.

So here I'm talking about the domain of integration.

We have a formula.

A formula has a left hand side and a right had side.

And each of them is an integral of something the

function over domain.

So i'd like to trace here on this blackboard what happens to

the domain and what happens to the boundary as we go from left

to right or from right to left.

And I observe that for the domain on the left we start

with an interval but on the the right we take the boundary

of this interval.

It's like going from the from the left hand side to the

right hand side for domains corresponds to taking

the boundary.

But as far as the integrand is concerned, on the right I

have some function f and on the left I have f prime.

So going from right to left I differentiate.

So this formula could be thought of as a trade-off

between taking the boundary and differentiation.

Taking the boundary at the level of domain, taking the

derivative or differentiating the integrand.

So this is the basic structure -- this is how I would like to

think about the fundamental theorem of calculus -- of

one variable calculus -- which is written here.

I would to first of all think of taking an integral as

combining or pairing two different types of objects.

One is algebraic, say a function, the other one

geometric, a domain.

That's the first thing and if I think about integration in this

way, then I clearly see what is the structure of this formula.

At the level of domains it's going to the boundary, at the

level of functions it's taking the derivative.

So this suggests that perhaps if I am on not on the line but

on the plane or in three dimensional space, if I'm doing

multivariable calculus, now I should also have a, formula

like this where on the right I would have some object

integrated and on the left I would have some

object integrated.

But the correspondents would be like in the fundamental theorem

of one variable calculus.

That is to say at the level of domains we go from the domain

to its boundary, and at the level of integrands we go from

omega, whatever it is, this we will have to discuss in more

detail, to its derivative of some sorts.

That will be the basic structure of the formulas

that we will derive.

And they will be the analogues of the fundamental theorem of

calculus and the will allow us to learn a lot of things

about integrals.

And we will also talk about various real-life applications

of these integrals.

What do these integrals mean?

And various applications of those integrals.

And such formulas will allow us to express integrals of one

kind in terms of integrals of the other kind.

That's basically the outline of what we're going to do next.

So now the question is to discuss what kind of objects

we should be integrating and over what?

The integrands and the domains of integration.

What are the most general objects which we should

introduce which we could then use to obtain a formula like

this and what is the meaning of the corresponding integrals.

These are the question that we will have to ask now.

So let's remember first of all what kind of integrals we

have learned up to now in multivariable calculus.

We have learned, in the second season so to speak, in the last

month or so, we've learned how to integrate functions --

integrals, double integrals of functions and triple

integrals of functions.

Over some regions in the two dimensional space on plane and

over solids, three dimensional regions in the three

dimensional space.

But in those cases we have integrated functions.

It turns out that this type with of integrals are not

sufficient to do justice to the kind of formulas

which I promised you.

We will have to generalize these integrals first.

And there are two types of limitations that we have now.

Two types of things that will have to relax and consider

more general integrals.

We will generalize these integrals in two ways.

First of all we will learn how to integrate more general

objects and functions.

We will integrate vector fields.

We will learn how to integrate vector field.s.

We will talk about what they are, vector fields.

And what is the meaning of the corresponding integral.

And that'll be the first kind of generalization

that we will make.

The second generalization is that we will consider

more general domains.

They will be roughly, the difference is that

they will be curved.

We will consider most general curved domains, whereas the

domains we have looked at up to now are flat domains.

Let me explain what I mean by this.

Flat versus curved.

You can just illustrate this.

The kind of regions that we consider here, the two

dimensional regions are -- it could be some region

on the plane like this.

And someone might say well this is curved.

It kind of looks curved.

That would be misleading because what is curved is not

the domain itself, which is the interior of this

curve, but the boundary.

In other words, in this subject, one of the most

important things is to distinguish domains

and their boundaries.

That's why I would like to emphasize it and I would

like to trace this boundary in a different color.

Yello.

This is a boundary and this is a domain.

These are two different things.

This is a domain.

For example the domain is two dimensional, it's on the plane.

But the boundary is one-dimensional.

And this is always the case.

The boundary of an n-dimensional domain is going

to n minus one-dimensional, you lose one dimension when you

pass through the boundary.

Same things happened here.

Here the domain was an interval.

It was one dimensional and the boundary was zero dimensional,

it consisted of two points.

Maybe I should make them yellow also to keep the analogy.

So the boundary here is curved.

It is curved, I wouldn't disagree with that.

But the domain itself is flat.

It's flat in the sense that if you look at any point here,

and you look at the small neighborhood of this point.

If you look at a small neighborhood of this point you

don't know what the boundary looks like, so you don't have

any illusions about the domain.

You just look at the very small part of this domain here.

And you see that it's just the same as a small part of the

plane around this point.

And the plane is flat. r2, the blackboard is flat.

So locally this domain everywhere looks like

the blackboard itself, so it is flat.

You should contrast that with the boundary itself.

The boundary is not flat.

A flat one-dimensional domain would be an integral

or a line segment.

So this will be flat.

For example this is flat, this is also flat.

So you could have, of course, the domain which is flat and

its boundary is also flat.

It migh have some corners.

So you see what I mean for the one-dimensional boundaries,

here it's curved, here it's flat.

Then you could ask, well give me an example of a two

dimensional curved doman and that's really easy,

think of a sphere.

A sphere is curved.

Because a sphere cannot be, it's two dimensional, it's the

same dimension as the plane, same dimension as the

blackboard, but you can't put the sphere inside

the blackboard.

I mean you can project that but you will have to make some

violent transformation to the object.

You will change its geometry.

So that's the point.

A curved two dimensional object cannot be fit in

a two-dimensional plane.

Anything that is two dimensional and that fits

in the two dimensional plane is flat.

There are a lot of two dimensional objects like a

sphere or a [UNINTELLIGIBLE]

and many other things, if you can imagine, like the surface

of the ocean, those are also two dimensional.

But there are not like the plane, so they are curved.

So far we have not really talked about integrating

over such domains and we'll need to do that here.

What's the boundary of a sphere?

That's a very good question.

So before answering this question let's talk about a

simpler question right but analogous question, in

dimension one lower.

There will be many analogies between objects of different

dimensions but always it's easier to look at the

situation where dimension is lower than higher.

So oftentimes is better to start with lower dimensions the

proceed to higher domensions.

So a good question here would be what about boundaries of

one dimensional domains.

So if you just have an interval like this, we

know what the boundary is.

It's very similar to what we discussed in one

variable calculus.

So if your domain is like this then the boundary

consists of two points.

But here's an example of a domain which does not have a

boundary or a more proper was to say it has an

empty boundary.

A circly.

That's a question that's similar to your question.

You asked me what is the boundary of a sphere.

So a simpler question would be what is the

boundary of a circle.

Boundary of a circle is empty.

So you see the difference between this and this.

In some sense you can think of you can obtain this by

combining these two end points and closing these

two end points.

It actually all make sense because the boundary, as we

discussed, in the boundary these points will enter

with different signs.

This one, if we orient it like this.

This one with sign plus and this one with sign minus.

When we close them they will sort of cancel each other.

That's consistent with our natural observation that a

circle has no boundary.

Likewise a sphere has no boundary or has empty boundary.

But if you take the upper hemisphere -- the upper

hemisphere has a boundary which is a circle.

Upper hemisphere is like a dome and the dome has a boundary

which is a circle.

There are many things that you need to be able to visualize

here and to understand and to see that there are certain

objects which do have boundaries.

There are objects which don't have boundaries.

You should always keep track of where you are.

If you are talking about the domain, the boundary inside

three dimensional space, two dimensional space.

There are many different options.

This I have now explained.

The second aspect is that we will integrate over more

general domains which in practical terms means that we

talk about single integrals.

Up to now we would only consider integrals like this.

In other words, we just integrate over an interval.

But now we will learn how to integrate over curved

domains like this.

Curved one-dimensional, so that means this yellow.

And it could be closed like this.

By the way, this has no boundary either, just

like the circle.

Neither of them has a boundary.

This one has a boundary.

And of course you could have a curved curve, a general

curve which has a boundary.

This minus this as a boundary.

So there's a whole variety.

There are closed curves without boundary there are curves

with the boundary.

The curves could be flat, like this.

Or the could be curved.

We'll have to learn how to integrate over all of them.

And that's just a single integral.

For one dimensional integrals.

For two dimensional integrals we will have to learn how to

integrate over general surfaces in r3.

For example over a sphere, or hemisphere, or over

[UNINTELLIGIBLE].

In case you get worried, that's it.

Because when we comes to three dimension, to have and a curved

three dimensional domain we have to go to four

dimension space.

And that's subject for the next course.

Season four.

We don't do it here.

We stop at three dimension.

We only consider in this class objects which can be embedded,

immersed in a two dimensional or three dimensional space.

And therefore as far as the three dimensional domains

are concerned, they're all going to be flat.

For example, the interior of the sphere or a ball is

three-dimensional and unlike this sphere, which is two

dimension and curved, the ball is three dimensional and flat

because it just sits in a three dimensional space

which is flat.

To have something curved which three dimensional you'd have to

introduce one more dimension, you'd have to think of -- but

it will be very difficult to visualize, so we're

not to do that.

So now we need to talk about the integrand, we need to talk

about the objects that we will integrate over those domains.

Question.

That's right.

I would take this, so the question is what would be the

analogue of pass through the boundary of this picture.

So on this picture on the left-hand side we would have

that a double integral over this domain and on the right

hand side of the formula we would have a single integral

over the boundary of this domain.

No inside will be on the left.

When I say domain I mean the interior.

That's the shaded area here, that's the domain.

It's two dimensional.

An example of a kind of formula which we will prove

will look as follows.

On the left we'll have a double integral over this, let's call

this r of something, which I'm not going to specify now, and

on the right-hand side we'll have the integral over this,

over the boundary of r of something else.

Any other questions?

Right, what did I mean when I said that the upper

hemisphere has a boundary which is a circle.

I don't have good visual aids, unfortunately, so I have to

just use my hands like this.

Think of an upper hemisphere.

Let me try to draw this.

It's going to look something like this.

That's the dome.

But it does have a boundary, which is this.

The circle at the bottom is its boundary. [UNINTELLIGIBLE]

The boundary means the following: the difference

between a point here and the point in the interior is the

following: if you have a point in the interior can go in

any direction you like.

There are two degrees of freedom, two independent

directions in which you can go.

By the way the same here if you take a point in the interior

you can go in any direction, but if you are on the boundary

you can also go in all directions but when I say in

all directions I mean to go in some directions and stay

within the domain.

If you want to stay within the domain, you can go inside, you

can go along the boundary, but you're prohibited

from going outside.

So some directions are prohibited.

That's what it means to be on the boundary.

Likewise if you look at the circle at the base of

this dome you see that you cannot go down.

This is prohibited.

You can only go up on the dome or you can go

along on the boundary.

That's the difference between these points.

But even if you have a point very close to the boundary you

can go down a little bit.

And other questions? [UNINTELLIGIBLE]

Not every, because see I have not put the integrand yet.

We'll have to learn that.

See the trick is that at the level of domains you take the

boundary but that the level of the integrand you take

the derivative this way.

So, in other words, what you will have here, schematically I

write it as a derivative of something, which is on

the right hand side.

So it's not the most general integral you see here.

It's the integral of something which is derivative of

something from the right.

So this is not a formula which allows you to express any

double integral as a single integral, but only allows you

to express a double integral of something which

is a derivative.

As a single integral of the quantity of which

it is a derivative.

But we'll talk about this more later so I don't wnt to spend

too much time making it precise at the moment.

What we need to do now is describe the kind of objects

that we'll have to integrate to really do justice to

this kind of formula.

And the point is that it's not enough to just integrate

functions. we have to include more general objects, which

are called vector fields.

And everybody knows what this concept is, maybe not the term.

It's actually a very intuitive concept which

I'm going to explain.

The way to think about is as follows: first of all

let's remember what is a vector to begin with.

What is a vector?

Everything I say can be applied in r2 and r3, on the plane and

in space, but it'll be easier to stay in r2, because there's

fewer components and so on.

Let me say it first in r2 and then we'll get

generalize into r3.

In r2, on the plane.

What is a vector in r2?

A vector in r2, as we know -- well first of all we can draw

it like this, it's an object which has not only magnitude

but also direction.

This is a vector.

Algebraically we are presented by a pair of numbers,

we write it as P,Q.

It's not just a single number, it can be represented but

a pair of numbers once you choose a coordinate system.

P,Q a and we also use a different notation Pi plus Qj,

you remember we've learned, i and j was our notation for the

unit vectors on the x-axis and along the y-axis.

That's a single vector, that's just one vector.

But suppose you would like to attach a vector to

each point on the plane.

Suppose we have attached the vector, a quantity like this,

but to point on the plane.

So what is this going to look like?

Say this point will have this vector and at this point we'll

have that vector and then this have this vector and this

point will have this vector.

So a priori they can all be different.

There should be no connection.

The simplest situation would be just that assigning the same

one to all points, but that's kind of boring so you can

assign them in a very complicated way.

This actually occurs in practice, in reality.

In many different ways.

The simplest way to think about it is wind maps.

It's funny because in the book, the picture which illustrates

vector fields is the picture of the wind map of San

Francisco Bay Area.

So I guess for good reason because our winds change and

have an interesting pattern.

Some years ago, I used to do a lot of windsurfing, and I

remember there was this web site where you could go and you

get a beautiful picture of the winds, real-time, all across

the San Francisco Bay.

If you want to wind surf.

Let's say that this is a bay, roughly.

This is San Francisco, this is the Golden Gate Bridge

This is San Francisco.

It's very schematic.

This is Berkeley.

I'm looking from the top.

It's important to say that, I'm looking from the air.

So there are many spots in the bay for wind surfing.

So let's say here, the Berkeley Marina, Emeryville, here you

have Treasure Island and so on.

If you look at this map you know right away that

the wind here is east point, eastward winds.

And then here will be north and here will be south.

That's the meaning -- you've all seen such maps.

You can imagine such maps.

Because the thing about wind it is that the wind does not just

have magnitude, the force of the wind, but also

has direction.

It can point in the northern direction, the east, the

west, south and so on.

Wind, at a given point, is a vector.

It's not a number, it's a vector.

But wind is everywhere, so therefore at each point you

have a vector, a wind vector.

So wind represents a vector field.

So this is a vector field.

So a vector field is an assignment of a vector to each

point on the plane, or maybe some given domain on the plane.

Not everywhere but I mean, of course -- say here I talk about

a particular domain which is on the map which is just San

Fransisco Bay area, I'm not talking about the entire United

States or the entire state of California.

So this is a vector field.

Is this clear?

Now let me contrast that to a function.

Up to now we've talked about functions we've been

integrating functions.

A function will be just -- a function means assigning

to each point not a vector but a number.

For example, think of barometric pressure.

At each point you have a barometric pressure and

barometric pressure is not a vector, it's a number.

So at each point you have a barometric pressure, at each

point you have a number, so you have a function.

What we normally call the function is assignment

over number of each point on the plane.

And vector field is almost like this, the only difference is

that we have assign to each point a number but a vector.

But in some sense, why not?

Okay so how would we represent such a vector field

algebraically?

Algebraically each of those guys, each of those vectors

would have two components but now these components

depend on the point.

So a point x,y on the plane would have a vector which

would have components P of x,y and Q of x,y.

So we will write P of x,y and Q of x,y and we will the square

brackets as we usually to.

Or we could also write it as P of x,y times i

plus Q of x,y times j.

Just like we did before.

PQ or Pi plus Qj.

The only difference is that now P and Q themselves

become functions.

So it's P of x,y Q of x,y.

And we will call this vector field f of x,y.

Oftentimes for practical purposes you could say that a

vector field is nothing but a pair of functions, so it's not

a single function, it's two functions, because a vector

has two components.

And now each component becomes a function of x,y.

But really reducing it to a pair of functions is not a good

idea because it kinds of strips the vector of its dramatic

meaning because a vector is much more than just

a pair of numbers.

First of all this pair of numbers can only be given once

we choose a coordinate system.

After all what are these numbers?

P and Q?

I have drawn them here.

Those numbers are the projections of the vector

onto the x and y-axis.

But if I choose a different coordinate system the vector

will stay the same, but the projections will be different.

So you have to remember this algebraic presentation really

depends on the choice of coordinates.

But the vector itself doesn't.

In other words, let's say here is a map of San

Francisco Bay Area.

San Francisco Bay Area doesn't know anything

about coordinates.

It exists without any coordinates.

Somebody will come and make the coordinate system like this,

but somebody else will come and make it like this.

Who knows?

Maybe for some purposes this will be more convenient.

The wind doesn't know about the coordinate system.

The wind will blow in this direction no matter what

coordinate system you prescribe.

So for wind is for real, it's a real dramatic object.

And the two numbers we assign, or two numbers which we used to

be represent this vector, is an abstraction, is a the way to

represent them algebraically.

This way of representing it algebraically depends on the

choice of coordinates, so that's why I'm saying that

reducing a vector to a pair of numbers, or reducing a vector

field to a pair of functions somehow doesn't do

justice to the object.

It sort of strips it of it dramatic meaning.

Nevertheless, in our calculations we will be working

with vector fields as though they were just represented by

these two function so for all practical purposes we just have

to integrate those functions or differentiate those

functions independently.

[UNINTELLIGIBLE]

Let's say at this point, this is point x,y, so

you have this vector.

So here's another point, which we'll have x,1

and y,1 coordinates.

At this point -- let me use a different chalk

to represent vectors.

So this vector goes like this, this is the vector

assigned to the point x,y.

The vector assigned to this point will be this vector,

so this vector will be P of x,1 y,1 comma Q of x,1 y,1.

A priori has nothing to do with this.

At each point you have a vector. [UNINTELLIGIBLE]

A function represents a bunch of numbers, so in other words,

this function is a number attached to each point x,y for

any x,y you have some number, and now you have these two

functions assembled into a vector.

So you have a vector each x,y.

The vector attached to x,y would be this vector, and that

vector attached to x1,y1 will be this vector, the vector

attached to x2, y2 will be this vector and so on.

It's represented by a formula.

Let's look at an example of a vector field.

The simplest example would be f of x,y is just i.

What does it mean?

That means that p in this case is 1, and q is zero.

From the point of view of this general formula, p times i plus

q times j, p is 1 and q is zero so I just write it as i.

What is the genetic of this formula?

It says that the vector attached to any point is

one and the same vector i.

So what does it look like?

Well at each point I just have vector i.

This is vector i.

A vector that is parallel to the x-axis and has magnitude 1.

So here I have this vector, and here I have this vector.

I cannot draw them all because I cannot account for all the

points on this blackboard, or even a small part of the

blackboard because I think there are too many of them.

So I'm just drawing a few of this.

This that's what's called a diagram, the vector field

diagram and it's not meant to represent the entire vector

field because inevitably no matter how may vectors I will

draw I will miss some points.

This is just meant to give you an idea as to what the

vector field looks like.

Of course, the point is that in this studies, in this course we

are going to consider the kind of vector fields that we will

consider will be nice, in the sense that the vector field in

general, the vectors will change from point to point, but

they will change it a continuous manner,

in a smooth manner.

So even if you know what the vectors look like at a certain

number of points, that should give you an idea as to what

vectors look like in between.

So this is the simples vector field it's called

a constant vector field.

Let's do something more complicated.

Question?

Yes.

If we say vector field on the plane.

That's right.

Very good question.

The question is do we have to assign a vector to every

point on the plane or not?

And it's the same kind of question as asking, suppose you

have a function does it mean that we have assigned a

number to each point.

And the answer is it depends.

By default if I just say this is a function on the line or

the plane, I'm saying that it is defined everywhere.

But we know that there are many interesting functions which

are not defined everywhere.

For example, even in a one dimensional context, say

functions f of x equals 1/x.

This is an important function and it's not defined at zero.

So this function assigns an number to every point x on the

line, except x equals zero.

Sometimes we only want to know the values of the function on

this certain domain, let's say on an interval from zero to

1 and we don't care what happens at other points.

Same goes with the vector fields.

If we say vector fields on the plane we are implicitly

saying that there's a vector attached to every point.

In many situations a formula that we write will not be

well defined at some points on the plane, right?

So then it means that the vector field is not defined or

it could be that we just just only care about the values

across a certain domain.

So if we only define it on that domain then we don't care

what happens outside.

Any other question?

So next example.

Let's say F of x,y equals x times i -- let me

use i -- xi plus yj.

So this is slightly more interesting.

This is certainly not a constant vector field because

now the function P of x,y is not a constant function

but it is function x.

And Q of x,y is a function y.

So what does it look like?

For example, let's say we take the point 1, zero.

If we take the point 1, zero.

I should put and arrow, when I talk about vectors I have to

put arrows on top that's the usual notation.

So let's say F at 1, zero.

This is the point 1, zero.

If I substitute 1, zero I get i, so here is

going to be vector i.

Just like in the previous example, but this is really

the only point where it is going to be a vector.

Let's take, for example, the zero, 1.

Zero, 1 we get j.

So this is a point zero, 1 and we get this vector and likewise

you could take negative 1, you could put plus minus 1 here and

this would be plus minus i, so this will go in this direction.

This will be negative i at this point.

And likewise here plus minus y will be plus minus j, so

this will be this vector.

So this already gives kind of tells you what the pattern is.

The pattern is that at each point, in fact, the vector

is going to go in the direction of the position

vector of that point.

For example, this is point 1, zero.

The position vector of this point is this vector.

And the value of our vector field is going to be

just that vector.

But it wouldn't be right to draw this vector starting at

this point, because we are drawing this picture with the

understanding that the vector that starts at this point is

the vector assigned at this point.

But, as we discussed many times, the same vector can

be presented in many different ways.

It can start with any point.

Here we start with the point to which it is attached.

This vector is attached to the point 1, zero

so we draw it here.

This vector is attached to the point zero, 1 as we did here.

So we do it here and more generally if we have a point

with coordinates x,y the vector attached to this point is going

to be xi plus wj, but that just happens to be, in this example,

it just happens to be the position vector of this point

which we normally draw like this.

So what should I draw then for this point?

I have to take this vector and transport it here so that

the initial point is here and not here.

I end up with -- transported here.

So that's the vector ij, this is vector xi plus ij, which is

prescribed to this point and it's the same as this vector.

So I can now kind of draw.

So you see they all move away from center, move

away from the origin.

At each point the direction is exactly away from the origin,

but the magnitude changes, the magnitude the is equal to the

distance from the origin, so if I go far away the magnitude

will be even higher.

If I am very close the magnitude is very small.

So that's the vector field that I get.

It's kind of an explosion.

It's not quite an explosion, if i have an explosion it actually

decays if you go farther away but here actually the

force increases.

The magnitude increases.

It's like the universe expanding.

That's right.

That's one of the scenarios.

We would like to think that it's expanding and it is

expanding now, but we don't know what will

happen eventually.

It might all collapse.

But not before the end of the semester, don't worry.

Unless some of those formulas are incorrect.

Let's suppose that you wanted the same kind of deal, you

wanted the vector field which also goes away from the origin

but you didn't want the magnitude to be higher and

higher, greater and greater.

Let's say you wanted the vector field where each vector

has magnitude 1.

This is very easy to do.

You could just convert each of those vectors into

a unit vector pointing in the same direction.

So third example would be, you would have to take x times i

plus y times j and divide by the magnitude.

And the magnitude is x squared plus y squared.

The square root of x squared plus y squared.

So now each of these vectors is a unit vector pointing in the

direction of the position vector like before.

So in other words the magnitude is not increasing but

stays fixed equal to 1.

Or if you want the magnitude vector field to decay you can

divide by a higher power of this.

So for instance you can also take F of x, y equals xi plus

yj but divide by -- actually you could divide by the square

of this, but it is a little better to the divide by the

cube and I will explain why. x squared plus y squared 3/2.

If we do that then the magnitude of this vector,

the length if you will, will be what?

When we take the magnitude we get square root of x squared

plus y squared in the numerator, we get --

actually, how do I want it?

I want to have twice the square of the length -- that's right.

No it was correct.

That's exactly what I want.

So the magnitude would then be 1 divided by x squared plus

y squared which is the square of the length.

Which is square of the x y.

So that corresponds to the vector field where the

magnitude is proportional or equal to the inverse square of

the length, of the distance.

And this is very typical for very important

forces of nature.

So this is going to be, the behavior of the vector field

is like the behavior of the force of gravity.

Or the electric force force, the electro-magnetic force.

The force of gravity is inverse proportion to the

square of the distance.

That's exactly what we get.

But because in the new numerator we have the position

vector itself, we divide by -- instead of dividing by the

second power we divide by the third power, so that when we

take the length there will be one power here, and

three powers here.

So on two powers on the left.

Because I would like to get the vector field, such that the

length of a vector is equal to 1, divided by the square

of the length of this.

Because this kind of behavior is characteristic of important

forces in nature, like the force of gravity.

If you have two masses, say sun and earth, two masses, two

objects, then the force of gravity is inversely

proportional to the square of the distance between them and

also proportional to the masses, or some coefficient

of proportionality.

Do we do what?

Do we cube every problem we do?

This is just one example.

I'm not saying that all vector fields look like this.

This is just one vector field.

Just trying to illustrate how you could produce new vector

fields by starting with some known vector fields.

So the next question is, so if that these are the vector

fields, next question is how do we have the integrate

these vector fields?

What I am going to explain it how to integrate vector

fields over curves.

That's right.

Why would we be interested in this.

That's the question.

This is a math class.

Of course, we wouldn't be doing that if weren't a good reason

for it and there is a good reason for it, which I am

going to explain now.

What I mean by this comment is that in math I'm not going to

say that we do useless stuff, on the contrary, we do

very useful stuff.

But sometimes it takes a while to really appreciate what

exactly this is good for.

You have to be patient.

But in this case actually you don't have to be patient

because I'm going to explain it right away.

So this is one of those cases.

So I would like to explain the real problem which will

actually lead us naturally to the notion of an integral of a

vector field over a curve.

And it's actually very easy to explain.

This has to do with calculating the work done by a force.

By the way, one important example of a vector field,

which I have kind of alluded to here, when I said the force of

gravity, I should say more forcefully, probably.

Which is that forces, in general, are represented by the

vector fields because a force also has a magnitude

and direction.

A force pushes you in a certain direction, not only, with

different levels of strength.

It also pushes you in a certain direction, pulls you in

a certain direction.

Like force of gravity.

Force of gravity pulls me down and not -- you know I'm not

flying out of this class, some of you may have wanted

but I'm staying here.

The only thing that happens is that this floor prevents me

from falling down, but if suddenly this floor would

disappear, I would go down.

I would go down, not up and not right, not left.

That means that the force of gravity, in this case, action

has a very specific direction, down.

Like the apple which fell on Newton's head.

So force has a certain direction.

Likewise electro-magnetic force, other forces, they are

represented by vector fields.

And if you have a force that oftentimes you may be

interested in the amount of work that a certain force, a

force vector field does in moving an object from

one point to another.

From point a to point b.

And the upshot is that this kind of quantity, the work done

by force is precisely represented by an integral over

a curve, where the curve would be the path along which the

force moved the object.

Let's say article, if it's an electromagnetic field.

Or if it's the force of gravity it would be any object.

But to explain why an integral, as always, we have to go back

to kind of the elementary situation where everything

is simplified.

Where the force is constant and the displacement is linear and

we look at the formula for the work and then we kind of

average it out over a general curve.

And that's how we get to the integral.

So the first question I'm going to ask is what is the amount of

work that is needed to move a particle from this point to

this point along a linear segment with some

displacement delta r.

If you have a constant vector field, if a force is constant,

forces everywhere are constant, like in my first

example where it was i.

What is the amount of work needed to move an article

from point a to point b.

And the constant force vector field, like this.

The value of the force is the same everywhere.

And actually it's very easy to illustrate this in

the following way.

I will do it not with a particle but with -- this

is what I'm talking about.

I'm moving this, unfortunately I have erased it, but I

hope you memorized it.

I'm moving this eraser from this point to this point

and I'm applying force.

Now the most efficient way to do it would be for the force to

be exactly parallel know the blackboard, but in reality I

don't do this, because I also want to -- I'm going to

push sort of against the blackboard.

So first of all I don't lose the eraser also

it erases better.

So in other words, oftentimes the displacement vector and

the force vector are not parallel to each other.

That's natural.

For example, I can go like -- there is a component of my

force which is moving it this way, but there is also

component which goes this way and keeps it close

to the blackboard.

I can do it like this, or I can do it like this.

A stupid thing to do would be to go like this.

Because then I wouldn't be able to move.

So in order to do this I have to have an non zero component

in this direction.

But the I can have some component in the

transverse direction.

The question is what is the work that has been done?

And the point is that what is essential for this process is

not that force itself, but only the horizontal component, the

component in the direction of displacement.

In other words, the work is equal to the displacement which

-- let's just write the word displacement times the

component of the force along the displacement vector.

In other words, what matters is how thoroughly I

erase the blackboard.

It doesn't matter how much I sweat doing this.

What I'm talking about is not how much work I have done,

but the question is what is the result?

And for the result it doesn't matter how much pressure I was

applying towards the board.

What matters is how much force I applied in the direction

of the displacement vector.

So that's why I get this formula.

But if so, it means that house algebraically I draw this is

the displacement vector, I have to draw it again -- and

so this is the force.

The force vector and this is displacement vector,

let's call it delta r.

And so the work done moving this particle is delta r times

f, but not f, but rather this is an angle theta, so this is

actually the length of this is f, a length of f times cosine

of the angle, times cosine of theta.

So I'm trying to say that this is just a dot

product between dr and F.

That's what I'm trying to say.

The work is a dot product of the displacement vector

and the force vector.

I this kind of simplified situation where displacement

is linear and the vector fields is constant.

If you want you can write it as F dr -- it doesn't matter.

Sorry, I should have written dr not delta r.

So that's that the idealized situation.

Now the general situation will be as follows: I will have

a more general curve, not necessarily a line segments so

it will be -- do it like this.

This will be some point a, and this b, and at each

point I will have some points like this.

So you see that the vector field changes.

It has different values at different points.

So let's suppose I would like to calculate the work done by

this force in moving this particle from this point to

this point along this curve.

So what should I do?

I should break it into small pieces, and this is a typical

thing, this should be almost reflex by now because this is

the kind of thing we always do when the talk about integrals.

We break it into small pieces and each small piece we

approximate by this idealized situation where thiss will be

-- displacement will be just linear.

And the vector field will be constant because if the vector

field varies in a continuous was, in a smooth way, more

precisely if it is differentiable.

If it's a good approximation on each, on a small subsegment

here, to think that the vector field doesn't changed so much.

So it's a good approximation to the replace it by the value at

any given point within this small integral.

The work will be equal to -- the total work be the sum of

the work on each of the small intervals, and this will be the

sum of F evaluated at some points along this, within

this interval dot delta ri.

so you know what will happen when this partition

becomes finer and finer.

The sum will converge to an integral.

This is very reminiscent of all kinds of integrals

that we've done up to now.

We've done double integrals, we've done integrals, we've

had triple integrals.

It's always the same idea.

If you have an expression like this, where you have a sum over

a small -- over this kind of partition then in the limit you

just replace this by an integral and this is valid if

the quantities involved satisfy some natural conditions

like being differentiable.

so the result of this is an integral of F dr -- I would

like to say from a to b, but it would be a little bit, it would

not be a good idea to write it like this.

When we were in one variable calculus, there was only one

way connect point a with point b, because we were going

along the single line that we are given.

Here we are on the plane, and there are many different ways

to connect these point and, of course, the answer will depend

on which way I'm going.

The longer the path, the longer the work, the larger the

work that will be done.

So instead I will write it like this, over C.

And I don't mean a, b, c.

I just want to denote the curve by letters C.

A and B denote the end points but C denotes

this entire curve.

It is important, however, to remember the orientation

of this curve.

We go from A to B and not from B to A.

If we were going from B to A, the calculation

would be different.

Let's talk about this formula.

So this was to convince you that an integral of a vector

field over a curve is important.

It gives you, the point is like the work done by a

force on a certain particle.

But now I have to explain in more detail what exactly

this integral stands for.

So I would like to explain it in more detail and also give

you a recipe on how to calculate it.

Let me write it down one more time.

F dr over c.

So for the dot product we have a nice formula in terms of the

components of the vector.

So let me remind you that f is a vector field

which looks like this.

It's P of x,y times i plus Q of x,y times j.

What is dr?

dr is the vector which is dx time i plus dy times j.

Let's take the dot product of these two.

The dot product is going to be P of xy dx plus Q of xy dy.

In retrospect the result looks very natural, because in single

variable calculus the only quantities that we were

allowed to integrate where integrals like this.

F dx, F of x dx.

First of all, there was a function which depended on only

one variable because it's single variable calculus, but

also there was only one dx.

We could only integrate against dx because there

were no other variables.

Now we are on the plane.

So there is dx and there is dy.

So we could integrate a quantity of the form P times

dx and we could integrate a quantity of the

form Q times dy.

And this is the most general expression which we could

actually integrate over a curve in two dimensional space.

Now we got this by thinking in terms of the work done by force

and by writing as a dot product of the force vector

field in dr.

But you see the end result is actually very nice and you we

could have actually written this end result from

the beginning.

As the most natural expression really that we could think of

integrating over a given curve.

Again because you have dx and dy and N of them could come

with some factor, some function, which will be some

function P and some function Q.

So now the only thing that remains is we need to explain

how to actually compute this and this is where the methods

which we developed at the very beginning of this course are

going to become very useful.

I'm talking about, of course, parametric curves.

To compute this integral, we need to parametrize our curve.

In other words, the curve was like this, but with

parametrizing, which means that we actually identify it with

a certain interval.

An interval on some auxiliary line with a coordinate

which we usually call t.

So we identified this curve on the plane with a straight

object, with a line segment on this auxiliary line.

Identify this point with this, and we identify this point with

this, and the point in the middle I identify

in a certain way.

So that there is a 1:1 correspondence between points

here and points here.

Algebraically we write it like this, we say x is f of t and y

is g of t along this curve, and we put some limits

on t, from A to B.

Here are the end points were actually points on the plane,

that's why I called them capital A and capital B because

actually capital A was actually a pair of numbers.

It was some, which actually now I can write as

f of a and g of a.

And this b is f of b and g og b.

Let's substitute this in this formula, so what do we get?

We will get integral now just over this integral, which is

great because this is already kind of an integral

which we did in single variable calculus.

Before it was an integral over this curve but now we write

this integral over this curve as an integral over the line

segment by using this parameterization.

We have to substitute, instead of x and y we have to

substitute f of t and g of t.

And then we have to write dx in terms of t.

But what is dx? dx is, of course, just f prime dt, just

by usual formula. dx is f prime of t dt, and dy

is g prime of t dt.

So I replace this dx by f prime of t dt.

And then I write the second one which is Q of f of t g

of t times g prime of t dt, then close the bracket.

Or, if you will, you could write it as a

sum of two integrals.

This is the first integral and this is the second one.

But you see I have, by using the parametric form for this

curve I have been able to the rewrite this integral which

involved x and y as an integral in one auxiliary

variable, namely t.

We have one minute to do an example of this.

So let's say compute the integral of z dx

plus x dy plus y dz.

On purpose I'm doing it in three dimensional space just to

show you that actually everything works out way in the

same way in the three dimensional space as in a

two dimensional space.

The difference is that now you have three variables, x, y, z

and likewise you have dx, dy, dz.

And so, let's say you have a curve in which x is t squared,

y is t cubed, z is t squared.

And t goes from zero to 1.

So the first the term -- so likewise put dz you will have

some h prime of t dt where h is the formula for

[UNINTELLIGIBLE].

So this will be an integral from zero to 1, z is equal to t

squared, dx is equal to 2t times dt, sorry it was z dx, x

is equal to t squared, dy is equal to 3t squared and finally

here I have t cubed and dz is 2t dt.

So you end up with a very simple integral

in one variable.

We'll stop here and we'll continue on Thursday.