Uploaded by UCBerkeley on 08.12.2009

Transcript:

Hey guys, how are you doing?

Are you ready for your exams?

No?

Yeah, it was much nicer at the end of August, you

know, the semester was

just beginning, and the weather was so nice.

Now, it's winter time in Berkeley, which means

like 60 degrees, 55.

So, today is the review lecture.

OK.

In the last few lectures, I gave you an overview from a

kind of theoretical point of view of what this

course is about.

And, in particular, this last chapter that we have studied.

And today, I would like to focus more on the

practical aspects of it.

And, basically on the question of what do you need to know to

do well on the final exam, OK.

So, I'll go over that most important points of the

material which we'll focus on in the final exam.

Which, again, will be on December 19.

And, I'm not going to go over all the information again,

everything is available online, here, and also on B space.

I will also have last minute office hours on the day

before, on the 18th.

In case, you know, Sunday don't see the difference between curl

and divergence, and you panic the night before, you can come

to see me on the 18th and we'll sort it out, I'll

sort it out for you.

OK.

Also, I want to tell you one more time, that the exam

will be at Hearst Gym in three different rooms.

OK.

And, the room is determined by your section.

It's determined by who your GSI is.

And, your GSIs have already announced the rooms for you,

but you can find all this information again

here and here.

OK?

But you should be aware of where your exam room is.

It's important that you are at the right place.

You don't end up in -- you know, it's Hearst Gym, so it's

big, and it's easy to get lost.

So be sure you know which room you should show up at.

OK.

Any questions about the organizational

aspect of the final?

Okay, good.

So, now let's talk about the material.

I already said that this exam will focus on the material

after the second midterm, what we call the third season.

And, but it doesn't mean that you should forget what happened

in the first two seasons, meaning that certainly

problems, solving all the problems, will require skills

which you learned before.

You should not forget all this stuff, right.

Especially multiple integrals are very important part of

this, of this material.

So, I think it's a good idea to go over that old material.

Perhaps not devote as much time as the material of the last

chapter, which studied since the last exam.

But, you should probably review it nonetheless.

Any questions from this part of the audience?

So, I would like to break it into the material.

So, I will only talk about the material since the last exam.

But again, don't take it as meaning that you should

not know anything about what happened before.

You should, but the focus will be on this material.

OK.

That's why I'm going over this material now.

Alright, so the first thing I would like to talk

about is integrals.

In this part of the course, we have talked about different

kinds of integrals.

And, I would like to summarize again what those integrals are,

and what kind of information you need to be able

to compute them.

So, the first thing I want to focus is just the integrals.

What kind of integrals.

Well, we integrate over curves, and we integrate over surfaces.

So, there are two types of integrals.

First of all, depending on the dimension of the domain.

Let me talk first about integrals over curves.

So here, we have two type of integrals also.

Type I integral is an integral of a function, so this an

integral that looks like this fds, where f is a

function over curve.

So, a curve here could be over -- could be a curve on the

plane that is R2 or in space.

On the plane or in space.

But, the way we handle this integral is very similar.

Now, how do we compute such an integral.

To compute such an integral, the first step is to

perameterize your curve.

Perameterize the curve.

So, which by the way, was the very first subject that we

learned in this course.

So, this is a good example of what I mean when I say, you

know, revisit, review the material from the

previous chapters.

Because, certainly to be able to do such an integral, you

have to know how to parameterize curves.

And, that's something we learned at the very beginning.

OK.

So, parameterizing the curve means introducing

an auxiliary parameter, which we usually call

t, but you may call it something else if you like.

In writing each of the coordinates as functions of

this auxiliary parameter, so, you would have, x is a function

of this paremeter t, y is a function of this parameter

t, and z is a function of this parameter t.

That would be in the case when the curve is in space.

If it's on the plane, you would only have x and y of course.

Right.

And once you do that, this is integral, and then you have to

say, that t, what is the range for this auxiliary parameter t,

and the range will be between some numbers a and b.

So, then this integral will be equal to the integral of f,

where instead of x, y and z, you insert those three

functions: x of t, y of t, and z of t.

Right?

And, you integrate from a to b.

But, in addition, you also have to insert a factor

which has to do with the arc length of the curve.

And that would be the following: x prime of t

squared, plus y prime of t squared, plus z prime of t

squared, and finally dt.

And again, if you are on the plane, you just don't have the

z variable, you just erase this, and you get the

formula for the plane.

Now, what does this represent.

This represents a mass of say, a thin wire in the shape of

this curve, if is a density function, is mass

density function.

Also, remember that if f is equal to 1.

If f is equals to 1 in other words, you just

have this square root.

This integral represents simply the arc length.

So, this integral measures the total mass or total length of

this object of this curve.

That's what this integral is about.

It's very easy to distinguish this integral from the next

type of integral, because you see that in this integral,

you integrate functions.

Whereas, the second type of integral, you integrate

vector fields.

So, the second type you have an integral with vector field,

rather than a function.

F is a vector field.

So, as such it has components: i, in front of i you have p, in

front of j you have q, then you have a third component r.

If this is a vector field in space.

If it's on the plane, again, you only have two components.

OK.

So, vector field is more than a function.

It has three components in space, two components

on the plane.

And, this integral can also be written as Pdx

plus Qdy plus Rdz.

This is the same, because the R here stands for dz equals

dxi plus dyj plus dzk.

And, so if you take the dot product of this two, you

end up with this form.

So these are two different ways to write a line

integral of a vector field.

Either this way or that way.

It's the same.

Okay, in the case of a line integral of a vector field

there is an additional complication, which we don't

have for integrals of this type.

Integral of this type, it's enough to just write this.

You have to specify the curve, you have to specify the

function, and that's it.

But, in the case of a line integral of a vector field,

you have to specify additional piece of data.

So, which I kind of, I wanted to emphazie.

Here you need orientation.

Of course, when I say you need to specify, what I really mean

is that I need to specify.

In other words, when I, if I give you problem, compute this

line integral, I have to tell you with respect to

which orientation.

OK.

If I don't tell you, it means that the problem

is not well posed.

Unless, you know, there's something said, which

should enable you to find the orientation.

In other words, this is not well defined, if you just have

the vector field and the curve.

You have to also have orientation on this curve.

What do I mean by orientation.

A curve is going to be, to look like this.

So it goes from some point a to point b.

And, orientation means the direction.

Which way do with traverse this curse.

So, there are two possibly orientations.

Here is one orientation, and here's another orientation.

You can go from a to b, you

can go from b to a.

And, of course, the answer, if you choose the first

orientation, is going to be just the negative of the answer

if you use, choose the second orientation.

But, it's a different answer, unless the integral is just 0,

you will get a different answer depending on which

orientation you choose.

So, orientation has to be specified, and you have to

follow that orientation.

You have to be careful when you set up such an integral, to

make sure that you are computing it in the

right direction.

You are doing the integral, oriented in the right way.

Of course, here I have drawn a curve, which has two endpoints.

There are also curves which do not have endpoints,

closed curves.

Here's an example of a closed curve.

In the case of a closed curve -- it also has two orientation,

which we usually call clockwise and counterclockwise, right.

So, counterclockwise is this one.

There is also a somewhat misleading terminology in

the book, which is called positive orientation.

Which I will prefer not to use because it's a matter

of convention, right.

So, let's just call it counterclockwise

or a clockwise.

In the book, counterclockwise is also called positive.

But, I think there's potential for misunderstanding if

you use that terminology.

So, let's just stick to clockwise, counterclockwise.

I think it's fairly, it's very self-explanatory.

OK.

Now, let's suppose you have this.

You have chosen such an orientation.

What will be, what will, how to compute this integral.

So, you use again parameterization like this, but

now it's important to make sure that your orientation, that

you've chosen, agrees with the orientation on the

auxiliary of parameter t.

This t is actually, just belongs to the one-dimensional

space to the line.

And, the line is oriented, because points on the line

correspond to numbers, and numbers are oriented, we can

say which numbers is bigger between any two numbers.

So, when we write it like this, it means that we identified

the curve with the interval from a to b.

And, the interval from a to b is always oriented from

a to b, like this, right.

So, let's suppose that under this parameterization, you

choose orientation like this.

So, it goes from this point to this point.

And, this point is t equals a, and this point is t equals b.

OK.

Let's suppose that.

In this case, this integral is computed as an integral from

a to b of P x prime, t dt.

Well, let's just write like this.

Plus Q y prime plus R z prime dt.

Here is a possible trick question.

Let's suppose I will ask you on the exam to compute such a line

integral, where the curve is oriented in a different way.

So, if you don't pay attention, you would say, okay.

You wound find your parameterization, where

again, let's say t equals a corresponds to this and t

equals b corresponds to this.

Let's suppose.

And you, look if you don't pay attention to this stuff, to the

orientation, you might want to write it again like this.

That would be wrong, you see.

Because, you have to make sure that with respect to that

orientation, which you are told to use, the parameterization is

such that, you know, the lowest one is the initial point, the

lowest value a is initial point and the largest value b

corresponds to the endpoint.

In this picture, that's the case.

And, that's why we end up with this integral from a to b.

In other words, the integral that I got with respect to

t has this orientation.

Which comes from the orientation of the curve.

But, if the orientation is like this, you have to go from t

equal b to t equal a, right.

Because, you have to go from this point to this point.

This point corresponds to t equals b, this point

corresponds to t equals a.

That's why, so in this case, then you should write it as

from b to a, and then the same thing.

Then of course what is integral from b to a, integral from b to

a, is the same as negative integral from a to b.

So, you will actually end up with minus this integral.

OK.

Is that clear?

Ask me if it's not clear.

This is a saddle point, which you have to remember.

Now, what does this integral represent?

Interpretation of this integral is that it represents work done

by force, this is work done by force F along c from the

point a to the point b.

You see, from this, if you think about this

interpretation, it becomes more clear why there's a sign,

because when you talk about the work done by force, the force

is moving an object or it may be resisting movement

of the object.

And, so it's important whether you go from a

to b or from b to a.

If the force, let's say the force comes from the wind, and

the wind is blowing this way, and so the wind is carrying

me this way, so the wind does perform a certain

amount of work, right.

And this work is positive.

But what if the wind is blowing this way, but I'm still,

I still insist on going there, okay.

Because I'm stubborn, so I go against the wind.

So, in that case, I do work not the wind, and so the wind is

resisting me, therefore the work of the wind is

negative, you see.

So, it's important to know not just the trajectory of the

object, but it's also important, you know, whether

the relative position of the force and the trajectory.

In other words, the orientation of the trajectory relative

to the direction of the vector field, you see.

So, it becomes more clear that if I go this way, if I go this

way against the wind, it's a negative work.

If I go this way with the wind, it would be positive, OK.

You have a question.

[INAUDIBLE].

That's right, that's right.

When I write it like this, so the question is about this PQR.

In this formula I spelled it out, I wrote f is a

function of x, y, z.

But now, x, y, z have become functions of t.

So, I substituted these three functions into f.

Likewise in this formula, I didn't write just this time,

but what I mean by this is, and the same, similar for Q and R.

OK.

So, the orientation is important, change of

orientation results in an appearance of a sign.

So, that's, that's pretty much all about the setup

of integrals for curves.

So, next we move on to integrals over surfaces.

Integrals over surfaces.

So, what do we need to know here.

Here again there are two types of integrals that

we are interested in.

And, there is the a lot of similarity between the first

type for surfaces and the first type for curves, and the second

type for surfaces and the second type for curves.

So the first, integrals of the first type over surfaces are

integrals are integrals of functions.

So, here again f is a function.

In this case, you have a surface m and it

is embedded in R3.

You could also have a surface embedded in R2, that would

be just a region on the plane, like this, right.

But that's, so that's integrating over such regions

with the subject of the earlier chapter of course, chapter 15.

They said that's the usual doubling integrals.

So, here the novelty is when you look at surfaces in R3,

which we dno not see it in R2, which cannot fit in a plane.

But only fit in a three-dimensional space, like

a sphere or part of a sphere.

So, what I'm talking about is something this, like

a cap, a cap like this.

That will be m.

So, what does, how do I compute this.

Again, to compute, I have to parameterize my surface.

Parameterize.

Because it is now two-dimensional, I have to use

two auxiliary parameters, which we usually call u and v.

And, so we write, x is some function of u and v, y is some

function of u and v and z is some function of u and v; u and

v run over points on some domain already on a plane.

So, what I'm doing is I'm trying to identify a curve

surface like this, like this m with a flat surface like

this, which I call d.

So each point here will correspond to a point here.

And, it's a one-to-one correspondence.

That's what this parameterization means.

Once you have this parameterization, you have a

formula for this integral.

To write this formula, I will use the following notation.

I will call R of u,v the vector field, which is obtained from

this parameterization by interpreting each of the three

functions x, y, and z as components of a three vector.

So, you'll have x of u,v i plus y of u,v j plus z of u,v k.

This is vector R.

And so I can differentiate this vector.

For example, R sub u will simply mean x sub u times

i plus y sub u times j plus z sub u times k.

So, again, this stands for partial derivative of

x with respect to u.

This is a partial derivative, partial derivative

with respect to u.

So, this is another example of something we use in

this part of the course.

We're building on some material that we've learned before.

Because, of course we learned partial derivatives

earlier, right.

Before the second exam.

So again, you have to know, for example, how to compute

partial derivatives.

Without knowing how to compute partial derivatives, you will

not be able to compute such intervals.

Okay, so once we have this, we also have R of v,

define in a similar way.

You just put derivatives with respect to v instead of u.

Then the formula is the following.

I actually made, I only drew one sign of integration, I

should have drawn two signs of integration.

For double integrals, we draw two signs.

Well, it's not a big deal.

I will not deduct points for this, since I've

made the same mistake.

I'll go easy on you if you put one integration sign instead

of two for double integrals.

But, for triple integrals, you should put more than one.

For double integrals, I will be lenient on double integrals

in terms of how many integral signs you put.

OK.

Let me put both two signs here though.

So, integral over this m, you'll be able to write as

a double integral over d.

Where here, you'll put f where again you put x of

uv, y of uv, and z or uv.

Then you will put R sub u cross R sub v.

This is a cross product, again something we learned before.

And, then you take the magnitude of this vector, dA.

dA is the usual measure of integration on uv plane.

So this term, ru cross rv is just like this term, which

you have four integrals of functions over curves.

The first type of it.

It's an analog, it's an analog of that.

That ru cross rv absolute value is an analog of this.

OK.

So, now the question really becomes how do

you parameterize it.

Once you parameterize it, once you parameterize it, it's very

easy, because you just need to compute derivatives, and you

take the cross-product, and you end up with just an

ordinary double integral.

But the question is how do you compute this.

So, well there are at least two special cases,

which you should know.

The first special case is when your surface

is part of a graph.

It's part of a graph of a function.

In this case, that formula simplifies it.

And in fact, it's a good idea to remember, to

remember this formula.

Or, you will actually have a cheat sheet, so you can just

write it on the cheat sheet.

You don't have to remember.

Okay, so special cases.

The first special case is when the surface, M, is

part of the graph of some function g of x, y.

So, in this case, what we can do is say that x is u, y

is v and z is g of u,v.

Or we can stop pretending that u and v are two additional

auxiliary variables, and just call them what they are.

I mean, let's just call this auxiliary variables x and y, so

then parameterization will be in terms of x and y, and it

will have x is x, y is y and z is g of x,y.

So, in this case actually, you can compute this cross-product.

This has been done.

I did it in class, I wrote this formula in class,

the computation explained in the book.

You must have done it many times from homework.

So, you might as well just remember this formula, and the

formula is like this it's minus dg over dx minus dg over dy, so

this i and this is k if I remember it correctly.

Please correct me if I'm wrong.

I think it's correct.

This formula you just get by applying, by applying

that general formula in this particular case.

It's just that when you compute the cross-product so you have

to compute this, you know, the determinant of a 3 by 3 matrix.

It's just that the matrix becomes very simple, simplifies

it so it's easy to compute.

I'm sorry?

[INAUDIBLE]

I'm sorry.

That's right.

I did not close the parenthesis, for

which I apologize.

Ok.

Please tell me because we want our world-wide audience to see

only the correct formulas.

And now, if you have this, you can compute it's

magnitude very easily.

And, so it's just going to be the sum, square root of the sum

of squares of this components.

So, it's like this.

OK.

So, we actually have not just the magnitude, but we actually

have a formula for the cross-product, which I

will use in a minute.

OK.

That is the first special case.

There is a second special case, when your surface

is part of a sphere.

The second special case is m is part of the sphere, of

a sphere of some radius.

Let's call this radius R.

So, this radius R is just a number.

So, you got a sphere, which is of this radius, and your m

could be just the entire sphere .

It can be the entire sphere.

Or, it could be a hemisphere, or it could be part of a

sphere, which is within, confined in a cone, like an

ice cream cone, right, but that's just the surface

of the ice cream.

So, in each of these cases, a good way to parameterize

such a surface is to use spherical coordinates.

Spherical coordinates we learned before.

Spherical coordinates give us an alternative coordinate

system for the entire three-dimensional space.

And, you got three coordinates, we got rho, phi, and theta.

But now, we are on a sphere.

We are on sphere of radius R.

So, from the point of view of spherical coordinates, this

coordinate, rho, is fixed.

Rho, the coordinate of spherical [UNINTELLIGIBLE],

is equal to R.

But, we still have two remaining spherical

coordinates, which we can use to parameterize our surface.

Which is good, because surface is two-dimensional, so it

requires two independent coordinates.

And phi and theta give us these two coordinates.

So, in other words, u and v here are phi and theta.

For example, if it's the entire sphere, you will have theta

going from 0 to 2 pi, and phi going from 0 to pi, which is

the full range of fi and theta.

If it's the upper hemisphere, again theta will go from 0 to

2 pi, and phi will go from 0 to pi over 2.

And here, theta is from 0 to 2 pi, and phi, well it depends

how much ice cream you get.

So, it's in your interest to have it as large

as possible, I guess.

Although, then it becomes more difficult to hold it.

So, let's just settle on the middle ground, pi over 4.

OK.

So,that's how you, this is an example of parameterization.

And now, in this case, there is a formula for, let me erase

this, for R, u cross Rv.

But, now we don't have u and v, we have phi and theta.

And, so R phi cross R theta, in this case, and I'll just write

the formula for the magnitude.

It's actually what you would expect it to be.

It's exactly the formula for the Jacobian, except that you

have to set rho equals R.

R squared times sin phi

[INAUDIBLE]

Right, so the question is about the cone.

How do I measure phi.

Phi is measured from the z axis, right.

So, 0 to pi over 4 would be, this kind of range.

Yep, it works.

Although, I have to say in this picture it looks a little

bit less than pi over 4.

I don't want to short change you with ice cream.

But, this is just an approximate picture.

Not the best day to eat ice cream at Berkeley though,

perhaps earlier in the semester will be more appropriate.

So, that's a formula for this.

And, let's go back to this double integral.

So, that's all you need.

You need this guy, you need your function, where you stick

those expressions, coming from your parameterization, and

then you end up with a simple double interval.

So you compute it.

What is the interpretation of this.

What is the meaning of this.

The meaning, just like in a case of Type I intervals for

curves is that it represents something that has to do

with the area or mass.

Something when you average density over your surface.

So the typical example is that this represents mass of say

aluminum foil in the shape of M, where f, this function

f, which appears in that integral, f is the mass

density function.

Now, it doesn't have to be mass, it could be any quantity

which you can measure, which can have a density.

For instance, charge.

This foil could be charged.

But, in a complicated way, so that some parts are

more charged than others.

So, you have to integrate, you have to average them out.

And, so you can have a, f may be the charge density function,

not necessarily mass density function.

So, tehn you will be calculating total charge.

By the way, if it is mass, you could also talk about the

momenta of the mass, which means that you would be

multiplying under this integral f by x or y or z, right.

So, you can also find the center of mass for this

membrane, for this surface.

So, if somebody, if a question is phrased in the following

way, find the charge of an aluminum foil in the shape of

such-and-such surface, if charged density function is

such-and-such function.

Then, you immediately know that you have to do this integral.

So that's the integral of the first kind or Type I

integral over surfaces.

And next we go to Type II integrals.

Type II integrals are integrals for vector fields.

In this case, we are given a vector field F.

And, we are computing this, which is oftentimes called,

sometimes called surface integral of a vector field, but

oftentimes it's also called flux, flux of F across M.

So, if you see the word flux it refers to the integral

of the second type.

And, I explained what it means, it means the rate of flow,

for example, of a fluid through this membrane.

If F is a velocity vector field.

So, this integral actually was defined originally as follows.

We said that we have to take the dot product of F with the

unit normal vector field along M.

And, this will be a function, and then we take its integral

as a Type I integral.

So, this definition already makes clear that we need

orientation, we need a choice of n.

So just like integrals of second type the curves are not

really well defined if you just specify the vector

field and the surface.

Integrals of the second type are also not well defined if

you just specify vector field and surface.

You have to specify orientation on your surface.

So, it's exactly the same.

We need orientation.

And, orientation should be given to you if you have

to solve a problem.

If you are given a problem where you have to compute flux

of a vector field across a given surface, the problem has

to stay what is orientation, or there should be a way to

find which orientation it's supposed to be.

Now, again there are two different types of orientation,

there are two different orientations, in case the

surface is orientable, but of course we will be dealing

with orientable surfaces.

I explained to you a Mobius strip.

Remember a Mobius strip is an example of a surface

which is not orientable.

But, we will be working with orientable surfaces.

So, in orientable surfaces we will have two different types,

two different orientations, which, they will

differ by a sign.

So, how do you find out which one, which orientation to use.

So, let me first of all you rewrite this to

make it computable as a double integral.

So, again I'm assuming that my surface is parameterize

in this way, by this functions of u and v.

OK.

And in this case, this double integral will be written as

follows: the double integral over this domains D, which you

have there, and here you have F dot ru cross rv.

I mean, in principle you could try to compute this integral by

using this n, and there are different ways, there are

different ways to try to find this normal vector, OK,

a unit normal vector.

So, you could actually apply this formula, then compute this

integral as a Type I integral.

You could compute as a Type I integral, but I think, I

suggest, I strongly suggest, that you use it as, that you

use this formula for it, OK.

This, because, the point is that there are many ways to

find a normal vector after.

For instance, you know, we talked about normal

vectors for graphs and functions for example.

But what we need to do after this, you have to first

of all normalize it.

You have to get a normal vector of unit plane.

You have to normalize it.

And, then in addition, you will have to compute this

ds, which itself is a rather complicated tasks, right.

You have to compute this ds, which we here compute as the

magnitude of our u crossed rv.

The nice thing about this formula is that this formula is

obtained from this formula, where instead of n you take

this ru crossed rv, which we normalized by dividing

by its magnitude.

So, what happens is that there is some consolation.

This norm, this magnitude get canceled by the same magnitude

in the denominator.

So, the formula actually simplifies.

That's why, I think that this formula is nicer, so it is easy

to get confused if you use a different type of formula.

I mean you can do it at your own risk if you like, if you

know what I mean, but I just think this is the

best way to do it.

In the framework, for the kind of problems that we

are dealing with here.

So, here is the formula, but how do you see if this is the

right answer, and not its negative.

You see, so this is a good question.

How to see that you get the right orientation

in this formula.

In other words, you could have a problem, say where you are

asked to compute, to use orientation, to compute with

respect to the orientation, upward orientation.

So let's say you are asked to use upward orientation.

How do you know that this formula gives you the integral

with the upward orientation and not the downward orienation.

For example, right.

So, first of all what do we mean by upward orientation.

It means that you have to choose n of the form something

times i plus something times j, where this doesn't matter, plus

something, which is positive, times k.

This is upward.

And if I put less than 0 here, it will be downward.

So, that's the terminology.

Upward vector is a vector whose k component or z

component is positive.

And, downward vector is a vector which has a

negative z component.

OK.

So, let's say you're ask to compute the integral over a

surface, assuming the upward orientation, and you have this

formula, and the natural question is does this formula

give me the upper orientation or it gives me a

downward orientation.

We have to check that, because if it does give me the

upward orientation, I will get the right answer.

If it gives me downward orientation, then I would get

the minus sign, I will get negative of the right answer.

So, it means I will get the wrong answer.

If I get the wrong answer, it means that I don't get the full

score on this problem, right.

So, how do you see that.

Well, here's an example.

Let's look at some special cases where this actually can

be computed more explicitly.

And, the first case is when m is part of the graph.

So, for example, let's look at the case when, for example,

M is part of a graph of the function g of x,y.

In this case, we have already computed our x cross our y.

It is written on this formula.

This is our x cross our y.

So, let's me just copy this formula here.

So, this formula, in this formula the coefficient in

front of i and the coefficient in front of j depends on this

function that you have, of which your, you know, which

gives you this graph, right.

So this coefficient depends.

But this coefficient is always equal to 1.

So this coefficient is 1.

So, it's positive.

So, that means if you are using this formula, for rx cross ry,

right, if you are using this formula for rx cross ry, then

you are getting orientation, or you are computing, this formula

is computing the integral with respect to the upward

orientation, you see.

So, this is an important point that upwardly it could be, this

gives you a normal vector, not normalized, not [? unit ?], but

it gives you a normal vector.

And, by looking at this vector, you can find out with

respect to which orientation you are now computing.

And then you have to match that with what you are asked to do.

In this case, rx cross ry.

So, you have to look more closely at this vector, and

see, what does it look like [UNINTELLIGIBLE]?

In this particular case, the simplest information we can

infer by this vector is the fact that coefficient

in front of k is 1.

So, it is upward, it's always upward.

So, now you match, you say OK so what am I asked to do.

I'm asked to do double integral, surface integral

with respect to the upward orientation.

Oh, is my orientation upward.

Yes it is.

So, this is the right formula: F dot rx cross ry is

the right formula.

It is the right formula.

For upward orientation.

But, what if I want to trick you and I ask you on the test,

I ask you to compute a double integral, compute a surface

integral with respect to the downward orientation.

If you don't pay attention, you just, you have copied this

formula on your cheat sheet right, and you copied this

formula on your cheat sheet, and, so you think you

are in good shape.

You just applied this formula and you get an answer, but you

get the wrong answer, because you did not pay attention

to what was asked.

You were asked to compute with respect to downward

orientation, but you computed with respect

to upward orientation.

Just simply, because the formula you are using

always gives you upward.

But, I have the right to ask you to compute with respect

to downward orientation.

You know, both integrals makes sense.

It's my right and my prerogative, in fact, to ask

you to define any, to ask you to compute with respect to any

orientation that I like.

So, in that case, you have to be clever, and you

have to put a minus sign.

So, this is the right formula.

This is for downward orientation.

Because putting this minus sign, you can interpret it

as putting the minus sign in front of rx cross ry.

If I put it in front of rx cross ry, I will get plus here,

plus here and most important, I get minus 1 here.

If I get minus 1 here, I got orientation vector

which is downward.

So, that matches what I am asked to do, right.

To do the downward orientation.

So, that's why this is the correct formula.

Is this clear?

Ask me.

Go ahead.

[INAUDIBLE]

Right, I'll give another example in a minute for when

its not upward or downward.

[INAUDIBLE]

Very good.

What if the coefficient is 0.

So, OK, in this case is it never happens, right.

Because in this case, it's just always like this.

So, here is the next level of trick question.

Which is that, suppose that some ru cross rv, OK, but

for some other u and v, and suppose it just happens

that the component of the k factor is 0, right.

In this case, I should not be asking you to compute in

respect to upward or downward orientation, because if it's k

component is 0, it means it's parallel to the x, y plane.

So, it's neither upward nor downward.

And, saying that compute with respect to upward orientation

would be misleading.

In other words, ambiguous.

It will not give you any information.

Because, neither of the two vectors is upward

or downward, right.

So, in this case, you actually have to use another way

to describe it, OK.

So what are other ways to say, we have to describe in words,

you know, there are two possibilities.

You have to be able to distinguish between them.

If the k component is non-0, you can distinguish between

the sign of k, right.

Positive sign or negative sign.

But, if it is 0, for example, you can talk about the sign of

the coefficient in front j.

Even in this case, here is another trick

question I could ask.

Instead of saying compute this integral, even in this case

when it's part of a graph, instead of saying, it's the

upward or downward, I could say compute it with respect to the

orientation which is to the left.

So, what does it mean.

What should you do if you see that.

Don't panic.

If the orientation is described in terms of left, right, it

means that we're talking about the y coordinate, right.

Instead of the z coordinate, which would be k and I've got

the y coordinate, which would be j.

So, to the left would mean that the coefficient is negative,

and to the right would mean that the coefficient in front

of j is positive, right.

Now, of course in this case it's not clear, it depends on

the function g, sometimes, just the fact that there is a minus

sign here, doesn't mean anything because a function

could be negative.

So, the total sign would be positive, and so, right.

But, in the particular example, which you are computing, you

should be able to see if it's positive or negative

over that range.

So, what I'm saying is, what you are interested in is just

the piece of, of a graph.

So, you are working with a particular domain D in uv, and

you have to see whether over this domain certain coefficient

is positive or negative, right.

That's all it is.

And, for example, suppose you find that ru cross rv, let me

now switch to the more general setting where you actually have

coordinates u and v and its not necessarily graph of a

function, but more general surface.

So, you could have ru cross rv is equal to some, you know,

some -- A of u,v i plus B of u,v j plus C of u,v k.

This is what you will compute by taking this partial and

taking the cross-product.

So, you'll get three functions, and then you

just have to look closely.

What do the it look like.

Well, let's say this is u squared plus v squared, OK.

So, it's not 1 but it's going to be positive.

Well, the only point where it will not be

positive will be 0,0.

So, let's suppose that your domain does not

include point 0,0.

So, then it's always positive.

If you look, if you see that, you know, that the coefficient

in front of j is positive, which means that the

orientation vector is pointing to the right.

Because, because this is x, well, of course when we say, we

say that, it's all relative to how we organize our

coordinate system.

What do we mean by left and right.

But there is a way which we're kind of used to, which would be

like x goes this way and y goes this way, and this

is a z,right.

So, so in this case to the right means that it's a

positive number in front of j.

To the left means it's a negative number.

So, in this case it's going to be a vector, not necessarily

like this, but maybe like this.

So, this is to the right, one and for the left it's this way.

Is that clear?

There is another way to describe orientation by using

the words outward and inward.

So, usually, this is often used spheres or for closed domains.

OK.

So, that's another way.

That should also be pretty self-explanatory.

So, at the end of the day, you have to compute the formula

by using ru cross rv.

But, the bottom line is what you get is either the correct

answer, or either the answer which, if you put in, if you

stick it here, you get the correct answer, or

it's negative.

So, it becomes fine-tuning issue.

Is it correct, or should I put a minus sign.

So, then you match the description of that

orientation, which is given in wards -- upward, downward,

left, right, to what you got.

And, usually it should be fairly easy to see.

And, then that's how you see whether you got

the right answer.

So, you use that formula without any modification, or

you put a negative sign.

That's all it is.

Any questions?

Yes.

[INAUDIBLE]

Right.

So what if the sign changes.

If the sign changes, of course then the description would not

be correct, I mean well define.

The description should be such that if it says to the right,

it means that that coefficient should stay positive or

negative, whatever, right.

Okay, alright.

So, that's about the set up of the integrals.

And now, I want to say a few words about the connection

between different integrals.

These are the formulas which we've learned.

So, this is 2, formulas.

So, we have essentially learned 3 different formulas.

We have the Fundamental Theorem for line integrals.

We have Stokes Formula, and then we have Divergence Theory.

OK.

So now, I have explained -- oh yeah.

Also, Green's Theorem is a special case of this, when m

actually belongs to the plane is a special case this.

So, that's why I didn't single it out.

So, I already explained the kind of general theory behind

this, and how you could think of all of this formulas, a

special cases of one and only formula.

what I would like to focus now is the practical

aspects of this.

How can you apply these formulas to computing

integrals of the type that we have just discussed.

So here, I would like to emphasize, first of all, the

point aspect of these formulas.

In these formulas, always, there is a trade-off between

the sort of a boundary, the sort of a geometric boundary

of a domain, and a sort of derivative of the office

that you are integrating.

It's a trade-off.

On the left, it's a general domain of integration, but

the integrand is special.

In the first formula, you're computing only line integrals

of vector fields of this form.

The gradients, the conservative vector fields.

In this formula, integrals only of curl, of something which is

a curl of another vector field.

Here, you compute a function, which is divergence

of some vector field.

So, that kind of derivative of other objects.

But, the domains are most general, most general curve,

most general surface, most general three-dimensional

solid.

On the other hand, on the right-hand side you have the

most general object, here a general function here a general

vector field, here also a general vector field.

But, your integrating over domains, which are kind of

derivatives of the domains on the left-hand side.

Where in the geometric world, derivative means boundary.

So, this is a very important point to remember when you

decide which formulsa is appropriate for this or that

exercise, or for this or that parameterization.

So on the right-hand side, you've got general integrand,

but special domain, mainly boundary.

Domain.

Here it's derivative, here it's boundary.

OK.

So, the most -- so don't try to use the formula, where for

example, where just on this grounds you can rule it out.

In other words, or don't use it in a straightforward way.

Use it in a more creative way.

So let me give you just two examples of how this

formulas could be used.

So, example 1.

So, let's suppose you have, you're asked to compute a line

integral of a vector field, but the curve over which you have

to compute is very complicated.

So, that's the first indication that you should replace

the domain of integration.

OK.

So, in this case, let's say you were asked to compute this.

So, the clever thing to do is to replace this domain by

this curve, which has the same endpoints, right.

And, then you can argue that this line integral is equal to,

let's call this c1 and let's call this c2, this line

integral is equal to integral over c2 over F dr plus the

double integral of curl F ds, where you integrate over the in

interior of this domain.

You see, you can use this formula.

Let's call it D or M, let's call it M.

OK.

So, the reason is the following that if you take this to the

left-hand side, you will get negative c2, right.

I'm writing it like this, because, I prefer to

say this is equal to this plus something.

But, I could put this integral to the left with minus sign,

which means I'm integrating over c1 minus c2, and then you

have to see that this is a boundary of M; c1 minus c2

would be going with this orientation, that would

be boundary of M.

Now, if I do it like this, this brings up another issue, which

is the issue of orientation in this formula.

So, I'm using this formula, but in both cases, I have integrals

which depend on orientation.

And, we have this rule for orientations.

OK.

Which we discussed already many times.

So, orientation on the left and on the right have to match in

this way where if you walk on the boundary on the

surface in your left.

So, the way I did it right now would mean that the orientation

on this surface would have to go that way.

OK.

So, here it would be the surface where orientation is

going through the black board.

The usual orientation, which we, you normally consider,

would be this orientation towards us, right.

And, that orientation, which we would get

with counterclockwise.

But, here I have chosen orientation which goes

clockwise, well because I also put minus c2

so it reverse this.

And so this will be the orientation, let's call it

away from us, not towards us, but way from us.

Because, if somebody would be walking, if their feet are on

the blackboard, but their head points that way, so they would

be walking in that classroom.

They'll be like Spider Man, and they would be walking also on

this bizarre world, which is behind this blackboard, which

we don't even know what's going on over there, right.

But, that person would be walking like this.

The surface will be on their left.

So, normally if I were to walk like this, I would have to walk

counterclockwise for the surface to be on my left.

So, if I wanted to have orientation here, towards us, I

would have to, for this formula to be true, I would have to

choose up as the orientation.

But, so I'm putting sort of several things into example,

because I'm out of time essentially, but I'm trying

to sort of pack a lot of information into

this one example.

So, you should really look at it more carefully, and

different aspects of this, OK.

The orientations have to be compatible.

And there is one more thing which I'm doing here, which is

that I'm replacing a complicate path by an easy path, a simple

path, and the trade-off is that this guy is equal to this one.

But, there's a price to pay, and the price is the double

integral over this membrane, over this sort of film.

Think of it sort of like a film that you get from soap.

It's like a frame and there is a kind of soap.

So, you have to integrate over that, the curl.

The best possible scenario, where this works,

is when curl is 0.

If curl is 0 there is no price to pay, and, in fact, this

complicated integral over this complicated curve is equal, is

just straight equal to the integral here.

But, sometimes it's not 0, sometimes it's not 0, but

it could be very simple.

So, it's not such a big price to pay sometimes.

You see what I mean.

So, you have to, you have to see different options, the

different options that you have.

So, this is one example, where I am utilizing

formula number 2.

And, let me give you one more example, which is

kind of similar, but for formal number three.

So, maybe I should emphasize here, that in this formulas,

orientations have to be compatible.

This is important.

I'm not going to explain one more time, because we spent a

lot of time talking about this in the previous few lectures.

So, example number 2.

Example number 2, is kind of a similar trade-off, but where

you have to compute a double integral.

So let's say you have this upper hemisphere, and then you

also have this disc, this base of this upper hemisphere.

And, together they form the boundary of the

upper half of this bowl.

So, let's call this B, upper half of a bowl.

So, the last formula will tell you that the integral of

divergence of some vector field E over this B is equal to the

flux through the surface, through the boundary.

And, the boundary now consists of tow things.

Let's call this M1 and let's call this M2.

So, the formula then will be that it's double integral

over M1 plus double integral the over M2.

But, I have to say with respect to which orientation, and the

rule in the divergence theorem is that the orientation is

outward, which here would mean like this, and here

would mean like this.

Here it's outward, actually is upward on top part, and on

the bottom outward is actually downward.

But, both are outward.

So, that's what I mean here and here.

So, how can we utilize this formula.

For example, what you can do is you can express this as the

difference between this and this.

So, this way you can express a complicated integral over

something very curved, this sphere, or upper hemisphere.

In terms of two flat integrals, one of them is over a

three-dimensional solid, which actually may be quite simple --

depending on what -- I made a mistake like this. dV.

It could actually be 0, this divergence.

It could be 0.

So, then this actually drops out.

And, then you can express this guy as a negative

of this guy, right.

Or, even if it's non-0, it could be something very

simple, like 1, function 1 or something.

So, then it's easier to compute.

And, this one is certainly easy to compute easier to compute

than this one, because it's an integral over a flap region.

So, you have this trade-off again.

Similar to the trade-off discussed in the previous

example -- between an integral over one part of the boundary,

and the integral over the other part of the boundary, with a

surplus or price to pay coming from the integral

of the divergence.

So, these are very typical examples of the kind of things

that you need to know to be able to apply these formulas.

So, here's what we're going to do now.

I'm going to stop, and we're going to do a

course evaluations.

It will just take a few minutes.

Okay.

Come on guys.

Come on over, so we'll start distributing them.

And, we'll just take a few minutes, and after this,

after we're done with course the evaluations, I'll be

here for office hours.

And, so you can ask me questions about anything,

well related to the course.

Okay.

Are you ready for your exams?

No?

Yeah, it was much nicer at the end of August, you

know, the semester was

just beginning, and the weather was so nice.

Now, it's winter time in Berkeley, which means

like 60 degrees, 55.

So, today is the review lecture.

OK.

In the last few lectures, I gave you an overview from a

kind of theoretical point of view of what this

course is about.

And, in particular, this last chapter that we have studied.

And today, I would like to focus more on the

practical aspects of it.

And, basically on the question of what do you need to know to

do well on the final exam, OK.

So, I'll go over that most important points of the

material which we'll focus on in the final exam.

Which, again, will be on December 19.

And, I'm not going to go over all the information again,

everything is available online, here, and also on B space.

I will also have last minute office hours on the day

before, on the 18th.

In case, you know, Sunday don't see the difference between curl

and divergence, and you panic the night before, you can come

to see me on the 18th and we'll sort it out, I'll

sort it out for you.

OK.

Also, I want to tell you one more time, that the exam

will be at Hearst Gym in three different rooms.

OK.

And, the room is determined by your section.

It's determined by who your GSI is.

And, your GSIs have already announced the rooms for you,

but you can find all this information again

here and here.

OK?

But you should be aware of where your exam room is.

It's important that you are at the right place.

You don't end up in -- you know, it's Hearst Gym, so it's

big, and it's easy to get lost.

So be sure you know which room you should show up at.

OK.

Any questions about the organizational

aspect of the final?

Okay, good.

So, now let's talk about the material.

I already said that this exam will focus on the material

after the second midterm, what we call the third season.

And, but it doesn't mean that you should forget what happened

in the first two seasons, meaning that certainly

problems, solving all the problems, will require skills

which you learned before.

You should not forget all this stuff, right.

Especially multiple integrals are very important part of

this, of this material.

So, I think it's a good idea to go over that old material.

Perhaps not devote as much time as the material of the last

chapter, which studied since the last exam.

But, you should probably review it nonetheless.

Any questions from this part of the audience?

So, I would like to break it into the material.

So, I will only talk about the material since the last exam.

But again, don't take it as meaning that you should

not know anything about what happened before.

You should, but the focus will be on this material.

OK.

That's why I'm going over this material now.

Alright, so the first thing I would like to talk

about is integrals.

In this part of the course, we have talked about different

kinds of integrals.

And, I would like to summarize again what those integrals are,

and what kind of information you need to be able

to compute them.

So, the first thing I want to focus is just the integrals.

What kind of integrals.

Well, we integrate over curves, and we integrate over surfaces.

So, there are two types of integrals.

First of all, depending on the dimension of the domain.

Let me talk first about integrals over curves.

So here, we have two type of integrals also.

Type I integral is an integral of a function, so this an

integral that looks like this fds, where f is a

function over curve.

So, a curve here could be over -- could be a curve on the

plane that is R2 or in space.

On the plane or in space.

But, the way we handle this integral is very similar.

Now, how do we compute such an integral.

To compute such an integral, the first step is to

perameterize your curve.

Perameterize the curve.

So, which by the way, was the very first subject that we

learned in this course.

So, this is a good example of what I mean when I say, you

know, revisit, review the material from the

previous chapters.

Because, certainly to be able to do such an integral, you

have to know how to parameterize curves.

And, that's something we learned at the very beginning.

OK.

So, parameterizing the curve means introducing

an auxiliary parameter, which we usually call

t, but you may call it something else if you like.

In writing each of the coordinates as functions of

this auxiliary parameter, so, you would have, x is a function

of this paremeter t, y is a function of this parameter

t, and z is a function of this parameter t.

That would be in the case when the curve is in space.

If it's on the plane, you would only have x and y of course.

Right.

And once you do that, this is integral, and then you have to

say, that t, what is the range for this auxiliary parameter t,

and the range will be between some numbers a and b.

So, then this integral will be equal to the integral of f,

where instead of x, y and z, you insert those three

functions: x of t, y of t, and z of t.

Right?

And, you integrate from a to b.

But, in addition, you also have to insert a factor

which has to do with the arc length of the curve.

And that would be the following: x prime of t

squared, plus y prime of t squared, plus z prime of t

squared, and finally dt.

And again, if you are on the plane, you just don't have the

z variable, you just erase this, and you get the

formula for the plane.

Now, what does this represent.

This represents a mass of say, a thin wire in the shape of

this curve, if is a density function, is mass

density function.

Also, remember that if f is equal to 1.

If f is equals to 1 in other words, you just

have this square root.

This integral represents simply the arc length.

So, this integral measures the total mass or total length of

this object of this curve.

That's what this integral is about.

It's very easy to distinguish this integral from the next

type of integral, because you see that in this integral,

you integrate functions.

Whereas, the second type of integral, you integrate

vector fields.

So, the second type you have an integral with vector field,

rather than a function.

F is a vector field.

So, as such it has components: i, in front of i you have p, in

front of j you have q, then you have a third component r.

If this is a vector field in space.

If it's on the plane, again, you only have two components.

OK.

So, vector field is more than a function.

It has three components in space, two components

on the plane.

And, this integral can also be written as Pdx

plus Qdy plus Rdz.

This is the same, because the R here stands for dz equals

dxi plus dyj plus dzk.

And, so if you take the dot product of this two, you

end up with this form.

So these are two different ways to write a line

integral of a vector field.

Either this way or that way.

It's the same.

Okay, in the case of a line integral of a vector field

there is an additional complication, which we don't

have for integrals of this type.

Integral of this type, it's enough to just write this.

You have to specify the curve, you have to specify the

function, and that's it.

But, in the case of a line integral of a vector field,

you have to specify additional piece of data.

So, which I kind of, I wanted to emphazie.

Here you need orientation.

Of course, when I say you need to specify, what I really mean

is that I need to specify.

In other words, when I, if I give you problem, compute this

line integral, I have to tell you with respect to

which orientation.

OK.

If I don't tell you, it means that the problem

is not well posed.

Unless, you know, there's something said, which

should enable you to find the orientation.

In other words, this is not well defined, if you just have

the vector field and the curve.

You have to also have orientation on this curve.

What do I mean by orientation.

A curve is going to be, to look like this.

So it goes from some point a to point b.

And, orientation means the direction.

Which way do with traverse this curse.

So, there are two possibly orientations.

Here is one orientation, and here's another orientation.

You can go from a to b, you

can go from b to a.

And, of course, the answer, if you choose the first

orientation, is going to be just the negative of the answer

if you use, choose the second orientation.

But, it's a different answer, unless the integral is just 0,

you will get a different answer depending on which

orientation you choose.

So, orientation has to be specified, and you have to

follow that orientation.

You have to be careful when you set up such an integral, to

make sure that you are computing it in the

right direction.

You are doing the integral, oriented in the right way.

Of course, here I have drawn a curve, which has two endpoints.

There are also curves which do not have endpoints,

closed curves.

Here's an example of a closed curve.

In the case of a closed curve -- it also has two orientation,

which we usually call clockwise and counterclockwise, right.

So, counterclockwise is this one.

There is also a somewhat misleading terminology in

the book, which is called positive orientation.

Which I will prefer not to use because it's a matter

of convention, right.

So, let's just call it counterclockwise

or a clockwise.

In the book, counterclockwise is also called positive.

But, I think there's potential for misunderstanding if

you use that terminology.

So, let's just stick to clockwise, counterclockwise.

I think it's fairly, it's very self-explanatory.

OK.

Now, let's suppose you have this.

You have chosen such an orientation.

What will be, what will, how to compute this integral.

So, you use again parameterization like this, but

now it's important to make sure that your orientation, that

you've chosen, agrees with the orientation on the

auxiliary of parameter t.

This t is actually, just belongs to the one-dimensional

space to the line.

And, the line is oriented, because points on the line

correspond to numbers, and numbers are oriented, we can

say which numbers is bigger between any two numbers.

So, when we write it like this, it means that we identified

the curve with the interval from a to b.

And, the interval from a to b is always oriented from

a to b, like this, right.

So, let's suppose that under this parameterization, you

choose orientation like this.

So, it goes from this point to this point.

And, this point is t equals a, and this point is t equals b.

OK.

Let's suppose that.

In this case, this integral is computed as an integral from

a to b of P x prime, t dt.

Well, let's just write like this.

Plus Q y prime plus R z prime dt.

Here is a possible trick question.

Let's suppose I will ask you on the exam to compute such a line

integral, where the curve is oriented in a different way.

So, if you don't pay attention, you would say, okay.

You wound find your parameterization, where

again, let's say t equals a corresponds to this and t

equals b corresponds to this.

Let's suppose.

And you, look if you don't pay attention to this stuff, to the

orientation, you might want to write it again like this.

That would be wrong, you see.

Because, you have to make sure that with respect to that

orientation, which you are told to use, the parameterization is

such that, you know, the lowest one is the initial point, the

lowest value a is initial point and the largest value b

corresponds to the endpoint.

In this picture, that's the case.

And, that's why we end up with this integral from a to b.

In other words, the integral that I got with respect to

t has this orientation.

Which comes from the orientation of the curve.

But, if the orientation is like this, you have to go from t

equal b to t equal a, right.

Because, you have to go from this point to this point.

This point corresponds to t equals b, this point

corresponds to t equals a.

That's why, so in this case, then you should write it as

from b to a, and then the same thing.

Then of course what is integral from b to a, integral from b to

a, is the same as negative integral from a to b.

So, you will actually end up with minus this integral.

OK.

Is that clear?

Ask me if it's not clear.

This is a saddle point, which you have to remember.

Now, what does this integral represent?

Interpretation of this integral is that it represents work done

by force, this is work done by force F along c from the

point a to the point b.

You see, from this, if you think about this

interpretation, it becomes more clear why there's a sign,

because when you talk about the work done by force, the force

is moving an object or it may be resisting movement

of the object.

And, so it's important whether you go from a

to b or from b to a.

If the force, let's say the force comes from the wind, and

the wind is blowing this way, and so the wind is carrying

me this way, so the wind does perform a certain

amount of work, right.

And this work is positive.

But what if the wind is blowing this way, but I'm still,

I still insist on going there, okay.

Because I'm stubborn, so I go against the wind.

So, in that case, I do work not the wind, and so the wind is

resisting me, therefore the work of the wind is

negative, you see.

So, it's important to know not just the trajectory of the

object, but it's also important, you know, whether

the relative position of the force and the trajectory.

In other words, the orientation of the trajectory relative

to the direction of the vector field, you see.

So, it becomes more clear that if I go this way, if I go this

way against the wind, it's a negative work.

If I go this way with the wind, it would be positive, OK.

You have a question.

[INAUDIBLE].

That's right, that's right.

When I write it like this, so the question is about this PQR.

In this formula I spelled it out, I wrote f is a

function of x, y, z.

But now, x, y, z have become functions of t.

So, I substituted these three functions into f.

Likewise in this formula, I didn't write just this time,

but what I mean by this is, and the same, similar for Q and R.

OK.

So, the orientation is important, change of

orientation results in an appearance of a sign.

So, that's, that's pretty much all about the setup

of integrals for curves.

So, next we move on to integrals over surfaces.

Integrals over surfaces.

So, what do we need to know here.

Here again there are two types of integrals that

we are interested in.

And, there is the a lot of similarity between the first

type for surfaces and the first type for curves, and the second

type for surfaces and the second type for curves.

So the first, integrals of the first type over surfaces are

integrals are integrals of functions.

So, here again f is a function.

In this case, you have a surface m and it

is embedded in R3.

You could also have a surface embedded in R2, that would

be just a region on the plane, like this, right.

But that's, so that's integrating over such regions

with the subject of the earlier chapter of course, chapter 15.

They said that's the usual doubling integrals.

So, here the novelty is when you look at surfaces in R3,

which we dno not see it in R2, which cannot fit in a plane.

But only fit in a three-dimensional space, like

a sphere or part of a sphere.

So, what I'm talking about is something this, like

a cap, a cap like this.

That will be m.

So, what does, how do I compute this.

Again, to compute, I have to parameterize my surface.

Parameterize.

Because it is now two-dimensional, I have to use

two auxiliary parameters, which we usually call u and v.

And, so we write, x is some function of u and v, y is some

function of u and v and z is some function of u and v; u and

v run over points on some domain already on a plane.

So, what I'm doing is I'm trying to identify a curve

surface like this, like this m with a flat surface like

this, which I call d.

So each point here will correspond to a point here.

And, it's a one-to-one correspondence.

That's what this parameterization means.

Once you have this parameterization, you have a

formula for this integral.

To write this formula, I will use the following notation.

I will call R of u,v the vector field, which is obtained from

this parameterization by interpreting each of the three

functions x, y, and z as components of a three vector.

So, you'll have x of u,v i plus y of u,v j plus z of u,v k.

This is vector R.

And so I can differentiate this vector.

For example, R sub u will simply mean x sub u times

i plus y sub u times j plus z sub u times k.

So, again, this stands for partial derivative of

x with respect to u.

This is a partial derivative, partial derivative

with respect to u.

So, this is another example of something we use in

this part of the course.

We're building on some material that we've learned before.

Because, of course we learned partial derivatives

earlier, right.

Before the second exam.

So again, you have to know, for example, how to compute

partial derivatives.

Without knowing how to compute partial derivatives, you will

not be able to compute such intervals.

Okay, so once we have this, we also have R of v,

define in a similar way.

You just put derivatives with respect to v instead of u.

Then the formula is the following.

I actually made, I only drew one sign of integration, I

should have drawn two signs of integration.

For double integrals, we draw two signs.

Well, it's not a big deal.

I will not deduct points for this, since I've

made the same mistake.

I'll go easy on you if you put one integration sign instead

of two for double integrals.

But, for triple integrals, you should put more than one.

For double integrals, I will be lenient on double integrals

in terms of how many integral signs you put.

OK.

Let me put both two signs here though.

So, integral over this m, you'll be able to write as

a double integral over d.

Where here, you'll put f where again you put x of

uv, y of uv, and z or uv.

Then you will put R sub u cross R sub v.

This is a cross product, again something we learned before.

And, then you take the magnitude of this vector, dA.

dA is the usual measure of integration on uv plane.

So this term, ru cross rv is just like this term, which

you have four integrals of functions over curves.

The first type of it.

It's an analog, it's an analog of that.

That ru cross rv absolute value is an analog of this.

OK.

So, now the question really becomes how do

you parameterize it.

Once you parameterize it, once you parameterize it, it's very

easy, because you just need to compute derivatives, and you

take the cross-product, and you end up with just an

ordinary double integral.

But the question is how do you compute this.

So, well there are at least two special cases,

which you should know.

The first special case is when your surface

is part of a graph.

It's part of a graph of a function.

In this case, that formula simplifies it.

And in fact, it's a good idea to remember, to

remember this formula.

Or, you will actually have a cheat sheet, so you can just

write it on the cheat sheet.

You don't have to remember.

Okay, so special cases.

The first special case is when the surface, M, is

part of the graph of some function g of x, y.

So, in this case, what we can do is say that x is u, y

is v and z is g of u,v.

Or we can stop pretending that u and v are two additional

auxiliary variables, and just call them what they are.

I mean, let's just call this auxiliary variables x and y, so

then parameterization will be in terms of x and y, and it

will have x is x, y is y and z is g of x,y.

So, in this case actually, you can compute this cross-product.

This has been done.

I did it in class, I wrote this formula in class,

the computation explained in the book.

You must have done it many times from homework.

So, you might as well just remember this formula, and the

formula is like this it's minus dg over dx minus dg over dy, so

this i and this is k if I remember it correctly.

Please correct me if I'm wrong.

I think it's correct.

This formula you just get by applying, by applying

that general formula in this particular case.

It's just that when you compute the cross-product so you have

to compute this, you know, the determinant of a 3 by 3 matrix.

It's just that the matrix becomes very simple, simplifies

it so it's easy to compute.

I'm sorry?

[INAUDIBLE]

I'm sorry.

That's right.

I did not close the parenthesis, for

which I apologize.

Ok.

Please tell me because we want our world-wide audience to see

only the correct formulas.

And now, if you have this, you can compute it's

magnitude very easily.

And, so it's just going to be the sum, square root of the sum

of squares of this components.

So, it's like this.

OK.

So, we actually have not just the magnitude, but we actually

have a formula for the cross-product, which I

will use in a minute.

OK.

That is the first special case.

There is a second special case, when your surface

is part of a sphere.

The second special case is m is part of the sphere, of

a sphere of some radius.

Let's call this radius R.

So, this radius R is just a number.

So, you got a sphere, which is of this radius, and your m

could be just the entire sphere .

It can be the entire sphere.

Or, it could be a hemisphere, or it could be part of a

sphere, which is within, confined in a cone, like an

ice cream cone, right, but that's just the surface

of the ice cream.

So, in each of these cases, a good way to parameterize

such a surface is to use spherical coordinates.

Spherical coordinates we learned before.

Spherical coordinates give us an alternative coordinate

system for the entire three-dimensional space.

And, you got three coordinates, we got rho, phi, and theta.

But now, we are on a sphere.

We are on sphere of radius R.

So, from the point of view of spherical coordinates, this

coordinate, rho, is fixed.

Rho, the coordinate of spherical [UNINTELLIGIBLE],

is equal to R.

But, we still have two remaining spherical

coordinates, which we can use to parameterize our surface.

Which is good, because surface is two-dimensional, so it

requires two independent coordinates.

And phi and theta give us these two coordinates.

So, in other words, u and v here are phi and theta.

For example, if it's the entire sphere, you will have theta

going from 0 to 2 pi, and phi going from 0 to pi, which is

the full range of fi and theta.

If it's the upper hemisphere, again theta will go from 0 to

2 pi, and phi will go from 0 to pi over 2.

And here, theta is from 0 to 2 pi, and phi, well it depends

how much ice cream you get.

So, it's in your interest to have it as large

as possible, I guess.

Although, then it becomes more difficult to hold it.

So, let's just settle on the middle ground, pi over 4.

OK.

So,that's how you, this is an example of parameterization.

And now, in this case, there is a formula for, let me erase

this, for R, u cross Rv.

But, now we don't have u and v, we have phi and theta.

And, so R phi cross R theta, in this case, and I'll just write

the formula for the magnitude.

It's actually what you would expect it to be.

It's exactly the formula for the Jacobian, except that you

have to set rho equals R.

R squared times sin phi

[INAUDIBLE]

Right, so the question is about the cone.

How do I measure phi.

Phi is measured from the z axis, right.

So, 0 to pi over 4 would be, this kind of range.

Yep, it works.

Although, I have to say in this picture it looks a little

bit less than pi over 4.

I don't want to short change you with ice cream.

But, this is just an approximate picture.

Not the best day to eat ice cream at Berkeley though,

perhaps earlier in the semester will be more appropriate.

So, that's a formula for this.

And, let's go back to this double integral.

So, that's all you need.

You need this guy, you need your function, where you stick

those expressions, coming from your parameterization, and

then you end up with a simple double interval.

So you compute it.

What is the interpretation of this.

What is the meaning of this.

The meaning, just like in a case of Type I intervals for

curves is that it represents something that has to do

with the area or mass.

Something when you average density over your surface.

So the typical example is that this represents mass of say

aluminum foil in the shape of M, where f, this function

f, which appears in that integral, f is the mass

density function.

Now, it doesn't have to be mass, it could be any quantity

which you can measure, which can have a density.

For instance, charge.

This foil could be charged.

But, in a complicated way, so that some parts are

more charged than others.

So, you have to integrate, you have to average them out.

And, so you can have a, f may be the charge density function,

not necessarily mass density function.

So, tehn you will be calculating total charge.

By the way, if it is mass, you could also talk about the

momenta of the mass, which means that you would be

multiplying under this integral f by x or y or z, right.

So, you can also find the center of mass for this

membrane, for this surface.

So, if somebody, if a question is phrased in the following

way, find the charge of an aluminum foil in the shape of

such-and-such surface, if charged density function is

such-and-such function.

Then, you immediately know that you have to do this integral.

So that's the integral of the first kind or Type I

integral over surfaces.

And next we go to Type II integrals.

Type II integrals are integrals for vector fields.

In this case, we are given a vector field F.

And, we are computing this, which is oftentimes called,

sometimes called surface integral of a vector field, but

oftentimes it's also called flux, flux of F across M.

So, if you see the word flux it refers to the integral

of the second type.

And, I explained what it means, it means the rate of flow,

for example, of a fluid through this membrane.

If F is a velocity vector field.

So, this integral actually was defined originally as follows.

We said that we have to take the dot product of F with the

unit normal vector field along M.

And, this will be a function, and then we take its integral

as a Type I integral.

So, this definition already makes clear that we need

orientation, we need a choice of n.

So just like integrals of second type the curves are not

really well defined if you just specify the vector

field and the surface.

Integrals of the second type are also not well defined if

you just specify vector field and surface.

You have to specify orientation on your surface.

So, it's exactly the same.

We need orientation.

And, orientation should be given to you if you have

to solve a problem.

If you are given a problem where you have to compute flux

of a vector field across a given surface, the problem has

to stay what is orientation, or there should be a way to

find which orientation it's supposed to be.

Now, again there are two different types of orientation,

there are two different orientations, in case the

surface is orientable, but of course we will be dealing

with orientable surfaces.

I explained to you a Mobius strip.

Remember a Mobius strip is an example of a surface

which is not orientable.

But, we will be working with orientable surfaces.

So, in orientable surfaces we will have two different types,

two different orientations, which, they will

differ by a sign.

So, how do you find out which one, which orientation to use.

So, let me first of all you rewrite this to

make it computable as a double integral.

So, again I'm assuming that my surface is parameterize

in this way, by this functions of u and v.

OK.

And in this case, this double integral will be written as

follows: the double integral over this domains D, which you

have there, and here you have F dot ru cross rv.

I mean, in principle you could try to compute this integral by

using this n, and there are different ways, there are

different ways to try to find this normal vector, OK,

a unit normal vector.

So, you could actually apply this formula, then compute this

integral as a Type I integral.

You could compute as a Type I integral, but I think, I

suggest, I strongly suggest, that you use it as, that you

use this formula for it, OK.

This, because, the point is that there are many ways to

find a normal vector after.

For instance, you know, we talked about normal

vectors for graphs and functions for example.

But what we need to do after this, you have to first

of all normalize it.

You have to get a normal vector of unit plane.

You have to normalize it.

And, then in addition, you will have to compute this

ds, which itself is a rather complicated tasks, right.

You have to compute this ds, which we here compute as the

magnitude of our u crossed rv.

The nice thing about this formula is that this formula is

obtained from this formula, where instead of n you take

this ru crossed rv, which we normalized by dividing

by its magnitude.

So, what happens is that there is some consolation.

This norm, this magnitude get canceled by the same magnitude

in the denominator.

So, the formula actually simplifies.

That's why, I think that this formula is nicer, so it is easy

to get confused if you use a different type of formula.

I mean you can do it at your own risk if you like, if you

know what I mean, but I just think this is the

best way to do it.

In the framework, for the kind of problems that we

are dealing with here.

So, here is the formula, but how do you see if this is the

right answer, and not its negative.

You see, so this is a good question.

How to see that you get the right orientation

in this formula.

In other words, you could have a problem, say where you are

asked to compute, to use orientation, to compute with

respect to the orientation, upward orientation.

So let's say you are asked to use upward orientation.

How do you know that this formula gives you the integral

with the upward orientation and not the downward orienation.

For example, right.

So, first of all what do we mean by upward orientation.

It means that you have to choose n of the form something

times i plus something times j, where this doesn't matter, plus

something, which is positive, times k.

This is upward.

And if I put less than 0 here, it will be downward.

So, that's the terminology.

Upward vector is a vector whose k component or z

component is positive.

And, downward vector is a vector which has a

negative z component.

OK.

So, let's say you're ask to compute the integral over a

surface, assuming the upward orientation, and you have this

formula, and the natural question is does this formula

give me the upper orientation or it gives me a

downward orientation.

We have to check that, because if it does give me the

upward orientation, I will get the right answer.

If it gives me downward orientation, then I would get

the minus sign, I will get negative of the right answer.

So, it means I will get the wrong answer.

If I get the wrong answer, it means that I don't get the full

score on this problem, right.

So, how do you see that.

Well, here's an example.

Let's look at some special cases where this actually can

be computed more explicitly.

And, the first case is when m is part of the graph.

So, for example, let's look at the case when, for example,

M is part of a graph of the function g of x,y.

In this case, we have already computed our x cross our y.

It is written on this formula.

This is our x cross our y.

So, let's me just copy this formula here.

So, this formula, in this formula the coefficient in

front of i and the coefficient in front of j depends on this

function that you have, of which your, you know, which

gives you this graph, right.

So this coefficient depends.

But this coefficient is always equal to 1.

So this coefficient is 1.

So, it's positive.

So, that means if you are using this formula, for rx cross ry,

right, if you are using this formula for rx cross ry, then

you are getting orientation, or you are computing, this formula

is computing the integral with respect to the upward

orientation, you see.

So, this is an important point that upwardly it could be, this

gives you a normal vector, not normalized, not [? unit ?], but

it gives you a normal vector.

And, by looking at this vector, you can find out with

respect to which orientation you are now computing.

And then you have to match that with what you are asked to do.

In this case, rx cross ry.

So, you have to look more closely at this vector, and

see, what does it look like [UNINTELLIGIBLE]?

In this particular case, the simplest information we can

infer by this vector is the fact that coefficient

in front of k is 1.

So, it is upward, it's always upward.

So, now you match, you say OK so what am I asked to do.

I'm asked to do double integral, surface integral

with respect to the upward orientation.

Oh, is my orientation upward.

Yes it is.

So, this is the right formula: F dot rx cross ry is

the right formula.

It is the right formula.

For upward orientation.

But, what if I want to trick you and I ask you on the test,

I ask you to compute a double integral, compute a surface

integral with respect to the downward orientation.

If you don't pay attention, you just, you have copied this

formula on your cheat sheet right, and you copied this

formula on your cheat sheet, and, so you think you

are in good shape.

You just applied this formula and you get an answer, but you

get the wrong answer, because you did not pay attention

to what was asked.

You were asked to compute with respect to downward

orientation, but you computed with respect

to upward orientation.

Just simply, because the formula you are using

always gives you upward.

But, I have the right to ask you to compute with respect

to downward orientation.

You know, both integrals makes sense.

It's my right and my prerogative, in fact, to ask

you to define any, to ask you to compute with respect to any

orientation that I like.

So, in that case, you have to be clever, and you

have to put a minus sign.

So, this is the right formula.

This is for downward orientation.

Because putting this minus sign, you can interpret it

as putting the minus sign in front of rx cross ry.

If I put it in front of rx cross ry, I will get plus here,

plus here and most important, I get minus 1 here.

If I get minus 1 here, I got orientation vector

which is downward.

So, that matches what I am asked to do, right.

To do the downward orientation.

So, that's why this is the correct formula.

Is this clear?

Ask me.

Go ahead.

[INAUDIBLE]

Right, I'll give another example in a minute for when

its not upward or downward.

[INAUDIBLE]

Very good.

What if the coefficient is 0.

So, OK, in this case is it never happens, right.

Because in this case, it's just always like this.

So, here is the next level of trick question.

Which is that, suppose that some ru cross rv, OK, but

for some other u and v, and suppose it just happens

that the component of the k factor is 0, right.

In this case, I should not be asking you to compute in

respect to upward or downward orientation, because if it's k

component is 0, it means it's parallel to the x, y plane.

So, it's neither upward nor downward.

And, saying that compute with respect to upward orientation

would be misleading.

In other words, ambiguous.

It will not give you any information.

Because, neither of the two vectors is upward

or downward, right.

So, in this case, you actually have to use another way

to describe it, OK.

So what are other ways to say, we have to describe in words,

you know, there are two possibilities.

You have to be able to distinguish between them.

If the k component is non-0, you can distinguish between

the sign of k, right.

Positive sign or negative sign.

But, if it is 0, for example, you can talk about the sign of

the coefficient in front j.

Even in this case, here is another trick

question I could ask.

Instead of saying compute this integral, even in this case

when it's part of a graph, instead of saying, it's the

upward or downward, I could say compute it with respect to the

orientation which is to the left.

So, what does it mean.

What should you do if you see that.

Don't panic.

If the orientation is described in terms of left, right, it

means that we're talking about the y coordinate, right.

Instead of the z coordinate, which would be k and I've got

the y coordinate, which would be j.

So, to the left would mean that the coefficient is negative,

and to the right would mean that the coefficient in front

of j is positive, right.

Now, of course in this case it's not clear, it depends on

the function g, sometimes, just the fact that there is a minus

sign here, doesn't mean anything because a function

could be negative.

So, the total sign would be positive, and so, right.

But, in the particular example, which you are computing, you

should be able to see if it's positive or negative

over that range.

So, what I'm saying is, what you are interested in is just

the piece of, of a graph.

So, you are working with a particular domain D in uv, and

you have to see whether over this domain certain coefficient

is positive or negative, right.

That's all it is.

And, for example, suppose you find that ru cross rv, let me

now switch to the more general setting where you actually have

coordinates u and v and its not necessarily graph of a

function, but more general surface.

So, you could have ru cross rv is equal to some, you know,

some -- A of u,v i plus B of u,v j plus C of u,v k.

This is what you will compute by taking this partial and

taking the cross-product.

So, you'll get three functions, and then you

just have to look closely.

What do the it look like.

Well, let's say this is u squared plus v squared, OK.

So, it's not 1 but it's going to be positive.

Well, the only point where it will not be

positive will be 0,0.

So, let's suppose that your domain does not

include point 0,0.

So, then it's always positive.

If you look, if you see that, you know, that the coefficient

in front of j is positive, which means that the

orientation vector is pointing to the right.

Because, because this is x, well, of course when we say, we

say that, it's all relative to how we organize our

coordinate system.

What do we mean by left and right.

But there is a way which we're kind of used to, which would be

like x goes this way and y goes this way, and this

is a z,right.

So, so in this case to the right means that it's a

positive number in front of j.

To the left means it's a negative number.

So, in this case it's going to be a vector, not necessarily

like this, but maybe like this.

So, this is to the right, one and for the left it's this way.

Is that clear?

There is another way to describe orientation by using

the words outward and inward.

So, usually, this is often used spheres or for closed domains.

OK.

So, that's another way.

That should also be pretty self-explanatory.

So, at the end of the day, you have to compute the formula

by using ru cross rv.

But, the bottom line is what you get is either the correct

answer, or either the answer which, if you put in, if you

stick it here, you get the correct answer, or

it's negative.

So, it becomes fine-tuning issue.

Is it correct, or should I put a minus sign.

So, then you match the description of that

orientation, which is given in wards -- upward, downward,

left, right, to what you got.

And, usually it should be fairly easy to see.

And, then that's how you see whether you got

the right answer.

So, you use that formula without any modification, or

you put a negative sign.

That's all it is.

Any questions?

Yes.

[INAUDIBLE]

Right.

So what if the sign changes.

If the sign changes, of course then the description would not

be correct, I mean well define.

The description should be such that if it says to the right,

it means that that coefficient should stay positive or

negative, whatever, right.

Okay, alright.

So, that's about the set up of the integrals.

And now, I want to say a few words about the connection

between different integrals.

These are the formulas which we've learned.

So, this is 2, formulas.

So, we have essentially learned 3 different formulas.

We have the Fundamental Theorem for line integrals.

We have Stokes Formula, and then we have Divergence Theory.

OK.

So now, I have explained -- oh yeah.

Also, Green's Theorem is a special case of this, when m

actually belongs to the plane is a special case this.

So, that's why I didn't single it out.

So, I already explained the kind of general theory behind

this, and how you could think of all of this formulas, a

special cases of one and only formula.

what I would like to focus now is the practical

aspects of this.

How can you apply these formulas to computing

integrals of the type that we have just discussed.

So here, I would like to emphasize, first of all, the

point aspect of these formulas.

In these formulas, always, there is a trade-off between

the sort of a boundary, the sort of a geometric boundary

of a domain, and a sort of derivative of the office

that you are integrating.

It's a trade-off.

On the left, it's a general domain of integration, but

the integrand is special.

In the first formula, you're computing only line integrals

of vector fields of this form.

The gradients, the conservative vector fields.

In this formula, integrals only of curl, of something which is

a curl of another vector field.

Here, you compute a function, which is divergence

of some vector field.

So, that kind of derivative of other objects.

But, the domains are most general, most general curve,

most general surface, most general three-dimensional

solid.

On the other hand, on the right-hand side you have the

most general object, here a general function here a general

vector field, here also a general vector field.

But, your integrating over domains, which are kind of

derivatives of the domains on the left-hand side.

Where in the geometric world, derivative means boundary.

So, this is a very important point to remember when you

decide which formulsa is appropriate for this or that

exercise, or for this or that parameterization.

So on the right-hand side, you've got general integrand,

but special domain, mainly boundary.

Domain.

Here it's derivative, here it's boundary.

OK.

So, the most -- so don't try to use the formula, where for

example, where just on this grounds you can rule it out.

In other words, or don't use it in a straightforward way.

Use it in a more creative way.

So let me give you just two examples of how this

formulas could be used.

So, example 1.

So, let's suppose you have, you're asked to compute a line

integral of a vector field, but the curve over which you have

to compute is very complicated.

So, that's the first indication that you should replace

the domain of integration.

OK.

So, in this case, let's say you were asked to compute this.

So, the clever thing to do is to replace this domain by

this curve, which has the same endpoints, right.

And, then you can argue that this line integral is equal to,

let's call this c1 and let's call this c2, this line

integral is equal to integral over c2 over F dr plus the

double integral of curl F ds, where you integrate over the in

interior of this domain.

You see, you can use this formula.

Let's call it D or M, let's call it M.

OK.

So, the reason is the following that if you take this to the

left-hand side, you will get negative c2, right.

I'm writing it like this, because, I prefer to

say this is equal to this plus something.

But, I could put this integral to the left with minus sign,

which means I'm integrating over c1 minus c2, and then you

have to see that this is a boundary of M; c1 minus c2

would be going with this orientation, that would

be boundary of M.

Now, if I do it like this, this brings up another issue, which

is the issue of orientation in this formula.

So, I'm using this formula, but in both cases, I have integrals

which depend on orientation.

And, we have this rule for orientations.

OK.

Which we discussed already many times.

So, orientation on the left and on the right have to match in

this way where if you walk on the boundary on the

surface in your left.

So, the way I did it right now would mean that the orientation

on this surface would have to go that way.

OK.

So, here it would be the surface where orientation is

going through the black board.

The usual orientation, which we, you normally consider,

would be this orientation towards us, right.

And, that orientation, which we would get

with counterclockwise.

But, here I have chosen orientation which goes

clockwise, well because I also put minus c2

so it reverse this.

And so this will be the orientation, let's call it

away from us, not towards us, but way from us.

Because, if somebody would be walking, if their feet are on

the blackboard, but their head points that way, so they would

be walking in that classroom.

They'll be like Spider Man, and they would be walking also on

this bizarre world, which is behind this blackboard, which

we don't even know what's going on over there, right.

But, that person would be walking like this.

The surface will be on their left.

So, normally if I were to walk like this, I would have to walk

counterclockwise for the surface to be on my left.

So, if I wanted to have orientation here, towards us, I

would have to, for this formula to be true, I would have to

choose up as the orientation.

But, so I'm putting sort of several things into example,

because I'm out of time essentially, but I'm trying

to sort of pack a lot of information into

this one example.

So, you should really look at it more carefully, and

different aspects of this, OK.

The orientations have to be compatible.

And there is one more thing which I'm doing here, which is

that I'm replacing a complicate path by an easy path, a simple

path, and the trade-off is that this guy is equal to this one.

But, there's a price to pay, and the price is the double

integral over this membrane, over this sort of film.

Think of it sort of like a film that you get from soap.

It's like a frame and there is a kind of soap.

So, you have to integrate over that, the curl.

The best possible scenario, where this works,

is when curl is 0.

If curl is 0 there is no price to pay, and, in fact, this

complicated integral over this complicated curve is equal, is

just straight equal to the integral here.

But, sometimes it's not 0, sometimes it's not 0, but

it could be very simple.

So, it's not such a big price to pay sometimes.

You see what I mean.

So, you have to, you have to see different options, the

different options that you have.

So, this is one example, where I am utilizing

formula number 2.

And, let me give you one more example, which is

kind of similar, but for formal number three.

So, maybe I should emphasize here, that in this formulas,

orientations have to be compatible.

This is important.

I'm not going to explain one more time, because we spent a

lot of time talking about this in the previous few lectures.

So, example number 2.

Example number 2, is kind of a similar trade-off, but where

you have to compute a double integral.

So let's say you have this upper hemisphere, and then you

also have this disc, this base of this upper hemisphere.

And, together they form the boundary of the

upper half of this bowl.

So, let's call this B, upper half of a bowl.

So, the last formula will tell you that the integral of

divergence of some vector field E over this B is equal to the

flux through the surface, through the boundary.

And, the boundary now consists of tow things.

Let's call this M1 and let's call this M2.

So, the formula then will be that it's double integral

over M1 plus double integral the over M2.

But, I have to say with respect to which orientation, and the

rule in the divergence theorem is that the orientation is

outward, which here would mean like this, and here

would mean like this.

Here it's outward, actually is upward on top part, and on

the bottom outward is actually downward.

But, both are outward.

So, that's what I mean here and here.

So, how can we utilize this formula.

For example, what you can do is you can express this as the

difference between this and this.

So, this way you can express a complicated integral over

something very curved, this sphere, or upper hemisphere.

In terms of two flat integrals, one of them is over a

three-dimensional solid, which actually may be quite simple --

depending on what -- I made a mistake like this. dV.

It could actually be 0, this divergence.

It could be 0.

So, then this actually drops out.

And, then you can express this guy as a negative

of this guy, right.

Or, even if it's non-0, it could be something very

simple, like 1, function 1 or something.

So, then it's easier to compute.

And, this one is certainly easy to compute easier to compute

than this one, because it's an integral over a flap region.

So, you have this trade-off again.

Similar to the trade-off discussed in the previous

example -- between an integral over one part of the boundary,

and the integral over the other part of the boundary, with a

surplus or price to pay coming from the integral

of the divergence.

So, these are very typical examples of the kind of things

that you need to know to be able to apply these formulas.

So, here's what we're going to do now.

I'm going to stop, and we're going to do a

course evaluations.

It will just take a few minutes.

Okay.

Come on guys.

Come on over, so we'll start distributing them.

And, we'll just take a few minutes, and after this,

after we're done with course the evaluations, I'll be

here for office hours.

And, so you can ask me questions about anything,

well related to the course.

Okay.