

PROFESSOR: I would like to
review just briefly what

we discussed last time.

We gave very important results,
and that was Green's Theorem.

And I would like to
know if you remember

when I said about the
settling for this problem.

So we'll assume we have
a domain without a hole,

D. D is a domain
without a hole inside,

without punctures or holes.


There is a scientific name in
mathematics for such a domain.

This is going to be
simply connected.


And this is a difficult
topological theorem,

but this is what we expect, OK?

And what does it mean?

What does it mean?

It means that in the C being
a Jordan curve was what?

How?

This was continuous,
no self intersections.


In such a case, we set up
M and N to be C1 functions.


And then we proceed through
the path integral of C.

Do you like this kind of C,
or you prefer a straight C?

The path integral of C of M
of xy dx class, N of xy, dy,

everything is in plane.

I'm sorry that I
cannot repeat that,

but we discussed that
time, is in the plane of 2.

And then what do
you remember in terms

of how this path integral,
[INAUDIBLE] inside,

is connected to a double
integral over the whole domain.

In particular, do you remember
this is easy to memorize

but do you remember
what's inside?

Because for the final, you are
expected to know his result.

STUDENT: [INAUDIBLE]

PROFESSOR: N sub X.

STUDENT: Minus M sub Y.

PROFESSOR: Minus M
sub Y. [INAUDIBLE]

must M M and N M sub Y.
Here is the Y. Of course this

would be dA in plane,
and in the if you

want to represent this
in the general format,

the MdX minus the MdY.

Feel free to do that.

One was a correlary
or a consequence.

This theorem was that if I
were to take this big M to be

the minus Y as a function,
then this function N will

be plus X, what will I get?

I would get that minus
YdX plus NdY will be what?


STUDENT: [INAUDIBLE]

PROFESSOR: Two times, excellent.

You are very awake.

So I wanted to catch you.

I couldn't catch you.

I thought you would say
the A of the domain,

but you said it right.

You said Y is the
area of the domain.

You probably
already in your mind

did the math saying X sub X
is one, minus Y sub 1 is 1.

1 plus 1 is two, so the
two part [INAUDIBLE].

OK, so what did we do with it?

We just stared at it?

No.

We didn't just stare at it.

We did something nice
with it last time.

We proved that, finally,
that the area, this radius R

will be pi R
squared, and we also

proved that the area
[INAUDIBLE] is what?

I'm testing you to
see if you remember.

STUDENT: AB pi.

PROFESSOR: AB pi.

Very good.

Or pi AB.

It's more, I like it the
way you said it, AB pi,

because pi is a transcendental
number, and you go around

and it's like partly
variable to put at the end.

And the real numbers
that could be anything,

so [INAUDIBLE] they are the
semi axes of the ellipse.

So we gain new knowledge and
we are ready to move forward.

And we're going to move
forward to something

called section 13.5, which
is the surface integral.

We will come back
to Green's Theorem

because there are
generalizations

of the Green's Theorem to
more complicate the case.

But in order to
see those, we have

to learn a little bit more.

In mathematics, you need to
know many things, many pieces

of the puzzle, and then
you put them together

to get the whole picture.

All right, so what
is 13.5 about?

This is just review.

13.5, if should be looking like
a friend, old friend, to you.

And I'll show you
in a minute why this

is called the surface integral.


I saw that US natives
don't pronounce integral,

they pronounce innegral.

And everybody that I heard
in romance languagespeaking

countries like Spanish,
Italian, Portuguese,

they put the T there
out, very visibly.

So it doesn't matter.

Even some accent difference
in different parts

of the United States
pronounce it differently.

So what is the surface
integral about?

It's about integrating a smooth
function, not a vector value,

but a real value function.

Let's say you have G or XY being
a nice interglobal function

over some surfaces.

And you say, I'm
going to take it,

double integral, over S of GDS,
where DS will be area level.


I had a student one time who
looked at two different books

and said, I have a problem
with this, [INAUDIBLE].

In one book it shows a
big, fat snake over S.

And in another book, a
double integral over it,

and I don't know which one it
is because I don't understand.

No matter how you denote it,
it's still a double integral.

You know why?

Because it's an
integral over a surface.

The same thing, integral over
a surface or a domain plane,

or anything twodimensional
will be a double integral.

So it doesn't matter
how you denote it.

In the end, it's going
to be a double integral.

Now, what in the world
do we mean by that?

DS is an old friend of
yours, and I don't know

if you remember him at all.

He was infinitesimal element
on some curved or linear patch.

Imagine your favorite surface.

Let's assume it's a graph.

It doesn't have to be a graph,
but let's assume it's a graph.

And that's your
favorite surface S.

And then you draw
coordinate lines,

and you are looking at a patch.


And this patch looks small,
but it's not small enough.

I want this to be
infinitesimally small.

Imagine that these curvature
lines become closer and closer

to one another.

And then we look in the
directions of DX and DY,

and then you say, wait a
minute, I'm not in plane.

If I were in plane,
DA will be DX, DY.

If you work with [INAUDIBLE],
I will be DX with DY.

So we've matched
the orientation.

If you would change
DY, [INAUDIBLE]

put the minus in front.

But this happens
because thank God this

will be a rectangular 1 patch
in plane, in the plane of 2.

But what if you
were on the surface?

On the surface, you
don't have this animal.

You will have which animal
I'm testing your knowledge.

I'm doing review with you.

For sure, you will
see something that

involves the S in the final.

Have you started browsing
through those finals

I sent you?

Just out of curiosity.

And do they look awful to you?

They look awful to you.

Come on.

I'm going to work with
you on some of those.

I don't want you to
have I don't want

you to be afraid of this final.

Because compared to
other exams that you'll

have in other courses,
where a lot of memorization

is emphasized, this
should not be a problem.

So you could go over
the types of problems

that are significant
in this course,

you will not have any you
shouldn't have any problem.

And I sent you three samples.

Didn't I send you three
samples with solutions?

Those are going to help
you once you read the exam

and you can go ahead
and try the exam

or go ahead, read the solutions.

If I give you more of that, then
you should be doctors in those,

and you would be able to
solve them yourselves.

What about this one?

This is not DA.

It's a DA times something.

There is some factor in front
of that, and why is that?

In case of Z equals
F of X and Y,

you should know that by heart,
and I know that some of you

know it.

You just have to ring
the bell, and I'll

start ringing the bell.

Look at my first step.

And now you know, right?

STUDENT: [INAUDIBLE] 1

PROFESSOR: I start with 1.

You said it right.

1 plus

STUDENT: F of X.

PROFESSOR: F of X

STUDENT: F squared.

PROFESSOR: Squared plus

STUDENT: [INAUDIBLE]

PROFESSOR: SY squared.

So this what you're doing.

What are you going to do?

You're going to
do wait a minute.

This animal of mine,
that looks so scary,

this is nothing but what?

It's the same thing as, not
the picture, my picture.

It's going to be double integral
over a plane or domain D.

Well, I just said
goodbye to the picture,

but I find you are really smart.

I would have drawn the
[INAUDIBLE] of a picture here.

This is S and this
is D. What is D?

It's the projection,
projects the shadow.

The projection of S
on the plane XY when

I have to deal with a graph.

So when I have to deal with a
graph, my life is really easy.

And I said I'd get double
integral over D of G of God

knows what in the end will
be a function of X and Y. OK?

And here I'm going to have
square root of this animal.

Let me change it,
F sub X squared

like because in this
one it is like that.

Plus 1.

It doesn't matter
where I put the 1.

DXDY.

DXDY will be like the area of an
infinitesimally small rectangle

based on displacement
DX and displacement DY

and disintegration.

So this is DA.

Make the distinction
between the DA and the DX.

Can I draw the two animals?

Let me try again.

So you have the
direction of X and Y.

You have to be imaginative and
see that some coordinate lines

are [INAUDIBLE] for fixing Y.

When I fix Y, I sliced a
lot like that very nicely.

That's the same piece of
cheese that I've been dreaming

because I didn't have lunch.

I was too busy not
to have any lunch.

So you slice it
up like that where

Y equals constant to slice
it up like that for X

equals constant.

What you get are socalled
coordinate lines.

So the coordinate
lines are [INAUDIBLE].

Y equals my zeros, and
X equals the zeros.

And when they get to be
many dense and refined,

your curvilinear element is
this between two curves

like this two curves like that.

Shrunk in the limit.

It's an infinitesimal element.

This shadow is going
to be a rectangle.

Say that again, Magdalena.

This is not just
delta X and delta Y.

This is DX and DY
because I shrink them

until it become
infinitesimally small.

So you can imagine,
which one is bigger?

DS is bigger, or DA is bigger?

STUDENT: DS is bigger.

PROFESSOR: DS is bigger.

DS is bigger.

And can I see it's true?

Yes.

Because for God's sake, this
is greater than 1, right?

And if I multiply the
little orange area, by that,

I'm going to get this,
which is greater than 1.

They could be equal when
both would be plainer, right?

If you have a plane or surface
on top of a plane or surface,

then you have two
tiny rectangles

and you have like a prism
between them, goes down.

But in general, the
curve in your [INAUDIBLE]

here let me make
him more curvilinear.

He looks so so square.

But he's between two lines,
but he's a curvilinear.

Dinah says that he belongs to a
curved surface, not a flat one.

All right.

When he could be flat,
these guys go away.

Zero and zero.

And that would be it.

If somebody else, they
well, this is hard to imagine,

but what if it could
be a tiny this

would not be curvilinear, right?

But it would be something like
a rectangular patch of a plane.

You have a grid in that plane.

And then it's just DS
would be itself a rectangle.

When you project
that rectangle here,

it will still be a rectangle.

When we were little I mean,
little, we were in K12,

we're smart in math better
than other people in class

did you ever have to do
anything with the two areas?

I did.

This was the shadow.

The projection in this
was that [INAUDIBLE].

And do you know what
the relationship

would be if I have a plane.

I'm doing that for actually,
I'm doing that for Casey

because she has something
similar to that.

So imagine that
you have to project

a rectangle that's in plane to
a rectangle that is the shadow.

The rectangle is on the ground.

The flat ground.

What's the relationship
between the two ends?


STUDENT: [INAUDIBLE]

PROFESSOR: No matter
what it is, but assume

it's like a rectangle up
here and the shadow is also

a rectangle down here.

Obviously, the rectangle
down here, the shadow

will be much smaller than
this because this is oblique.

It's an oblique.

And assume that I
have this plane making

an angle, a fixed angle with
this laying on the table.

STUDENT: [INAUDIBLE]

PROFESSOR: Excellent.

STUDENT: cosine

PROFESSOR: Which one
is cosine of what?

So the S would be the
the equal sign of theta,

or the A will be the
S cosine of theta?

STUDENT: [INAUDIBLE] DA.

PROFESSOR: DA is the S
cosine of theta, a very smart

[INAUDIBLE].

How does she know [INAUDIBLE]?

STUDENT: Because it's
got to be less than one.

PROFESSOR: It's less
than one, right?

Cosine theta is
between zero and one,

so you think which one is less.

All right, very good.

So when you have a
simple example like that,

you were back to
your K12, and you

were happy I
just meant we were

avoiding three years of exams.

We only have [INAUDIBLE].

But now exams became
serious, and look.

This is curvilinear
elemental variant.

So let me write it how
people call the S's then.

Some people call it
curvilinear elemental variant.

Yeah?

Many engineers I
talk to do that.


Now, I think we should just
call it surface area element.


[? I'm ?] a physicist, so you
also say surface area element.

So I think we should just
learn each other's language.

We are doing the same things.

We just we have a language
barrier between it's

not writing interdisciplinary,
so if we could establish

a little bit more work in
common, because there are so

many applications to
engineering of this thing,

you have no idea yet.

OK, let's pick a problem like
the ones we wrote in the book,

and see how hard it gets.

It shouldn't get very hard.

I'll start with one, the
only one, that is naturally

coming to your mind
right now, which would

be the one where G would be 1.

Somebody has to tell
me what that would be.


So guys, what if G would be 1?

STUDENT: [INAUDIBLE]

PROFESSOR: Very good.

It would be the
A of the surface.

I'm going to look for
some simple application.

Nothing is simple.

Why did we make this problem,
this book, so complicated?

OK, it' s good.

We can pick I can make
up a problem like this one.


But I can do a better job.

I can give you an
better example.

I'm looking at the
example 1 in section 13.5.

I'll give you
something like that

if I were to write an exam 1.

I put on it something
like Z equals

X squared plus 1 squared.

You know is my favorite
eggshell which is a [INAUDIBLE].


And somebody says,
I'm not interested

in the whole surface,
which is infinitely large.

I'm only interested in a
piece of a surface that

is above the disk D of
center O and radius 1.

So say, what, Magdalena?

Say that I want just
that part of the surface

that he's sitting above the
disk of center O and radius 1.

And I want to know how to
set up the surface integral.

Set up main surface
area integral.


And of course, when you
first see that you freak out

for a second, and then you say,
no, no, that's not a problem.

I know how to do that.

So example 1 out of
this section would

be a double integral over your
S. You have to call it names.

S. 1 instead of G and DS.


But then you say wait a minute.

I know that is true, but I
have to change it accordingly.

The same thing is here.

So I'm going to have it
over D. And D is the shadow,

DS is the plane of what?

Of 1 times.

I know I'm silly saying 1
times, but that's what it is.

Square root of S of X squared
plus S of Y squared plus 1.

DS, DY or DA as Rachel
said, somebody said.

Aaron said.

I don't know, you just
whispered, I should say.


All right.

So first of all, this
looks a little bit bad.

It makes me a
little bit nervous.

But in the end, with your
help, I'm going to do it.

And I'm going to do it by
using what kind of coordinates?

I'm

STUDENT: [INAUDIBLE]

PROFESSOR: Former
coordinates of the Y and Z.

It would be a killer.

Double, double, square root
of 1 plus who's telling me

what's coming next?

STUDENT: 4X squared.

4X squared, excellent.

4R squared, you say.

STUDENT: [INAUDIBLE]

PROFESSOR: OK.

Let me write it with
X and Y, and then

realize that this is our square.

How about that?

And then I have DX,
DY over the domain D,

and now I finally become
smart and say I just

fooled around here.

I want to do it in four
coordinates finally.

And that means I'll say
zero to 2 pi for theta.

So that theta will be the
last of the [INAUDIBLE].

R will be from zero to 1.


And So what?

This is an ugly, fairly
ugly, I just [INAUDIBLE].

I don't know what
I'm going to do yet.

I reduced our confusion, right?

But I'm not done.

STUDENT: R.

PROFESSOR: R. Never forget it.

So if I didn't have this
R, I would be horrible.

Why would it be horrible?

Imagine you couldn't have the R.

STUDENT: [INAUDIBLE]

PROFESSOR: We have to look that
this thing in integral table

or some use the calculator,
which we are not allowed

to do in this kind of course.

So what do we do?

We say it's a new substitution.

I have an R. That's a blessing.

So U equals 4 squared plus 1.

DU equals 8R, DR. I
think R, DR is a block.

And I know that's what I'm
going to do is a U substitution.

And I'm almost there.


It's a pretty good
example, but the one

you have as a first example
in this section, 13.5,

it's a little bit
too computational.

It's not smart at all.

It has a similar function over a
rectangle, something like that.

But it's a little bit
too confrontational.

We are looking
for something that

is not going examples that are
going to be easy to do and not

involve too much heavy
competition by him, because you

do everything by him.

Not like you don't have
a calculator, et cetera.

And the exam is very
limited in time, DU over 8.

So you say OK, I'm
know what that is.

That's going to be the A of S.
And that is going to be 2 pi.

Why can't I be so confident
and pull 2 pi out?

STUDENT: [INAUDIBLE]

PROFESSOR: Because there
is no dependence on theta.

All right?

So I have that one.

And then you go all right,
integral, square of you

times the U over 8 1 over 8DU.

And I have to be careful
because when R is zero

if I put zero and 1 here
like some of my students,

I'm dead meat, because I'm going
to lose a lot of credit, right?

So I have to pay attention.

R is 0, and U equals?

STUDENT: 1.

PROFESSOR: 1.

R equals 1.

U equals 5.

And I worked this out
and I should be done.

And that's you should
expect something like that.

Nice, not computational,
you kind of looking.

What is integral of square of U?

STUDENT: [INAUDIBLE]

PROFESSOR: So you have
you do the three halves,

and you pull out the 2/3, right?

That's what you do.

And then you go between U equals
1 down, and U equals 5 up.

And it's like one of those
examples we worked before.

Remember, and more
important, you

had something like
that for surface area?

Oh, my god.

4 over 8.

How much is 4 over 8?

STUDENT: [INAUDIBLE]

PROFESSOR: One half.


Right?

So we will have 1 over
6, and write pi times 5

to the three halves minus 1.

So do I like it?

I would leave it like that.

I'm fine.

I'll forget about it.

I have people who care.

I don't care how some people
write it 5 with 5 minus 1

because they think
it looks better.

It doesn't.

That's the scientific
equation, and I'm fine with it.

Right?

OK.

So expect something like
maybe I'm talking too much,

but maybe it's a good thing
to tell you what to expect

because we have to [INAUDIBLE].

At the same time, we're
teaching new things

as staff instructors doing
review of what's important.

I'm thinking if I'm doing things
right and at the same pace,

I should be finished
with chapter 13

at the end of next week.

Because after 13.5,
we have 13.6 which

is a generalization
of Green's Theorem.

13.6 as you recall is
called Stokes' Theorem.

13.7 is also a generalization
of Green's Theorem.

And they are all related.

It's like the trinity
on [INAUDIBLE].

That's the Divergence Theorem.

That is the last section,
13.7, Divergence Theorem.

So if I am going
at the right pace,

by what is next
wee on Thursday?

The 23rd?

I should be more or less
done with the chapter.

And I'm thinking I have
all the time in the world

to review with you
from that moment on.

In which sense are
we going to review?

We are going to review
by solving past finals.

Right?

That's what we are that's
what I'm planning to do.

I'm going to erase this
and move on to something

more spectacular.

Many OK.

This second part that
I want to teach you

now about, many instructors
in regular courses

just skip it because they do
not want to teach you not you,

you are honor students.

But they don't want
to teach the students

about some more general
ways to look at a surface.

Remember, guys, a surface
that is written like that

is called a graph.

But not all the
surfaces were graphs.

And actually for a surface
S, what the most general way

to represent the presentation
would be a parameterization.


And I'll do a little bit
of a review for those.

R little R or big R
big R, because that's

the position vector the
way I serve it to you

on a plate, whether,
for curves in space.

I say that's R of P. And when
we moved on curves to surfaces,

I said you move your path
two directions of motion.

You have two what are
those called in mechanics?

Degrees of freedom.

So you have two
degrees of freedom

like latitude and longitude.

Then R belongs
the position vector

is a function of two variables,
and it belongs to R3,

because it's a vector in R3.

And want to have imagine
that my hand is a surface.

Well, OK.

This is the position vector, I'm
just kind of sweeping my hand,

going this way, one
degree of freedom.

Or going that way, the
other degree of freedom.

This is what
parameterization is.

So for a sphere, if you want to
parameterize the whole sphere

I'll be done in a second.

I need you to see
if you remember

how to parameterize a sphere.

I'm testing you.

I'm mean today.

So examples.

Example 1 is
parameterize a sphere.


Was it hard?

That was a long
time ago, my god.

X, Y, and Z are what?

Latitude from Santa Clause.

Always latitude from
the North Pole is 5.

Longitude is from zero to 5.

The meridian is zero to 5.

That was theta, the
parameter of theta.

R was the distance
from this to a point.

But R was allowed to be
from take many values.

Now if I'm moving on
a sphere of radius

A let me make
that radius a just

to make your life miserable.

Assume that A would
be a sample, A.

How am I going to write
that parameterization?

STUDENT: X equals
A plus [INAUDIBLE]?

PROFESSOR: A something,
A something, A something.

STUDENT: A [INAUDIBLE]

PROFESSOR: He is right.

I have to move on.

STUDENT: [INAUDIBLE]

PROFESSOR: Go slow.

So I have the last
one you were right,

Buddy, you have the
memory of a medical doctor

and some day you will
be a medical doctor.

Not everybody has a good memory.

So the way you can do that
is, wait a minute, this is pi,

right?

This [INAUDIBLE].

If you want the Z, you
start with that first.

And since Z is adjacent, you
go R, cosine, sine, phi equals

sine phi.

Now we started with X
because he's worked on this

and remembers everything.

He has it memorized.

Sine phi for both.

And times what in both cases?

He's just the guy
who's not here.

So sine phi.

It helps to memorize N
cosine theta, and sine theta.

Is that really easy to memorize?

So where phi was the
latitude from the North

Pole between zero
and phi, it theta

was the longitude excuse
me, guys longitude from zero

to 2 pi, all around one more.

So you say wait a
minute, Magdalena,

these are Euler's angle.

What do they call in mechanics?

I think they call
them Euler angles.

But anyway, for
phi theta, we call

them latitude and longitude.

I'll replace them, because look,
I want R to be in terms of U,V.

So in mathematics, it's
not about location.

We can call them
whatever we want.

Mathematics is about the freedom
to call people names no

to call things names
and people names

STUDENT: Could U not equal zero?

PROFESSOR: Who?

STUDENT: U.

PROFESSOR: Yes.

So U can

STUDENT: [INAUDIBLE]

PROFESSOR: yeah, but
why didn't I write zero?

Well

STUDENT: [INAUDIBLE]
makes sense.

PROFESSOR: because,
yeah, you can take both.

If I want to study
differentiability,

I usually have to take it less
than and less than and less

than and less than because we
studied differentiability on

[INAUDIBLE].

But right now, I can take them
from the North Pole itself

to the South Pole itself so.

I'm not deleting any meridian.

If I were suppose
I were to delete it.

By the way, what does this mean?

I'm just kidding.

I'll put it back.

But Alex had a smart
question over there,

and he made me thinking.

It's a dangerous thing
when people make you think.

So it goes from zero to 2 pi.

Why would that be?

Imagine you have all the
meridians in the world

except for one.

From the sphere, you cut it and
remove the Greenwich meridian,

the one that passes
through Greenwich Village.

The one not the one in New
York, the one next to London,

right?

So put it back.

Put that meridian back.

It's like you take an
orange, and you make a slice.

I am OK.

Stop with the fruit
because I'm hungry.

Now, example two.

Now, imagine another surface
area you're used to, the what?

The paraboloid is one of our
favorite guys this semester.

X squared plus Y squared.

What is the
parameterization of that?


Well, if I write it
like that, it's a graph.

But if I don't want to
write it as a graph,

I have to write
it as a parameter.

What am I going to do?

I really know X to be U, right?

That's the simplest
choice possible.

Y could be V. And then Z will
be U squared plus V squared.

And there I am.

[SNEEZE]

So I'm going to write bless
your heart, [INAUDIBLE].


V plus J plus U squared
plus V squared, K. So this

is the parameterization
of a paraboloid.

That one of them
there are infinitely

many the one that comes
to mind because it's

the easiest one to think about.

STUDENT: [INAUDIBLE].

PROFESSOR: Good.

For a minute, guys,
you didn't need me.

You didn't need me at all
to come up with those.

But maybe you would need me
to remember, or maybe not

to remind you of the helicoid.

Helicoid.

Did you go to the,
as I told you to go

to the [INAUDIBLE] spectrum
what was that called?

The

STUDENT: Science spectrum.

PROFESSOR: Science spectrum.

And dip into soap solution
the thingy was a metal

rod with a with a what?

With; a helix made
of metal so the soap

film would take which shape?

The shape of this spiral that's
going to go inside here, right?

That's called a helicoid.

OK.

All right.

You're not mad at me.

STUDENT: No.

PROFESSOR: OK, good.

So in this case, R
of UV will be what?

It was a long time ago, once
upon a time I gave it to you.

It's extremely hard to
memorize if you don't work

with it on a regular basis.

If it were a helix,
what would it be?

If it were a helix, it
would be R of T right?

It would be like equal
sign T, A sine T, BT.

Say it again, Magdalena,
that was a long time ago,

chapter 10.

Chapter 10.

Equal sign, T, A sine
T, MBT, standard helix.

This is not going to be that.

It's going to be U cosine B.
U sine B. Look at the picture.

And imagine that these guys
are extended to infinity.

It's not just the
stairs themselves,

or whatever they are.

There are infinite lines,
straight lines, and busy.

This is done.

NB is a positive constant.


But your parameters are
U and V. Any other guy

that comes to mind, I'm out
of imagination right now.

You can do a torus on the
fold that looks like a donut.

You will have two parameters.

Imagine a donut.

How do you I'm not
going to write that.

Eventually I could give you
that as an extra credit thing.

What are the two degrees of
freedom of moving on the donut,

assuming that you would
like to move in circles?


STUDENT: [INAUDIBLE]


PROFESSOR: Let me draw a
donut, because I'm hungry,

and I really I cannot help it.

I just have to this is
called a torus in mathematics.

And you'll have one degree
of freedom will be like this,

the other degree of
freedom will be like that.

This is U and B.
Instead of U and B,

mathematicians,
apologists, geometers,

they call those angles phi
and theta because they really

are between zero and 2 pi.

It has a rotation like
that along the donut.

You can cut, slice
the donut, or if they

don't put cheese filling in it.

That was a bad idea not
having anything to eat.

And the other angle will be your
2 pi along this little circle.

So you still have two degrees
of freedom on a donut.

It's a surface.

You can write the
parameterization.

Yes?

STUDENT: Why is a
pie this way around.

Why is it like [INAUDIBLE].

PROFESSOR: It
doesn't have to be.

STUDENT: Or is it just
kind of like [INAUDIBLE]?

PROFESSOR: That's
what they call it.

Yeah.

So they are between 2 and 2 pi.

While I erase or should
I enough expectation

in terms of parameterization,
I have to night

teach you something about that.

If somebody would say I'm
giving you a patch of a surface,

but that patch of a
surface is in a frame

it's a nice parameterization.

This is the P on the surface.


And you say, well,
the parameterization

is going to be R
of U and V equals

X of UVI plus Y of
UVJ plus Z of UVK.


And suppose that somebody says
this is you favorite test.


Find V. Well, that
would be absurd.

My god, how do we do that?

Find the flux
corresponding to do

we say restart
just a second just

to restart with applications.

[INAUDIBLE]


We don't say what kind of
vector field that it is,

but we will say plus
corresponding to the vector

field.

F [INAUDIBLE].


And this vector
field, I'll tell you

in a second what's expected
from this to be a vector field.

Through, on the surface, we
find on the surface yes.

Mathematicians say
define normal surface S.

But a physicist will
say flux through,

the flux corresponding
to F through the surface.


Yes.

So you'll say why would that
be, and what is the flux?

By definition, how
should we denote it?

Let's make a beautiful script
F. That's crazy, right?

And then it goes doubling
over the surface F test.

Is anybody mechanical
engineering here?

Do you know the flux formula?

It's going to be [INAUDIBLE]
over S of F, this magic thing.

Not DN, DS.

Do you know what N means?

What it is N for
mechanical engineering,

[INAUDIBLE] engineers?

N to would be the unit normal
vector field to the surface S.

How would you want
to imagine that?

You would have a surface, and
you have this velocity vectors

here at the bottom that goes
to S. And this field goes up.

You'll have a force and
acceleration, velocity,

you have everything
going this way.

And you want to find
out what happens.

You introduce this notion
of flux through the surface.

Another way to have a
flux through the surface

maybe through the same
surface but associated

through another
kind of concept

if there could be
something else.

In electromagnetism, F would be
something else, some other type

of vector field.

Yes, sir.

STUDENT: [INAUDIBLE].

PROFESSOR: So find out, by
the way until next time,

if you were an electrical
engineering major, what

would flux be for you guys?

Two surfaces, one would be the
meaning of the vector field

F for you, and
why would you care

about the electromagnetic
flux or something like that.

I don't want to talk
too much about it.

It's for you to do the
search and find out.

So suppose that
somebody gives you

this notion that says you
have a parameteric surface.

Give an application
of that and find out

how you're going
be deal with it.

I'll give you an example
that shouldn't be too hard.


I'll make up my own example.

And looks like example 6, but
it's going to be different.


Example.


Find the flux F if F will
be a simple function.

Let's say something equals X, I
plus Y,J Z, K at every point X,

Y at every point
of the space XYZ.

That means you could have this
vector field defined everywhere

in space in [INAUDIBLE].

But you only care about
this acting on the surface.

So it's acting on the surface.


And then what will the flux be?

On the surface, which surface?

My favorite one, Z equals
X squared plus Y squared.


First of all, you say
wait, wait, Magdalena,

do you want to do it like that?

Do you want to say F
over XY to be a graph?

Or do you want to consider it
as a parameterized surface?

And that means it's the same
thing, equivalent to or if

and only if, who tells me again
what R was for such a surface?

STUDENT: XI.

PROFESSOR: X is
U. Y is V, so U

STUDENT: [INAUDIBLE]

PROFESSOR: I, that
would be J, then good.

U squared plus U squared UK.

Well, when you say that,
we have first of all,

we have no idea what
the heck we need to do,

because do we want to do it
in this form like a graph?

Or do we want to do
it parameterized?

We have to set up
formulas for the flats.

It's not so easy.

So assume that we are brave
enough and we start everything.

I want to understand what
flux really is as an integral.

And let me set it up for the
first case, the case of Z

equals F of X and Y.
And I'm happy with it

because that's
the simplest case.

Who's going to teach
me what I have to do?

You are confusing.

I have double integral over S
minus theory of F in general.

This is a general
vector value field.


It could be anything.

Could be anything.

But then I have to [INAUDIBLE],
because N corresponds

to the normal to the surface.

So I it's not so easy, right?

I have to be a little bit smart.

If I'm not smart

STUDENT: [INAUDIBLE]


PROFESSOR: That you
are getting close.

So guys, the normal
twoway surface somebody

gave you a surface, OK?

And normal to a surface
is normal to the plane

the tangent plane
of the surface.

So how did we get that?

There were many ways to do it.

Either you write
the tangent plane

and you know it by heart
that was Z minus Z zero

equals what the heck was
that S of X times X minus X

equals plus X of Y
times Y minus Y zero.

And from here you collect
what do you collect?

You move everybody
it's a moving sale.

You move everybody to the
left hand side and that's it.

[INAUDIBLE] moving sale.

OK?

And everybody will be
giving you some components.

You're going to have minus S
of X S minus X zero minus S

of Y, Y minus Y zero, plus
1 this is really funny.

1 times Z minus Z, Z.

Your normal will
be given by what?

The normal how do
you collect the normal?

STUDENT: [INAUDIBLE]

PROFESSOR: Pi is A, B, C. A,
B, and C will be the normal.

Except it's not unitary.

And the mechanical engineer
tells you, yeah, you're

stupid well, they
never say that.

They will stay look, you have
to be a little more careful.

Not say they are equal.

What do they mean?

They say for us,
in fluid mechanics,

solid mechanics, when we write
N, we mean you mean vector.

You are almost there.

What's missing?

STUDENT: Magnitude. [INAUDIBLE].

PROFESSOR: Very
good, the magnitude.

So they will say, go ahead and
you [INAUDIBLE] the magnitude.

And you are lucky now that
you know what N will be.

On the other hand

STUDENT: [INAUDIBLE].

PROFESSOR: This is excellent.

The one on the bottom Alex
is thinking like in chess, two

or three moves ahead.

You should get two extra
credit points with that.

STUDENT: All right.

PROFESSOR: You already got it.

DS is 1 plus S of X squared
plus F of X squared.

The 1 on the bottom and the
1 on the top will simplify.

So say it again, Magdalena.

Let me write it down here.

1 S of X, minus S of Y
1 over all this animal,

S of X squared plus S
of Y squared plus 1.

This is the thinking
like the early element

times the early element
will be the same thing.

I'll write it twice even if you
laugh at me because we are just

learning together,
and now you finally

see everybody can
see that desimplifies.


So it's going to be easy to
solve this integral in the end,

right?

So let's do the
problem, finally.

I'm going to erase it.

Let's do this problem
just for us, at any point.

I didn't say where.

Over the same thing.

The DS was over V01.

So the picture is
the same as before.

The S will be the
nutshell, the eggshell

I don't know what it was
over the domain D plane.

The domain D plane
was D of zero 1.

And I say that I need
to use another color.

This it's going to be my
shell, my surface S. Z

equals X squared
plus [INAUDIBLE].

How do you compute the flux?

Well, this is that.

So if we have to be a
little bit careful and smart

and say double integral over
S, and now without rushing,

we have to do a good job.

First of all, how do
you do the dot product?

The dot product

STUDENT: [INAUDIBLE]

PROFESSOR: Right.

So first component
times first component,

a second component,
second component times

second component plus
that component times

third component.

So if 1 is X, F2 is 1.

Good.

Z, though, he's not free.

He's married.

Why is he married?

STUDENT: [INAUDIBLE].

PROFESSOR: Because he
depends on X and Y.

So Z was even here,
because I'm on the surface.

I don't care what F does
away from the surface,

but when he sticks
to the surface, when

he's origin is on
the surface, then he

has to listen to the surface.

And that Z is not independent.

The Z is X squared
by Y squared here.

In a bracket, we are
over the surface.

That product minus S
of X, minus S of Y. I

know you're going to
laugh at me because I

haven't written where they are.

But that's what I
need your help for.

DA.


Who are they?

Who is this guy?

STUDENT: The [INAUDIBLE].

PROFESSOR: What?

STUDENT: [INAUDIBLE].

PROFESSOR: Negative 2X.

Is it?

STUDENT: No.

PROFESSOR: How about this guy?

STUDENT: [INAUDIBLE].

PROFESSOR: Negative 2Y.

How about this guy?

I'm just kidding.

OK.

So finally we should be able
to compute this integral.

That looks awful.

Over D.

So instead of S, we
have the D, which

is the disk of
radius one in plane.

And we say, OK, I have,
oh my god, it's OK.

This times that is how much?

STUDENT: [INAUDIBLE].

PROFESSOR: Minus 2X squared.

Right?

There.

Take the green.

This times that is how much?

Minus the Y squared.

And this times that is finally
just X squared plus Y squared.

Very nice think.

I think that at
first, but now I see

that life is beautiful
again DX, DY

that I can go ahead and do it.

I can get a hold of this.

And inside that, what do I
what am I left with in the end?

STUDENT: [INAUDIBLE].

PROFESSOR: Minus 2
times this animal,

called X squared plus Y squared,
which is going to be R squared.

So the flux the flux for
this problem in the end

is going to be very
nice and sassy.

Look at that.

F would be

STUDENT: There
would not be any

STUDENT: [INAUDIBLE]


PROFESSOR: What?

STUDENT: You've got
minus 2 and the plus 1.

PROFESSOR: Oh, thank God.

Thank God you exist.

So I thought about
it before, but then I

said I don't know why.

I messed up.

So we have minus R squared.

Very good.

It's easy.

Times an R from the
Jacobian, DR is theta.

And theta is between 0 and 2 pi.

And R between 0 and 1.

And now I will need
a plumber to tell me

what I do the limits
of the integrals,

because I think I'm getting
a negative answer, so.


STUDENT: [INAUDIBLE].

PROFESSOR: I'll do
it, and then you

tell me why I got what I got.

I have a minus
pulled out by nature.

And then I have integral

STUDENT: R [INAUDIBLE].

PROFESSOR: R to the
fourth of a fourth.

Very good.

But you have your [INAUDIBLE] so
when I do between zero and 1

STUDENT: It's [INAUDIBLE].

PROFESSOR: 1 over
4 you are too

fast as 2 pi that's a
good thing minus pi over 2,

you said, Gus.

And I could see it
coming straight at me

and hit me between the eyes.

What is the problem.

Is there a problem?

Without an area as a flux, would
that say, what is the negative?

Yes.

How can I make it positive?

This is my question.

STUDENT: Change the direction.

PROFESSOR: Change
the direction of who?

STUDENT: The flux.

PROFESSOR: The flux.

I could change the direction.

So what is it that
doesn't match?

[INAUDIBLE]

If I want to keep the
flux will be the same.

When I can change the
orientation of the service.

And instead I get a minus then.

My N was it sticking
in oh, my god.

So is it sticking
in or sticking out?

Look at it.

Think about it.

I have minus the positive guy
minus another positive guy,

and 1 sticking out.

But it goes with
the holes inside.

This is the paraboloid
[INAUDIBLE].

If I have something I minus I
minus J, does it go out or in?

STUDENT: In.

PROFESSOR: It goes in.

It goes in, and it'll be up.

So it's going to be like
all these normals are

going to be like a vector
field like that, like amoebas.

But they are pointing
towards inside.

Do I like that?

Yes, because I'm a
crazy mathematician.

Does the engineer like that?

No.

Why?

The flux is pointing in or out?

The flux.

The flux.

The flux, the flux
is pointing out.

Are you guys with me?

X plus Y X plus I plus J.
It's like this pointing out.

So the flux get
out of the surface.

It's like to pour water
inside, and the water's

just a net not a net, but
like something that holds it in.

And like a

STUDENT: Like a [INAUDIBLE]?

PROFESSOR: pasta strainer.

And the water goes up
[SPRAYING NOISE], well,

like a jet.

Like that.

So that is your flux
going through the surface.

Are you happy that I took
the normal pointing inside?

No.

That was crazy.

So here comes you, the
mechanical engineer majoring

in solid or [INAUDIBLE]
and say Magdalena,

you should have taken
the outer normal,

because look at the
flux pointing out.

Take the outer of normal,
and things are going

to looks right and nice again.

So if I were to
change the normal,

I would put the
plus, plus, minus.

I'll take the outer normal.

And in the end I
get plus 5 over 2.

So no remark.

If I change N to minus N, this
would become the outer normal.

Then the flux would
become pi over 2. solar

flux depends on the what?

The match between
the flux, the angles,

sort of between the flux if
function, vector [INAUDIBLE]

function, and the normal
that I take to the surface.

Right?

I can change the normal and
I get the opposite answer.

In absolute values,
the same flux.

So flux should be equal
[INAUDIBLE] the absolute value.

Unlike the area that should
be always a positive number.

Volume, that should always
be a positive number.

So if I get a limited area,
that means I messed up.

If I get a negative
on all of them,

it means messed up in my
computation somewhere.

But that doesn't mean
I messed up here.

I just chose the other normal.

It's possible.

So the flux can be taken as
is and put in absolute value.

All right.

OK.

We have to think of it like
the surface, and stuff that

goes through surface
in electric circuits.

Can you do some research
for you about flux

and electrical engineering?

And next time somebody
tells me a story about it.

Who is again who is
electrical engineering major

here?

Oh, so five people.

You're going to get four
extra credit points.

You guys are jealous.

I'm going to give you four extra
credit points if in 10 minutes

you can tell us a little bit
about where flux can be seen.

Well, you don't have
to come to the board.

You can just talk to us
from outside if you want,

or down inside the classroom.

Tell us where the
notion of flux appears

in the electric
circuits and why it

would be important for
Calculus 3 as well.

OK.

Now a big question
before I let you go.

Can I have a flux
that corresponds

to a parameterization?

That is my big worry, that
I have to do that as well.

Eventually, could I
have solved this problem

if the surface that
is parameterized

was my friend
who was my friend?

I don't remember.

UI plus VJ plus U
squared plus you

gave it to me OK, that
was the previous example,

and that's the last
example on the board.

So you have double integral
of force field times NDS.

Now, what if I say I don't
want to do it like this Z

equals F of XY.

So I don't want to
do it like that.

I want to do it in
a different way.

That means you pulling out of
your brain some old memories.

F was F, right?

You need to leave F alone,
poor fellow, because he

has no better way to do it.

This is becoming
complicated, the [INAUDIBLE]

mechanical engineering.


And what's given to you
before, but you don't remember?

R was given to you
as position vector.

R sub U and R sub V,
you may not remember

that was a long time ago we
proved that R sub U and R sub

V were on the surface.

They are both tensions
of the surface.

It was a long time ago.

So the normal is
[INAUDIBLE], and that's

exactly what I wanted to
say the normal will be.

Not quite pressed product,
but just like before,

pressed product
divided by the norm,

because then the unit normal
vector has to be length 1.

So I have to divide
by the number.

[SNEEZE]

The DS

STUDENT: Thank you.

PROFESSOR: is going to
OK, now it's up to you guys.

You're smart.

You know what I want to say.

So I'll pretend that you
know what DS is in terms

of the parameterization.

What's coming?

We said that.

It was a long time ago.

You can guess it by
just being smart

STUDENT: [INAUDIBLE].

PROFESSOR: or you can

STUDENT: [INAUDIBLE].

PROFESSOR: Yes, exactly.

And you got another
one extra credit point.


STUDENT: [INAUDIBLE]

PROFESSOR: So since before,
they were simplified,

for god's sake.

Now we have the new kind
of writing area element DS.

They also have to simplify.

It wasn't hard to see.

So you could have
done it like that.

You could have done
it like that, how?

Somebody need to help me,
because I have no idea what

I'm going to do here.

Do we get the same thing or not?

This is the question.

And I'm going to
finish with that,

but I don't want to
go home I'm not

going to let you go home
until you finish this.

F was a simple,
beautiful vector field.

Given like that.

This is a force.

May the force be
with you like that.

But we changed it in U,V because
we are acting on the surface S,

what is the pressure
in V, right?

So you have UI plus VJ
plus you gave it to me

U squared plus V squared.

Am I right, or am
I talking nonsense?


All right.

So now again I have
to be seeing them.

Am I getting the same thing?

If I'm not getting
the same thing,

I can just go home and
get drunk and be sad.

But I have to get
the same thing.

Otherwise, there is something
wrong with my setup.

So I have to have U, V.
U squared plus V squared.

Close.

Dot product.

This guy over on top say what?

Magdalena, this guy over on top
has to be has to be a what?

Well, I didn't say what it was.

I should do it now.

Right?

So how will we do that?

We were saying R of
UV will be UI plus VJ

plus U squared plus V squared.

OK.

So R sub U will be
you teach me quickly,

and R sub [INAUDIBLE]
is voila.

STUDENT: [INAUDIBLE]

PROFESSOR: 1

STUDENT: [INAUDIBLE]

PROFESSOR: Plus zero
thank you plus 2U, OK.

0 plus 1J plus 2VK Am I done?

I'm done.

No, I'm not done.

What do I have to do?

Cross them.


Cross multiply IJK.

This looks nice.

Look, it's not so ugly.

I thought it would
be uglier, right?

OK.

What it is?

What it this thing?

STUDENT: [INAUDIBLE].

PROFESSOR: Minus the
U, I. Minus plus.

Minus, plus 1.

2V minus because it's

STUDENT: Minus.

PROFESSOR: minus in front.

Right.

So I'm alternating.

And 1K.

So again, I get minus
X of S minus XY and 1,

and again, I'm pointing
in, and that's bad.

So my normal will point
inside the surface

like needles that are
perpendicular to the surface

pointing inside.

But that's OK.

In the end, I take
everything in absolute value.

Right?


So again, I do the same math.

So I get minus I don't
want to do it anymore.

Minus 2A squared, minus 2B
squared, plus your squared,

plus this squared,
then you save me

and you said minus 2
squared [INAUDIBLE] squared.

DUDV.

But DUDV means that UV is a
pair, a point in this, guys.

UV.

It's a pair in the
disk of radius one.

So I'm getting exactly,
what exactly the same thing

as before.

Because this is
minus R squared, so I

get integral, integral, minus
R squared times R. DR, D theta.

From zero to 1,
from zero to 2 pi,

and I get the same
answer, which was?

STUDENT: [INAUDIBLE].

PROFESSOR: Minus what?

STUDENT: [INAUDIBLE]

PROFESSOR: Pi over

STUDENT: [INAUDIBLE].

PROFESSOR: You see?

I already forgot.

STUDENT: 2.

PROFESSOR: So what
matters is that we

take the flux in
absolute value because it

depends on the
orientation of the normal.

If we take the
normal [INAUDIBLE].

Please, one thing I want you
to do when you go home now,

open the book which
maybe you rarely do,

but now it's
really the material

became complicated enough.

We are not just doing math,
calculus, we are doing physics,

we are doing mechanics, we are
dealing with surface integrals

and flux.

I want you to open the book
at page I don't know.

At surface integrals
starts at page 1,063.

Section 13.5.

And it keeps going like that,
pretty pictures of surfaces

and fluxes and so on.

Vector fields.

And it keeps going like that.

But it doesn't cover anything
new except what I said today.

It's just that it shows
you examples that are not

as beautiful as the ones I
gave, but they are essentially

the same, only a little
bit nastier to complete.

So up to 1,072.

So that is what you're
going to do this weekend,

plus the homework.

Keep on the homework.

Now, if you get stuck
Saturday, Sunday,

whenever you try your homework
you get stuck, what do you do?

STUDENT: [INAUDIBLE]

PROFESSOR: You email me.

So you say what in the world
is going on with this problem

because I tried it seven
times and 88 times.

And then you got
the brownie points.

STUDENT: [INAUDIBLE]

PROFESSOR: [INAUDIBLE] problem.

STUDENT: [INAUDIBLE] by 32.

PROFESSOR: There
was a problem, guys.

There are not so many problems.

But the only part, serious
part that we would catch,

he found it first, and
he tried it 88 times.


I'll never forget you, though,
because you are unique,

and that I appreciated
that very much.

So doing this weekend, do
not hesitate to pester.

I will answer all the web
work problems you have.

I want you to do well.

Next week is the last
week on new theory,

and then we start
working for the final,

so by the time of the final,
you'll be [INAUDIBLE].

STUDENT: [INAUDIBLE]?

PROFESSOR: Yes, sir.

Oh, I appreciated
that you did that.

STUDENT: [INAUDIBLE]


PROFESSOR: Again,
I forgot these.


With the extra points you
got, you shouldn't care.
