TTU Math2450 Calculus3 Secs 13.6 - 13.7

Stokes' Theorem and Divergence Theorem

PROFESSOR: Do you have
any kind of questions?

There were a few questions
about the homework.

Casey, you have that problem
we need to the minus D?

STUDENT: Yeah.

PROFESSOR: The minus D?

Let's do that in class, because
there were several people who

faced that problem.

You said you faced it, and
you got it and can I cheat?

Can I take your work so I
can present it at the board?

I'm serious about it.

STUDENT: OK, um.

PROFESSOR: So I know
we've done this together.

I don't even
remember the problem.

How was it?

Homework problem.

STUDENT: She knows it.

[INAUDIBLE]

STUDENT: But she knows it.

PROFESSOR: They'll work with you
to be minus M. Can you tell me,

Casey?

Can I tell the statement?

STUDENT: Well, a black
guy's [INAUDIBLE].

PROFESSOR: If you find it, give
it to me and I'll give you $2.

STUDENT: It's your problem.

[INAUDIBLE]

PROFESSOR: How is it's this one.

STUDENT: No.

PROFESSOR: Oh, no.

It's not this one.

STUDENT: Can I have it from you?

I won't give you anything.

STUDENT: Um, it's doable.

[INAUDIBLE]

PROFESSOR: So can somebody with
me now, that's my handwriting.

STUDENT: Yeah, I know.

It is weird.

PROFESSOR: OK.

All right.

So the problem says--

STUDENT: Do you have
a problem with me?

PROFESSOR: Any--

STUDENT: And then we're all
here for her. [INAUDIBLE].

Doesn't it feel like
[INAUDIBLE] kind of a bit?

PROFESSOR: It's the one
that has X of D equals

into the minus the cosign D.

STUDENT: Oh, [INAUDIBLE].

PROFESSOR: Y of T, and you
go by your exclamation.

I understand that you
love this problem.

And so you've had this type
of pathing to grow compute.

The pathing to grow with respect
to the [INAUDIBLE] fellow man

well meant that
in life is slowly

because nobody
[INAUDIBLE] with you.

And to go over C
of the integer will

be a very nice friend of
yours, [INAUDIBLE] explain it,

but of course, they are both
functions of T in general,

and you will have
the DS element,

and what does this mean?

S is [INAUDIBLE].

It means that you are
archic element should

be expressed in terms of what?

Who in the world is
the archeling infinite

decimal element.

It's the speed times the t.

STUDENT: Say it again?

PROFESSOR: It's the speed.

STUDENT: And what was the speed?

R in front of T. [INAUDIBLE].

Right?

So you will have to
transform this path integral

into an integral, respected
T, where T takes values

from a T0 to a T1.

And I don't want to give
you your notebook back.

STUDENT: It's OK.

PROFESSOR: OK, now I'll do
the same thing all over again,

and you control me and then
if I do something wrong,

you've done me.

And what were the--
what was the path?

Specified as what?

STUDENT: XYZ? [INAUDIBLE].

PROFESSOR: Yeah, the path was--

STUDENT: [INAUDIBLE].

PROFESSOR: T equals
from zero to pi over 2.

I have to write it down.

So let us write the--
[INAUDIBLE] are from the T.

The speed square root of is
from the T squared plus Y

prime of T squared, because
the sampling occurred.

Before we do that, we have
to go ahead and compute

X prime and Y prime.

And of course
that's product rule,

and I need a better marker.

STUDENT: [INAUDIBLE].

PROFESSOR: Yes, sir?

STUDENT: Do you think
the arc too is really

taken as an arc [INAUDIBLE]?

This

PROFESSOR: This is the--

STUDENT: Because we take-- I
would consider it as a path

function that looks like
an arc, or, like, thinking

that it's missing one
rule, and that's about it.

That's fine.

PROFESSOR: No, no, no
no, no, no, no, no.

no.

OK, let me explain.

So suppose you are
[INAUDIBLE] arc in plane

and this is your r of t.

STUDENT: Oh, OK.

PROFESSOR: And that's
called the position vector,

and that's x of t, y of t.

OK.

What is your velocity vector?

Velocity vector would be
in tangent to the curve.

Suppose you go in this
direction, counterclockwise,

and then our prime of
t will be this guy.

And it's gonna be
x prime, y prime.

And we have to
find its magnitude.

And its magnitude
will be this animal.

So the only thing here
is tricky because you

will have to do this
carefully, and there

will be a simplification
coming from the plus

and minus of the binomial.

So a few people missed it
because of that reason.

So let's see what
we have-- minus

e to the minus t, first
prime, times second one

prime plus the first one
prime times the second prime.

Good.

We are done with this first guy.

The second guy will be minus
e to the minus t sin t.

Why do I do this?

Because I'm afraid that
this being on the final.

Well, it's good practice.

You may expect
something a little bit

similar to that, so why don't we
do this as part of our review,

which will be a very good idea.

We are gonna do lots of
review this week and next

week already, because
the final is coming close

and you have to go over
everything that you've covered.

Let's square them,
and add them together.

OK.

When we add them together,
this guy, the speed [INAUDIBLE]

is going to-- bless you.

It is going-- it's not going
to bless, it's going-- OK,

you are being blessed, and
now let's look at that.

You have e to the
minus 2t cosine squared

and e to the minus
2t sine squared,

and when you add
those parts, the sine

squared plus cosine
squared stick together.

They form a block called 1.

Do you guys agree with me?

So what we have as the
first result of that

would be this guy.

But then, when you
take twice the product

of these guys in the
binomial formula,

and twice the product of these
guys, what do you notice?

We have exactly the
same individuals inside,

but when you do twice the
product of these two red ones,

you have minus, minus, plus.

But when you do twice the
product of these guys,

you have minus, plus, minus.

So they will cancel out.

The part in the middle
will cancel out.

And finally, when I square
this part and that part,

what's going to happen them?

And I'm gonna shut up because I
want you to give me the answer.

Square this animal, square
this animal, add them together,

what do you have?

STUDENT: Squared, squared
total-- [INTERPOSING VOICES]

PROFESSOR: T to the minus 2t,
so exactly the same as this guy.

So all I know under
the square root,

I'm gonna get square root of
2 times e to the minus 2t.

Which is e to the minus
t square root of 2.

Am I right, [INAUDIBLE]
that's what we got last time?

All right.

So I know who this will be.

I don't know who this will be,
but I'm gonna need your help.

Here I write it, x squared
of t plus y squared of t

in terms of t, squaring them
and adding them together.

It's gonna be again a piece of
cake, because you've got it.

How much is it?

I'm waiting for you to tell me.

This is this one.

[INAUDIBLE]

E to the?

STUDENT: Minus t

PROFESSOR: And anything else?

STUDENT: Was it 2?

[INAUDIBLE]

PROFESSOR: Why it times 2?

STUDENT: Times 2
in the last one.

Because we had an e to the minus
2t plus an e to the minus 2t.

PROFESSOR: So I took
this guy and squared it,

and I took this guy
and squared it--

STUDENT: No, we don't
have [INAUDIBLE].

PROFESSOR: And I sum them up.

And I close the issue.

Unless I have sine squared plus
cosine squared, which is 1,

so we adjust it to the minus 2t.

Agree with me?

All right, now we have
all the ingredients.

Do we have all the
ingredients we need?

We have this, we have
that, we have that.

And we should just go ahead
and solve the problem.

So, integral from 0 to pi over
2, this friend of yours, e

to the minus 2t
plus [INAUDIBLE].

The speed was over there, e to
the minus t times square root

of 2.

That was the speed.

[INAUDIBLE] magnitude, dt.

Is this what we got?

All right.

Now, we are almost
done, in the sense

that we should wrap things up.

Square root 2 gets out.

And then integral of it to the
minus 3t from zero to pi over 2

is our friend.

We know how to deal with him.

We have dt.

So when you integrate
that, what do you have?

Let me erase--

STUDENT: Negative square
root 2 over 3 [INAUDIBLE] 3t.

PROFESSOR: Right.

So let me erase this part.

So we have-- first we
have to copy this guy.

Then we have e to the
minus 3t divided by minus 3

because that is
the antiderivative.

And we take that into t equals
zero and t equals pi over 2.

Square root of 2 says I'm
going out, and actually minus 3

says also I'm going out.

So he doesn't want to be
involved in this discussion.

[INAUDIBLE]

Now, e to the minus 3 pi over
2 is the first thing we got.

And then minus e to the 0.

What's e to the 0?

STUDENT: 1.

PROFESSOR: 1.

So in the end, you have
to change the sign.

You have root 2 over 3
times bracket notation when

you type this in
WeBWorK because based

on your syntax, if your syntax
is bad, you are-- for example,

here we have to put
^ minus 3 pi over 2.

Are you guys with me?

Do you understand the words
coming out of my mouth?

So here you have to
type the right syntax,

and you did, and you got--

STUDENT: And I
didn't [INAUDIBLE]

but I need to type
the decimal answer.

In terms of decimal places.

PROFESSOR: This is a problem.

It shouldn't be like that.

Sometimes unfortunately--
well, fortunately it rarely

happens that WeBWorK program
does not take your answer

in a certain format.

Maybe the pi screws
everything up.

I don't know.

But if you do this
with your calculator,

eventually, you can What was
the approximate answer you got,

[INAUDIBLE]?

STUDENT: 0.467

PROFESSOR: And blah, blah, blah.

I don't know.

I think WeBWorK only cares
for the first two decimals

to be correct.

As I remember.

I don't know.

Now I have to ask
the programmer.

So this would be
approximately-- you

plug in the approximate answer.

I solved the problem,
so I should give myself

the credit, plus
a piece of candy,

but I hope I was able to
save you from some grief

because you have so much review
going on that you shouldn't

spend time on problems
that you headache

for computational reasons.

Actually, I have
computational reasons,

because we are not androids
and we are not computers.

What we can do is
think of a problem

and let the software
solve the problem for us.

So our strength does not consist
in how fast we can compute,

but on how well we can solve a
problem so that the calculator

or computer can carry on.

All right.

I know I covered
up to 13.6, and let

me remind you what we covered.

We covered some beautiful
sections that were called 13.4.

This was Green's theorem.

And now, I'm really proud
of you that all of you

know Green's theorem very well.

And the surface
integral, which was 13.5.

And then I promised
you that today we'd

move on to 13.6, which
is Stokes' theorem,

and I'm gonna do that.

But before I do that, I want
to attract your attention

to a fact that this is a bigger
result that incorporates 13.4.

So Stokes' theorem
is a more general

result. So let me make a
diagram, like a Venn diagram.

This is all the cases
of Stokes' theorem,

and Green's is one of them.

And this is something you've
learned, and you did very well,

and we only considered
this theorem

on a domain that's
interconnected.

It has no holes in it.

Green's theorem
can be also taught

on something like a
doughnut, but that's not

the purpose of this course.

You have it in the book.

It's very sophisticated.

13.6 starts at-- oh, my
God, I don't know the pages.

And being a co-author
of this book

means that I should
remember the pages.

All right, there it is.

13.6 is that page 1075.

OK, and let's see what
this theorem is about.

I'm gonna state it as
first Stokes' theorem,

and then I will see why Green's
theorem is a particular case.

We don't know yet why that is.

Well, assume you
have a force field,

may the force be with you.

This is a big vector-valued
function over a domain in R 3

that includes a surface s.

We don't say much
about the surface S

because we try to
avoid the terminology,

but you guys should
assume that this

is a simply connected
surface patch with a boundary

c, such that c is
a Jordan curve.

We use the word Jordan curve
as a boundary of the surface,

but we don't say
simply connected.

And I'm going to ask
you, what in the world

do we mean when we
simply connected?

I've used this before.

I just want to test your
memory and attention.

Do you remember what that meant?

I have some sort of little
hill, or s could be a flat disc,

or it could be a
patch of the plane,

or it could be just any
kind of surface that

is bounded by Jordan curve c.

What is a Jordan curve?

But what can we say about c?

So c would be nice
piecewise continuous-- we

assumed it continuous actually.

Most cases--

STUDENT: It has to connect
to itself, doesn't it?

PROFESSOR: No
self-intersections.

So we knew that from before,
but what does it mean,

simply connected for us?

I said it before.

I don't know how
attentive you were.

Connectedness makes
you think of something.

No holes in it.

So that means no holes.

No punctures.

No holes, no punctures.

So why-- don't draw it.

I will draw it so you can laugh.

Assume that the dog
came here and took

a bite of this surface.

And now you have a hole in it.

Well, you're not supposed
to have a hole in it,

so tell the dog to go away.

So you're not gonna have
any problems, any puncture,

any hole, any problem with this.

Now the surface is assumed
to be a regular surface,

and we've seen that before.

And since it's a regular
surface, that means

it's immersed in
the ambient space,

and you have an N orientation.

Orientation which is the
unit normal to the surface.

Can you draw it, Magdalena?

Yes, in a minute,
I will draw it.

At every point you
have an N unit normal.

What was the unit
normal for you when

you parametrize the surface?

That was the stick that
has length 1 perpendicular

to the tangent length, right?

So if you wanted to
do it for general R,

you would take those R sub u or
sub v the partial [INAUDIBLE].

And draw the cross
product, and this

is what I'm trying to do now.

And just make the length be 1.

So if the surface is
regular, I can parametrize it

as [INAUDIBLE] will exist
in that orientation.

I want something more.

I want N orientation to be
compatible to the direction

of travel on c-- along c.

Along, not on, but
on is not bad either.

So assume that this
is a hill, and I'm

running around the boundary.

Look, I'm just running around
the boundary, which is c.

Am I running in a
particular direction that

tells you I'm a mathematician?

It tells you that I'm a weirdo.

Yes.

So in what kind of
direction am I running?

Counterclockwise.

Why?

Because I'm a nerd.

Like Sheldon or something.

So let's go around,
and so what does

it mean I am compatible
with the orientation?

Think of the right hand rule.

Or forget about right hand
rule, I hate that word.

Let's think faucet.

So if your motion
is along the c,

so that it's like you are
unscrewing the faucet,

it's going up.

That should mean
that your orientation

n should go up, or, not
down, in the other direction.

So if I take c to be my
orientation around the curve,

then the orientation of
the surface should go up.

Am I allowed to go around the
opposite direction on the c.

Yes I am.

That's, how it this called,
inverse trigonometric,

or how do we call such a thing.

STUDENT: Clockwise?

PROFESSOR: Clockwise,
you guessed it.

OK, clockwise.

If I would go
clockwise in plane,

then the N should
be pointing down.

So it should be oriented
just the opposite way

on the surface S.

All right, that's sort
of easy to understand now

because most of
you are engineers

and you deal with this
kind of stuff every day.

What is Stokes' theorem?

Stokes' theorem says well, in
that case, the path integral

over c of FdR, F dot dR.

What the heck is this?

I'm not gonna finish the
sentence, because I'm mean.

There is a sentence there,
an equation, but I'm mean.

So I'm asking you first,
what in the world is this?

F is the may the
force be with you.

R is the vector position.

What is this animal?

The book doesn't tell you.

This is the work that
you know so well.

All right.

So you may hear math majors
saying they don't care.

They don't care because they're
not engineers or physicists,

but work is very important.

The work along the curve
will be equal to-- now

comes the beauty--
the beautiful part.

This is a double integral
over the surface [INAUDIBLE]

with respect to the
area element dS.

Oh, guess what, you wouldn't
know unless somebody taught you

before coming to
class, this is going

to be curl F. What is curl?

It's a vector.

So I have to do dot product
with another vector.

And that vector is N.

Some people read the book
ahead of time, which is great.

I would say 0.5% or less
of the students read ahead

in a textbook.

I used to do that
when I was young.

I didn't always have
the time to do it,

but whenever I had the
possibility I did it.

Now a quiz for you.

No, don't take any sheets
out, but a quiz for you.

Could you prove
to me, just based

on this thing that looks
weird, that Green's theorem is

a particular case of this?

So prove-- where
should I put it?

That was Stokes' theorem.

Stokes' theorem.

And I'll say
exercise number one,

sometimes I put this
in the final exam,

so I consider this
to be important.

Prove that Green's theorem is
nothing but a particular case

of Stokes' theorem.

And I make a face in the sense
that I'm trying to build trust.

Maybe you don't trust me.

But I-- let's do this together.

Let's prove together
that this is what it is.

Now, the thing is, if I were
to give you a test right now

on Green's theorem,
how many of you would

know what Green's theorem said?

So I'll put it here
in an-- open an icon.

Imagine this would be
an icon-- or a window,

a window on the computer screen.

Like a tutorial reminding
you what Green's theorem was.

So Green's theorem said what?

We have to-- bless you.

So Zander started
the theorem by-- we

have a domain D that was
also simply connected.

What does it mean?

No punctures, no holes.

No holes.

Even if you have a
puncture that's a point,

that's still a hole.

You may not see it, but if
anybody punctured the portion

of a plane, you are in trouble.

So there are no such things.

And c is a Jordan curve.

And then you say, OK,
how is it, how was F?

F was a c 1.

What does it mean
that if is a c 1?

F is a vector-valued function
that's differentiable,

and its derivatives
are continuous,

partial derivatives.

And so you think F of xy will
be M of xy I plus n of xyj

is a vector field, so
it's a multivariable,

so I have two variables.

OK?

So you think, OK, I
know what this is.

Like, this would be a force.

If this were a
force, I would get

the vector-- the work--
how can I write this again?

We didn't write it like that.

We wrote it as Mdx
plus Ndy, which

is the same thing as before.

Why?

Because Mr. F is MI plus NJ.

Not k, Magdalena.

You were too nice, but you
didn't want to shout at me.

And dR was what? dR was
dxI plus dyJ, right?

So when you do this, the
product which is called work,

the integral will
read Mdx plus Ndy,

and this is what it
was in Green's theorem.

And what did we claim it was?

Now, you know it, because
you've done a lot of homework.

You're probably sick and tired
of Green's theorem and you say,

I understand that work-- a
path integral can be expressed

as a double integral some way.

Do you know this by heart?

You proved this to me last
time you know it by heart.

That was N sub x minus M sub y.

And we memorized it-- dA
over this is a planar domain.

It's a domain in
plane d, [INAUDIBLE].

I said it, but I
didn't write it down.

So double integral over d,
N sub x minus M sub y dA.

We've done that.

That was section-- what
section was that, guys?

13.4.

Yeah, for sure, you will have a
problem on the final like that.

Do not expect lots of problems.

Do not expect 25 problems.

You will not have the time.

So you will have
some 15, 16 problems.

This will be one of them.

You mastered this
Green's theorem.

When you sent me
questions from WeBWorK

I realized that you were able
to solve the problems where

these would be
easy to manipulate,

like constants and so on.

That's a beautiful case.

There was one that gave
[INAUDIBLE] a headache,

and then I decided--
number 22, right?

Where this was more complicated
as an integrant in y, and guys,

your domain was like that.

And then normally to integrate
with respect to y and then x,

you would have had to split this
integral into two integrals--

one over a part of the
triangle, the other one

over part of the triangle.

So the easier way
was to do it how?

To do it like that, with
horizontal integrals.

And we've done that.

I told you-- I gave
you too much, actually,

I served it to you on a
plate, the proof-- solution

of that problem.

But you have many others.

Now, how do we prove that
this individual equation that

looks so sophisticated
is nothing

but that for the case
when S is a planar patch?

If S is like a hill,
yeah, then we believe it.

But what if S is the
domain d in plane Well,

then this S is exactly this d.

So it reduces to d.

So you say, wait a
minute, doesn't it

have to be curvilinear?

Nope.

Any surface that is bounded
by c verifies Stokes' theorem.

Say it again, Magdalena.

Any surface S that
is regular, so I'm

within the conditions
of the theorem, that

is bounded by a Jordan curve,
will satisfy the theorem.

So let's see what I've become.

That should became
a friend of yours,

and we already know
who this guy is.

So the integral FdR is your
friend integral Mdx plus Ndy

that's staring at you over c.

It's an integral over one
form, and it says that's work.

And the right-hand side, it's
a little bit more complicated.

So we have to think.

We have to think.

It's not about computation,
it's about how good we

are at identifying everybody.

If I go, for this particular
case, S is d, right?

Right.

So I have a double integral
over D. Sometimes you ask me,

but I saw that over a domain
that's a two-dimensional domain

people wrote only one
snake, and it looks fat,

like somebody fed
the snake too much.

Mathematicians are lazy people.

They don't want to write always
double snake, triple snake.

So sometimes they say,
I have an integral

over an n-dimensional domain.

I'll make it a fat snake.

And that should be enough.

Curl F N-- we have
to do this together.

Is it hard?

I don't know.

You have to help me.

So what in the world was that?

I pretend I forgot everything.

I have amnesia.

STUDENT: [INAUDIBLE].

PROFESSOR: Yeah, so
actually some of you

told me by email
that you prefer that.

I really like it
that you-- maybe I

should have started a
Facebook group or something.

Because instead of the
personal email interaction

between me and you,
everybody could see this.

So some of you tell me, I
like better this notation,

because I use it
in my engineering

course, curl F. OK, good, it's
up to you what you want to use.

d/dx, d/dy-- I mean it.

In principle, in r3,
but I'm really lucky.

Because in this case, F
is in r2, value in r2.

STUDENT: You mean d/dz?

PROFESSOR: Huh?

STUDENT: d/dx, d/dy, d/dz.

PROFESSOR: I'm sorry.

You are so on the ball.

Thank you, Alexander.

STUDENT: No, I thought I had
completely misunderstood--

PROFESSOR: No, no, no,
no, I wrote it twice.

So M and N and 0, M is a
function of x and y only.

N of course-- do I
have to write that?

No, I'm just being silly.

And what do I get in this case?

STUDENT: [INAUDIBLE].

PROFESSOR: I times this
guy-- how much is this guy?

STUDENT: 0.

PROFESSOR: 0.

Why is that 0?

Because this contains no z,
and I prime with respect to z.

So that is nonsense,
0i minus 0j.

Why is that?

Because 0 minus something
that doesn't depend on z.

So plus, finally-- the
only guy that matters there

is [INAUDIBLE], which
is this, which is that.

So because I have
derivative of N

with respect to h
minus derivative of M

with respect to y.

And now I stare at it,
and I say, times k.

That's the only guy that's
not 0, the only component.

Now I'm going to go
ahead and multiply this

in the top product with him.

But we have to be smart
and think, N is what?

STUDENT: [INAUDIBLE].

PROFESSOR: It's
normal to the surface.

But the surface is
a patch of a plane.

The normal would be trivial.

What will the normal be?

The vector field of all
pencils that are k-- k.

It's all k, k everywhere.

All over the domain is k.

So N becomes k.

Where is it?

There, N becomes k.

So when you multiply
in the dot product

this guy with this
guy, what do you have?

STUDENT: [INAUDIBLE].

PROFESSOR: N sub x
minus M sub y dA.

QED-- what does it mean, QED?

$1, which I don't
have, for the person

who will tell me what that is.

Latin-- quod erat demonstrandum,
which was to be proved, yes?

So I'm done.

When people put QED, that means
they are done with the proof.

But now since mathematicians
are a little bit illiterate,

they don't know much about
philosophy or linguistics.

Now many of them,
instead of QED,

they put a little square box.

And we do the same in our books.

So that means I'm
done with the proof.

Let's go home, but not that.

So we proved that for the
particular case of the planar

domains, Stokes' theorem
becomes Green's theorem.

And actually this is the curl.

And this-- well, not the curl.

But you have the curl of F
multiplied with dot product

with k and this green
fellow is exactly

the same as N sub
x minus N sub y

smooth function,
real value function.

All right, am I done?

Yes, with this Exercise 1,
which is a proof, I'm done.

You haven't seen many
proofs in calculus.

You've seen some from
me that we never cover.

We don't do epsilon delta in
regular classes of calculus,

only in honors.

And not in all the honors
you've seen some proofs

with epsilon delta.

You've seen one or two
proofs from me occasionally.

And this was one simple proof
that I wanted to work with you.

Now, do you know if
you're ever going

to see proofs in math
classes, out of curiosity?

US It depends how much
math you want to take.

If you're a math major, you
take a course called 3310.

That's called
Introduction to Proofs.

If you are not a math
major, but assume

you are in this
dual program-- we

have a beautiful and tough dual
major, mathematics and computer

science, 162 hours.

Then you see everything
you would normally

see for an engineering major.

But in addition, you
see a few more courses

that have excellent proofs.

And one of them is linear
algebra, Linear Algebra 2360.

We do a few proofs--
depends who teaches that.

And in 3310 also
you see some proofs

like, by way of contradiction,
let's prove this and that.

OK, so it's sort of fun.

But we don't attempt long
and nasty, complicated proofs

until you are in graduate
school, normally.

Some of you will do
graduate studies.

Some of you-- I
know four of you--

want to go to medical school.

And then many of you hopefully
will get a graduate program

in engineering.

OK, let's see another
example for this section.

I don't particularly
like all the examples

we have in the book.

But I have my favorites.

And I'm going to go
ahead and choose one.

There is one that's a
little bit complicated.

And you asked me about it.

And I wanted to
talk about this one.

Because it gave several
of you a headache.

There is Example 1, which
says-- what does it say?

Evaluate fat integral
over C of 1 over 2 i

squared dx plus zdy plus xdz
where C is the intersection

curve between the plane x plus
z equals 1 and the ellipsoid x

squared plus 2y
squared plus z squared

equals 1 that's oriented
counterclockwise as viewed

from the above picture.

And I need to draw the picture.

The picture looks really ugly.

You have this ellipsoid.

And when you draw
this intersection

between this plane and the
ellipsoid, it looks horrible.

And the hint of
this problem-- well,

if you were to be given
such a thing on an exam,

the hint would be
that a projection--

look at the picture.

The projection of the curve of
intersection on the ground--

ground means the
plane on the equator.

How shall I say that?

The x, y plane is this.

It looks horrible.

And it looks like an egg.

It's not supposed
to be an egg, OK?

It's a circle.

I'm sorry if it
looks like an egg.

OK, and that would be the
only hint you would get.

You would be asked to
figure out this circle

in polar coordinates.

And I'm not sure if all of
you would know how to do that.

And this is what worried me.

So before we do everything,
before everything,

can we express this
in polar coordinates?

How are you going to set
up something in r theta

for the same domain
inside this disc?

STUDENT: [INAUDIBLE].

PROFESSOR: So if we
were, for example,

to say x is r cosine
theta, can we do that?

And i to be r sine
theta, what would

we get instead of this equation?

Because it looks horrible.

We would get-- this equation,
let's brush it up a little bit

first.

It's x squared plus y squared.

And that's nice.

But then it's minus twice--
it's just x plus 1/4

equals 1/4, the heck with it.

My son says, don't say "heck."

That's a bad word.

I didn't know that.

But he says that he's being
told in school it's a bad word.

So he must know what
he's talking about.

So this is r squared.

And x is r cosine theta.

Aha, so there we almost
did it in the sense

that r squared equals
r cosine theta is

the polar equation,
equation of the circle

in polar coordinates.

But we hate r.

Let's simplify by an r.

Because r is positive--
cannot be 0, right?

It would be a point.

So divide by r and get
r equals cosine theta.

So what is r equals
cosine theta?

r equals cosine theta
is your worst nightmare.

So I'm going to make a face.

That was your worst
nightmare in Calculus II.

And I was just talking to a
few colleagues in Calculus II

telling me that the
students don't know that,

and they have a big
hard time with that.

So the equation of this circle
is r equals cosine theta.

So if I were to
express this domain,

which in Cartesian
coordinates would be written--

I don't know if you want to--
as double integral, We'd?

Have to do the
vertical strip thingy.

But if I want to do it
in polar coordinates,

I'm going to say,
I start-- well,

you have to tell
me what you think.

We have an r that
starts with the origin.

And that's dr.
How far does r go?

For the domain inside, r goes
between 0 and cosine theta.

STUDENT: Why were you
able to divide by r

if it could have equaled 0?

PROFESSOR: We already did.

STUDENT: Yes, but
then you just said

you could only do that
because it never equaled 0.

PROFESSOR: Right, and for
0 we pull out one point

where we take the
angle that we want.

We will still get
the same thing.

STUDENT: [INAUDIBLE].

PROFESSOR: No, r will be any--

STUDENT: Oh, I see.

PROFESSOR: Yeah,
so little r, what

is the r of any
little point inside?

The r of any little
point inside is

between 0 and N cosine theta.

Cosine theta would be the r
corresponding to the boundary.

Say it again-- so every
point on the boundary

will have that r
equals cosine theta.

The points inside the domain--
and this is on the circle,

on C. This is the circle.

Let's call it C ground.

That is the C.

So the r, the points
inside have one property,

that their r is between
0 and cosine theta.

If I take r theta
with this property,

I should be able to
get all the domain.

But theta, you have to be a
little bit careful about theta.

STUDENT: It goes from pi
over 2 to negative pi over 2.

PROFESSOR: Actually, yes.

So you have theta will be
between minus pi over 2

and pi over 2.

And you have to think
a little bit about how

you set up the double integral.

But you're not there yet.

So when we'll be there
at the double integral

we will have to think about it.

What else did I want?

All right, so did I give
you the right form of F?

Yes.

I'd like you to compute curl
F and N all by yourselves.

So compute.

This is going to be F1.

This is going to be F2.

This is going to be F3.

And I'd like you to
realize that this

is nothing but integral over
C F dR. So who is this animal?

This is the work, guys.

All right, so I should
be able to set up

some integral, double
integral, over a surface

where I have curl F times N dS.

So what I want you to
do is simply-- maybe

I'm a little bit too lazy.

Take the curl of F and
tell me what it is.

Take the unit normal vector
field and tell me what it is.

And then we will
figure out the rest.

So you say, wait a
minute, Magdalena, now

you want me to look at this
Stokes' theorem over what

surface?

Because C is the red boundary.

So you want me to look at
this surface, right, the cap?

So the surface could be the cap.

But what did I tell you before?

I told you that Stokes' theorem
works for any kind of domain

that is bounded by the
curve C. So is this the way

you're going to do it-- take
the cap, put the normals,

find the normals, and do all
the horrible computation?

Or you will simplify
your life and understand

that this is exactly the same
as the integral evaluated

over any surface bounded by C.

Well, this horrible thing
is going to kill us.

So what's the simplest
way to do this?

It would be to do it
over another surface.

It doesn't matter
what surface you have.

This is the C. You can take any
surface that's bounded by C.

You can take this balloon.

You can take this one.

You can take the
disc bounded by C.

You can take any surface
that's bounded by C.

So in particular,
what if you take

the surface inside
this red disc,

the planar surface
inside that red disc?

OK, do you see it?

OK, that's going to
be part of a plane.

What is that plane?

x plus z equals 1.

So you guys have to
tell me who N will be

and who the curl will be.

And let me show you again
with my hands what you have.

You have a surface that's
curvilinear and round

and has boundary C.

The boundary is C. You
have another surface that's

an ellipse that has
C as a boundary.

And this is sitting in a plane.

And I want-- it's very hard
to model with my hands.

But this is it.

You see it?

You see it?

OK, when you project
this on the ground,

this is going to become that
circle that I just erased,

so this and that.

We have a surface integral.

Remember, you have dS here up,
and you have dA here down--

dS here up, dA here down.

So that shouldn't be
hard to do at all.

Now what is N?

N, for such an individual,
will be really nice and sassy.

x plus z equals 1.

So what is the normal
to the plane x plus z?

[INAUDIBLE]

So who is this normal for
D curl F times [INAUDIBLE]

but N d-- I don't know,
another S, S tilde.

So for this kind of surface,
I have another dS, dS tilde.

So who's going to
tell me who N is?

Well, it should be
x plus z equals 1.

What do we keep?

What do we throw away?

The plane is x plus z equals 1.

What's the normal?

So the plane is x plus
0y plus 1z equals 1.

What's the normal to the plane?

STUDENT: Is it i plus k
over square root of 2?

PROFESSOR: i plus k,
very good, but why

does Alexander say the
over square root of 2?

Because it says,
remember guys, that that

has to be a unit normal.

We cannot take i plus k based
on being perpendicular to the x

plus z.

Because you need
to normalize it.

So he did.

So he got i plus k
over square root of 2.

How much is curl F?

You have to do
this by yourselves.

I'll just give it to you.

I'll give you three
minutes, and then I'll

check your work based on
the answers that we have.

And in the end, I'll have to do
the dot product and keep going.

Is it hard?

I should do it
along with you guys.

I have i jk d/dx, d/dy, d/dz.

Who were the guys? y
squared over 2 was F1.

z was F2.

x was F3.

And let's see what you got.

I'm checking to see if
you get the same thing.

Minus psi is the first guy.

[INAUDIBLE] the next one?

STUDENT: Minus j.

PROFESSOR: Minus j.

STUDENT: Minus yk.

PROFESSOR: yk.

And I think that's
what it is, yes.

So when you do the integral,
what are you going to get?

I'm going to erase this here.

You have your N.
And your N is nice.

What was it again, Alexander?

i plus k over square
root of 2, right?

So let's write
down the integral W

will be-- double
integral over the domain.

Now, in our case, the domain
is this domain, this one here.

Let's call it-- do you want to
call it D or D star or D tilde?

I don't know what.

Because we use to call
the domain on the ground

D. Let's put here D star.

So over D star, and the cap
doesn't exist in your life

anymore.

You said, bye-bye bubble.

I can do the whole
computation on D star.

I get the same answer.

So you help me right?

I get minus 1 times
1 over root 2.

Am I right?

A 0 for the middle term, and
a minus y times 1 over root 2,

good-- this is the
whole thing over here.

My worry is about dS star.

What was dS star?

dS star is the area
limit for the plane-- are

limit for how can I call this?

For disc, for D star, not for D.

It's a little bit complicated.

D is a projection.

So who reminds me how we did it?

dS star was what times dA?

This is the surface area.

And if you have a surface that's
nice-- your surface is nice.

STUDENT: It's area, so r?

PROFESSOR: What was this
equation of this surface

up here?

This is the ellipse that goes
projected on the surface.

STUDENT: Cosine of theta.

PROFESSOR: The equation
of the plane, see?

The equation of the plane.

So I erased it.

So was it x plus z equals 1?

STUDENT: Yes.

PROFESSOR: So z
must be 1 minus x.

So this is going to be the
square root of 1 plus-- minus 1

is the first partial.

Are you guys with me?

Partial with respect to x of
this guy is minus 1 squared

plus the partial of
this with respect to y

is missing 0 squared.

And then comes
dA, and who is dA?

dA is dxdy in the floor plane.

This is the [INAUDIBLE] that
projects onto the floor.

Good, ds star is going to
be then square root 2dA.

Again, the old
trick that I taught

you guys is that
this will always

have to simplify with
[INAUDIBLE] on the bottom

of the N. Say what?

Magdalena, say it again.

Square root of 2DA, this
is that magic square root

of 1 plus [INAUDIBLE].

This guy, no matter what
exercise you are doing,

will always simplify with
the bottom of N [INAUDIBLE],

so you can do this
simplification

from the beginning.

And so in the end, what
are you going to have?

You're going to have
W is minus y minus 1

over the domain D in the
plane that this will claim.

The square root of
[INAUDIBLE], and then you'll

have dA, which is dxdy

OK, at this point suppose
that you are taking the 5.

And this is why I
got to this point

because I wanted
to emphasize this.

Whether you stop here
or you do one more step,

I would be happy.

Let's see what I mean.

So you would have minus who
is y r cosine theta minus 1.

dA will become instead
of dxdy, you have--

STUDENT: [INAUDIBLE].

PROFESSOR: r, very
good. r dr is theta.

So you're thinking--

STUDENT: [INAUDIBLE].

PROFESSOR: --well,
so you're thinking--

I'm looking here what we have--
r was from 0 to cosine theta,

and theta is from
minus [INAUDIBLE].

Please stop here, all right?

So in the exam, we will not
expect-- on some integrals who

are not expected to
go on and do them,

which they set up the
integral and leave it.

Yes, sir?

STUDENT: Why did you throw
r cosine theta for y?

PROFESSOR: OK, because
let me remind you,

when you project the image
of this ellipse on the plane,

we got this fellow, which
is drawn in the book

as being this.

So we said, I want to
see how I set this up

in [INAUDIBLE] coordinates.

The equation of the
plane of the circle

was r equals cosine theta,
and this was calculus too.

That's why we
actually [INAUDIBLE].

So if somebody
would ask you guys,

compute me instead of an
area over the domain, what

if you compute for me the
linear area of the domain?

How would you do that?

Well, double integral of 1
or whatever-that-is integral

of r drd theta,
instead of 1, you

can have some other ugly
integral looking at you.

I put the stop here.

Theta is between minus
pie over 2n pi over 2

because I'm moving from here
to here, from here to here, OK?

Nr is between 0 and the margin.

Who is on the margin?

I started 0.

I ended cosine theta.

I started 0, ended cosine theta.

Cosine theta happens
online for the boundary,

so that's what you do.

Do we want you to do that?

No, we want you to leave it.

Yes?

STUDENT: He was asking why you
had a negative y minus 1 r sine

theta, not r cosine theta.

PROFESSOR: You are so right.

I forgot that x was r cosine
theta, and y was r sine theta.

You are correct.

And you have the group
good observation.

So r was [INAUDIBLE]
cosine theta.

And x was r cosine theta.

y was r sine theta.

Very good.

OK, so if you get something
like that, we will now

want you to go on, we
will want you to stop.

Let me show you one
where we wanted to go on,

and we indicate it
like this, example 3.

So here, we just dont'
want you to show some work,

we wanted to actually
get the exact answer.

And I'll draw the picture,
and don't be afraid of it.

It's going to look
a little bit ugly.

You have the surface
Z equals 1 minus x

squared minus 2y squared.

And you have to evaluate
over double the integral

of the surface S.
This is the surface.

Let me draw the surface.

We will have to understand
what kind of surface that is.

Double integral of curl F
[INAUDIBLE] dS evaluated

where F equals xI plus y squared
J plus-- this looks like a Z e

to the xy.

It's very tiny.

I bet you won't see it.

[INAUDIBLE] xy k and S. Is
that part of the surface?

Let me change the marker so
the video can see better.

Z-- this is a bad marker.

Z equals-- what was it, guys?

1 minus x squared minus 2y
squared with Z positive or 0.

And the [? thing ?] is
I think we may give you

this hint on the exam.

Think of the Stokes theorem
and the typical-- think

of the Stokes theorem
and the typical tools.

You have learned them.

OK, what does it mean?

We have like an
eggshell, which is coming

from the parabola [INAUDIBLE].

This parabola [INAUDIBLE]
is S minus x squared

minus 2y squared,
and we call that S,

but you see, we have
two surfaces that are

in this picture bounded by c.

The other one is the domain
D, and it's a simple problem

because your domain D is
sitting on the xy plane.

So it's a blessing that you
already know what D will be.

D will be those pairs
xy with what property?

Can you guys tell
me what D will be?

Z should be 0, right?

If you impose it
to be 0, then this

has to satisfy x squared
plus 2y squared less than

or equal to 1.

Who is the C?

C are the points
on the boundary,

which means exactly x squared
plus 2y squared is equal to 1.

What in the world is this curve?

STUDENT: [INAUDIBLE].

PROFESSOR: It's an ellipse.

Is it an ugly ellipse?

Uh, not really.

It's a nice ellipse.

OK, what do they give us?

They give us xy
squared and Z times

e to the xy, so this
is F1, F2, and F3.

So the surface itself
is just the part

that corresponds to Z
positive, not all the surface

because the whole surface
will be infinitely large.

It's a paraboloid that keeps
going down to minus infinite,

so you only take this part.

It's a finite patch that I stop.

So this is a problem
that's amazingly simple

once you solve it one time.

You don't even have to show your
work much in the actual exam,

and I'll show you why.

So Stokes theorem tells
you what in this case?

Let's review what
Stokes theorem says.

Stokes theorem says, OK,
you have the work performed

by the four steps that's
given to you as a vector value

function along the path
C, which is given to you

as this wonderful ellipse.

Let me put C like I did it
before, C. This is not L,

it's C, which what is that?

It's the same as
double integral over S,

the round paraboloid [INAUDIBLE]
like church roof, S curl F

times N dS.

But what does it
say, this happens

for any-- for every, for
any, do you know the sign?

STUDENT: [INAUDIBLE].

PROFESSOR: Surface
is bounded by C.

And here is that winking
emoticon from-- how

is that in Facebook?

Something like that?

A wink would be a good
hint on the final.

What are you going to do
when you see that wink?

If it's not on the
final, I will wink at you

until you understand
what I'm trying to say.

It means that you can change
the surface to any other surface

that has the boundary
C. What's the simplest

surface you may think of?

STUDENT: [INAUDIBLE].

PROFESSOR: The D.
So I'm going to say,

double integral over D. Curl
left, God knows what that is.

We still have to
do some work here.

I'm making a sad
face because I really

wanted no work whatsoever.

N becomes-- we've done this
argument three times today.

STUDENT: It's k.

PROFESSOR: It's a k.

That is your blessing.

That's what you have
to indicate on the exam

that N is k when I look
at the plane or domain.

STUDENT: And dS is DA.

PROFESSOR: And dS is dA.

It's much simpler than
before because you

don't have to project.

You are already projecting.

You are all to the floor.

You are on the ground.

What else do you have to do?

Not much, you just have to be
patient and compute with me

something I don't like to.

Last time I asked you
to do it by yourself,

but now I shouldn't be lazy.

I have to help you.

You have to help me.

i j k z dx z dx z dz of what?

x y squared and
this horrible guy.

What do we get?

Well, it's not so
obvious anymore.

STUDENT: [INAUDIBLE].

PROFESSOR: It's Z prime this
guy with respect to y Zx,

very good.

The x into the xy times i, and
I don't care about the rest.

Why don't I care about the rest?

Because when I prime y squared
with respect to Z goes away.

So I'm done with the first term.

I'm going very slow as you
can see, but I don't care.

So I'm going to erase more.

Next guy, minus and then
we'll make an observation.

The same thing here,
I go [INAUDIBLE].

So I have x to the
Z Zy e to the xy.

Are you guys with me, or
am I talking nonsense?

So what am I saying?

I'm saying that I expand with
respect to the j element here.

I have a minus because
of that, and then I

have the derivative of
this animal with respect

to x, which is Zy into
the x y j, correct?

STUDENT: Yes.

PROFESSOR: Finally, last but
not least, and actually that's

the most important guy, and
I'll tell you in a second why.

What is the last guy?

STUDENT: [INAUDIBLE].

PROFESSOR: 0.

So one of you will
hopefully realize what

I'm going to ask you right now.

No matter what I got here,
this was-- what is that called?

Work that is not necessary,
it's some stupid word.

So why is that not necessary?

Why could I have said star
i plus start j-- God knows

what that is-- plus 0k.

Because in the end, I have to
multiply that product with k,

so no matter what we do
here, and we sweat a lot.

And so no matter
what we put here

it would not have made a
difference because I have

to take this whole curl and
multiply as a dot with k,

and what matters is
only what's left over.

So my observation is this
whole thing is how much?

STUDENT: 0.

PROFESSOR: 0, thank God.

So the answer is 0.

And we've given
this problem where

the answer is 0 about four
times on four different finals.

The thing is that many
students won't study,

and they didn't know the trick.

When you have a surface like
that, that bounds the curve

C. Instead of doing
Stokes over the surface,

you do Stokes over the domain
and plane, and you'll get zero.

So poor kids, they
went ahead and tried

to compute this from
scratch for the surface,

and they got nowhere.

And then I started the fights
with, of course, [INAUDIBLE],

but they don't want to
give them any credit.

And I wanted to give
them at least some credit

for knowing the
theorem, the statement,

and trying to do something for
the nasty surface, the roof

that is a paraboloid.

They've done something,
so in the end,

I said I want to
do whatever I want,

and I gave partial credit.

But normally, I was told
not to give partial credit

for this kind of a thing because
the whole key of the problem

is to be smart,
understand the idea,

and get 0 without doing any
work, and that was nice.

Yes, sir?

STUDENT: Does that mean
that all we would really

needed to do compute
the curl is the k part?

Because if k would
have been something,

then there would have
been a dot on it.

PROFESSOR: Exactly,
but only if-- guys,

no matter what, if we give you,
if your surface has a planer

boundary-- say it again?

If your surface,
no matter what it

is-- it could look like
geography-- if your surface has

a boundary in the plane xy
like it is in geography,

imagine you have a
hill or something,

and that's the sea level.

And around the hill you have
the rim of the [INAUDIBLE].

OK, that's your planar curve.

Then you can reduce
to the plane,

and all the arguments
will be like that.

So the thing is you get 0
when the curl has 0 here,

and there is [INAUDIBLE].

Say it again?

When the F is given to you
so that the last component

of the curl is zero, you
will get 0 for the work.

Otherwise, you can get something
else, but not bad at all.

You can get something
that-- let's do

another example like that where
you have a simplification.

I'm going to go ahead
and erase the whole--

STUDENT: So, let's say if I
knew the [INAUDIBLE] equal to 0,

so I--

PROFESSOR: Eh, you cannot know
unless you look at the F first.

You see--

STUDENT: Let's say that
I put the F on stop,

and I put the equation,
which is F d r,

and I put the curl
F [INAUDIBLE], so

and then I said--
I looked at it.

I said, oh, it's a 0.

PROFESSOR: If you
see that's a big 0,

you can go at them
to 0 at the end.

STUDENT: OK.

PROFESSOR: Because the
dot product between k,

that's what matters.

The dot product between k and
the last component of the curl.

And in the end,
integral of 0 is 0.

And that is the lesson.

STUDENT: We should
also have N equal to k

if we don't have that.

PROFESSOR: Yeah, so I'm saying
if-- um, that's a problem.

This is not going to happen, but
assume that somebody gives you

a hill that looks like that,
and this is not a planar curve.

This would be a really
nasty curve in space.

You cannot do that anymore.

You have to apply [INAUDIBLE]
for the general surface.

But if your boundary sees a
planar boundary [INAUDIBLE],

then you can do that,
and simplify your life.

So let me give you
another example.

This time it's not going to be--
OK, you will see the surprise.

And you have a sphere, and
you have a spherical cap,

the sphere of radius R, and
this is going to be, let's say,

R to be 5.

And this is z equals 3.

You have the surface.

Somebody gives
you the surface S.

That is the spherical cap
of the sphere x squared

plus y squared plus z squared
equals 25 above the plane z

equals 3.

Compute double
integral of F times--

how did we phrase this
if we phrase it as a--

STUDENT: Curl FN?

PROFESSOR: No, he said, curl FN.

I'm sorry, if we rephrase
it as work curl FN

over S, whereas this
is the spherical cap.

This is S.

So you're going to
have this on the final.

First thing is, stay calm.

Don't freak out.

This is a typical--
you have to say, OK.

She prepared me well.

I did review, [INAUDIBLE].

For God's sake, I'm
going to do fine.

Just keep in mind that
no matter what we do,

it's not going to involve
a heavy computation like we

saw in that horrible
first example

I gave you-- second
example I gave you.

So the whole idea is to make
your life easier rather than

harder.

So what's the
first thing you do?

You take curl F, and you want
to see what that will be.

i j k is going to
be d dx, d dy, d dz.

And you say, all
right, then I'll

have x squared yz xy
squared z and xy z squared.

And then you say, well,
this look ugly, right?

That's what you're going to say.

So what times i minus what times
j plus what times k remains up

to you to clue the computation,
and you say, wait a minute.

The first minor is it math?

No, the first minor-- minor is
the name of such a determinant

is just a silly path.

So you do x yz squared
with respect to y,

it's xz squared minus prime
with respect to z dz xy squared.

Next guy, what do we have?

Who tells me?

He's sort of significant
but not really--

STUDENT: yz squared?

PROFESSOR:yz squared, good.

It's symmetric in a way.

x squared y, right guys?

Are you with me?

STUDENT: Mh-hmm.

PROFESSOR: And for
the k, you will have?

STUDENT: y squared z.

PROFESSOR: y squared z.

STUDENT: And x squared z.

PROFESSOR: z squared z because
if you look at this guy--

so we [INAUDIBLE] again.

And you say, well, I have
derivative with respect

to x is y squared z.

The derivative with
respect to y is

x squared y squared z and
then minus x squared z.

Then I have-- [INAUDIBLE].

I did, right?

So this is squared.

What matters is that I
check what I'm going to do,

so now I say, my
c is a boundaries.

That's a circle, so the
meaning of this integral given

by Stokes is actually
a path integral

along the c at the
level z equals 3.

I'm at the third floor looking
at the world from up there.

I have the circle on the
third floor z equals 3.

And then I say, that's going
to be F dot dR God knows what.

That was originally the work.

And Stokes theorem
says, no matter

what surfaces
you're going to take

to have a regular surface
without controversy,

without holes that bounded
family by the circles c,

you're going to be in business.

So I say, the heck with
the S. I want the D,

and I want that D to be colorful
because life is great enough.

Let's make it D.

That D has what meaning?

z equals 3.

I'm at the level
three, but also x

squared plus y squared must
be less than or equal to sum.

Could anybody tell
me what that is?

STUDENT: 16.

PROFESSOR: So how do I know?

I will just plug in a 3 here.

3 squared is 9.

25 minus 9, so I get x squared
plus y squared equals 16.

So from here to here,
how much do I have?

STUDENT: 4.

PROFESSOR: 4, right?

So that little radius of that
yellow domain, [INAUDIBLE].

OK, so let's write
down the thing.

Let's go with D. This domain
is going to be called D.

And then I have this curl F.

And who is N?

N is k.

Why?

Because I'm in a
plane that's upstairs,

and I have dA because
whether the plane is

upstairs or downstairs
on the first floor,

dA will still be dxdy.

OK, so now let's compute
what we have backwards.

So this times k will
give me double integral

over D of y squared times.

Who is z?

I'm in a domain, d,
where z is fixed.

STUDENT: 3.

PROFESSOR: z is 3.

Minus x squared times 3.

And--

STUDENT: dx2.

PROFESSOR: I just came
up with this problem.

If I were to write
it for the final,

I would write it even simpler.

But let's see, 3 and
3, and then nothing.

And then da, dx dy, right?

Over the d, which
is x squared plus y

squared this is
[INAUDIBLE] over 16.

How do I solve such a integral?

I'm going to make it nicer.

OK.

How would I solve
such an integral?

Is it a painful thing?

STUDENT: [INAUDIBLE].

PROFESSOR: Well,
they're coordinates.

And somebody's going to help me.

And as soon as we are
done, we are done.

3 gets out.

And instead of x squared
plus y squared is then 16,

I have r between 0 and 4,
theta between 0 and 2pi.

I have to take
advantage of everything

I've learned all the semester.

Knowledge is power.

What's missing?

r.

A 3 gets out.

And here I have to be just
smart and pay attention

to what you told me.

Because you told me, Magdalena,
why is our sine theta not

our cosine theta?

[INAUDIBLE]

This is r squared, sine
squared theta minus r

squared cosine squared theta.

So what have I taught
you about integrals

that can be expressed as
products of a function of theta

and function of r?

That they have a
blessing from God.

So you have 3 integral
from 0 to 2pi minus 1.

I have my plan when
it comes to this guy.

Because it goes on my nerves.

All right.

Do you see this?

[INAUDIBLE] theta.

STUDENT: That's the [INAUDIBLE].

[INTERPOSING VOICES]

PROFESSOR: Do you know
what I'm coming up with?

[INTERPOSING VOICES]

PROFESSOR: Cosine
of a double angle.

Very good.

I'm proud of you guys.

If I were to test--
oh, there was a test.

But [INAUDIBLE] next
for the whole nation.

Only about 10% of the
students remembered that

by the end of the
calculus series.

But I think that's not--
that doesn't show weakness

of the [INAUDIBLE] programs.

It shows a weakness in
the trigonometry classes

that are either missing from
high school or whatever.

So you know that
you want in power.

Now, times what
integral from 0 to 4?

STUDENT: r squared.

Or r cubed.

PROFESSOR: r cubed, which again
is wonderful that we have.

And we should be able to
compute the whole thing easily.

Now if I'm smart,

STUDENT: [INAUDIBLE].

PROFESSOR: How can we see?

STUDENT: Because the cosine to
the integral is sine to theta.

And [INAUDIBLE].

PROFESSOR: Right.

So the sine to theta, whether
I put it here or here,

is still going to be 0.

The whole thing will be 0.

So I play the game.

Maybe I should've given
such a problem when we

wrote this edition of the book.

I think it's nicer than
the computational one

you saw before.

But I told you this trick so
you remember it for the final.

And you are to promise
that you'll remember it.

And that was the whole
essence of understanding

that the Stokes' theorem can
become Green's theorem very

easily when you work
with a surface that's

a domain in plane, a
planar domain. [INAUDIBLE].

Are you done with this?

OK.

So you say, OK, so what else?

This was something
that's sort of fun.

I understand it.

Is there anything left
in this whole chapter?

Fortunately or unfortunately,
there is only one section left.

And I'm going to
go over it today.

STUDENT: Can I ask you a quick
question about [INAUDIBLE] 6--

PROFESSOR: Yes, sire.

STUDENT: --before you move on?

PROFESSOR: Move on?

STUDENT: I was an idiot.

PROFESSOR: No, you are not.

STUDENT: And when I
was writing these down,

I missed the variable.

So I have the
integral of fdr over c

equals double integral
over f, curl f dot n.

PROFESSOR: ds.

STUDENT: I didn't
write down what c was.

I didn't write down
what this c was.

PROFESSOR: The c was
the whatever boundary

you had there of the surface s.

And that was in
the beginning when

we defined the sphere, when
we gave the general statement

for the function.

So I'm going to try
and draw a potato.

We don't do a very
good job in the book

drawing the solid body.

But I'll try and draw
a very nice solid body.

Let's see.

You have a solid body.

Imagine it as a potato,
topologically a sphere.

It's a balloon that you blow.

It's a closed surface.

It closes in itself.

And we call that r in the book.

It's a solid region enclosed
by the closed surfaces.

Sometimes we call
such a surface compact

for some topological reasons.

Let's put s.

s is the boundary of r.

We as you know our old friend
to be a vector value function.

And again, if you
like a force field,

think of it as a force field.

Now, I'm not going to
tell you what it is.

It's [INAUDIBLE] function
differential [INAUDIBLE]

the partial here is continuous.

The magic thing is that this
surface must be orientable.

And if we are going to immerse
it, it's a regular surface.

Then of course, n exists.

And your [INAUDIBLE], guys,
doesn't have to be outwards.

It could be inwards [INAUDIBLE].

Let's make the convention that n
will be outwards by convention.

So we have to have an agreement
like they do in politics,

between Fidel Castro and Obama.

By convention, whether
we like it or not,

let's assume the normal
will be pointing out.

Then something magic happens.

And that magic thing, I'm not
going to tell you what it is.

But you should tell
me if you remember

what the double integral was
in this case, intolerance

of physics.

Shut up, Magdalena.

Don't tell them everything.

Let people remember
what this was.

So what is the second term?

This is the so-called
famous divergence theorem.

So this is the divergence.

If you don't remember
that, we will review it.

dV is the volume integral.

I have a [INAUDIBLE] integral
over the solid potato,

of course.

What is this animal [INAUDIBLE]?

OK.

Take some milk and strain
it and make cheese.

And you have that kind
of piece of cloth.

And you hang it.

And the water goes through
that piece of cloth.

[INAUDIBLE] have this
kind of suggestive image

should make you
think of something we

talked about before.

Whether that was fluid
dynamics or electromagnetism,

[INAUDIBLE], this
has the same name.

f is some sort of field,
vector [INAUDIBLE] field.

N is the outer
normal in this case.

What is the meaning of that, for
a dollar, which I don't have?

It's a four-letter word.

It's an F word.

STUDENT: Flux.

PROFESSOR: Very good.

I'm proud of you.

Who said it first?

Aaron said it first?

I owe you a dollar.

You can stop by my office.

I'll give you a dollar.

STUDENT: He said it
five minutes ago.

PROFESSOR: So the flux-- He did?

STUDENT: Yeah, he did.

Silently.

PROFESSOR: Aaron
is a mindreader.

OK.

So the flux in the
left-hand side.

This thing you don't
know what it is.

But it's some sort of potato.

What is the divergence
of something?

So if somebody gives you the
vector field F1, [INAUDIBLE],

where these are functions
of xyz, [INAUDIBLE].

What is the divergence
of F by definition?

Remember section 13.1?

Keep it in mind for the final.

So what do we do?

Differentiate the
first component respect

to x plus differentiate
the second component

respect to y plus differentiate
the third component

respect to c, sum them up.

And that's your divergence.

OK?

How do engineers
write divergence?

Not like a mathematician
or like a geometer.

I'm doing differential geometry.

How do they write?

STUDENT: Kinds of [INAUDIBLE].

PROFESSOR: [INAUDIBLE] dot if.

This is how engineers
write divergence.

And when they write curl,
how do they write it?

They write [INAUDIBLE]
cross product.

Because it has a meaning.

If you think about
operator, you have ddx

applied to F1, ddy applied
to F2, ddz applied to F3.

So it's like having the dot
product between ddx, ddy,

ddz operators, which would
be the [INAUDIBLE] operator

acting on F1, F2, F3.

So you go first first,
plus second second,

plus third third, right?

It's exactly the same idea that
you inherited from dot product.

Now let's see the last two
problems of this semester.

except for step the review.

But the review's another story.

So I'm going to pick one
of your favorite problems.

OK.

Example one, remember
your favorite tetrahedron.

I'm going to erase it.

Instead of the potato, you can
have something like a pyramid.

And you have example one.

Let's say, [INAUDIBLE]
we have that.

Somebody gives you the F.

I'm going to make
it nice and sassy.

Because the final is coming
and I want simple examples.

And don't expect anything
[INAUDIBLE] really nice

examples also on the final.

Apply divergence
theorem in order

to compute double integral
of F dot n ds over s,

where s is the surface of the
tetrahedron in the picture.

And that's your
favorite tetrahedron.

We've done that like
a million times.

Somebody gave you a--
shall I put 1 or a?

1, because [INAUDIBLE] is
[INAUDIBLE] is [INAUDIBLE].

So you have the plane
x plus y plus z.

Plus 1 you intersect
with the axis'.

The coordinates, you take
the place of coordinates

and you form a tetrahedron.

Next tetrahedron is a
little bit beautiful

that it has 90 degree
angles at the vertex.

And it has a name, OABC.

OABC is the tetrahedron.

And the surface of
the tetrahedron is s.

How are you going
to do this problem?

You're going to say, oh
my god, I don't know.

It's not hard.

STUDENT: It looks like
you're going to use

the formula you just gave us.

PROFESSOR: The
divergence theorem.

STUDENT: And the divergence
for that is really easy.

It's just a constant.

PROFESSOR: Right.

And we have to give a name to
the tetrahedron, [INAUDIBLE]

T, with the solid tetrahedron.

And its area, its
surface is this.

Instead of a potato, you
have the solid tetrahedron.

So what do you write?

Exactly what [INAUDIBLE] told
you, triple integral over T.

Of what?

The divergence of F, because
that's the divergence theorem,

dv.

Well, it should be easy.

Because just as you
said, divergence of F

would be a constant.

How come?

Differentiate this
with respect to x, 2.

This with respect to y, 3.

This with respect to z, 5.

Last time I checked this was
10 when I was [INAUDIBLE].

So 10 says I'm going for a walk.

And then triple integral of
the volume of 1dv over T,

what is this?

[INTERPOSING VOICES]

PROFESSOR: Well, because I
taught you how to cheat, yes.

But what if I were to ask
you to express this as--

STUDENT: [INAUDIBLE]?

PROFESSOR: Yeah.

Integrate one at a time.

So you have 1dz, dy, dx--
I'm doing review with you--

from 0 to 1 minus x
minus y from 0 to--

STUDENT: [INTERPOSING VOICES]

PROFESSOR: --1 minus
x from zero to 1.

And how did I teach
you how to cheat?

I taught you that in
this case you shouldn't

bother to compute that.

Remember that you
were in school and we

learned the volume
of a tetrahedron

was the area of the base
times the height divided by 3,

which was one half
times 1 divided by 3.

So you guys right.

The answer is 10 times 1 over 6.

Do I leave it like that?

No.

Because it's not nice.

So the answer is 5/3.

Expect something like
that on the final,

something very similar.

So you'll have to apply
the divergence theorem

and do a good job.

And of course, you
have to be careful.

But it shouldn't be hard.

It's something that
should be easy to do.

Now, the last problem of the
semester that I want to do

is an application of
the divergence theorem

is over a cube.

So I'm going to erase
it, the whole thing.

And I'm going to
draw a cube, which is

an open-topped box upside down.

Say it again, an open-topped
box upside down, which

means somebody gives
you a cubic box

and tells you to
turn it upside down.

And you have from
here to here, 1.

All the dimensions
of the cube are 1.

The top is missing, so
there's faces missing.

The bottom face is missing.

Bottom face is missing.

Let's call it-- you know,
what shall we call it?

F1.

Because it was the top,
but now it's the bottom.

OK?

And the rest are F2, F3, F4,
F5, and F6, which is the top.

And I'm going to erase.

And the last thing before this
section is to do the following.

What do I want?

Evaluate the flux double
integral over s F dot n ds.

You have to evaluate that.

For the case when F-- I usually
don't take the exact data

from the book.

But in this case, I want to.

Because I know you'll read it.

And I don't want you to have any
difficulty with this problem.

I hate the data myself.

I didn't like it very much.

It's unit cube, OK.

So x must be between 0 and 1.

y must be between 0
and 1 including them.

But z-- attention guys-- must
be between 0 [INAUDIBLE],

without 0.

Because you remove the
face on the ground.

z is greater than 0 and
less than or equal to 1.

And do we want anything else?

No.

That is all.

So let's compute
the whole thing.

Now, assume the box
would be complete.

If the box were
complete, then I would

have the following, double
integral over all the

faces F2 union with
F3 union with F4 union

with F5 union with--
oh my God-- F6 of F

dot 10 ds plus double integral
over what's missing guys, F1?

Of n dot n ds-- F,
Magdalena, that's the flux.

F dot n ds.

If it were complete,
that would mean

I have the double integral
over all the six faces.

In that case, this sum would
be-- I can apply finally

the divergence theorem.

That would be triple
integral of-- God

knows what that is-- divergence
of F dv over the cube.

What do you want us
to call the cube?

STUDENT: C.

PROFESSOR: C is usually what
we denote for the curve.

STUDENT: How about q?

PROFESSOR: Beautiful.

Sounds like.

Oh, I like that. q.

q is the cube inside
the whole thing.

Unfortunately, this
is not very nicely

picked just to make
your life miserable.

So you have dv x over y.

There is no j, at least that.

ddz minus this way.

As soon as we are
done with this,

since I gave you no break,
I'm going to let you go.

So what do we have?

y minus 2z.

Does it look good?

No.

Does it look bad?

No, not really bad either.

If I were to solve
the problem, I

would have to say triple,
triple, triple y minus 2z.

And now-- oh my
God-- dz, dy, dx.

I sort of hate when a little bit
of computation 0 to 1, 0 to 1,

0 to 1.

But this is for
[INAUDIBLE] theorem.

Is there anybody
missing from the picture

so I can reduce it
to a double integral?

STUDENT: x.

PROFESSOR: x is missing.

So I say there is no x inside.

I go what is integral
from 0 to 1 of 1dx?

STUDENT: 1

PROFESSOR: 1.

So I will rewrite it
as integral from 0 to1,

integral from 0 to1,
y minus 2z, dz, dy.

Is this hard?

Eh, no.

But it's a little bit obnoxious.

When I integrate with
respect to z, what do I get?

Yz minus z squared.

No, not that-- between z equals
0 1 down, z equals 1 up to z

equals 0 down dy.

So z goes from 0
to 1 [INAUDIBLE].

When z is 0 down,
I have nothing.

STUDENT: Yeah, 1
minus-- y minus 1.

PROFESSOR: 1. y
minus 1, not so bad.

Not so bad, dy.

So I get y squared
over 2 minus y.

Between 0 and 1, what do I get?

STUDENT: Negative 1/2.

PROFESSOR: Negative 1/2.

All right.

Let's see what we've got here.

Yeah.

They got [INAUDIBLE].

And now I'm asking you
what's going to happen.

Our contour is the
open-topped box upside down.

This is what we need.

This is what we--

STUDENT: Couldn't you
just the double integral?

PROFESSOR: We just have
to compute this fellow.

We need to compute that fellow.

So how do we do that?

How do we do that?

STUDENT: What is the
problem asking for again?

PROFESSOR: So the problem
is asking over this flux,

but only over the box'
walls and the top.

The top, one, two, three,
four, without the bottom,

which is missing.

In order to apply
divergence theorem,

I have to put the bottom back
and have a closed surface that

is enclosing the whole cube.

So this is what I want.

This is what I know.

How much is it guys?

Minus 1/2.

And this is, again, what I need.

Right?

That's the last thing
I'm going to do today.

[INTERPOSING VOICES]

STUDENT: F times k da.

PROFESSOR: Let's compute it.

k is a blessing, as
you said, [INAUDIBLE].

It's actually minus k.

Why is it minus k?

Because it's upside down.

And it's an altered normal.

STUDENT: Oh, it is the
[INAUDIBLE] normal.

OK.

Yeah.

That's right.

PROFESSOR: [INAUDIBLE].

So minus k.

But it doesn't [INAUDIBLE].

The sign matters.

So I have to be careful.

F is-- z is 0, thank God.

So that does away.

So I have x y i dot
product with minus k.

What's the beauty of this?

0.

STUDENT: 0.

PROFESSOR: Yay.

0.

So the answer to this
problem is minus 1/2.

So the answer is minus 1.

And we are done with
the last section

of the book, which is 13.7.

It was a long way.

We came a long way
to what I'm going

to do next time and
the times to come.

First of all, ask me from now
on you want a break or not.

Because I didn't give
you a break today.

We are not in a hurry.

I will pick up exams.

And I will go over
them together with you.

And by the time we
finish this review,

we will have solved two or
three finals completely.

We will be [INAUDIBLE].

And so the final is on the 11th,
May 11 at 10:30 in the morning.

I think.

STUDENT: It's the 11.

The 12 is [INAUDIBLE].

The 12th is the other class.

STUDENT: Yeah.

I'm positive.

PROFESSOR: We are
switching the two classes.

STUDENT: [INAUDIBLE].

PROFESSOR: And it's May 11
at 10:30 in the morning.

On May 12, there are other
math courses that have a final.

But fortunately for
them, they start at 4:30.

I'm really blessed that I
don't have that [INAUDIBLE].

They start at 4:30,
and they end at 7:00.

Can you imagine how
frisky you feel when you

take that final in the night?

Good luck with the homework.

Ask me questions about the
homework if you have them.