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 in-negral.
And everybody that I heard in romance language-speaking
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 two-dimensional 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 so-called 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 K-12,
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 K-12, 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 two-way 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 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.