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?
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.