MAGDALENA TODA: I'm starting early, am I?
It's exactly 12:30.
The weather is getting better, hopefully,
and not too many people should miss class today.
Can you start an attendance sheet for me [INAUDIBLE]?
I know I can count on you.
OK.
I have good markers today.
I'm going to go ahead and talk about 12.3,
double integrals in polar coordinates.
These are all friends of yours.
You've seen until now only double integrals that
involve the rectangles, either a rectangle, we saw [INAUDIBLE],
and we saw some type of double integrals,
of course that involved x and y, so-called type
1 and type 2 regions, which were--
so we saw the rectangular case.
You have ab plus cd, a rectangle.
You have what other kind of a velocity [INAUDIBLE]
over the the main of the shape x between a and be and y.
You write wild and happy from bottom to top.
That's called the wild-- not wild, the vertical strip
method, where y will be between the bottom function
f of x and the top function f of x.
And last time I took examples where
f and g were both positive, but remember, you don't have to.
All you have to have is that g is always greater than f,
or equal at some point.
And then what else do we have for these cases?
These are all continuous functions.
What else did we have?
We had two domains.
Had one and had two.
Where what was going on, we have played a little bit
around with y between c and d limits with points.
These are horizontal, so we take the domain
as being defined by these horizontal strips between let's
say a function.
Again, I need to rotate my head, but I didn't do my yoga today,
so it's a little bit sticky.
I'll try.
x equals F of y, and x equals G of y, assuming, of course,
that f of y is always greater than or equal to g of y,
and the rest of the apparatus is in place.
Those are not so hard to understand.
We played around.
We switched the integrals.
We changed the order of integration from dy dx
to dx dy, so we have to change the domain.
We went from vertical strip method
to horizontal strip method or the other way around.
And for what kind of example, something
like that-- I think it was a leaf like that,
we said, let's compute the area or compute
another kind of double integral over this leaf in two
different ways.
And we did it with vertical strips,
and we did the same with horizontal strips.
So we reversed the order of integration,
and we said, I'm having the double integral over domain
of God knows what, f of xy, continuous function,
positive, continuous whenever you want, and we said da.
We didn't quite specify the meaning of da.
We said that da is the area element,
but that sounds a little bit weird, because it makes
you think of surfaces, and an area element
doesn't have to be a little square in general.
It could be something like a patch on a surface, bounded
by two curves within your segments in each direction.
So you think, well, I don't know what that is.
I'll tell you today what that is.
It's a mysterious thing, it's really beautiful,
and we'll talk about it.
Now, what did we do last time?
We applied the two theorems that allowed
us to do this both ways.
Integral from a to b, what was my usual [? wrist ?] is down,
f of x is in g of x, right?
dy dx.
So if you do it in this order, it's
going to be the same as if you do it in the other order.
ab are these guys, and then this was cd on the y-axis.
This is the range between c and d in altitudes.
So we have integral from c to d, integral from,
I don't know what they will be.
This big guy I'm talking-- which one is the one?
This one, that's going to be called x equals f of y,
or g of y, and let's put the big one G and the smaller one,
x equals F of y.
So you have to [? re-denote ?] these functions,
these inverse functions, and use them as functions of y.
So it makes sense to say-- what did we do?
We first integrated respect to x between two functions of y.
That was the so-called horizontal strip method, dy.
So I have summarized the ideas from last time
that we worked with, generally with corners x and y.
We were very happy about them.
We had the rectangular domain, where x was between ab
and y was between cd.
Then we went to type 1, not diabetes, just type 1 region,
type 2, and those guys are related.
So if you understood 1 and understood the other one,
and if you have a nice domain like that,
you can compute the area or something.
The area will correspond to x equals 1.
So if f is 1, then that's the area.
That will also be a volume of a cylinder based
on that region with height 1.
Imagine a can of Coke that has height 1,
and-- maybe better, chocolate cake,
that has the shape of this leaf on the bottom,
and then its height is 1 everywhere.
So if you put 1 here, and you get the area element,
and then everything else can be done
by reversing the order of integration if f is continuous.
But for polar coordinates, the situation
has to be reconsidered almost entirely, because the area
element, da is called the area element for us,
was equal to dx dy for the cartesian coordinate case.
And here I'm making a weird face, I'm weird, no?
Saying, what am I going to do, what is this
going to become for polar coordinates?
And now you go, oh my God, not polar coordinates.
Those were my enemies in Calc II.
Many people told me that.
And I tried to go into my time machine
and go back something like 25 years ago
and see how I felt about them, and I remember that.
I didn't get them from the first 48 hours
after I was exposed to them.
Therefore, let's do some preview.
What were those polar coordinates?
Polar coordinates were a beautiful thing,
these guys from trig.
Trig was your friend hopefully.
And what did we have in trigonometry?
In trigonometry, we had a point on a circle.
This is not the unit trigonometric circle,
it's a circle of-- bless you-- radius r.
I'm a little bit shifted by a phase of phi 0.
So you have a radius r.
And let's call that little r.
And then, we say, OK, how about the angle?
That's the second polar coordinate.
The angle by measuring from the, what
is this called, the x-axis.
Origin, x-axis, o, x, going counterclockwise,
because we are mathemeticians.
Every normal person, when they mix into a bowl,
they mix like that.
Well, I've seen that most of my colleagues--
this is just a psychological test, OK?
I wanted to see how they mix when
they cook, or mix up-- most of them
mix in a trigonometric sense.
I don't know if this has anything to do with the brain
connections, but I think that's [? kind of weird. ?]
I don't have a statistical result, but most of the people
I've seen that, and do mathematics, mix like that.
So trigonometric sense.
What is the connection with the actual Cartesian coordinates?
D you know what Cartesian comes from as a word?
Cartesian, that sounds weird.
STUDENT: From Descartes.
MAGDALENA TODA: Exactly.
Who said that?
Roberto, thank you so much.
I'm impressed.
Descartes was--
STUDENT: French.
MAGDALENA TODA: --a French mathematician.
But actually, I mean, he was everything.
He was a crazy lunatic.
He was a philosopher, a mathematician,
a scientist in general.
He also knew a lot about life science.
But at the time, I don't know if this is true.
I should check with wiki, or whoever can tell me.
One of my professors in college told me that at that time,
there was a fashion that people would
change their names like they do on Facebook nowadays.
So they and change their name from Francesca
to Frenchy, from Roberto to Robby, from-- so
if they would have to clean up Facebook and see
how many names correspond to the ID, I think less than 20%.
At that time it was the same.
All of the scientists loved to romanize their names.
And of course he was of a romance language,
but he said, what if I made my name a Latin name,
I changed my name into a Latin name.
So he himself, this is what my professor told me, he
himself changed his name to Cartesius.
"Car-teh-see-yus" actually, in Latin, the way it should be.
OK, very smart guy.
Now, when we look a x and y, there
has to be a connection between x, y as the couple, and r theta
as the same-- I mean a couple, not the couple,
for the same point.
Yes, sir?
STUDENT: Cartesius.
Like meaning flat?
The name?
MAGDALENA TODA: These are the Cartesian coordinates,
and it sounds like the word map.
I think he had meant
STUDENT: Because the meaning of carte--
STUDENT: But look, look.
Descartes means from the map.
From the books, or from the map.
So he thought what his name would really mean,
and so he recalled himself.
There was no fun, no Twitter, no Facebook.
So they had to do something to enjoy themselves.
Now, when it comes to these triangles,
we have to think of the relationship between x, y
and r, theta.
And could somebody tell me what the relationship between x, y
and r, theta is?
x represents
STUDENT: R cosine theta.
STUDENT: r cosine theta, who says that?
Trigonometry taught us that, because that's
the adjacent side over the hypotenuse for cosine.
In terms of sine, you know what you have,
so you're going to have y equals r sine theta,
and we have to decide if x and y are allowed
to be anywhere in plane.
For the plane, I'll also write r2.
R2, not R2 from the movie, just r2 is the plane,
and r3 is the space, the [? intriguing ?]
space, three-dimensional one.
r theta, is a couple where?
That's a little bit tricky.
We have to make a restriction.
We allow r to be anywhere between 0 and infinity.
So it has to be a positive number.
And theta [INTERPOSING VOICES] between 0 and 2 pi.
STUDENT: I've been sick since Tuesday.
MAGDALENA TODA: I believe you, Ryan.
You sound sick to me.
Take your viruses away from me.
Take the germs away.
I don't even have the-- I'm kidding,
Alex, I hope you don't get offended.
So, I hope this works this time.
I'm making a sarcastic-- it's really,
I hope you're feeling better.
I'm sorry about that.
So you haven't missed much.
Only the jokes.
So x equals r cosine theta, y equals r sine theta.
Is that your favorite change that
was a differential mapping from the set x,
y to the set r, theta back and forth.
And you are going to probably say, OK
how do you denote such a map?
I mean, going from x, y to r, theta and back,
let's suppose that we go from r, theta to x, y,
and that's going to be a big if.
And going backwards is going to be the inverse mapping.
So I'm going to call it f inverse.
So that's a map from a couple to another couple of number.
And you say, OK, but why is that a map?
All right, guys, now let me tell you.
So x, you can do x as a function of r, theta, right?
It is a function of r and theta.
It's a function of two variables.
And y is a function of r and theta.
It's another function of two variables.
They are both nice and differentiable.
We assume not only that they are differentiable,
but the partial derivatives will be continuous.
So it's really nice as a mapping.
And you think, could I write the chain rule?
That is the whole idea.
What is the meaning of differential?
dx differential dy.
Since I was chatting with you, once, [? Yuniel ?],
and you asked me to help you with homework,
I had to go over differential again.
If you were to define, like Mr. Leibniz did,
the differential of the function x with respect
to both variables, that was the sum, right?
You've done that in the homework,
it's fresh in your mind.
So you get x sub r, dr, plus f x sub what?
STUDENT: Theta.
MAGDALENA TODA: Sub theta d-theta.
And somebody asked me, what if I see skip the dr?
No, don't do that.
First of all, WeBWorK is not going to take the answer.
But second of all, the most important stuff
here to remember is that these are small, infinitesimally
small, displacements.
Infinitesimally small displacements in the directions
x and y, respectively.
So you would say, what does that mean, infinitesimally?
It doesn't mean delta-x small.
Delta-x small would be like me driving 7 feet, when
I know I have to drive fast to Amarillo to be there in 1 hour.
Well, OK.
Don't tell anybody.
But, it's about 2 hours, right?
So I cannot be there in an hour.
But driving those seven feet is like a delta x.
Imagine, however, me measuring that speed of mine
in a much smaller fraction of a second.
So shrink that time to something infinitesimally small,
which is what you have here.
That kind of quantity.
And dy will be y sub r dr plus y sub theta d-theta.
And now, I'm not going to go by the book.
I'm going to go a little bit more
in depth, because in the book-- First of all, let me tell you,
if I went by the book, what I would come with.
And of course the way we teach mathematics
all through K-12 and through college is swallow this theorem
and believe it.
So practically you accept whatever we give you
without controlling it, without checking if we're right,
without trying to prove it.
Practically, the theorem in the book
says that if you have a bunch of x,
y that is continuous over a domain, D,
and you do change the variables over--
STUDENT: I forgot my glasses.
So I'm going to sit very close.
MAGDALENA TODA: What do you wear?
What [INAUDIBLE]?
STUDENT: I couldn't tell you.
I can see from here.
MAGDALENA TODA: You can?
STUDENT: Yeah.
My vision's not terrible.
MAGDALENA TODA: All right. f of x, y da.
If I change this da as dx dy, let's say,
to a perspective of something else
in terms of polar coordinates, then
the integral I'm going to get is over the corresponding domain D
star, whatever that would be.
Then I'm going to have f of x of r theta, y of r theta,
everything expressed in terms of r theta.
And instead of the a-- so we just
feed you this piece of cake and say, believe it,
believe it and leave us alone.
OK?
That's what it does in the book in section 11.3.
So without understanding why you have to-- instead of the r
d theta and multiply it by an r.
What is that?
You don't know why you do that.
And I thought, that's the way we thought it
for way too many years.
I'm sick and tired of not explaining why
you multiply that with an r.
So I will tell you something that's quite interesting,
and something that I learned late in graduate school.
I was late already.
I was in my 20s when I studied differential forms
for the first time.
And differential forms have some sort
of special wedge product, which is very physical in nature.
So if you love physics, you will understand more or less
what I'm talking about.
Imagine that you have two vectors, vector a and vector b.
For these vectors, you go, oh my God.
If these would be vectors in a tangent plane to a surface,
you think, some of these would be
tangent vectors to a surface.
This is the tangent plane and everything.
You go, OK, if these were infinitesimally
small displacements-- which they are not, but assume they would
be-- how would you do the area of the infinitesimally small
parallelogram that they have between them.
This is actually the area element right here, ea.
So instead of dx dy, you're not going to have dx dy,
you're going to have some sort of, I don't know,
this is like a d-something, d u, and this
is a d v. And when I compute the area of the parallelogram,
I consider these to be vectors, and I
say, how did we get it from the vectors
to the area of the parallelogram?
We took the vectors, we shook them off.
We made a cross product of them, and then we
took the norm, the magnitude of that.
Does this makes sense, compared to this parallelogram?
Yeah.
Remember, guys, this was like, how big
is du, a small infinitesimal displacement,
but that would be like the width, one of the dimensions.
There's the other of the dimension of the area element
times-- this area element is that tiny pixel that
is sitting on the surface in the tangent plane, yeah?
Sine of the angle between the guys.
Oh, OK.
So if the guys are not perpendicular to one another,
if the two displacements are not perpendicular to one another,
you still have to multiply the sine of theta.
Otherwise you don't get the element
of the area of this parallelogram.
So why did the Cartesian coordinates not pose a problem?
For Cartesian coordinates, it's easy.
It's a piece of cake.
Why?
Because this is the x, this is the y, as little tiny measures
multiplied.
How much is sine of theta between Cartesian coordinates?
STUDENT: 1.
MAGDALENA TODA: It's 1, because its 90 degrees.
When they are orthogonal coordinates,
it's a piece of cake, because you have 1 there,
and then your life becomes easier.
So in general, what is the area limit?
The area limit for arbitrary coordinates--
So area limit for some arbitrary coordinates
should be defined as the sined area.
And you say, what do you mean that's a sined area,
and why would you do that.?
Well, it's not so hard to understand.
Imagine that you have a convention, and you say,
OK, dx times dy equals negative dy times dx.
And you say, what, what?
If you change the order of dx dy,
this wedge stuff works exactly like the-- what is that called?
Cross product.
So the wedge works just like the cross product.
Just like the cross product.
In some other ways, suppose that I am here, right?
And this is a vector, like an infinitesimal displacement,
and that's the other one.
If I multiply them one after the other,
and I use this strange wedge [INTERPOSING VOICES] the area,
I'm going to have an orientation for that tangent line,
and it's going to go up, the orientation.
The orientation is important.
But if dx dy and I switched them,
I said, dy, swap with dx, what's going to happen?
I have to change to change to clockwise.
And then the orientation goes down.
And that's what they use in mechanics when it comes
to the normal to the surface.
So again, you guys remember, we had 2 vector products,
and we did the cross product, and we got the normal.
If it's from this one to this one,
it's counterclockwise and goes up,
but if it's from this vector to this other vector,
it's clockwise and goes down.
This is how a mechanical engineer
will know how the surface is oriented
based on the partial velocities, for example
He has the partial velocities along a surface,
and somebody says, take the normal, take the unit normal.
He goes, like, are you a physicist?
No, I'm an engineer.
You don't know how to take the normal.
And of course, he knows.
He knows the convention by this right-hand rule,
whatever you guys call it.
I call it the faucet rule.
It goes like this, or it goes like that.
It's the same for a faucet, for any type of screw,
for the right-hand rule, whatever.
What else do you have to believe me are true?
dx wedge dx is 0.
Can somebody tell me why that is natural to introduce
such a wedge product?
STUDENT: Because the sine of the angle between those is 0.
MAGDALENA TODA: Right.
Once you flatten this, once you flatten the parallelogram,
there is no area.
So the area is 0.
How about dy dy sined area?
0.
So these are all the properties you
need to know of the sine area, sined areas.
OK, so now let's see what happens
if we take this element, which is a differential,
and wedge it with this element, which is also a differential.
OK.
Oh my God, I'm shaking only thinking about it.
I'm going to get something weird.
But I mean, mad weird.
Let's see what happens.
dx wedge dy equals-- do you guys have questions?
Let's see what the mechanics are for this type of computation.
I go-- this is like a-- displacement wedge
this other displacement.
Think of them as true vector displacements,
and as if you had a cross product, or something.
OK.
How does this go?
It's distributed.
It's linear functions, because we've
studied the properties of vectors,
this acts by linearity.
So you go and say, first first, times plus first times
second-- and times is this guy, this weirdo--
plus second times first, plus second times second,
where the wedge is the operator that
has to satisfy these functions.
It's similar to the cross product.
OK.
Then let's go x sub r, y sub r, dr
wedge dr. Oh, let's 0 go away.
I say, leave me alone, you're making my life hard.
Then I go plus x sub r-- this is a small function.
y sub theta, another small function.
What of this displacement, dr d theta.
I'm like those d something, d something,
two small displacements in the cross product.
OK, plus.
Who is telling me what next?
STUDENT: x theta--
MAGDALENA TODA: x theta yr, d theta dr. Is it fair?
I did the second guy from the first one with the first guy
from the second one.
And finally, I'm too lazy to write it down.
What do I get?
STUDENT: 0.
MAGDALENA TODA: 0.
Why is that?
Because d theta, always d theta is 0.
It's like you are flattening-- there is no more parallelogram.
OK?
So the two dimensions of the parallelogram become 0.
The parallelogram would become [? a secant. ?]
What you get is something really weak.
And we don't talk about it in the book,
but that's called the Jacobian.
dr d theta and d theta dr, once you introduce the sine area,
you finally understand why you get this r here,
what the Jacobian is.
If you don't introduce the sine area,
you will never understand, and you cannot explain it
to anybody, any student have.
OK, so this guy, d theta, which the r is just
swapping the two displacements.
So it's going to be minus dr d theta.
Why is that, guys?
Because that's how I said, every time I swap two displacements,
I'm changing the orientation.
It's like the cross product between a and b,
and the cross product between b and a.
So I'm going up or I'm going down, I'm changing orientation.
What's left in the end?
It's really just this guy that's really weird.
I'm going to collect the terms.
One from here, one from here, and a minus.
Go ahead.
STUDENT: Do the wedges just cancel out?
MAGDALENA TODA: This was 0.
This was 0.
And this dr d theta is nonzero, but is the common factor.
So I pull him out from here.
I pull him out from here.
Out.
Factor out, and what's left is this guy over here
who is this guy over here.
And this guy over here with a minus
who gives me minus d theta yr.
That's all.
So now you will understand why I am going to get r.
So the general rule will be that the area element dx
dy, the wedge sined area, will be--
you have to help me with this individual,
because he really looks weird.
Do you know of a name for it?
Do you know what this is going to be?
Linear algebra people, only two of you.
Maybe you have an idea.
So it's like, I take this fellow,
and I multiply by that fellow.
Multiply these two.
And I go minus this fellow times that fellow.
STUDENT: [INAUDIBLE]
MAGDALENA TODA: It's like a determinant of something.
So when people write the differential system,
[INTERPOSING VOICES] 51, you will understand
that this is a system.
OK?
It's a system of two equations.
The other little, like, vector displacements,
you are going to write it like that.
dx dy will be matrix multiplication dr d theta.
And how do you multiply x sub r x sub theta?
So you go first row times first column give you that.
And second row times the column gives you this.
y sub r, y sub theta.
This is a magic guy called Jacobian.
We keep this a secret, and most Professors don't even
cover 12.8, because they don't want to tell
people what a Jacobian is.
This is little r.
I know you don't believe me, but the determinant of this matrix
must be little r.
You have to help me prove that.
And this is the Jacobian.
Do you guys know why it's called Jacobian?
It's the determinant of this matrix.
Let's call this matrix J. And this
is J, determinant of [? scripture. ?]
This is called Jacobian.
Why is it r?
Let's-- I don't know.
Let's see how we do it.
This is r cosine theta, right?
This is r sine theta.
So dx must be what x sub r?
X sub r, x sub r, cosine theta.
d plus.
What is x sub t?
x sub theta.
I need to differentiate this with respect to theta.
STUDENT: It's going to be negative r sine theta.
MAGDALENA TODA: Minus r sine theta, very good.
And d theta.
Then I go dy was sine theta-- dr,
I'm looking at these equations, and I'm
repeating them for my case.
This is true in general for any kind of coordinates.
So it's a general equation for any kind of coordinate,
two coordinates, two coordinates,
any kind of coordinates in plane,
you can choose any functions, f of uv, g of uv,
whatever you want.
But for this particular case of polar coordinates
is going to look really pretty in the end.
What do I get when I do y theta?
r cosine theta.
Am I right, guys?
Keen an eye on it.
So this will become-- the area element will become what?
The determinant of this matrix.
Red, red, red, red.
How do I compute a term?
Not everybody knows, and it's this times
that minus this times that.
OK, let's do that.
So I get r cosine squared theta minus minus plus r sine
squared theta.
dr, d theta, and our wedge.
What is this?
STUDENT: 1.
MAGDALENA TODA: Jacobian is r times 1,
because that's the Pythagorean theorem, right?
So we have r, and this is the meaning of r, here.
So when I moved from dx dy, I originally had the wedge
that I didn't tell you about.
And this wedge becomes r dr d theta,
and that's the correct way to explain
why you get the Jacobian there.
We don't do that in the book.
We do it later, and we sort of smuggle through.
We don't do a very thorough job.
When you go into advanced calculus,
you would see that again the way I explained it to you.
If you ever want to go to graduate school,
then you need to take the Advanced Calculus I, 4350
and 4351 where you are going to learn about this.
If you take those as a math major or engineering major,
it doesn't matter.
When you go to graduate school, they
don't make you take advanced calculus again
at graduate school.
So it's somewhere borderline between senior year
and graduate school, it's like the first course you would take
in graduate school, for many.
OK.
So an example of this transformation
where we know what we are talking about.
Let's say I have a picture, and I
have a domain D, which is-- this is x squared
plus y squared equals 1.
I have the domain as being [INTERPOSING VOICES].
And then I say, I would like-- what would I like?
I would like the volume of the-- this
is a paraboloid, z equals x squared plus y squared.
I would like the volume of this object.
This is my obsession.
I'm going to create a vase some day like that.
So you want this piece to be a solid.
In cross section, it will just this.
In cross section.
And it's a solid of revolution.
In this cross section, you have to imagine
revolving it around the z-axis, then creating a heavy object.
From the outside, don't see what's inside.
It looks like a cylinder.
But you go inside and you see the valley.
So it's between a paraboloid and a disc,
a unit disc on the floor.
How are we going to try and do that?
And what did I teach you last time?
Last time, I taught you that-- we have to go over a domain D.
But that domain D, unfortunately,
is hard to express.
How would you express D in Cartesian coordinates?
You can do it.
It's going to be a headache.
x is between minus 1 and 1.
Am I right, guys?
And y will be between-- now I have two branches.
One, and the other one.
One branch would be square-- I hate square roots.
I absolutely hate them.
y is between 1 minus square root x squared,
minus square root 1 minus x squared.
So if I were to ask you to do the integral like last time,
how would you set up the integral?
You go, OK, I know what this is.
Integral over D of f of x, y, dx dy.
This is actually a wedge.
In my case, we avoided that.
We said dh.
And we said, what is f of x, y?
x squared plus y squared, because I
want everything that's under the graph, not above the graph.
So everything that's under the graph.
F of x, y is this guy.
And the I have to start thinking,
because it's a type 1 or type 2?
It's a type 1 the way I set it up,
but I can make it type 2 by reversing
the order of integration like I did last time.
If I treat it like that, it's going
to be type 1, though, right?
So I have to put dy first, and then
change the color of the dx.
And since mister y is the purple guy,
y would be going between these ugly square roots that
to go on my nerves.
And then x goes between minus 1 and 1.
It's a little bit of a headache.
Why is it a headache, guys?
Let's anticipate what we need to do if we do it like last time.
We need to integrate this ugly fellow in terms of y,
and when we integrate this in terms of y, what do we get?
Don't write it, because it's going to be a mess.
We get x squared times y plus y cubed over 3.
And then, instead of y, I have to replace those square roots,
and I'll never get rid of the square roots.
It's going to be a mess, indeed.
And I may even-- in general, I may not even
be able to solve the integral, and that's
a bit headache, because I'll start
crying, I'll get depressed, I'll take Prozac, whatever
you take for depression.
I don't know, I never took it, because I'm never depressed.
So what do you do in that case?
STUDENT: Switch to polar.
MAGDALENA TODA: You switch to polar.
So you use this big polar-switch theorem, the theorem that
tells you, be smart, apply this theorem,
and have to understand that the D, which was this expressed
in [INTERPOSING VOICES] Cartesian coordinates
is D. If you want express the same thing as D star,
D star will be in polar coordinates.
You have to be a little bit smarter, and say r theta,
where now you have to put the bounds that limit--
STUDENT: r.
MAGDALENA TODA: r from?
STUDENT: 0 to 1.
MAGDALENA TODA: 0 to 1, excellent.
You cannot let r go to infinity, because the vase is
increasingly.
You only needs the vase that has the radius 1 on the bottom.
So r is 0 to 1, and theta is 0 to 1 pi.
And there you have your domain this time.
So I need to be smart and say integral.
Integral, what do you want to do first?
Well, it doesn't matter, dr, d theta, whatever you want.
So mister theta will be the last of the two.
So theta will be between 0 and 2 pi, a complete rotation.
r between 0 and 1.
And inside here I have to be smart
and see that life can be fun when
I work with polar coordinates.
Why?
What is the integral?
x squared plus y squared.
I've seen him somewhere before when
it came to polar coordinates.
STUDENT: R squared.
STUDENT: That will be r squared.
MAGDALENA TODA: That will be r squared.
r squared times-- never forget the Jacobian,
and the Jacobian is mister r.
And now I'm going to take all this integral.
I'll finally compute the volume of my vase.
Imagine if this vase would be made of gold.
This is my dream.
So imagine that this vase would have,
I don't know what dimensions.
I need to find the volume, and multiply it
by the density of gold and find out-- yes, sir?
STUDENT: Professor, like in this question, b time is dt by dr,
but you can't switch it--
MAGDALENA TODA: Yes, you can.
That's exactly my point.
I'll tell you in a second.
When can you replace d theta dr?
You can always do that when you have something under here,
which is a big function of theta times
a bit function of r, because you can treat them differently.
We will work about this later.
Now, this has no theta.
So actually, the theta is not going
to affect your computation.
Let's not even think about theta for the time being.
What you have inside is Calculus I. When you have a product,
you can always switch.
And I'll give you a theorem later.
0 over 1, r cubed, thank God, this
is Calc I. Integral from 0 to 1, r
cubed dr. That's Calc I. How much is that?
I'm lazy.
I don't want to do it.
STUDENT: 1/4.
MAGDALENA TODA: It's 1/4.
Very good.
Thank you.
And if I get further, and I'm a little bi lazy, what do I get?
1/4 is the constant, it pulls out.
STUDENT: So, they don't--
MAGDALENA TODA: So I get 2 pi over 4, which is pi over 2.
Am I right?
STUDENT: Yeah.
MAGDALENA TODA: So this constant gets out,
integral comes in through 2 pi.
It will be 2 pi, and this is my answer.
So pi over 2 is the volume.
If I have a 1-inch diameter, and I
have this vase made of gold, which is a piece of jewelry,
really beautiful, then I'm going to have pi over 2 the volume.
That will be a little bit hard to see
what we have in square inches.
We have 1.5-something square inches, and then--
STUDENT: More.
MAGDALENA TODA: And then multiply
by the density of gold, and estimate,
based on the mass, how much money that's going to be.
What did I want to tell [? Miteish? ?]
I don't want to forget what he asked me, because that
was a smart question.
When can we reverse the order of integration?
In general, it's hard to compute.
But in this case, I'm you are the luckiest person
in the world, because just take a look at me.
I have, let's see, my r between 0 and 2 pi,
and my theta between 0 and 2 pi, and my r between 0 and 1.
Whatever, it doesn't matter, it could be anything.
And here I have a function of r and a function g of theta only.
And it's a product.
The variables are separate.
When I do-- what do I do for dr or d theta?
dr. When I do dr-- with respect to dr,
this fellow goes, I don't belong in here.
I'm mister theta that doesn't belong in here.
I'm independent.
I want to go out.
And he wants out.
So you have some integrals that you got out a g of theta,
and another integral, and you have f of r dr in a bracket,
and then you go d theta.
What is going to happen next?
You solve this integral, and it's going to be a number.
This number could be 8, 7, 9.2, God knows what.
Why don't you pull that constant out right now?
So you say, OK, I can do that.
It's just a number.
Whatever.
That's going to be integral f dr, times
what do you have left when you pull that out?
A what?
STUDENT: Integral.
MAGDALENA TODA: Integral of G, the integral of g of theta,
d theta.
So we just proved a theorem that is really pretty.
If you have to integrate, and I will try to do it here.
No--
STUDENT: So essentially, when you're integrating with respect
to r, you can treat any function of only theta as a constant?
MAGDALENA TODA: Yeah.
I'll tell you in a second what it means, because--
STUDENT: Sorry.
MAGDALENA TODA: You're fine.
Integrate for domain, rectangular domains,
let's say u between alpha, beta, u between gamma,
delta, then what's going to happen?
As you said very well, integral from-- what
do you want first, dv or du?
dv, du, it doesn't matter.
v is between gamma, delta.
v is the first guy inside, OK.
Gamma, delta.
I should have cd.
It's all Greek to me.
Why did I pick that three people?
If this is going to be a product of two functions, one is in u
and one is in v. Let's say A of u and B of v,
I can go ahead and say product of two constants.
And who are those two constants I was referring to?
You can do that directly.
If the two variables are separated through a product,
you have a product of two separate variables.
A is only in u, it depends only on u.
And B is only on v. They have nothing to do with one another.
Then you can go ahead and do the first integral with respect
to u only of a of u, du, u between alpha, beta.
That was your first variable.
Times this other constant.
Integral of B of v, where v is moving,
v is moving between gamma, delta.
Instead of alpha, beta, gamma, delta,
put any numbers you want.
OK?
This is the lucky case.
So you're always hoping that on the final,
you can get something where you can separate.
Here you have no theta.
This is the luckiest case in the world.
So it's just r cubed times theta.
But you can still have a lucky case
when you have something like a function of r
times a function of theta.
And then you have another beautiful polar
coordinate integral that you're not going
to struggle with for very long.
OK, I'm going to erase here.
For example, let me give you another one.
Suppose that somebody was really mean to you,
and wanted to kill you in the final,
and they gave you the following problem.
Assume the domain D-- they don't even tell you what it is.
They just want to challenge you--
will be x, y with the property that x squared plus y
squared is between a 1 and a 4.
Compute the integral over D of r [? pan ?] of y over x and da,
where bi would be ds dy.
So you look at this cross-eyed and say, gosh,
whoever-- we don't do that.
But I've seen schools.
I've seen this given at a school, when they covered
this particular example, they've covered
something like the previous one that I showed you.
But they never covered this.
And they said, OK, they're smart,
let them figure this out.
And I think it was Texas A&M. They gave something like that
without working this in class.
They assumed that the students should
be good enough to figure out what
this is in polar coordinates.
So in polar coordinates, what does the theorem say?
We should switch to a domain D star that corresponds to D.
Now, D was given like that.
But we have to say the corresponding D
star, reinterpreted in polar coordinates,
r theta has to be also written beautifully out.
Unless you draw the picture, first of all, you cannot do it.
So the prof at Texas A&M didn't even say, draw the picture,
and think of the meaning of that.
What is the meaning of this set, geometric set,
geometric locus of points.
STUDENT: You've got a circle sub-
MAGDALENA TODA: You have concentric circles,
sub-radius 1 and 2, and it's like a ring, it's an annulus.
And he said, well, I didn't do it.
I mean they were smart.
I gave it to them to do.
So if the students don't see at least an example like that,
they have difficulty, in my experience.
OK, for this kind of annulus, you
see the radius would start here, but the dotted part
is not included in your domain.
So you have to be smart and say, wait a minute, my radius
is not starting at 0.
It's starting at 1 and it's ending at 2.
And I put that here.
And theta is the whole ring, so from 0 to 2 pi.
Whether you do that over the open set,
that's called annulus without the boundaries,
or you do it about the one with the boundaries,
it doesn't matter, the integral is not going to change.
And you are going to learn that in Advanced Calculus, why
it doesn't matter that if you remove the boundary,
you put back the boundary.
That is a certain set of a measure 0 for your integration.
It's not going to change your results.
So no matter how you express it-- maybe
you want to express it like an open set.
You still have the same integral.
Double integral of D star, this is
going to give me a headache, unless you help me.
What is this in polar coordinates?
STUDENT: [INAUDIBLE]
MAGDALENA TODA: I know when-- once I've
figured out the integrand, I'm going
to remember to always multiply by an r,
because if I don't, I'm in big trouble.
And then I go dr d theta.
But I don't know what this is.
STUDENT: r.
MAGDALENA TODA: Nope, but you're-- so r cosine theta is
x, r sine theta is y.
When you do y over x, what do you get?
Always tangent of theta.
And if you do arctangent of tangent, you get theta.
So that was not hard, but the students did
not-- in that class, I was talking
to whoever gave the exam, 70-something percent
of the students did not know how to do it,
because they had never seen something similar,
and they didn't think how to express this theta in r.
So what do we mean to do?
We mean, is this a product?
It's a beautiful product.
They are separate variables like [INAUDIBLE] [? shafts. ?] Now,
you see, you can separate them.
The r is between 1 and 2, so I can do-- eventually I
can do the r first.
And theta is between 0 and 2 pi, and as I taught you
by the previous theorem, you can separate the two integrals,
because this one gets out.
It's a constant.
So you're left with integral from 0 to 2 pi theta d
theta, and the integral from 1 to 2 r dr. r dr theta d theta.
This should be a piece of cake.
The only thing we have to do is some easy Calculus I.
So what is integral of theta d theta?
I'm not going to rush anywhere.
Theta squared over 2 between theta equals 0 down
and theta equals 2 pi up.
Right?
STUDENT: [INAUDIBLE]
MAGDALENA TODA: Yeah.
I'll do that later.
I don't care.
This is going to be r squared over 2 between 1 and 2.
So the numerical answer, if I know
how to do any math like that, is going to be--
STUDENT: 2 pi squared.
MAGDALENA TODA: 2 pi squared, because I
have 4 pi squared over 2, so the first guy
is 2 pi squared, times-- I get a 4 and 4 minus 1-- are
you guys with me?
So I get a-- let me write it like that.
4 over 2 minus 1 over 2.
What's going to happen to the over 2?
We'll simplify.
So this is going to be 3 pi squared.
Okey Dokey?
Yes, sir?
STUDENT: How did you split it into two integrals, right here?
MAGDALENA TODA: That's exactly what I taught you before.
So if I had not taught you before,
how did I prove that theorem?
The theorem that was before was like that.
What was it?
Suppose I have a function of theta, and a function of r,
and I have d theta dr. And I think
this weather got to us, because several people have
the cold and the flu.
Wash your hands a lot.
It's full of-- mathematicians full of germs.
So theta, you want theta to be between whatever you want.
Any two numbers.
Alpha and beta.
And r between gamma, delta.
This is what I explained last time.
So when you integrate with respect to theta first inside,
g of r says I have nothing to do with these guys.
They're not my type, they're not my gang.
I'm going out, have a beer by myself.
So he goes out and joins the r group,
because theta and r have nothing in common.
They are separate variables.
This is a function of r only, and that's
a function of theta only.
This is what I'm talking about.
OK, so that's a constant.
That constant pulls out.
So in the end, what you have is that constant that pulled out
is going to be alpha, beta, f of beta d theta as a number, times
what's left inside?
Integral from gamma to delta g of r
dr. So when the two functions F and G are functions of theta,
respectively, r only, they have nothing to do with one another,
and you can write the original integral
as the product of integrals, and it's really a lucky case.
But you are going to encounter this lucky case many times
in your final, in the midterm, in-- OK, now thinking of what
I wanted to put on the midterm.
Somebody asked me if I'm going to put-- they looked already
at the homework and at the book, and they asked me,
are we going to have something like the area of the cardioid?
Maybe not necessarily that-- or area
between a cardioid and a circle that intersect each other.
Those were doable even with Calc II.
Something like that, that was doable with Calc II,
I don't want to do it with a double integral in Calc III,
and I want to give some problems that are relevant to you guys.
The question, what's going to be on the midterm?
is not-- OK, what's going to be on the midterm?
It's going to be something very similar to the sample
that I'm going to write.
And I have already included in that sample
the volume of a sphere of radius r.
So how do you compute out the weight-- exercise 3 or 4,
whatever that is-- we compute the volume of a sphere using
double integrals.
I don't know if we have time to do this problem, but if we do,
that will be the last problem-- when you ask you teacher,
why is the volume inside the sphere, volume of a ball,
actually.
Well, the size-- the solid ball.
Why is it 4 pi r cubed over 2?
Your, did she tell you, or she told
you something that you asked, Mr. [? Jaime ?], for example?
They were supposed to tell you that you can prove that
with Calc II or Calc III.
It's not easy.
It's not an elementary formula.
In the ancient times, they didn't
know how to do it, because they didn't know calculus.
So what they tried to is to approximate it and see
how it goes.
Assume you have the sphere of radius r,
and r is from here to here, and I'm
going to go ahead and draw the equator, the upper hemisphere,
the lower hemisphere, and you shouldn't help me,
because isn't enough to say it's twice the upper hemisphere
volume, right?
So if I knew the-- what is this called?
If I knew the expression z equals
f of x, y of the spherical cap of the hemisphere,
of the northern hemisphere, I would be in business.
So if somebody even tries-- one of my students,
I gave him that, he didn't know polar coordinates very well,
so what he tried to do, he was trying to do,
let's say z is going to be square root of r
squared minus z squared minus y squared over the domain.
So D will be what domain? x squared
plus y squared between 0 and r squared, am I right guys?
So the D is on the floor, means x
squared plus y squared between 0 and r squared.
This is the D that we have.
This is D So twice what?
f of x, y.
The volume of the upper hemisphere
is the volume of everything under this graph, which
is like a half.
It's the northern hemisphere.
dx dy, whatever is dx.
So he tried to do it, and he came up
with something very ugly.
Of course you can imagine what he came up with.
What would it be?
I don't know.
Oh, God.
x between minus r to r.
y would be between 0 and-- you have to draw it.
STUDENT: It's going to be 0 or r.
STUDENT: Yeah.
STUDENT: Oh, no.
MAGDALENA TODA: So x is between minus r--
STUDENT: It's going to be as a function of x.
MAGDALENA TODA: And this is x.
And it's a function of x.
And then you go square root r squared minus x squared.
It looks awful in Cartesian coordinates.
And then for f, he just plugged in that thingy,
and he said dy dx.
And he would be right, except that I
would get a headache just looking
at it, because it's a mess.
It's a horrible, horrible mess.
I don't like it.
So how am I going to solve this in polar coordinates?
I still have the 2.
I cannot get rid of the 2.
How do I express-- in polar coordinates,
the 2 would be one for the upper part, one for the lower part--
How do I express in polar coordinates the disc?
Rho or r.
r between 0 to R, and theta, all the way from 0 to 2 pi.
So I'm still sort of lucky that I'm in business.
I go 0 to 2 pi integral from 0 to r,
and for that guy, that is in the integrand,
I'm going to say squared of z.
z is r squared minus-- who is z squared plus y squared
in polar coordinates?
r squared. very good. r squared.
Don't forget that instead of dy dx,
you have to say times r, the Jacobian, dr d theta.
Can we solve this, and find the correct formula?
That's what I'm talking about.
We need the u substitution.
Without the u substitution, we will be dead meat.
But I don't know how to do u substitution,
so I need your help.
Of course you can help me.
Who is the constant?
R is the constant.
It's a number.
Little r is a variable.
STUDENT: r squared is going to be the u.
MAGDALENA TODA: u, very good.
r squared minus r squared.
How come this is working so well?
Look why du will be constant prime 0 minus 2rdr.
So I take this couple called rdr, this block,
and I identify the block over here.
And rdr represents du over minus 2, right?
So I have to be smart and attentive,
because if I make a mistake at the end, it's all over.
So 2 tiomes integral from 0 to 2 pi.
I could get rid of that and say just 2 pi.
Are you guys with me?
I could say 1 is theta-- as the product, go out-- times--
and this is my integral that I'm worried about, the one only
in r.
Let me review it.
This is the only one I'm worried about.
This is a piece of cake.
This is 2, this is 2 pi.
This whole thing is 4 pi a.
At least I got some 4 pi out.
What have I done in here?
I've applied the u substitution, and I
have to be doing a better job.
I get 4 pi times what is it after u substitution.
This guy was minus 1/2 du, right?
This fellow is squared u, [? squared ?]
squared u as a power.
STUDENT: u to the 1/2.
MAGDALENA TODA: u to the one half.
And for the integral, what in the world do I write?
STUDENT: r squared--
MAGDALENA TODA: OK.
So when little r is 0, u is going to be r squared.
When little r is big R, you get 0.
Now you have to help me finish this.
It should be a piece of cake.
I cannot believe it's hard.
What is the integral of 4 pi?
Copy and paste.
Minus 1/2, integrate y to the 1/2.
STUDENT: 2/3u to the 3/2.
MAGDALENA TODA: 2/3 u to the 3/2, between u equals 0 up,
and u equals r squared down.
It still looks bad, but--
STUDENT: You've got a negative sign.
MAGDALENA TODA: I've got a negative sign.
STUDENT: Where is it--
MAGDALENA TODA: So when I go this minus that,
it's going to be very nice.
Why?
I'm going to say minus 4 pi over 2 times 2 over 3.
I should have simplified them from the beginning.
I have minus 5 pi over 3 times at 0 I have 0.
At r squared, I have r squared, and the square root
is r, r cubed.
r cubed.
Oh my God, look how beautiful it is.
Two minuses in a row.
Multiply, give me a plus.
STUDENT: This is the answer.
MAGDALENA TODA: Plus.
4 pi up over 3 down, r cubed.
So we proved something that is essential,
and we knew it from when we were in school,
but they told us that we cannot prove it,
because we couldn't prove that the volume of a ball was 4 pi r
cubed over 3.
Yes, sir?
STUDENT: Why are the limits of integration reversed?
Why is r squared on the bottom?
MAGDALENA TODA: Because first comes little r, 0,
and then comes little r to be big R. When I plug them
in in this order-- so let's plug them in first,
little r equals 0.
I get, for the bottom part, I get u equals r squared,
and when little r equals big R, I
get big R squared minus big R squared equals 0.
And that's the good thing, because when
I do that, I get a minus, and with the minus I already had,
I get a plus.
And the volume is a positive volume, like every volume.
4 pi [INAUDIBLE].
So that's it for today.
We finished 12-- what is that?
12.3, polar coordinates.
And we will next time do some homework.
Ah, I opened the homework for you.
So go ahead and do at least the first 10 problems.
If you have difficulties, let me know on Tuesday,
so we can work some in class.
STUDENT: [? You do ?] so much.
STUDENT: So, I went to the [INAUDIBLE], and I asked them,
[INTERPOSING VOICES]
[SIDE CONVERSATION]
STUDENT: Can you imagine two years
of a calculus that's the equivalent to [? American ?]
and only two credits?
MAGDALENA TODA: Because in your system,
everything was pretty much accelerated.
STUDENT: Yeah, and they say, no, no, no--
I had to ask him again.
[INAUDIBLE] calculus, in two years,
that is only equivalent to two credits.
I was like--
MAGDALENA TODA: Anyway, what happens
is that we used to have very good evaluators
in the registrar's office, and most of those people retired
or they got promoted in other administrative positions.
So they have three new hires.
Those guys, they don't know what they are doing.
Imagine, you would finish, graduate, today,
next week, you go for the registrar.
You don't know what you're doing.
You need time.
Yes?
STUDENT: I had a question about the homework.
I'll wait for [INAUDIBLE].
MAGDALENA TODA: It's OK.
Do you have secrets?
STUDENT: No, I don't.
MAGDALENA TODA: Homework is due the 32st.
STUDENT: No, I had a question from the homework.
Like I had a problem that I was working on, and I was like
MAGDALENA TODA: From the homework.
OK You can wait.
You guys have other, more basic questions?
[INTERPOSING VOICES]
MAGDALENA TODA: There is only one meeting.
Oh, you mean-- Ah.
Yes, I do.
I have the following three-- Tuesday,
Wednesday, and Friday- no, Tuesday, Wednesday,
and Thursday.
On Friday we can have something, some special arrangement.
This Friday?
OK, how about like 11:15.
Today, I have-- I have right now.
2:00.
And I think the grad students will come later.
So you can just right now.
And tomorrow around 11:15, because I have meetings
before 11 at the college.
STUDENT: Do you mind if I go get something to eat first?
Or how long do you think they'll be in your office?
MAGDALENA TODA: Even if they come,
I'm going to stop them and talk to you,
so don't worry about it.
STUDENT: Thank you very much.
I'll see you later.
STUDENT: I just wanted to say I'm sorry for coming in late.
I slept in a little bit this morning--
MAGDALENA TODA: Did you get the chance to sign?
STUDENT: Yes.
MAGDALENA TODA: There is no problem.
I'm--
STUDENT: I woke up at like 12:30-- I woke up at like 11:30
and I just fell right back asleep, and then I got up
and I looked at my phone and it was 12:30,
and I was like, I have class right now.
And so what happened was like--
MAGDALENA TODA: You were tired.
You were doing homework until late.
STUDENT: --homework and like, I usually
am on for an earlier class, and I
didn't go to bed earlier than I did last night,
and so I just overslept.
MAGDALENA TODA: I did the same, anyway.
I have similar experience.
STUDENT: You have a very nice day.
MAGDALENA TODA: Thank you.
You too.
So, show me what you want to ask.
STUDENT: There it was.
I looked at that problem, and I thought,
that's extremely simple, acceleration--
MAGDALENA TODA: Are they independent, really?
STUDENT: Huh?
MAGDALENA TODA: Are they-- b and t are independent?
I need to stop.
STUDENT: But I didn't even bother.