TTU Math2450 Calculus3 Sec 12.3

Double Integral in polar coordinates

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.

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.