TTU Math2450 Calculus3 Secs 12.1-12.2

Double Integral over a rectangular region and Double Integral over a non rectangular region

PROFESSOR: I have
some assignments

that I want to give you back.

And I'm just going
to put them here,

and I'll ask you to pick them
up as soon as we take a break.

There are explanations there
how they were computed in red.

If you have questions,
you can as me

so I can ask my grader about it.

Now, I promised you that
I would move on today,

and that's what I'm going to do.

I'm moving on to something
that you're gong to love.

[? Practically ?] chapter 12
is integration of functions

of several variables.

And to warn you
we're going to see

how we introduce introduction
to the double integral.

But you will say, wait a minute.

I don't even know if I
remember the simple integral.

And that's why I'm here.

I want to remind you what the
definite integral was both

as a formal definition-- let's
do it as a formal definition

first, then come up with a
geometric interpretation based

on that.

And finally write
down the definition

and the fundamental
theorem of calculus.

So assume you have a
function that's continuous.

Continuous over a certain
integral of a, b interval in R.

And you know that
in that case, you

can "define the
definite integral of f

of x from or between a and b."

And as the notation is denoted,
by integral from a to b f of x

dx.

Well, how do we define this?

This is just the notation.

How do we define it?

We have to have a set up, and
we are thinking of a x, y frame.

You have a function,
f, that's continuous.

And you are thinking,
oh, wait a minute.

I would like to be
able to evaluate

the area under the integral.

And if you ask your teacher
when you are in fourth grade,

your teacher will say, well,
I can give you some graphing

paper.

And with that
graphing paper, you

can eventually approximate
your area like that.

Sort of what you get here is
like you draw a horizontal

so that the little part
above the horizontal

cancels out with the little
part below the horizontal.

So more or less,
the pink rectangle

is a good approximation
of the first slice.

But you say yeah, but the first
slice is a curvilinear slice.

Yes, but we make it
like a stop function.

So then you say, OK,
how about this fellow?

I'm going to approximate
it in a similar way,

and I'm going to have a bunch
of rectangles on this graphing

paper.

And I'm going to
compute their areas,

and I'm going to come up
with an approximation,

and I'll give it to my
fourth grade teacher.

And that's what we
did in fourth grade,

but this is not fourth grade.

And actually, it's
very relevant to us

that this has
applications to our life,

to our digital world,
that people did not

understand when Riemann
introduced the Riemann sum.

They thought, OK, the idea
makes sense that practically we

have a huge picture
here, and I'm

taking a and b and a function
that's continuous over a and b.

And then I say I'm
going to split this

into a equidistant intervals.

I don't know how
many I want, but let

me make them eight of them.

I don't know.

They have to have
the same length.

And I'll call this delta x.

It has to be the same.

And, you guys, please forgive
me for the horrible picture.

They don't look like
the same step, delta x,

but it should be the same.

In each of them I
arbitrarily, say it again,

Magdalena, arbitrarily pick
x1 star, and another point,

x2 star wherever I want inside.

I'm just getting [INAUDIBLE].

X4 star, and this is x8 star.

But let's say that in general
I don't know they are 8.

They could be n.

xn star.

And passing to the
limit with respect

to n going to infinity,
what am I going to get?

Well, in the first
cam I'm going up,

and I'm hitting
at what altitude?

I'm hitting at the altitude
called f of x1 star.

And that's going to be the
height of this-- what is this?

Strip?

Right?

Or rectangle.

OK.

And I'm going to do
the same with green

for the second rectangle.

I'll pick x2 star, and
then that doesn't work.

And I'll take this.

Let's see if I can do
the light green one,

because spring is here.

Let's see.

That's beautiful.

I go up.

I hit here at x2 star.

I get f of x2 star.

And so on and so forth.

Until I get to, let's say,
the last of the Mohicans.

This will be xn minus
1, and this is going

to be xn star, the purple guy.

And this is going
to be the height

of that last of the Mohicans.

So when I compute the sum, I
call that approximating sum

or Riemann approximating sum,
because Riemann had nothing

better to do than invent it.

He didn't even know
that we are going

to get pixels that are in
larger and larger quantities.

Like, we get 3,000 by 900.

He didn't know we are going to
have all those digital gadgets.

But passing to the
limit practically should

be easier to understand
for teenagers now

age, because it's like
making the number of pixels

larger and larger, and the
pixels practically invisible.

Remember, I mean, I don't
know, those old TVs,

color TVs where you could
still see the squares?

STUDENT: Mm-hm.

PROFESSOR: Well, yeah.

When you were little.

But I remember them
much better than you.

And, yes, as the number
of pixels will increase,

that means I'm taking the limit
and going larger and larger.

That means
practically limitless.

Infinity will give
me an ideal image.

My eye will be as if I could see
the image that's a curvilinear

image as a real person.

And, of course, the
quality of our movies

really increased a lot.

And this is what I'm
trying to emphasize here.

So you have f of x1 star delta
x plus the last rectangle

area, f of xn star delta x.

Well, as a mathematician,
I don't write it like that.

How do I write it
as a mathematician?

Well, we are funny people.

We like Greek.

It's all Greek to me.

So we go sum and from-- no. k
from 1 to n, f of x sub k star.

So I have k from 1 to n exactly
an rectangles area to add.

And this is going to
be [INAUDIBLE], which

is the same everywhere.

In that case, I made
the partition is equal.

So practically I have
the same distance.

And what is this
limit? [? Lim ?]

is going to be exactly integral
from a to b of f of x dx.

And I make a smile here,
and I say I'm very happy.

This is as a meaning is
the area under the graph.

If-- well, I didn't
say something.

If I want it to be
positive, otherwise it's

getting not to be the
area under the graph.

The integral will still
be defined like that.

But what's going to happen if
I have, for example, half of it

above and half of it below?

I'm going to get this,
and I'm going to get that.

And when I add them, I'm going
to get a negative answer,

because this is a negative
area, and that's a positive area

and they try to
annihilate each other.

But this guy under
the water is stronger,

like an iceberg that's
20% on tip of the water,

80% of the iceberg
is under the water.

So the same thing.

I'm going to get a negative
answer in volume [INAUDIBLE].

OK.

Now, we remember that
very well, but now we

have to generalize this
thingy to something else.

And I will give you
a curvilinear domain.

Where shall I erase?

I don't know.

Here.

What if somebody gives you
the image of a potatoe-- well,

I don't know.

Something.

A blob.

Some nice curvilinear domain--
and says, you know what?

I want to approximate the area
of this image, curvilinear

image, to the best
of my abilities.

And compute it, and eventually I
have some weighted sum of that.

So if one would have
to compute the area,

it wouldn't be so hard,
because we would say,

OK, I have to
"partition this domain

into small sections using
a rectangular partition

or square partition."

And how?

Well, I'm going to--
you have to imagine

that I have a bunch
of a grid, and I'm

partitioning the whole thing.

And you say, wait a minute.

Wait a minute.

It's not so easy.

I mean, they are not all
the same area, Magdalena.

Even if you tried to make these
equidistant in both directions,

look at this guy.

Look at that guy.

He's much bigger than that.

Look at this small
guy, and so on.

So we have to imagine that we
look at the so-called normal

the partition.

And let's say in the normal,
or the length of the partition,

is denoted like that.

We have to give that a meaning.

Well, let's say "this
is the highest diameter

for all subdomains
in the picture."

And you say, wait a minute.

But these subdomains
should have names.

Well, they don't have names,
but assume they have areas.

This would be-- I have to
find a way to denote them

and be orderly.

A1, A2, A3, A4, A5, AN,
AM, AN, stuff like that.

So practically I'm looking
at the highest diameter.

When I have a domain, I
look at the largest instance

inside that domain.

So what would be the diameter?

The largest distance between
two points in that domain.

I'll call that the diameter.

OK.

I want that diameter to
go got 0 in the limit.

So I want this partition
to go to 0 in the limit.

And that means I'm
"shrinking" the pixels.

"Shrinking" in
quotes, the pixels.

How would I mimic
what I did here?

Well, it would be
easier to get the area.

In this case, I would have
some sort of A sum limit.

I'm sorry.

The curvilinear
area of the domain.

Let's call it-- what
do you want to call it?

D for domain--
inside the domain.

OK?

This whole thing would be what?

Would be limit of summation of,
let's say, limit of what kind?

k from 1 to n.

Limit n goes to infinity.

K from 1 to n of
these tiny A sub k's,

areas of the subdomain.

Wait a minute.

But you say, but what if
I want something else?

Like, I'm going to
build some geography.

This is the domain.

That's something like
on a map, and I'm

going to build a
mountain on top of it.

I'll take some Play-Do,
I'll take some Play-Do,

and I'm going to
model some geography.

And you say, wait a minute.

Do you make mountains?

I'm afraid to make Rocky
Mountains, because they

may have points where the
function is not smooth.

If I don't have
derivative at the peak,

them I'm in trouble, in general.

Although you say,
well, but the function

has to be only continuous.

I know.

I know.

But I don't want any kind
of really nasty singularity

where I can have a
crack in the mountain

or a well or
something like that.

So I assume the
geography to be smooth,

the function of
[INAUDIBLE] is continuous,

and the picture
should look something

like-- let's see
if I can do that.

The projection, the
shadow of this geography,

would be the domain, [? D. ?]
And this is equal, f of x what?

You say, what?

Magdalena, I don't understand.

The exact shadow of this fellow
where I have the sun on top

here-- that's the sun.

Spring is coming-- the shade
is the plain, or domain, x, y.

I take all my points in x, y.

I mean, I take really
all my points in x, y,

and the value of the altitude
on this geography at the point

x, y would be z
equals f of x, y.

And somebody's asking me, OK,
if this would be a can of Coke,

it would be easy to
compute the volume, right?

Practically you have a
constant altitude everywhere,

and you have the area of
the base times the height,

and that's your volume.

But what if somebody asks you to
find the volume under the hat?

"Find the volume
undo this graph."

STUDENT: I would take it
more as two functions.

So the top line would
be the one function,

and the bottom line would
be another function.

So if you take the volume of the
top function minus the volume

of the bottom
function, it'd give you

the total volume of the object.

PROFESSOR: And actually,
I want the total volume

above the sea level.

So I'm going to--
sometimes I can take it up

to a certain level where-- let's
say the mountain is up to here,

and I want it only up to here.

So I want everything,
including the-- the walls

would be cylindrical.

STUDENT: Yeah.

PROFESSOR: If I
want all the volume,

that's going to be
a little bit easier.

Let's see why.

I will have limit.

The idea is, as you
said very well, limit.

n goes to infinity.

A sum k from 1 to n.

And what kind of
partition can I build?

I'll take the
line, and I'll say,

I'll build myself
a partition with a,

let's say, the
typical domain, AK.

I have A1, A2 A3, A4, AK, AN.

How may of those little domains?

AN.

That will be all the
little subdomains

inside the green curve.

The green loop.

In that case, what do I do?

For each of these guys, I go
up, and I go, oh, my god, this

looks like a skyscraper,
but the corners,

when I go through
this surface, are

in the different dimensions.

What am I going to do?

That forces me to
build a skyscraper

by thinking I take a
point in the domain,

I go up until that hits the
surface, pinches the surface,

and this is the
altitude that I'm going

to select for my skyscraper.

And here I'm going to have
another skyscraper, and here

another one and another one,
so practically it's dense.

I have a skyscraper next
to the other or a less like

[INAUDIBLE].

Not so many gaps
in certain areas.

So I'm going to say
f of x kappa star.

Now those would be the
altitudes of the buildings.

Magdalena, you don't
know how to spell.

Altitudes of the buildings.

What are they?

Parallel [INAUDIBLE] by P's.

Can you say parallel by P?

OK.

[INAUDIBLE] what.

Ak where Ak will be the basis
of the area of the basis.

is of my building.

OK.

The green part will
be the flat area

of the floor of the skyscraper.

Is this hard?

Gosh, yes.

If you want to do it by
hand and take the limit

you would really kill
yourself in the process.

This is how you introduce it.

You can prove this limit exists,
and you can prove that limits

exist and will be the volume of
the region under the geography

z equals f of x,y and
above the sea level.

The seal level
meaning z equals z.

STUDENT: What's under a of k?

PROFESSOR: Ak.

STUDENT: What is [INAUDIBLE]

PROFESSOR: Volume of the region.

STUDENT: Oh, I know,
like what under it?

PROFESSOR: Here?

STUDENT: No, up.

PROFESSOR: Here?

STUDENT: Yes.

PROFESSOR: Area of the
basis of a building.

STUDENT: Oh, the basis.

PROFESSOR: So practically
this green thingy

is a basis like the base rate.

How large is the basement
of that building.

Ak.

Now how am I going
to write this?

This is something new.

We have to invent a notion
for it, and since it's Ak,

looks more or less like
a square or a rectangle.

You think, well, wouldn't--
OK, if it's a rectangle,

I know I'm going to get
delta x and delta y right?

The width times the height,
whatever those two dimensions.

It makes sense.

But what if I have
this domain that's

curvilinear or that
domain or that domain.

Of course, the diameter
of such a domain

is less than the diameter of the
partition, so I'm very happy.

The highest diameter,
say I can get it here,

and this is shrinking
to zero, and pixels

are shrinking to zero.

But what am I going to
do about those guys?

Well, you can assume that
I am still approximating

with some squares and as
the pixels are getting

to be many, many,
many more, it doesn't

matter that I'm doing this.

Let me show you what I'm doing.

So on the floor, on the-- this
is the city floor, whatever.

What we do in practice,
we approximate that

like on the graphing paper
with tiny square domains,

and we call them delta Ak will
be delta Sk times delta Yk,

and I tried to make it a uniform
partition as much as I can.

Now as the number of
pixels goes to infinity

and those pixels will
become smaller and smaller,

it doesn't there that the actual
contour of your Riemann sum

will look like graphing paper.

It will get refined, more
refined, more refined, smoother

and smoother, and
it's going to be

really close to the ideal
image, which is a curve.

So as that end goes
to infinity, you're

not going to see this-- what is
this called-- zig zag thingy.

Not anymore.

The zig zag thingy will go into
the limit to the green curve.

This is what the
pixels are about.

This is how our
life changed a lot.

OK?

All right.

Now good.

How am I going
compute this thing?

Well, I don't know, but let
me give it a name first.

It's going to be double
integral over-- what

do want the floor to be called?

We called d domain before.

What should I call this?

Big D. Not round.

Over D. That's the
floor, the foundation

of the whole city-- of
the whole area of the city

that I'm looking at.

Then I have f of xy,
da, and what is this?

This is exactly that.

It's the limit of sum of the--
what is the difference here?

You say, wait a
minute, Magdalena,

but I think I don't
understand what you did.

You tried to copy the
concept from here,

but you forgot you have a
function of two variables.

In that case, this mister,
whoever it is that goes up

is not xk, it's XkYk.

So I have two variables--
doesn't change anything

for the couple.

This couple represents a
point on the skyscraper

so that when I go up, I hit the
roof with this exact altitude.

So what is the double integral
of a continuous function

f of x and y, two variables,
with respect to area level.

Well, it's going to be just
the limit of this huge thing.

In fact, it's how
do we compute it?

Let's see how we
compute it in practice.

It shouldn't be a big deal.

What if I have a
rectangular domain,

and that's going to
make my life easier.

I'm going to have a
rectangular domain in plane,

and which one is the x-axis?

This one.

From A to B, I have the x
moving between a and Mr. y

says, I'm going to
be between c and d.

C is here, and d is here.

So this is going to be
the so-called rectangle

a, b cross c, d meaning
the set of all the pairs--

or the couples xy-- inside
it, what does it mean?

x, y you playing with
the property there.

X is between a and b, thank god.

It's easy.

And y must be between
c and d, also easy.

A, b, c, d are fixed real
numbers in this order.

A is less than b, and c is less.

And we have this
geography on top,

and I will tell you
what it looks like.

I'm going to try and draw
some beautiful geography.

And now I'm thinking
of my son, who is 10.

He played with this kind of toy
that was exactly this color,

lime, and it had needles.

Do you guys remember that toy?

I am sure you're young
enough to remember that.

You have your palm like that,
and you see this square thingy,

and it's all made
of needles that

look like thin,
tiny skyscrapers,

and you push through and all
those needles go up and take

the shape of your hand.

And of course, he would
put it on his face,

and you could see
his face and so on.

But what is that?

That's exactly the Riemann
sum, the Riemann approximation,

because if you think of
all those needles or tiny--

what are they, like
the tiny skyscrapers--

the sum of the them approximates
the curvilinear shape.

If you put that over your face,
your face is nice and smooth,

curvilinear except for
a few single areas,

but if you actually
look at that needle

thingy that is
giving the figure,

you recognize the figure.

It's like a pattern recognition,
but it's not your face.

I mean it is and it's not.

It's an approximation of
your face, a very rough face.

You have to take that
rough model of your face

and smooth it out.

How?

By passing to the
limit, and this is what

animation is doing actually.

On top of that you want this
to have some other properties--

illumination of some sort--
light coming from what angle.

That is all rendering
techniques are actually

applied mathematics.

In animation, the
people who programmed

Toy Story-- that
was a long time ago,

but everything that
came after Toy Story 2

was based on mathematical
rendering techniques.

Everything based on
the notion of length.

All right.

So the way we compute
this in practice

is going to be very simple,
because you're going to think,

how am I going to do the
rectangle for the rectangles?

That'll be very easy.

I split the rectangle perfectly
into other tiny rectangle.

Every rectangle will
have the same dimension.

Delta x and delta y.

Does it makes sense?

So practically when
I go to the limit,

I have summation f
of xk star, yk star

inside the delta x
delta y delta Magdalena,

the same kind of displacement
when I take k from 1 to n,

and I pass to the limit
according to the partition,

what's going to happen?

These guys, according
to Mr. Linux,

will go to be infinitesimal
elements, dx, dy.

This whole thing will
go to double integral

of f of x and y,
and Mr. y says, OK

it's like you want him to
integrate him one at a time.

This is actually something that
we are going to see in a second

and verify it.

X goes between a and b,
and y goes between c and d,

and this is an application
of a big theorem called

Fubini's Theorem that
says, wait a minute,

if you do it like this over
a rectangle a,b cross c,d,

you're double integral can
be written as three things.

Double integral over your
square domain f of x,y dA,

or you integral from c to
d, integral from a to b,

f of x,y dx dy, or you
can also swap the order,

because you say, well, you can
do the integration with respect

to y first.

Nobody stops you
from doing that,

and y has to be
between what and what?

STUDENT: C and d.

PROFESSOR: C and d, thank you.

And then whatever you get,
you get to integrate that

with respect to x from a to b.

So no matter in what
order you do it,

you'll get the same thing.

Let's see an easy example,
and you'll say, well,

start with some [INAUDIBLE]
example, Magdalena,

because we are just
starting, and that's

exactly what I'm going to.

I will just misbehave.

I'm not going to go by the book.

And I will say I'm going
by whatever I want to go.

X is between 0, 2, and
y is between 0 and 2

and 3-- this is 2, this
is 3-- and my domain

will be the rectangle
0, 2 times 0, 3.

This is neat on the floor.

Compute the volume of the
box of basis d and height 5.

Can I draw that?

It gets out of the picture.

I'm just kidding.

This is 5, and that's
sort of the box.

And you say, wait a minute, I
know that from third grade--

I mean, first grade, whenever.

How do we do that?

We go 2 units times
3 units that's

going to be 6 square inches
on the bottom of the box,

and then times 5.

So the volume has to be
2 times 3 times 5, which

is 30 square inches.

I don't care what it is.

I'm a mathematician, right?

OK.

How does somebody who just
learned Tonelli's-- Fubini

Tonelli's Theorem
do the problem.

That person will
say, wait a minute,

now I know that the
function is going to be z

equals f of xy, which in
this case happens to be cost.

According to what you told us,
the theorem you claim Magdalena

proved to this theorem,
but there is a sketch

of the proof in the book.

According to this,
the double integral

that you have over the
domain d, and this is dA.

DA will be called element of
area, which is also dx dy.

This can be solved in
two different ways.

You take integral
from-- where is x going?

Do we want to do it
first in x or in y?

If we put dy dx, that means
we integrate with respect

to y first, and y
goes between 0 and 3,

so I have to pay attention
to the limits of integration.

And then x between
0 and 2 and again

I have to pay attention to
the limits of integration

all the time and,
here, who is my f?

Is the altitude 5 that's
constant in my case?

I'm not worried about it.

Let me see if I get 30?

I'm just checking if this
theorem was true or is just

something that you cannot apply.

How do you integrate
5 with respect to y?

STUDENT: 5y.

PROFESSOR: 5y, very good.

So it's going to be 5y between
y equals 0 down and y equals 3

up, and how much
is that 5y, we're

doing y equals 0 down
and y equals 3 up,

what number is that?

STUDENT: 25.

PROFESSOR: What?

STUDENT: 25.

PROFESSOR: 25?

STUDENT: One [INAUDIBLE] 15.

PROFESSOR: No, you did--
you are thinking ahead.

So I go 5 times 3 minus
5 times 0 equals 15.

So when I compute this
variation of 5y between y

equals 3 and y equals
0, I just block in

and make the difference.

Why do I do that?

It's the simplest application
of that FT, fundamental theorem.

The one that I did not
specify in [INAUDIBLE].

I should have specified when
I have a g function that

is continuous between
alpha and beta, how do we

integrate with respect to x?

I get the antiderivative
of rule G. Let's call

that big G. Compute
it at the end points,

and I make the difference.

So I compute the
antiderivative at an endpoint--

at the other endpoint-- then I'm
going to make the difference.

That's the same thing I do
here, so 5 times 3 is 15,

5 times 0 is 0,
15 minus 0 is 15.

I can keep moving.

Everything in the
parentheses is the number 15.

I copy and paste, and that
should be a piece of cake.

What do I get?

STUDENT: 15.

PROFESSOR: I get 15
times x between 0 and 2.

Integral of 1 is x.

Integral of 1 is x
with respect to x,

so I get 15 times 2, which
is 30, and you go, duh,

[INAUDIBLE].

That was elementary mathematics.

Yes, you were lucky you
knew that volume of the box,

but what if somebody gave
you a curvilinear area?

What if somebody gave you
something quite complicated?

What would you do?

You have know calculus.

That's your only chance.

If you don't calculus,
you are dead meat.

So I'm saying, how
about another problem.

That look like it's
complicated, but calculus

is something
[INAUDIBLE] with that.

Suppose that I have a square
in the plane between-- this

is x and y-- do you
want square 0,1 0,1

or you want minus 1
to 1 minus 1 to 1.

It doesn't matter.

Well, let's take minus
1 to 1 and minus 1 to 1,

and I'll try to draw
as well as I can,

which I cannot but it's OK.

You will forgive me.

This is the floor.

If I were just a
little tiny square

in this room plus the
equivalent square in that room

and that room and that room.

This is the origin.

Are you guys with me?

So what you're
looking at right now

is this square foot
of carpet that I have,

but I have another one here and
another one behind the wall,

and so do I everything in mind?

X is between minus 1 and 1,
y is between minus 1 and 1.

And somebody gives you z
to be a positive function,

continuous function, which
is x squared plus y squared.

And you go, already.

Oh, my god.

I already have this
kind of hard function.

It's not a hard thing to do.

Let's draw that.

What are we going to get?

Your favorite [INAUDIBLE]
that goes like this.

And imagine what's
going to happen

with this is like a vase.

Inside, it has this
circular paraboloid.

But the walls of this vase are--
I cannot draw better than that.

So the walls of this
vase are squares.

And what you have inside is
the carved circular paraboloid.

Now I'm asking
you, how do I find

volume of the body under and
above D, which is minus 1,

1, minus 1, 1.

It's hard to draw that, right?

It's hard to draw.

So what do we do?

We start imagining things.

Actually, when you cut with
a plane that is y equals 1,

you would get a parabola.

And so when you look at what the
picture is going to look like,

you're going to have
a parabola like this,

a parabola like that,
exactly the same,

a parallel parabola like this
and a parabola like that.

Now I started drawing better.

And you say, how did you
start drawing better?

Well, with a little
bit of practice.

Where are the maxima
of this thing?

At the corners.

Why is that?

Because at the corners,
you get 1, 1 for both.

Of course, to do the absolute
extrema, minimum, maximum,

we would have to go back to
section 11.7 and do the thing.

But practically, it's easy
to see that at the corners,

you have the height 2 because
this is the point 1, 1.

And the same height, 2 and 2
and 2, are at every corner.

That would be the
maximum that you have.

So you have 1 minus 1 and so
on-- minus 1, 1, and minus 1,

minus 1, who is behind
me, minus 1, minus 1.

That goes all the way to 2.

So it's hard to do an
approximation with a three

dimensional model.

Thank god there is calculus.

So you say integral of x
squared plus y squared,

as simple as that, da over the
domain, D, which is minus 1,

1, minus 1, 1.

How do you write it
according to the theorem

that I told you
about, Fubini-Tonelli?

Then you have integral integral
x squared plus y squared dy dx.

Doesn't matter which
one I'm taking.

I can do dy dx.

I can do dx dy.

I just have to pay
attention to the endpoints.

Lucky for you the
endpoints are the same.

y is between minus 1 and 1.

x is between minus 1 and 1.

I wouldn't known how to compute
the volume of this vase made

of marble or made
of whatever you

want to make it unless I knew
to compute this integral.

Now you have to help me
because it's not hard

but it's not easy either, so we
need a little bit of attention.

We always start from the
inside to the outside.

The outer person has to be just
neglected for the time being

and I focus all my attention
to this integration.

And when I integrate
with respect to y,

y is the variable for me.

Nothing else exists
for the time being,

but y being a variable,
x being like a constant.

So when you integrate x
squared plus y squared

with respect to y, you have
to pay attention a little bit.

It's about the same if you
had 7 squared plus y squared.

So this x squared
is like a constant.

So what do you get inside?

Let's apply the fundamental
theorem of calculus.

STUDENT: x squared y.

PROFESSOR: x squared y.

Excellent.

I'm very proud of you.

Plus?

STUDENT: y cubed over 3.

PROFESSOR: y cubed over three.

Again, I'm proud of you.

Evaluated between y equals
minus 1 down, y equals 1 up.

And I will do the math later
because I'm getting tired.

Now let's do the math.

I don't know what
I'm going to get.

I get minus 1 to 1, a
big bracket, and dx.

And in this big bracket, I
have to do the difference

between two values.

So I put two parentheses.

When y equals 1, I
get x squared 1--

I'm not going to write
that down-- plus 1 cubed

over 3, 1/3.

I'm done with evaluating
this sausage thingy at 1.

It's an expression
that I evaluate.

It could be a lot longer.

I'm not planning to give you
long expressions in the midterm

because you're going to
make algebra mistakes,

and that's not what I want.

For minus 1, what do we
have Minus x squared.

What is y equals minus
1 plugged in here?

Minus 1/3.

I have to pay attention.

You realize that if I mess
up a sign, it's all done.

So in this case, I say, but
this I have minus, minus.

A minus in front of
a minus is a plus,

so I'm practically doubling
the x squared plus 1/3

and taking it
between minus 1 and 1

and just with respect to x.

So you say, wait a minute.

But that's easy.

I've done that when
I was in Calc 1.

Of course.

This is the nice part that
you get, a simple integral

from the ones in Calc 1.

Let's solve this one and find
out what the area will be.

What do we get?

Is it hard?

No.

Kick Mr. 2 out.

He's just messing
up with your life.

Kick him out.

2, out.

And then integral of
x squared plus 1/3

is going to be x
cubed over 3 plus--

STUDENT: x over 3.

PROFESSOR: x over 3, very good.

Evaluated between x equals
minus 1 down, x equals 1 up.

Let's see what we get.

2 times bracket.

I'll put a parentheses
for the first fractions,

and another minus, and
another parentheses.

What's the first edition
of fractions that I get?

1/3 plus 1/3.

I'll put 2/3 because I'm lazy.

Then minus what?

STUDENT: Minus 1/3.

PROFESSOR: Minus 1/3
minus 1/3, minus 2/3.

And now I should be able to
not beat around the bush.

Tell me what the answer
will be in the end.

STUDENT: 8/3.

PROFESSOR: 8/3.

Does that make sense?

When you do that in
math, you should always

think-- one of the famous
professors at Harvard

was saying one time
she asked the students,

how many hours of
life do we have have

in one day, blah, blah, blah?

And many students
came up with 36, 37.

So always make sure that the
answer you get makes sense.

This is part of a cube, right?

It's like carved in a
cube or a rectangle.

Now, what's the height?

If this were to go
up all the way to 2,

it would be 2, 2, and 2.

2 times 2 times 2 equals 8,
and what we got is 8 over 3.

Now, using our imagination,
it makes sense.

If I got a 16, I
would say, oh my god.

No, no, no, no.

What is that?

So a little bit, I would think,
does this make sense or not?

Let's do one more,
a similar one.

Now I'm going to count
on you a little bit more.

STUDENT: Professor,
did you calculate that

by just doing a quarter, and
then just multiplying it by 4?

Because then that
would just leave us

with zeroes [INAUDIBLE].

PROFESSOR: You mean in
that particular figure?

Yeah.

STUDENT: Yeah, because it
was perfectly [INAUDIBLE].

PROFESSOR: Yeah.

It's nice.

It's a little bit related
to some other problems that

come from pyramids.

By the way, how can you compute
the volume of a square pyramid?

Suppose that you have
the same problem.

Minus 1 to 1 for x and y.

Minus 1 to 1, minus 1 to 1.

Let's say the pyramid would
have the something like that.

What would be the volume
of such a pyramid?

STUDENT: [INAUDIBLE].

PROFESSOR: The height
is h for extra credit.

Can you compute the
volume of this pyramid

using double integrals?

Say the height is h and the
bases is the square minus 1,

1, minus 1, 1.

I'm sure it can be
done, but you know--

now I'm testing what you
remember in terms of geometry

because we will deal
with geometry a lot

in volumes and areas.

So how do you do that
in general, guys?

STUDENT: 1/3 [INAUDIBLE].

PROFESSOR: 1/3 the
height times the area

of the bases, which is what?

2 times 2.

2 times 2, 3, over 3, 4/3 h.

Can you prove that
with calculus?

That's all I'm saying.

One point extra credit.

Can you prove that
with calculus?

Actually, you would have
to use what you learned.

You can use Calc 2 as well.

Do you guys remember
that there were

some cross-sectional areas, like
this would be made of cheese,

and you come with a vertical
knife and cut cross sections.

They go like that.

But that's awfully hard.

Maybe you can do it differently
with Calc 3 instead of Calc 2.

Let's pick one from
the book as well.

OK.

So the same idea of using
the Fubini-Tonelli argument

and have an iterative-- evaluate
the following double integral

over the rectangle
of vertices 0, 0--

write it down-- 3,
0, 3, 2, and 0, 2.

So on the bases, you have a
rectangle of vertices 3, 0, 0,

0, 3, 2, and 0, 2.

And then somebody
tells you, find us

the double integral
of 2 minus y da

over r where r represents the
rectangle that we talked about.

This is exactly [INAUDIBLE].

And the answer we
should get is 6.

And I'm saying on top of
what we said in the book,

can you give a geometric
interpretation?

Does this have a
geometric interpretation

you can think of or not?

Well, first of all,
what is this animal?

According to the Fubini
theorem, this animal

will have to be-- I have
it over a rectangle,

so assume x will be
between a and b, y

will be between c and d.

I have to figure
out who those are.

2 minus y and dy dx.

Where is y between?

I should draw the
picture for the rectangle

because otherwise, it's
not so easy to see.

I have 0, 0 here, 3, 0 here, 3,
2 over here, shouldn't be hard.

So this is going to be 0, 2.

That's the y-axis and
that's the x-axis.

Let's see if we can see it.

And what is the meaning
of the 6, I'm asking you?

I don't know.

x should be between
0 and 3, right?

y should be between
0 and 2, right?

Now you are experts in this.

We've done this twice, and
you already know how to do it.

Integral from 0 to 3.

Then I take that,
and that's going

to be 2y minus y
squared over 2 between y

equals 0 down and
y equals 2 up dx.

That means integral from 0
to 3, bracket minus bracket

to make my life easier, dx.

Now, there is no x, thank god.

So that means I'm going
to have a constant

minus another constant, which
means I go 4 minus 4 over 2.

2, right?

The other one, for 0, I get 0.

I'm very happy I get 0
because in that case,

it's obvious that I get
2 times 3, which is 6.

So I got what the book
said I'm going to get.

But do I have a geometric
interpretation of that?

I would like to see
if anybody can--

I'm going to give you a
break in a few minues--

if anybody can think of a
geometric interpretation.

What is this f of xy if I were
to interpret this as a graph?

x equals f of x and y.

Is this--

STUDENT: 2 minus y.

PROFESSOR: So z equals 2
minus y is a plane, right?

STUDENT: Yes, but then you have
the parabola is going down.

PROFESSOR: And how do I get
to draw this plane the best?

Because there are
many ways to do it.

I look at this wall.

The y-axis is this.

The z-axis is the vertical line.

So I'm looking at this plane.

y plus z must be equal to 2.

So when is y plus z equal to 2?

When I am on a
line in the plane.

I'm going to draw that line
with pink because I like pink.

This is y plus z equals 2.

And imagine this line will be
shifted by parallelism as it

comes towards you on all these
other parallel vertical planes

that are parallel to the board.

So I'm going to have an
entire plane like that,

and I'm going to stop here.

When I'm in the plane
that's called x equals 3--

this is the plane
called x equals

3-- I have exactly this
triangle, this [INAUDIBLE].

It's in the plane
that faces me here.

I don't know if
you realize that.

I'll help you make a
house or something nice.

I think I'm getting hungry.

I imagine this again as
being a piece of cheese,

or it looks even like a piece
of cake would be with layers.

So our question is, if
we didn't know calculus

but we knew how to draw
this, and somebody gave you

this at the GRE
or whatever exam,

how could you have done
it without calculus?

Just by cheating and
pretending, I know how to do it,

but you've never done a
double integral in your life.

So I know it's a volume.

How do I get the volume?

What kind of geometric
body is that?

STUDENT: A triangle.

STUDENT: It's a
triangular prism.

PROFESSOR: It's a
triangular prism.

Good.

And a triangular prism
has what volume formula?

STUDENT: Base times height.

PROFESSOR: Base
times the height.

And the height has what area?

Let's see.

The base would be that, right?

And the height would be 3.

Am I right or not?

The height would be 3.

This is not--

STUDENT: It's 2.

Yeah.

STUDENT: No, it's 3.

DR. MAGDALENA TODA:
From here to here?

STUDENT: 3.

DR. MAGDALENA TODA: It's 3.

So how much is that?

How much-- OK.

From here to here is 2.

From here to here,
it's how much?

STUDENT: The height
is only-- I see--

STUDENT: It's also 2.

DR. MAGDALENA TODA: It's
also 2 because look at that.

It's an isosceles triangle.

This is 45 to 45.

So this is also 2.

2 to-- that's 90
degrees, 45, 45.

OK.

So the area of the shaded purple
triangle-- how much is that?

STUDENT: 2.

DR. MAGDALENA TODA: 2.

2 times 2 over 2.

2 times 3 equals 6.

I don't need calculus.

In this case, I
don't need calculus.

But when I have those
nasty curvilinear

z equals f of x, y, complicated
expressions, I have no choice.

I have to do the
double integral.

But in this case, even if
I didn't know how to do it,

I would still get the 6.

Yes, sir?

STUDENT: What if we
did that on the exam?

DR. MAGDALENA TODA:
Well, that's good.

I will then keep it in mind.

Yes.

It doesn't matter to me.

I have other colleagues who
really care about the method

and start complaining.

I don't care how you
get to the answer

as long as you got
the right answer.

Let me tell you my logic.

Suppose somebody hired you
thinking you're a good worker,

and you're smart and so on.

Would they care how you got to
the solution of the problem?

As long as the problem
was solved correctly, no.

And actually, the elementary
way is the fastest

because it's just 10 seconds.

You draw.

You imagine.

You know what it is.

So your boss will want you to
find the fastest way to provide

the correct solution.

He's not going to
care how you got that.

So no matter how
you do it, as long

as you've got the right
answer, I'm going to be happy.

I want to ask you to please
go to page 927 in the book

and read.

It's only one page.

That whole end section, 12.1.

It's called an informal
argument for Fubini's theorem.

Practically, it's a proof of
Fubini's theorem, page 927.

And then I'm going to go
ahead and start the homework

four, if you don't mind.

I'm going to go into WeBWork
and give you homework four.

And the first few
problems that you

are going to be
expected to solve

will be out of 12.1,
which is really easy.

I'll give you a
few minutes back.

And we go on with 12.2,
and it's very similar.

You're going to like that.

And then we'll go home or
wherever we need to go.

So you have a few
minutes of a break.

Pick up your extra credits.

I'll call the names.

Lily.

You got a lot of points.

And [INAUDIBLE].

And you have two separate ones.

Nathan.

Nathan?

Rachel Smith.

Austin.

Thank you.

Edgar.

[INAUDIBLE]

Aaron.

Andre.

Aaron.

Kasey.

Kasey came up with
a very good idea

that I will write
a review sample.

Did I promise that?

A review sample for the midterm.

And so I said yes.

Karen and Matthew.

Reagan.

Aaron.

When you submitted,
you submitted.

Yeah.

And [INAUDIBLE].

here.

And I'm done.

STUDENT: Did we
turn in [INAUDIBLE]?

DR. MAGDALENA TODA:
Yes, absolutely.

Now once we go over
12.2, you will say, oh,

but I understand
the Fubini theorem.

I didn't know whether
there's room for Fubini,

because once I cover the more
general case, which is in 12.2,

you are going to understand
Why Fubini-Tonelli

works for rectangles.

So if I think of a domain
that is of the following form,

in the x, y plane, I go x
is between and and b, right?

That's my favorite x.

So I take the pink
segment, and I

say, everything that
happens-- it's going

to happen on top of this world.

I have, let's say,
two functions.

To make my life easier, I'll
assume both of them [INAUDIBLE]

one bigger than the other.

But in case they are
not both positive,

I just need f to be bigger
than g for every point.

And the same argument
will function.

This is f, continuous positive.

Then g, continuous
positive but smaller

in values than this one.

Yes, sir?

STUDENT: [INAUDIBLE]
12.2 that we're starting?

DR. MAGDALENA TODA: 12.2.

And you are more organized
than I am, and I appreciate it.

So integration over a
non-rectangular domain.

And we call this a
type one because this

is what many books are using.

And this is that x is
between two fixed end points.

But y is between two
variable end points.

So what's going to happen to y?

y is going to take
values between the lower,

the bottom one, which is
g of x, and the upper one,

which is f of x.

So this is how we
define the domain that's

shaded by me with black
shades, vertical strips here.

This is the domain.

Now you really do
not need to prove

that double integral over
1 dA over-- let's call

the domain D-- is what?

Integral between f of x
minus g of x from a to b dx.

And you say, what?

Magdalena, what are
you trying to say?

OK.

Let's go back and
say, what if somebody

would have asked you the
same question in calculus 2?

Saying, guys I have a
question about the area

in the shaded strip,
vertical strip thing.

How are we going
to compute that?

And you would say,
oh, I have an idea.

I take the area under the graph
f, and I shade that in orange.

And I know what that is.

So you would say, I
know what that is.

That's going to be what?

Integral from a to be f of x dx.

Let's call that A1, right?

A1.

Then you go, minus the area
with-- I'm just going to shade

that, brown strips under g.

g of x dx.

And call that A2.

A1 minus A2.

We know both of these
formulas from where?

Calc 1 because that's where
you learned about the area

under the graph of a curve.

This is the area under
the graph of a curve f.

This is the area under
the graph of the curve g.

The black striped area
is their difference.

All right.

And so how much is that?

I'm sorry I put the wrong thing.

a, b.

That's going to be
integral from a to b.

Now you say, wait,
wait, wait a minute.

Based on what?

Based on some sort of
additivity property

of the integral of one
variable, which says integral

from a to b of f plus g.

You can have f plus, minus g.

It doesn't matter.

dx.

You have integral from a to b f
dx plus integral from a to b g

dx.

It doesn't matter what.

You can have a linear
combination of f and g.

Yes, Matthew?

MATTHEW: So this is
just for the domain?

So if you put it,
that would be down.

So there might be
another formula up here

that would be curved surface.

And this is the bottom,
so you're using integral

to find the base,
and then you're

going to plug that integral
into the other integral.

DR. MAGDALENA TODA: So I'm
just using the property that's

called linearity of
the simple integral,

meaning that if I have even
a linear combination like af

plus bg, then a-- I have not a.

Let me call it big A and
big B. Big A Af integral

of f plus big B integral of g.

You've learned that in Calc 2.

I'm doing this to apply it for
these areas that are subtracted

from one another.

If I were to add, as you
said, I would put something

on top of that.

And then it would be like
a superimposition onto it.

So I have integral from a to
b of f of x minus g of x dx.

And I claim that
this is the same

as double integral of the
1dA over the domain D.

How can you write
that differently?

I'll tell you how you
write that differently.

Integral from a to b of
integral from-- what's

the bottom value of Mr. Y?

So Mr. X knows what he's doing.

He goes all the way from a to b.

The bottom value of y is g of x.

You go from the bottom value
of y g of x to the upper value

f of x.

And then you here put 1 and dy.

Is this the same thing?

You say, OK, I know this one.

I know this one from calc 2.

But Magdalena, the one
you gave us is new.

It's new and not new, guys.

This is Fubini's
theorem but generalized

to something that depends on x.

So how do I do that?

Integral of 1dy.

That's what?

That's y measured between two
values that don't depend on y.

They depend only on x, g of x on
the bottom, f of x on the top.

So this is exactly the
integral from a to b.

In terms of the
round parentheses,

I put-- what is y between
f of x and g of x?

f of x minus g of x dx.

So it is exactly the
same thing from Calc 2

expressed as a double integral.

All right.

Now This is a type one
region that we talked about.

A type two region is a
similar region, practically.

What you have to keep
in mind is they're both

given here as examples.

But the technique is
absolutely the same.

If instead of
taking this picture,

I would take y to move
between fixed values,

like y has to be between
c and d-- this is my y.

These are the fixed values.

And then give me
some nice colors.

This curve and
that curve-- OK, I

have to rotate my head because
then this is going to be x.

This is going to be y.

And the blue thingy has
to be a function of y.

x is a function of y.

So how do I call that?

I have x or whatever
equals big F of y.

And here in the red one, I
have x equals big G of y.

And how am I going to
evaluate the striped area?

Of course striped because I
have again y is between c and d.

And what's moving is Mr. X.

And Mr. X refuses to
have fixed variables.

Now he goes, I move from
the bottom, which is G of y,

to the top, which is F of y.

How am I going to write
the double integral

over this domain of
1dA, where dA is dxdy.

Who's going to tell me?

Similarly, the same
reasoning as for this one.

I'm going to have the
integral from what to what

of integral from what to what?

Who comes first, dx or dy?

STUDENT: dx.

DR. MAGDALENA TODA:
dx, very good.

And dy at the end.

So y will be between
c and d, and x

is going to be between
G of y and F of y.

And here is y.

How can I rewrite this integral?

Very easily.

The integral from c to
d of the guy on top,

the blue guy, F of y, minus the
guy on the bottom, G of y, dy.

Some people call the
vertical stip method

compared to the horizontal
strip method, where

in this kind of
horizontal strip method,

you just have to view
x as a function of y

and rotate your head and apply
the same reasoning as before.

It's not a big deal.

You just need a little
bit of imagination,

and the result is the same.

An example that's
not too hard-- I

want to give you
several examples.

We have plenty of time.

Now it says, we have
a triangular region.

And that is enclosed by lines
y equals 0, y equals 2x,

and x equals 1.

Let's see what that means
and be able to draw it.

It's very important to be
able to draw in this chapter.

If you're not, just
learn how to draw,

and that will give
you lots of ideas

on how to solve the problems.

Chapter 12 is included
completely on the midterm.

So the midterm is
on the 2nd of April.

For the midterm, we have chapter
10, those three sections.

Then we have chapter
11 completely,

and then we have chapter 12
not completely, up to 12.6.

All right.

So what did I say?

I have a triangular region that
is obtained by intersecting

the following lines.

y equals 0, x equals
1, and y equals 2x.

Can I draw them and
see how they intersect?

It shouldn't be a big problem.

This is a line that
passes through the origin

and has slope 2.

So it should be
very easy to draw.

At 1, x equals 1, the y will
be 2 for this line of slope 2.

So I'll try to draw.

Does this look double to you?

So this is 2.

This is the point 1, 2.

And that's the line y equals 2x.

And that's the line y equals 0.

And that's the line x equals 1.

So can I shade this triangle?

Yeah, I can eventually,
depending on what they ask me.

What do they ask me?

Find the double
integral of x plus y dA

with respect to the area element
over T, T being the triangle.

So now I'm going to ask,
did they say by what method?

Unfortunately, they say,
do it by both methods.

That means both by x
intregration first and then

y integration and
the other way around.

So they ask you to change
the order of the integration

or do what?

Switch from vertical
strip method

to horizontal strip method.

You should get the same answer.

That's a typical
final exam problem.

When we test you, if
you are able to do this

through the vertical
strip or horizontal

strip and change the
order of integration.

If I do it with the
vertical strip method,

who comes first,
the dy or the dx?

Think a little bit.

Where do I put d--
Fubini [INAUDIBLE]

comes dy dx or dx dy?

STUDENT: dy.

PROFESSOR: dy dx.

So VSM.

You're going to laugh.

It's not written in the book.

It's like a childish name,
Vertical Strip Method,

meeting integration
with respect to y

and then with respect to x.

It helped my students
through the last decade

to remember about
the vertical strips.

And that's why I say something
that's not using the book, VSM.

Now, I have integral from-- so
who is Mr. X going from 0 to 1?

He's stable.

He's happy.

He's going between
two fixed values.

y goes between the
bottom line, which is 0.

We are lucky.

It's a really nice problem.

Going to y equals 2x.

So it's not hard at all.

And we have to integrate
the function x plus y.

It should be a piece of cake.

Let's do this together because
you've accumulated seniority

in this type of problem.

What do I put inside?

What's integral of x
plus y with respect to y?

Is it hard?

xy plus-- somebody tell me.

STUDENT: y squared.

PROFESSOR: y squared
over 2, between y

equals 0 on the bottom,
y equals 2x on top.

I have to be smart and
plug in the values y.

Otherwise, I'll never make it.

STUDENT: Professor?

PROFESSOR: Yes, sir?

STUDENT: Why did you take
2x as the final value

because you have a
specified triangle.

PROFESSOR: Because y
equals 2x is the expression

of the upper function.

The upper function is
the line y equals 2x.

They provided that.

So from the bottom function
to the upper function,

the vertical strips go
between two functions.

So when I plug in
here y equals 2x,

I have to pay attention
to my algebra.

If I forget the 2, it's all
over for me, zero points.

Well, not zero points,
but 10% credit.

I have no idea what I would
get, so I have to pay attention.

2x times x is 2x squared
plus 2x all squared-- guys,

keep an eye on me--
4x squared over 2.

I put the first value
in a pink parentheses,

and then I move on to
the line parentheses.

Evaluate it at 0.

That line is very lucky.

I get a 0 because y
equals 0 will give me 0.

What am I going to get here?

2x squared plus 2x squared.

Good.

What's 2x squared
plus 2x squared?

4x squared.

So a 4 goes out.

Kick him out.

Integral from 0
to 1 x squared dx.

Integral of x squared is?

Integral of x squared is?

STUDENT: x cubed over 3.

PROFESSOR: x cubed over 3.

And if you take it
between 1 and 0, you get?

STUDENT: 1.

PROFESSOR: 1/3.

1/3 times 4 is 4/3.

Suppose this is going to
happen on the midterm,

and I'm asking you to do it
reversing the integration

order.

Then you are going to check
your own work very beautifully

in the sense that
you say, well, now

I'm going to see if I made
a mistake in this one.

What do I do?

I erase the whole thing, and
instead of vertical strips,

I'm going to put
horizontal strips.

And you say, well, life is a
little bit harder in this case

because in this
case, I have to look

at y between fixed
values, y between 0 and 1.

So y is between 0 and 1--
0 and 2, fixed values.

And Mr. X says, I'm going
between two functions of y.

I don't know what those
functions of y are.

I'm puzzled.

You have to help
Mr. X know where

he's going because his life
right now is a little bit hard.

So what is the
function for the blue?

Now he's not blue anymore.

He's brown.

x equals 1.

So he knows what
he's going to be.

What is the x function
for the red line

that [INAUDIBLE] asked about?

STUDENT: y over 2.

PROFESSOR: x must be y over 2.

It's the same thing, but I have
to express x in terms of y.

So I erase and I say
x equals y over 2.

Same thing.

So x has to be between what and
what, the bottom and the top?

Well, I turn my head.

The top must be x equals 1,
and the bottom one is y over 2.

That's the bottom one,
the bottom value for x.

Now wish me luck because I
have to get the same thing.

So integral from 0 to 2 of
integral from y over 2 to 1.

Changing the order
of integration

doesn't change the
integrand, which is exactly

the same function, f of xy.

This is the f function.

Then what changes?

The order of integration.

So I go dx first,
dy next and stop.

I copy and paste the outer
ones, and I focus my attention

to the red parentheses
inside, which I'm

going to copy and paste here.

I'll have to do some
math very carefully.

So what do I have?

I have x plus y integrated
with respect to x.

If I rush, it's a bad thing.

STUDENT: So that
would be x squared.

PROFESSOR: x squared.

STUDENT: Over 2.

PROFESSOR: Over 2.

STUDENT: Plus xy.

PROFESSOR: Plus xy taken
between the following.

When x equals 1,
I have it on top.

When x equals y over 2,
I have it on the bottom.

OK.

This red thing, I'm a
little bit too lazy.

I'll copy and paste
it separately.

For the upper part, it's
really easy to compute.

What do I get?

When x is 1, 1/2, 1/2
plus when x is 1, y.

Minus integral of--
when x is y over 2,

I get y squared over
4 up here over 2.

So I should get y
squared over 8 plus--

I've got an x equals y over 2.

What do I get?

y squared over 2.

Is this hard?

It's very easy to make an
algebra mistake on such

a problem, unfortunately.

I have y plus 1/2 plus what?

What is 1/2 plus 1/8?

STUDENT: 5/8.

PROFESSOR: 5 over 8
with a minus y squared.

So hopefully I did this right.

Now I'll go, OK, integral from
0 to 2 of all of this animal, y

plus 1/2 minus 5
over 8, y squared.

What happens if I don't
get the right answer?

Then I go back and
check my work because I

know I'm supposed to get 4/3.

That was easy.

So what is integral of this
sausage, whatever it is?

y squared over 2 plus y
over 2 minus 5 over 8--

oh my god-- 5 over 8, y
cubed over 3, between 2 up

and 0 down.

When I have 0 down,
I plug y equals 0.

It's a piece of cake.

It's 0.

So what matters is
what I get when I plug

in the value 2 instead of y.

So what do I get?

4 over 2 is 2, plus 2 over 2
is 1, minus 2 cubed, thank god.

That's 8.

8 simplifies with 8 minus 5/3.

So I got 9/3 minus 5/3,
and I did it carefully.

I did a good job.

I got the same thing, 4/3.

So no matter which
method, the vertical strip

or the horizontal strip
method, I get the same thing.

And of course, you'll
always get the same answer

because this is what the Fubini
theorem extended to this case

is telling you.

It doesn't matter the
order of integration.

I would advise you to go
through the theory in the book.

They teach you more about
area and volume on page 934.

I'd like you to read that.

And let's see what I want to do.

Which one shall I do?

There are a few examples
that are worth it.

I'll pick the one that gives
people the most trouble.

How about that?

I take the few examples that
give people the most trouble.

One example that popped up on
almost each and every final

in the past 13 years
that involves changing

the order of integration.

So example problem on changing
the order of integration.

A very tricky, smart
problem is the following.

Evaluate integral from 0
to 1, integral from x to 1,

e to the y squared dy dx.

I don't know if you've
seen anything like that

in AP Calculus or Calc 2.

Maybe you have, in which case
your professor probably told

you that this is nasty.

You say, in what
sense is it nasty?

There is no expressible
anti-derivative.

So this cannot be expressed in
terms of elementary functions

explicitly.

It's not that there
is no anti-derivative.

There is an anti-derivative--
a whole family, actually--

but you cannot express them in
terms of elementary functions.

And actually, most functions
are not so bad in real world,

in real life.

Now, could you compute, for
example, integral from 1 to 3

of e to the t squared dt?

Yes.

How do you do that?

With a calculator.

And what if you don't have one?

You go to the lab over there.

There is MATLAB.

MATLAB will compute it for you.

How does MATLAB know
how to compute it

if there is no way to
express the anti-derivative

and take the value of the
anti-derivative between b

and a, like in the fundamental
theorem of calculus?

Well, the calculator or the
computer program is smart.

He uses numerical analysis
to approximate this type

of integral.

So he's fooling you.

He's just playing smarty pants.

He's smarter than
you at this point.

OK.

So you cannot do this by hand,
so this order of integration is

fruitless.

And there are people who
tried to do this on the final.

Of course, they
didn't get anywhere

because they couldn't
integrate it.

The whole idea of this one
is to-- some professors

are so mean they
don't even tell you,

hint, change the
order of integration

because it may work
the other way around.

They just give it to you, and
then people can spend an hour

and they don't get anywhere.

If you want to be mean to a
student, that's what you do.

So I will tell
you that one needs

to change the order of
integration for this.

This is the function.

We keep the function, but let's
see what happens if you draw.

The domain will be
x between 0 and 1.

This is your x value.

y will be between x and 1.

So it's like you have a square.

y equals x is your
diagonal of the square.

And you go from--
more colors, please.

You go from y equals x on the
bottom and y equals 1 on top.

And so the domain is
this beautiful triangle

that I make all in line
with vertical strips.

This is what it means,
vertical strips.

But if I do horizontal strips, I
have to change the color, blue.

And for horizontal
strips, I'm going

to have a different problem.

Integral, integral dx dy.

And I just hope to god
that what I'm going to get

is doable because if
not, then I'm in trouble.

So help me on this one.

If y is between what and what?

It's a square.

It's a square, so this will
be the same, 0 to 1, right?

STUDENT: Yep.

PROFESSOR: But Mr. X?

How about Mr. X?

STUDENT: And then it
will be between 1 and y.

PROFESSOR: Between--
Mr. X is this guy.

And he doesn't go between 1.

He goes between the
sea level, which is

x equals 0, to x equals what?

STUDENT: [INAUDIBLE].

PROFESSOR: Right?

So from x equals 0
through x equals y.

And you have the same individual
e to the y squared that before

went on your nerves.

Now he's not so bad, actually.

Why is he not so bad?

Look what happens in
the first parentheses.

This is so beautiful
that it's something

you didn't even hope for.

So we copy and paste
it from 0 to 1 dy.

These guys stay
outside and they wait.

Inside, it's our
business what we do.

So Mr. X is independent
from e to the y squared.

So e to the y squared pulls out.

He's a constant.

And you have integral
of 1 dx between 0 and y.

How much is that?

1.

x between x equals
0 and x equals y.

So it's y.

So I'm being serious.

So I should have said y.

Now, if your professor would
have given you, in Calc 2,

this, how would
you have done it?

STUDENT: U-substitution.

PROFESSOR: U-substitution.

Excellent.

What kind of
u-substitution [INAUDIBLE]?

STUDENT: y squared equals u.

PROFESSOR: y squared
equals u, du equals 2y dy.

So y dy together.

They stick together.

They stick together.

They attract each
other as magnets.

So y dy is going to be
1/2 du-- 1/2 pulls out--

integral e to the u du.

Attention.

When y is moving
between 0 and 1,

u is moving also
between 0 and 1.

So it really should
be a piece of cake.

Are you guys with me?

Do you understand what I did?

Do you understand the words
coming out of my mouth?

It's easy.

Good.

So what is integral
of e to the u du?

e to the u between
1 up and 0 down.

So e to the u de to the 1
minus e to the 0 over 2.

That is e minus 1 over 2.

I could not have solved this
if I tried it by integration

with y first and then x.

The only way I
could have done this

is by changing the
order of integration.

So how many times have I seen
this in the past 12 years

on the final?

At least six times.

It's a problem that
could be a little bit

hard if the student has
never seen it before

and doesn't know what to
do [? at that point. ?]

Let's do a few more
in the same category.

STUDENT: Professor?

PROFESSOR: Yes?

STUDENT: Where did this shape--
where did this graph come from?

Were we just saying
it was with the same--

PROFESSOR: OK.

I read it from here.

So this and that are the key.

This is telling me x is between
0 and 1, and at the same,

time y is between x and 1.

And when I read this
information on the graph,

I say, well, x is
between 0 and 1.

Mr. Y has the freedom to go
between the first bisector,

which is that, and the
cap, his cap, y equals 1.

So that's how I got
to the line strips.

And from the line strips, I said
that I need horizontal strips.

So I changed the
color and I said

the blue strips go between x.

x will be x equals
0 and x equals y.

And then y between 0
and 1, just the same.

It's a little bit tricky.

That's why I want to do one or
two more problems like that,

because I know that I remember
20-something years ago,

I myself needed a little
bit of time understanding

the meaning of reversing
the order of integration.

STUDENT: Does it matter
which way you put it?

PROFESSOR: In this case, it's
important that you do reverse.

But in general, it's
doable both ways.

I mean, in the other problems
I'm going to give you today,

you should be able
to do either way.

So I'm looking for a problem
that you could eventually

do another one.

We don't have so many.

I'm going to go ahead and
look into the homework.

Yeah.

So it says, you
have this integral,

the integral from 0
to 4 of the integral

from x squared to 4y dy dx.

Draw, compute, and also
compute with reversing

the order of integration
to check your work.

When I say that,
it sounds horrible.

But in reality, the
more you work on

that one, the more familiar
you're going to feel.

So what did I just say?

Problem number 26.

You have integral
from 0 to 4, integral

from x squared to 4x dy dx.

Interpret geometrically,
whatever that means,

and then compute the
integral in two ways,

with this given order
integration, which

is what kind of strips, guys?

Vertical strips.

Or reversing the
order of integration.

And check that the answer is the
same just to check your work.

STUDENT: So first--

PROFESSOR: First you draw.

First you draw because
if you don't draw,

you don't understand what
the problem is about.

And you say, wait a minute.

But couldn't I go ahead
and do it without drawing?

Yeah, but you're not
going to get too far.

So let's see what kind
of problem you have.

y and x.

y equals x squared is a what?

It's a pa--

STUDENT: Parabola.

PROFESSOR: Parabola.

And this parabola should
be nice and sassy.

Is it fat enough?

I think it is.

And the other one will
be 4x, y equals 4x.

What does that look like?

It looks like a line passing
through the origin that

has slope 4, so the
slope is really high.

STUDENT: Just straight.

PROFESSOR: y equals 4x
versus y equals x squared.

Now, do they meet?

STUDENT: Yes.

PROFESSOR: Yes.

Exactly where do they meet?

Exactly here.

STUDENT: 4.

PROFESSOR: So 4x equals x
squared, where do they meet?

They meet at-- it has
two possible roots.

One is x equals
0, which is here,

and one is x equals
4, which is here.

So really, my graph looks
just the way it should look,

only my parabola is
a little bit too fat.

This is the point of
coordinates 4 and 16.

Are you guys with me?

And Mr. X is moving
between 0 and 4.

This is the maximum
level x can get.

And where he stops here
at 4, a miracle happens.

The two curves intersect each
other exactly at that point.

So this looks like a
leaf, a slice of orange.

Oh my god.

I don't know.

I'm already hungry so I cannot
wait to get out of here.

I bet you're hungry as well.

Let's do this problem
both ways and then go

home or to have
something to eat.

How are you going to advise
me to solve it first?

It's already set
up to be solved.

So it's vertical strips.

And I will say
integral from 0 to 4,

copy and paste the outer part.

Take the inner part, and do the
inner part because it's easy.

And if it's easy, you tell
me how I'm going to do it.

Integral of 1 dy is y.

y measured at 4x is 4x,
and y measured at x squared

is x squared.

Oh thank god.

This is so beautiful
and so easy.

Let's integrate again.

4 x squared over 2 times x cubed
over 3 between x equals 0 down

and x equals 4 up.

What do I get?

I get 4 cubed over 2
minus 4 cubed over 3.

This 4 cubed is an obsession.

Kick him out.

1/2 minus 1/3.

How much is 1/2 minus 1/3?

My son knows that.

STUDENT: 1/6.

PROFESSOR: OK.

1/6, yes.

So we simply take it.

We can leave it like that.

If you leave it like that on
the exam, I don't mind at all.

But you could always put
64 over 6 and simplify it.

Are you guys with me?

You can simplify
it and get what?

32 over 3.

Don't give me decimals.

I'm not impressed.

You're not supposed
to use the calculator.

You are supposed to leave
this is exact fraction

form like that, irreducible.

Let's do it the
other way around,

and that will be the
last thing we do.

The other way around means
I'll take another color.

I'll do the horizontal stripes.

And I will have to rewrite
the meaning of these two

branches of functions with
x expressed in terms of y.

That's the only thing
I need to do, right?

So what is this?

If y is x squared, what is x?

STUDENT: Root y.

PROFESSOR: The inverse
function. x will be root of y.

You said very well.

So I have to write.

In [INAUDIBLE], I
have what I need

to have for the line
horizontal strip method.

And then for the other one,
x is going to be y over 4.

So what do I do?

So integral, integral, a
1 that was here hidden,

but I'll put it because
that's the integral.

And then I go dx dy.

All I have to care about is the
endpoints of the integration.

Now, pay attention a little
bit because Mr. Y is not

between 0 and 4.

I had very good
students under stress

in the final putting 0 and 4.

Don't do that.

So pay attention to the
limits of integration.

What are the limits?

0 and--

STUDENT: 16.

PROFESSOR: 16.

Very good.

And x will be between root
y-- well, which one is on top?

Which one is on the bottom?

Because if I move my head,
I'll say that's on top

and that's on the bottom.

STUDENT: The right side
is always on the top.

PROFESSOR: So the one that
looks higher is this one.

This is more than
that in this frame.

So square of y is on top and
y over 4 is on the bottom.

I should get the same answer.

If I don't, then I'm in trouble.

So what do I get?

Integral from 0 to 16.

Tonight, when I
go home, I'm going

to cook up the homework
for 12.1 and 12.1 at least.

I'll put some problems
similar to that

because I want to emphasize
the same type of problem

in at least two or three
applications for the homework

for the midterm.

And maybe one like that will
be on the final as well.

It's very important for
you to understand how,

with this kind of
domain, you reverse

the order of integration.

Who's helping me here?

Root y.

What is root y
when-- y to the 1/2.

I need to integrate.

So I need minus y over 4 and dy.

Can you help me integrate?

STUDENT: [INAUDIBLE].

PROFESSOR: 2/3 y
to the 3/2 minus--

STUDENT: y squared.

PROFESSOR: y squared
over 8, y equals 0

on the bottom, piece of cake.

That will give me 0.

I'm so happy.

And y equals 16 on top.

So for 16, I have 2/3.

And who's telling me what else?

STUDENT: 64.

PROFESSOR: 64.

4 cubed.

I can leave it 4 cubed if I want
to minus another-- well here,

I have to pay attention.

So I have 16 here.

I got square root of
16, which is 4, cubed.

Here, I put minus 4
squared, which was there.

How do you want me to
do this simplification?

STUDENT: [INAUDIBLE].

PROFESSOR: I can
do 4 to the fourth.

Are you guys with me?

I can put, like you
prefer, 16 squared over 8.

Is it the same answer?

I don't know.

Let's see.

This is really 4 to the 4,
so I have 4 times 4 cubed.

4 cubed gets out and
I have 2/3 minus 1/2.

And how much is that?

Again 1/6.

Are you guys with me?

1/6.

So again, I get 4 cubed
over 6, so I'm done.

4 cubed over 6 equals 32 over 3.

I am happy that
I checked my work

through two different methods.

I got the same answer.

Now, let me tell you something.

There were also times
when on the midterm

or on the final, due to
lack of time and everything,

we put the following
kind of problem.

Without solving this integral--
without solving-- indicate

the corresponding integral
with the order reversed.

So all you have to
do-- don't do that.

Just from here,
write this and stop.

Don't waste your time.

If you do the whole thing,
it's going to take you

10 minutes, 15 minutes.

If you do just reversing
the order of integration,

I don't know what it takes, a
minute and a half, two minutes.

So in order to save
time, at times,

we gave you just don't
solve the problem. reverse

the order of integration.

One last one.

One last one.

But I don't want to finish it.

I want to give you
the answer at home,

or maybe you can finish it.

It should be shorter.

You have a circular parabola,
but only the first quadrant.

So x is positive.

STUDENT: Question.

PROFESSOR: I don't know.

I have to find it.

Find the volume.

Example 4, page 934.

Find the volume
of the solid bound

in the above-- this is a
little tricky-- by the plane z

equals y and below
in the xy plane

by the part of the disk
in the first quadrant.

So z equals y means this
is your f of x and y.

So they gave it to you.

But then they say, but
also, in the xy plane,

you have to have the part of
the disk in the first quadrant.

This is not so easy.

They draw it for you to
make your life easier.

The first quadrant is that.

How do you write the unit
circle, x squared equals 1,

x squared plus y squared
less than or equal to 1,

and x and y are both positive.

This is the first quadrant.

How do you compute?

So they say compute the
volume, and I say just

set up the volume.

Forget about computing it.

I could put it in the
midterm just like that.

Set up an integral
without solving it

that indicates the volume
under z equals f of xy-- that's

the geography of z-- and above
a certain domain in plane,

above D in plane.

So you have, OK, what
this should teach you?

Should teach you that double
integral over d f of xy da

can be solved.

Do I ask to be solved?

No.

Why?

Because you can finish
it later, finish at home.

Or maybe, I don't even want
you to compute on the final.

So how do we do that?

f is y.

Would I be able to choose
whichever order integration I

want?

It shouldn't matter which order.

It should be more
or less the same.

What if I do dy dx?

Then I have to do the Fubini.

But it's not a
rectangular domain.

Aha.

So Magdalena, be a
little bit careful

because this is going to
be two finite numbers,

but these are functions.

STUDENT: It will
be an x function.

PROFESSOR: So the x
is between 0 and 1,

and that's going to be z.

You do vertical strips.

That's a piece of cake.

But if you do the
vertical strips,

you have to pay attention to
the endpoints for x and y,

and one is easy.

Which one is trivial?

STUDENT: Zero.

PROFESSOR: The bottom one, zero.

The one that's nontrivial
is the upper one.

STUDENT: There will be 1 minus--

STUDENT: Square root
of 1 minus y squared.

PROFESSOR: Very good.

Square root of 1
minus y squared.

So if I were to go one more step
further without solving this,

I'm going to ask you, could
this be solved by hand?

Well, so you have
it in the book--

STUDENT: Professor, should be
a [INAUDIBLE] minus x squared?

PROFESSOR: Oh yeah.

1 minus x squared.

Excuse me.

Didn't I write it?

Yeah, here I should have written
y equals square root of 1

minus x squared.

So when you do it-- thank you
so much-- you go integrate,

and you have y squared over 2.

And you evaluate
between y equals 0

and y equals square
root 1 minus x squared,

and then you do the [INAUDIBLE].

In the book, they
do it differently.

They do it with respect to
dx and dy and integrate.

But it doesn't
matter how you do it.

You should get the same answer.

All right?

[INAUDIBLE]?

STUDENT: [INAUDIBLE]
in that way,

doesn't the square root work out
better because there's already

a y there?

PROFESSOR: In the other case--

STUDENT: Doing dy dx.

PROFESSOR: Yeah,
in the other way,

it works a little
bit differently.

You can do
u-substitution, I think.

So if you do it the other
way, it will be what?

Integral from 0 to
1, integral form 0

to square root of 1
minus y squared, y dx dy.

And what do you do in this case?

You have integral from 0 to 1.

Integral of y dx is going
to be y is a constant.

x between the two values will
be simply 1 minus y squared dy.

So you're right.

Matthew saw that,
because he's a prophet,

and he could see
two steps ahead.

This is very nice
what you observed.

What do you do?

You take a u-substitution
when you go home.

You get u equals
1 minus y squared.

du will be minus 2y
dy, and you go on.

So in the book, we got 1/3.

If you continue
with this method,

I think it's the same answer.

STUDENT: Yeah.

I got 1/3.

PROFESSOR: You got 1/3.

So sounds good.

We will stop here.

You will get homework.

How long should I
leave that homework on?

Because I'm thinking maybe
another month, but please

don't procrastinate.

So let's say until
the end of March.

And keep in mind
that we have included

one week of spring
break here, which you

can do whatever you want with.

Some of you may be in Florida
swimming and working on a tan,

and not working on homework.

So no matter how, plan ahead.

Plan ahead and you will do well.

31st of March for
the whole chapter.