TTU Math2450 Calculus3 Sec 12.5

Triple Integrals

MAGDALENA TODA:
According to my watch,

we are right on time to start.

I may be one minute
early, or something.

Do you have questions out of the
material we covered last time?

What I'm planning on
doing-- let me tell you

what I'm planning to do.

I will cover triple
integrals today.

And this way, you
would have accumulated

enough to deal with most of
the problems in homework four.

You have mastered the
double integration by now,

in all sorts of coordinates,
which is a good thing.

Triple integrals
are your friend.

If you have understood
the double integration,

you will have no
problem understanding

the triple integrals.

The idea is the same.

You look at different
domains, and then you

realize that there are
Fubini-Tunelli type of results.

I'm going to present
one right now.

And there are also regions
of a certain type, that

can be treated
differentially, and then you

have cases in which reversing
the order of integration

for those triple integrals
is going to help you a lot.

OK, 12.5 is the name of the
section, triple integrals.

So what should you imagine?

You should imagine
that somebody gives you

a function of three variables.

Let's call that-- it doesn't
often have a name as a letter,

but let's call it w.

Being a function of three
coordinates, x, y, and z,

where x, y, z is in R3.

And we have some assumptions
about the domain D

that you are working
over, and you

have D as a closed-bounded
domain in R3.

Examples that you're going
to do use frequently.

Frequently used.

A sphere, a ball,
actually, because in here,

if a sphere is together with
a shell, it is the ball.

Then you have some types of
polyhedra in r3 of all types.

And by that, I mean
the classical polyhedra

whose sides are just polygons.

But you will also have some
curvilinear polyhedra, as well.

What do I mean?

I mean, we've seen that already.

For example, somebody
give you a graph

of a function, g of x and y, a
continuous function, and says,

OK, can you estimate the
volume under the graph?

Right?

And until now, we treated
this volume under the graph

as double integral of g of x, y,
continuous function over d, a,

where d, a was dx dy.

And we said double integral
over the projected domain

in the plane.

That's what I have.

But can I treat it, this
volume, can I treat it

as a triple integral?

This is the question,
and answer is--

so can I make three snakes?

The answer is yes.

And the way I'm
going to define that

would be a triple
integral over a 3D domain.

Let's call it curvilinear d in
r3, which is the volume-- which

is the body under the graph
of this positive function,

and above the projected
area in plane.

So it's going to be a
cylinder in this case.

And I'll put here 1 dv.

And dv is a mysterious element.

That's the volume element.

And I will talk a little
bit about it right now.

So what can you imagine?

They give you a way to
look at it in the book.

I mean, we give you a way
to look at it in the book.

It's not very thorough
in explanations,

but it certainly gives you the
general idea of what you want,

what you need.

So somebody gives you a potato.

It doesn't have to
be this cylinder.

It's something beautiful, a
body inside a compact surface.

Let's say there are
no self-intersections.

You have some compact surface,
like a sphere or a polyhedron,

assume it's simply
connected, and it doesn't

have any self-intersections.

So a beautiful
potato that's smooth.

If you imagine a potato
that has singularities,

like most potatoes have
singularities, boo-boos,

and cuts, so that's bad.

So think about some
regular surface

that's closed, no
self-intersections,

and that is a potato
that is [INAUDIBLE].

Oh let's call it--
p for potato, no,

because I have got to
use p for the partition.

So let me call it
D from 3D domain,

because it's a three-dimensional
domain, enclosed

by a curve, enclosed
by a compact surface.

So think potato.

What do we do in
terms of partitions?

Those pixels were pixels for
the 2D world in flat line.

But now, we don't
have pixels anymore.

Yes we do.

I was watching lots of
sci-fi, and the holograms

have the
three-dimensional pixels.

I'm going to try and
make a partition.

It's going to be a hard way
to partition this potato.

But you have to imagine you
have a rectangular partition,

so every little pixel
will be a-- is not cube.

It has to be a little
tiny parallelepiped.

So a 3D pixel, let me
put pixel in quotes,

because this is
kind of the idea.

Well have what
kind of dimensions?

We'll have three
dimensions, right?

Three dimensions, a delta xk,
a delta yk, and a delta zk

for the pixel number k.

That's pixel number k.

We have to number them,
see how many they are.

Where k is from 1 to n, and
is the total number of 3D

pixels that I'm covering
the whole thing.

So don't think graphing
paper, anymore,

because that's outdated.

Don't even think of 2D image.

Think of some hologram,
where you cover everything

with tiny, tiny,
tiny 3D pixels so

that in the limit, when you pass
to the limit, with respect to n

and going to infinity,
the discrete image,

you're going to have something
like a diamond shaped thingy,

will convert to
the smooth potato.

So as n goes to
infinity, that surface

made of tiny, tiny squares
will convert to the data.

So how do you actually find
a triple snake integral f

of x, y, z over the
domain D, dx dy dz.

OK, this theoretically
should be what?

Think pixels.

Limit as n goes to infinity
of the sum of the--

what do I need to do?

Think the whole
partition into pixels.

How many pixels? n pixels
total is called a p.

Script p.

And the normal p will be the
highest diameter of-- highest

diameter among all pixels.

And you're going to
say, Oh, what the heck?

I don't understand it.

I have these three-dimensional
cubes, or three-dimensional

barely by p that get tinier,
and tinier, and tinier.

What in the world is going to
be a diameter of such a pixel?

Well, you have to
take this pixel

and magnify it so we can look
at it a little bit better.

What do we mean by
diameter of this pixel?

Let's call this pixel k.

The maximum of all the distances
you can compute between two

arbitrary points inside.

Between two arbitrary
points inside,

you have many [INAUDIBLE].

So the maximum of the
distance between points,

let's call them r and
q inside the pixel.

So if it were for me
to ask you to find

that diameter in this
case, what would that be?

STUDENT: It's the diagonal.

MAGDALENA TODA: It's the
diagonal between this corner

and the opposite corner.

So this would be the highest
distance inside this pixel.

If it's a cube-- you see
that I wanted a cube.

If it's a parallelepiped,
it's the same idea.

So I have that opposite
corner distance kind of thing.

OK.

So I know what I want.

I want n to go to infinity.

That means I'm going to
have the p going to 0.

The length of the
highest diameter

will go shrinking to 0.

And then I'm going to say
here, what do I have inside?

F of some intermediate point.

In every pixel, I take a point.

Another pixel, another
point, and so on.

So how many such
points do I have?

n, because it's the
number of pixels.

So inside, let's call
this as pixel p k.

What is the little point that
I took out of the [INAUDIBLE]

inside the cube, or
inside the pixel?

Let's call that mister x k
star, y k star, x k star.

Why do we put a star?

Because he is a star.

He wants to be number one
in these little domains,

and says I'm a star.

So we take that intermediate
point, x k star,

y k star, x k star.

Then we have this function
multiplied by the delta v k.

somebody tell me what
this delta v k will mean,

because ir really looks weird.

And then k will be from 1 to n.

So what do you do?

You sum up all these
weighted volumes.

This is a weight.

So this is a volume.

All these weighted volumes.

We sum them up for all
the pixels k from 1 to n.

We are going to get
something like this cover

with tiny parallelepipeds
in the limit,

as the partitions' [? norm ?]
go to 0, or the number of pixels

goes to infinity.

This discrete
surface will converge

to the beautiful smooth
potato, and give you

a perfect linear image.

Actually, if we saw a
hologram, this is what it is.

Our eyes actually
see a bunch of I

tiny-- many, many, many,
many, millions of pixels

that are cubes in 3D.

But it's an optical illusion.

We see, OK, it's a curvilinear,
it's a smooth body of a person.

It's not smooth at all.

If you get closer and
closer to that diagram

and put your eye
glasses on, you are

going to see, oh, this
is not a real person.

It's made of pixels
that are all cubes.

just the same, you see your
digital image of your picture

on Facebook, whatever it is.

If you would be able
to be enlarge it,

you would see the pixels, being
little tiny squares there.

The graphical
imaging has improved.

The quality of our digital
imaging has improved a lot.

But of course, 20 years ago,
when you weren't even born,

we could still see the pixels
in the photographic images

in a digital camera.

Those tiny first cameras,
what was that, '98?

STUDENT: Kodak.

MAGDALENA TODA: Like AOL
cameras that were so cheap.

The cheaper the camera,
the worse the resolution.

I remember some resolutions
like 400 by 600.

STUDENT: Black and white.

MAGDALENA TODA: Not
black and white.

Black and white
would have been neat.

But really nasty in the sense
that you had the feeling

that the colors
were not even-- they

were blending into each
other, because the resolution

was so small.

So it was not at all
pleasing to the eye.

What was good is that
any kind of defects you

would have, something
like a pimple

could not be seen in that,
because the resolution was

so slow that you couldn't
see the boo-boos,

the pimples, the defects
of a face or something.

Now, you can see.

With the digital cameras
we have now, we can do,

of course, Adobe
Photoshop, and all of us

will look great if we
photoshop our pictures.

OK.

So this is what it
is in the limit.

But in reality, in
the everyday reality,

you cannot take Riemann
sums like that--

this is a Riemann
approximating sum--

and then cast to the limit, and
get ideal curvilinear domains.

No.

You don't do that.

You have to deal
with the equivalent

of the fundamental
theorem of calculus

from Calc I, which is called the
Fubini-Tonelli type of theorem

in Calc III.

So say it again.

So the Fubini-Tunelli
theorem that you

learned for double
integrals over a rectangle

can be generalized to the
Fubini-Tonelli theorem

over a parallelepiped.

And it's the same
thing, practically,

as applying the fundamental
theorem of calculus in Calc I.

So somebody says, well, let me
start with a simple example.

I give you a-- you
will say, Magdalena,

you are offending us.

This is way too easy.

What do you think, that we
cannot understand the concept?

I'll just try to start with
the simplest possible example

that I think of.

So x is between a and b.

In my case, they will
be positive numbers,

because I want everything
to be in the first octant.

First octant means x positive,
y positive, and z positive

all together.

To make my life easier,
I take that example,

and I say, I know the
numbers for x, y, z.

I would like you to
compute two integrals.

One would be the
volume of this object.

Let's call it body.

It's not a dead body,
it's just body in 3D.

The volume of the
body, we say it

like a mathematician, V of B.
What is that by definition?

Who's going to tell me?

Triple snake.

Don't say triple
snake to other people,

because other professors
are more orthodox than me.

They will laugh-- they
will not joke about it.

So triple integral over the
body of-- to get the volume,

the weight must be 1.

f integral must be 1, and
then you have exactly dV.

How can I convince
you what we have here,

in terms of Fubini-Tonelli?

It's really beautiful.

B is going to be a, b segment
cross product, c, d segment,

cross product.

What is the altitude?

E, f, e to f.

Interval e to f
means the height.

So length, width, height.

This is the box, or carry-on,
or USPS parcel, or whatever

box you want to measure.

So how am I going to set up
the Fubini-Tonelli integral?

a to b, c to d, e to f, 1.

And now, who counts first?

dz, dy, dx.

So it is like the equivalent
of the vertical strip

thingy in double corners,
double integrals.

Yes, sir?

STUDENT: Professor,
why did you use 1 dV?

Why 1?

MAGDALENA TODA: OK.

You'll see in a second.

This is the same
thing we do for areas.

So when you compute an
area-- very good question.

If you use 1 here, and
you put delta [? ak ?]

that is the graphing paper area.

It's going to be
all the tiny areas,

summed up, sum of all
the delta [? ak ?] which

means this little pixel,
plus this little pixel,

plus this little pixel,
plus this little pixel,

plus 1,000 pixels all together
will cover up the area.

If you have the
volume of a potato,

a body that is alive,
but shouldn't move.

OK, it should stay in one place.

Then, to compute the
volume of the potato,

you have to say, the
potato, the smooth potato,

is the limit of the sum of
all the tiny cubes of potato,

if you cut the potato in many
cubes, like you cut cheese.

They got a bunch of cheddar
cheese into small cubes,

and they feed us with crackers
and wine-- OK, no comments.

So you have delta vk, you
have 1,000 little cubes,

tiny, tiny, tiny,
like that Lego.

OK, forget about the cheese.

The cheese cubes
are way too big.

So imagine Legos that
are really performing

with millions of little pieces.

Have you seen the exhibit, Lego
exhibit with almost invisible

Legos at the Civic Center?

They have that art festival.

How many of you go
to the art festival?

Is it every April?

Something like that.

So imagine those little
tiny Legos, but being cubes

and put together.

This is what it is.

So f Vy.

Now, can we verify
the volume of a box?

It's very easy.

What do we do?

Well, first of all, I
would do it in a slow way,

and you are going to
shout at me, I know.

But I'll tell you why
you need to bear with me.

So integral of 1 dz goes first.

That's z between f and e.

So it's f minus e, am I right?

You say, duh, that's
to easy for me.

I'm know it's too
easy for me, but I'm

going somewhere with it.

dy dx.

The one inside, f minus e
is a constant, pulls out,

completely out of the product.

And then I have integral from
a to b of-- what is that left?

1 dy, y between d and c.

So d minus c, right?

d minus c dx.

And so on and so
forth, until I get it.

If minus c times d
minus c times b minus a,

and goodbye, because this
is the volume of the box.

It's the height.

This is the height.

No, excuse me, guys.

The height is-- this
one is the height.

This is the width, and this is
the length, whatever you want.

All right.

How could I have done it if
I were a little bit smarter?

STUDENT: You could have just
put it in three integrals.

MAGDALENA TODA:
Right Hey, I have

a theorem, just like
before, which says

three integrals in a product.

This is what Matt
immediately remembered.

We had two integrals
in a product last time.

So what have we proved
in double integrals

remains valid in
triple integrals

if we have something like that.

So I'm going the same theorem.

It's in the book.

We have a proof.

So you have integral from
a to b, c to d, e to f.

And then, some guys that you
like, f of x, times g of y,

times h of z.

Functions of x, y, z,
separated variables.

So f, a function of
x only, g a function

of y only, h a
function of z only.

This is the complicated case.

And then I have

STUDENT: dz, dy, dx.

MAGDALENA TODA: dz, dy, dx.

Excellent.

Thanks for whispering,
because I was a little bit

confused for a second.

So, just as Matt said,
go ahead and observe

that you can treat them one
at a time like you did here,

and integrate one at a time, and
integrate again, and pull out

a constant, integrate
again, pull out a constant.

But practically this is
exactly the same as integral

from a to b of f of x
alone, dx, times integral

from c to d, g of y alone,
dy, and times integral from e

to f of h of z, dz, and close.

So you've seen the version
of the double integral,

and this is the same result
for triple integrals.

And it's practically--
what is the proof?

You just pull out one
at a time, so the proof

is that you start working and
say, mister z counts here,

and he's the only
one that counts.

These guys get out for a
walk one at a time outside

of the first integral inside.

And then, integral
of h of z, dz,

over the corresponding domain,
will be just a constant, c1,

that pulls out.

And that is that--
c1 that pulls out.

Ans since you pull them out
in this product one at a time,

that's what you get.

I'm not going to give you this
as an exercise in the midterm

with a proof, but this is
one of the first exercises

I had as a freshman
in my multi--

I took it as a freshman,
as multivariable calculus.

And it was a pop quiz.

My professor just came
one day, and said, guys,

you have to try to do this
[? before ?] by yourself.

And some of us did,
some of us didn't.

To me, it really
looked very easy.

I was very happy to prove it,
in an elementary way, of course.

OK.

So how hard is it
to generalize, to go

to non-rectangular domains?

Of course it's a pain.

It's really a pain,
like it was before.

But you will be able to
figure out what's going on.

In most cases,
you're going to have

a domain that's really
not bad, a domain that

has x between fixed values.

For example y between
your favorite guys,

something like f of x and
g of x, top and bottom.

That's what you had
for double integral.

Well, in addition,
in this case, you

will have z between--
let's make this guy

big F and big G,
other functions.

This is going to be
a function of x, y.

This is going to be
a function of x, y,

and that's the
upper and the lower.

And find the triple integral
of, let's say 1 over d dV

will be a volume of the potato.

Now, I'm sick of potatoes,
because they're not

my favorite food.

Let me imagine I'm making a
tetrahedron, a lot of cheese.

I'm going to draw this same
tetrahedron from last time.

So what did we do last time?

We took a plane
that was beautiful,

and we said let's
cut with that plane.

This is the plane we are
cutting the cheese with.

It's a knife.

x plus y plus z equals 1.

Imagine that there's
an infinite knife that

comes into the frame.

Everything is cheese.

The space, the universe is
covered in solid cheese.

So the whole thing,
the Euclidean space

is covered in cheddar cheese.

That's all there.

From everywhere, you
come with this knife,

and you cut along
this plane-- hi

let's call this
[? high plane. ?]

And then you cut the x
plane along the x, y plane,

y, z plane and x, z plane.

What are these called?

Planes of coordinates.

And what do you obtain?

Then, you throw
everything away, and you

maintain only the
tetrahedron made of cheese.

Now, you remember
what the corners were.

This is 0, 0, 0.

It's a piece of cake.

But I want to know the vertices.

And you know them, and I
don't want to spend time

discussing why you know them.

So

STUDENT: 0--

MAGDALENA TODA: 1, 0, 0.

Thank you.

Huh?

STUDENT: 0, 1, 0.

MAGDALENA TODA: Yes.

And 0, 0, 1.

All right.

Great.

The only thing is, if we see
the cheese being a solid,

we don't see this part, the
three axes of corners behind.

so I'm going to make them
dotted, and you see the slice,

here, it has to
be really planar.

And you ask yourself,
how do you set up

the triple integral
that represents

the volume of this object?

Is it hard?

It shouldn't be hard.

You just have to think what
the domain will be like,

and you say the domain is
inside the tetrahedron.

Do you want d or t?

T from tetrahedron.

It doesn't matter.

We have a new name.

We get bored of all sorts
of names and notations.

We change them.

Mathematicians have imagination,
so we change our notations.

Like we cannot change
our identities,

and we suffer because of that.

So you can be a nerd
mathematician imagining

you're Spiderman,
and you can take,

give any name you want, and
you can adopt a new name,

and this is behind
our motivation

why we like to change names
and change notation so much.

OK?

So we have triple integral of
this T. All right, of what?

1 dV.

Good.

Now we understand
what we need to do,

just like [? Miteish ?]
asked me why.

OK, now we know this is going
to be a limit of little cubes.

If were to cover
this piece of cheese

in tiny, tiny,
infinitesimally small cubes.

But now we know a
method to do it.

So according to--
Fubini-Tonelli type of result.

We would have a between--
no, x-- is first, dz.

z is first, y is moving
next, x is moving last.

z is constrained to
move between a and b.

But in this case, a and b should
be prescribed by you guys,

because you should think
where everybody lives.

Not you, I mean the coordinates
in their imaginary world.

The coordinates
represent somebodies.

STUDENT: 0.

MAGDALENA TODA: x, 0 to 1.

How should I give you
a feeling for that?

Just draw this line.

This red segment between 0 to 1.

That expresses everything
instead of words

into pictures, because every
picture is worth 1,000 words.

y is married to
x, unfortunately.

y cannot say, oh, I am y,
I'm going wherever I want.

He hits his head against
this purple line.

He cannot go beyond
that purple line.

He's constrained, poor y.

So he says, I'm moving.

I'm mister y.

I'm moving in this direction,
but I cannot go past the purple

line in plane here.

I need you, because
if you go, I'm lost.

y is between 0 and--

STUDENT: 1 minus x.

MAGDALENA TODA: 1 minus x.

Excellent Roberto.

How did we think about this?

The purple line has
equation-- how do you

get to the equation of the
purple line, first of all?

In your imagination,
your plug in z equals 0.

So the purple line would
be x plus y equals 1.

And so mister y will
be 1 minus x here.

That's how you got it.

And finally, z is that--
mister z foes from the floor

all the way--
imagine somebody who

is like-- z is a
helium balloon, and he

is left-- you let him
go from the floor,

and he goes all the
way to the ceiling.

And the ceiling is not
flat like our ceiling.

The ceiling is
this oblique plane.

So z is going to hit his head
against the roof at some point,

and he doesn't know
where he is going

to hit his head, unless you
tell him where that happens.

So he knows he leaves at 0,
and he's going to end up where?

STUDENT: 1 minus y minus x.

MAGDALENA TODA: Excellent.

1 minus x minus y.

How do we do that?

We pull z out of that, and
say, 1 minus x minus y.

So that is the equation of
the shaded purple plane,

and this is as
far as you can go.

You cannot go past the
roof of your house,

which is the purple plane,
the purple shaded plane.

So here you are.

Is this hard?

No.

In many problems on the
final and on the midterm,

we tell you, don't even
think about solving that,

because we believe you.

Just set up the integral.

I might give you something
like that again, just

set up the integral and
you have to do that.

But now, I would like
to actually work it out,

see how hard it is.

So is this hard
to work this out?

I have to do it one at
a time, because you see,

I don't have fixed endpoints.

I cannot say, I'm applying the
problem with the integral if f

times the integral of g, times--
so I have to integrate one

at a time, because I don't
have fixed endpoints.

And the integral of 1dz is
z between that and that.

So z, 1 minus x minus y
will be what's left over,

and then I have dy,
and then I have dx.

And at this point it
looks horrible enough,

but we have to pray
that in the end

it's not going to be so hard,
and I'm going to keep going.

So we have integral from 0 to 1.

We have integral
from 0 to 1 minus x..

I'll just copy and paste it.

Which is integral from 0 to 1.

Now I have to think,
and that's dangerous.

I have 1 minux x
with respect to y.

This is going to be ugly.

That's a constant with
respect to y, and times y,

minus-- integrate with
respect to y, y is [? what? ?]

y squared over 2.

Between y equals 0 down.

That's going to save my
life, because for y equals 0,

0 is going to be a
great simplification.

And for y equals
1 minus x on top,

hopefully it's not going
to be the end of the world.

It looks ugly now, but
I'm an optimistic person,

so I hope that this is
going to get better.

And I can see it's
going to get better.

So I have integral
from here to 1.

And now I say, OK, let me think.

Life is not so bad.

Why?

1 minus x, 1 minus x
is 1 minus x squared.

I could think faster, you
could think faster than me,

but I don't want to rush.

1 minus x squared over 2.

So it's not bad at all.

Look, I'm getting this guy who
is beautiful in the end, when

I'm going to
integrate, and you have

to keep your fingers
crossed for me,

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

So I get integral from 0 to 1,
1/2 out, 1 minus x squared dx.

Is this bad?

Can you do this by
yourself without my help?

What are you going to do?

x squared minus 2x plus 1.

That's the square.

STUDENT: Why not just change it?

MAGDALENA TODA: Huh?

STUDENT: Why not just change it?

MAGDALENA TODA: You
can do it in many ways.

You can do whatever you want.

I don't care.

I want you to the right
answer one way or another.

So I'm going to clean a
little bit around here.

It's dirty.

You do it.

You have one minute
and a half to finish.

And tell me what you get.

STUDENT: 1 minus x
cubed over six negative.

MAGDALENA TODA: No, no.

In the end is the number.

What number?

But you have to go slow.

I need three people to
give me the same answer.

Because then it's like in that
proverb, if two people tell

you drunk, you go to bed.

I need three people to tell
me what the answer is in order

to believe them.

Three witnesses.

STUDENT: 1 [? by ?] 6.

MAGDALENA TODA: Who got
1 over 6, raise hand?

Wow, guys, you're fast.

Can you raise hands again?

OK, being fast doesn't
mean you're the best,

but I agree you do a very good
job, all of you in general.

So I believe there were
eight people or nine people.

1 over 6.

Now, how could I have cheated
on this problem on the final?

STUDENT: It's a
[? junction ?] from this--

MAGDALENA TODA: Right.

In this case, being a volume,
I would have been lucky enough,

and say, it is the
volume of a tetrahedron.

I go, the tetrahedron
has area of the base 1/2,

the height is 1.

1/2 times 1 divided by 3 is 1/6.

And just pretend on the
final that I actually

computed everything.

I could have done that, from
here jump to here, or from here

jump straight to here.

And ask you, how did you
get from here to here?

And you say, I'm a genius.

Could I not believe you?

I have to give you full credit.

However, what would you
have done if I said compute,

I don't know, something
worse, something

like triple integral of x,
y, z over the tetrahedron 2.

In that case, you cannot cheat.

You're not lucky
enough to cheat.

You're lucky enough
to cheat when

you have a volume
of a prism, you

have a volume of-- and volume
means this should be the number

1 here, number 1.

So if you have number 1, here,
or I ask you for the volume,

and it's a prism, or
tetrahedron, or sphere,

or something, go
ahead and cheat,

and pretend that you're
actually solving the integral.

Yes, sir.

STUDENT: What would that
represent, geometrically,

the triple integral of x, y, z?

MAGDALENA TODA: It's a
weighted triple integral.

I'm going to give
you examples later.

When you have mass and momentum,
when you compute the center

map, or you compute the
mass, and somebody give you

densities.

Let me get -- If you have a
triple integral over row at x,

y, z, this could be it,
but I [? recall ?] it row

for a reason, not just for fun.

And here, dx, dy, dz.

Very good question, and
it's very insightful.

For a physicist or
engineer, the guy

needs to know why we take
this weighted [? integral. ?]

If row is the
density of an object,

if it's everywhere the same, if
row is a homogeneous density,

for that piece of cheddar
cheese-- Oh my God

I'm so hungry-- row
would be constant.

If it's a quality cheddar
made in Vermont in the best

factory, whatever, row would
be considered to be a constant,

right?

And in that case, what happens?

If it's a constant,
it's a gets out,

and then you have row
times triple integral 1

dV, which is what?

The volume.

And then the volume times the
density of the piece of cheese

will be?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: The mass
of the piece of cheese,

in kilograms, because I
think in kilograms because I

can eat more.

OK?

Actually, no, I'm just kidding.

You guys have really-- I
mean, 2 pounds and 1 kilogram

is not the same thin.

Can somebody tell me why?

I mean, you know it's not
the same thing because,

the approximation.

But I'm claiming you cannot
compare pounds with kilograms

at all.

STUDENT: Pounds is
a measure of weight,

whereas kilograms is
a measure of mass.

MAGDALENA TODA: Excellent.

Kilogram is a measure
of mass, pound

is a measure of the
gravitational force.

It's a force measure.

So OK.

Which reminds me,
there was-- I don't

know if you saw this short
movie for 15 minutes that

got an award the
previous Oscar last year,

and there was an old lady
telling another old lady

in Great Britain, get
2 pounds of sausage.

And the other one says,
I thought we got metric,

because we are in
the European Union.

And she said, then get me
just the one meter of sausage,

or something.

So it was funny.

So it can be mass.

But what if this
density is not the same?

This is exactly why we
need to do the integral.

Imagine that the
density is-- we have

a piece of cake with layers.

And again, you see
how hungry I am.

So you have a layer, and
then cream, or whipped cream,

or mousse, and another
layer, and another mousse.

The density will vary.

But then there are bodies in
physics where the density is

even a smooth function.

It doesn't matter that you have
such a discontinuous function.

What would you do?

You just split.

You have triple row 1 for the
first layer, then triple row 2

for the second later,
the layer of mousse,

and then let's
say it's tiramisu,

you have another layer, row
three, dV3 for the top layer

of the tiramisu.

STUDENT: Can any row
be kept constant?

MAGDALENA TODA: So
these are discontinuous.

They are all constant, though.

That would be the
great advantage,

because presumably mousse would
have the constant density,

the dough has a constant,
homogeneous density, and so on.

But what if the density
varies in that body from point

to point?

Then nobody can do
it by approximation.

You'd say volume, mass 1 plus
mass 2 plus mass 3 plus mass 1.

You have to have a triple
integral where this row varies,

constantly varies.

And for an engineer,
that would be a puzzle.

Poor engineers says,
oh my God, the density

is different from
one point to another.

I have to find an
approximated function

for that density moving from one
point to another on that body.

And then the only way to do it
would be to solve an integral.

Imagine that somebody--
now it just occurred,

I never thought
about it-- we would

be measured in terms of
this type of integral.

Of course, people would be able
to measure mass right away.

But then, if you were
to know the density--

you cannot even know the density
at every point of the body.

It varies a lot, so
every point of our bodies

has a different
material and a density.

OK.

STUDENT: Tiramisu. [INAUDIBLE]

MAGDALENA TODA: Huh?

STUDENT: So you
use the tiramasu,

you're making me hungry.

MAGDALENA TODA:
Yeah, because now,

OK take your mind
off the tiramisu.

Think about an exam.

Then you don't--

STUDENT: Now I'm sick.

MAGDALENA TODA: Exactly.

Now you need something
against nausea.

Let's see what else
is interesting to do.

I'll give you ten minutes.

How much did I steal from you?

I stole constantly about
five minutes of your breaks

for the last few Tuesdays.

STUDENT: So the integral--

MAGDALENA TODA: The
integral of that.

I think I would be fair
to give you 10 minutes

as a gift today to compensate.

OK, so remind me to let
you go 10 minutes early.

Especially since
spring break is coming.

We have a 3D application.

We have several 3D applications.

Let me see which one
I want to mimic first.

Yeah.

I'm going to pick my favorite,
because I just want to.

So imagine you
have a disc that is

x squared plus y squared
equals 1 would be the circle.

That's the unit
disc on the floor.

And then I have the plane
x plus y plus z equals 8.

Then I'm going to
draw that plane.

I'll try my best.

It's similar to two
examples from the book,

but I did not want to
repeat the ones in the book

because I want you to
actually read them.

That's kind of the idea.

So you have this
picture, and you

realize that we had that
in the first octant before.

So I say, I don't
want the volume

of the body over the whole disc,
only over the part of the disc

which is in the first octant.

So I say, I want this domain
D, which is going to be what?

x squared plus y squared
less than or equal to 1

in plane, with x
positive, y positive.

Do you know what we call
that in trigonometry?

Does anybody know what we
call this in trigonometry?

Let me put the points
while you think.

Hopefully, you are
thinking about this.

This is 1 in x-axis.

1, 0, 0, and this is
0, 1, 0, and this is y.

If I were to go up
until I meet the plane,

what point would this--

STUDENT: [INAUDIBLE]

MAGDALENA TODA: What point
would this-- on the thing.

STUDENT: 1, 0, 7
and then 0, 1, 7.

MAGDALENA TODA: 1, 0, 7.

This would be you said 0, 1, 6.

And this would be 1, 0, 7.

How did you think about this?

How do you know?

STUDENT: Y plus z--

MAGDALENA TODA: Because z,
because it's on the y-axis,

and since you are on the
x-axis here, y has to be 0.

So you're right.

Very good.

Excellent.

Now I'm going to
say, I'd like to know

the-- compute the volume of
the body that is bounded above

from above by x plus y plus
z equals 8, who's projection

on the floor is the
domain D. And I'll say

volume of the cylindrical body.

So how could you obtain
such a, again-- No,

this is Murphy's Law.

OK, how could you obtain such
an object, such a cylinder?

STUDENT: Take a pencil,
and cut it into fourths.

MAGDALENA TODA: Huh?

STUDENT: Take like a cylindrical
pencil and cut it into fourths.

MAGDALENA TODA: Take a
salami, a piece of salami.

Cut that piece of salami into
four, into four quarters.

And then we take, we slice,
and we slice like that.

So we have something like--

STUDENT: I tried to think
of a non- food example.

MAGDALENA TODA: --a quarter.

How can I draw this?

OK, this is what it means.

You don't see this one.

You don't see this part.

You don't see this part.

This is curved.

And here, instead of cutting
with another perpendicular

plane, along the
salami-- so this

is the axis of the salami--
instead of taking the knife

and cutting like that, I'm
cutting an oblique plane,

and this is what this
oblique plane will do.

STUDENT: If you cut
that way, then you

would have only squares.

MAGDALENA TODA: Hmm?

So I'm going to have some
oblique-- I cannot draw better.

I don't know how to draw better.

So it's going to be an
oblique cut in the salami.

Let's think how we
do this problem.

Elementary, it will
be a piece of cake--

it would be a piece of--

STUDENT: A piece of salami.

MAGDALENA TODA: No.

It wouldn't be apiece of salami.

STUDENT: It could be done.

MAGDALENA TODA: How could we do
that quickly with the Calculus

III we know?

STUDENT: Find the
triple integral.

Oh, you want us to do
the double integral?

MAGDALENA TODA: Double, triple,
I don't know what to do.

What do you think is best?

Let's do that triple
integral first,

and you'll see that it's the
same thing as double integral.

Triple integral over B, the
body of the salami, 1 dV.

How can we set it up?

Well, this is a
little bit tricky.

It's going to be like that.

We can say, I have a double
integral over my domain,

D. When it comes to the z,
mister z has to be first.

So mister z says, I'm first.

I know where I'm going.

You guys, x and y
are bound together,

mired in the element
of area of the circles.

This is like dx dy.

But I am independent from you.

I am z.

So I'm going all the way
from the floor to what?

You taught me that.

8 minus x minus y, and 1.

This is the way to do
it as a triple integral,

but then Alex will
say, I could have

done this as a double integral.

Let me show you how.

I could have done it over
the domain D in plane.

Put the function,
8 minus x minus y

is [? B and ?] z from
the very beginning,

because that's my altitude
function, f of x and y.

So then I say dx dy, dx
dy, it doesn't matter.

That's the only theory element.

Fine.

It's the same thing.

This is what I wanted
you to observe.

Whether you view it like the
triple integral like that,

or you view it as the
double integral like that,

it's the same thing.

This is not a headache.

The headache is coming next.

This is not a headache.

So you can do it in two ways.

And I'd like to
look at the-- check

the two methods of doing this.

And set up the integrals
without solving them.

Can you read my mind?

Do you realize what I'm asking?

Imagine that would
be on the midterm.

What do you think I'm
asking, the two methods?

This can be interpreted
in many ways.

There are two methods.

I mean, one method by
doing it with Cartesian

coordinates x and y.

The other method is switching
to polar coordinates

and set up the integral
without solving.

And you say, why not solving?

Because I'm going to cheat.

I'm going to use a
TI-92 to solve it,

or I'm going to use
a Matlab or Maple.

If it looks a little
bit complicated,

then I don't want
to spend my time.

Actually, engineers,
after taking Calc III,

they know a lot.

They understand a lot
about volumes, areas.

But do you think if you work on
a real-life problem like that,

that your boss will
let you waste your time

and do the integral by hand?

STUDENT: No.

MAGDALENA TODA: Most integrals
are really complicated

in everyday life.

So what you're
going to do is going

to be a scientific software,
like Matlab, which is primarily

for engineers, Mathematica,
which is similar to Matlab,

but is mainly for
mathematicians.

It was invented
at the University

of Illinois Urbana-Champaign,
and they're still

very proud of it.

I prefer Matlab
because I feel Matlab

is stronger, has higher
capabilities than Mathematica.

You can use Maple.

Maple lets you set up the
endpoints even as functions.

And then it's user
friendly, you type in this,

you type in the endpoints.

It has little windows, here.

You don't need to
know any programming.

It's made for people who
have no programming skills.

So it's going to show
a little window on top,

here, here, here and here.

You [? have ?] those,
and you press Enter,

and it's going to spit
the answer back at you.

So this is how
engineers actually

solve the everyday integrals.

Not by hand.

I want to be able to
set it up in both ways

before I go home or
eat something, right?

So we don't have to spend
a lot of time on it.

But if you want to tell me
how I am going to set it up,

I would be very grateful.

So this is Cartesian,
and this is polar.

All right.

Who helps me?

In Cartesian-- which
one do you prefer?

I mean, it doesn't matter.

You guys are good
and smart, and you'll

figure out what I need to do.

If I want to do it in terms
of vertical strip-- so

for vertical strip
method-- first

I integrate with respect to
y, and then with respect to x.

And maybe, to test
your understanding,

let me change the
order of integrals

and see how much you
understood from that last time.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: So x is
between what and what?

STUDENT: 0 and 1.

MAGDALENA TODA: Look
at this picture.

I have to reproduce
this picture like that.

0 to 1, says Alex,
and he's right.

And why will he decide against--

STUDENT: 1 minus x squared.

MAGDALENA TODA: --square
root 1 minus x squared.

So we know very well
what we are going to do,

what Maple is
going to do for us.

1 square root 1 minus x squared.

And then what do I put here?

8 minus x minus y.

Can I do it by hand?

Yes, I guarantee to you
I can do it by hand.

Let me tell you why.

Because when we integrate
with respect to y, I get xy.

So I get xy, and y will be
plugged in 1 minus x squared.

How am I going to solve
an integral like this?

I can the first one with a
table, the second one with a u

substitution.

On the last one is a
little bit painful.

I'm going to have
y squared over 2--

STUDENT: That's
the easiest part.

MAGDALENA TODA: According
to Alex, yes, you're right.

Maybe that is the easiest.

STUDENT: That's the [INAUDIBLE]
part you can integrate--

MAGDALENA TODA: And I can
integrate one at a time,

and I'm going to
waste all my time.

So if I want to be an
efficient engineer,

and my boss is waiting for
the end-of-the-day project,

of course I'm not going
to do this by hand.

How about the other integral?

Same integral.

Same idea, y between 0 and 1.

And x between 0 and

STUDENT: Square root
of 1 minus y squared.

MAGDALENA TODA: Square
root of 1 minus y squared.

Because I'll do this guy
with horizontal strips,

and forget about
the vertical strips.

And here's the y-- I rotate
my head and it cracks,

so that means that
I need some yoga.

y is between 0 and 1.

Or gymnastics.

So x is between
0 and square root

1 minus y squared. [INAUDIBLE].

And I'll leave it
here on the meter.

And I'm going to make a
sample like I promised.

OK, good.

How would you do this to set up
the polar coordinate integral?

And that is why Alex said
maybe that's a pain because

of a reason.

And he's right, it's a little
bit painful to solve by hand.

But again, once you
switch to polar,

you can solve it with a
calculator or a computer

software, scientific
software in no time.

In Maple, you just have
to plug in the numbers.

You cannot plug in theta,
I think, as a symbol.

I'm not sure.

But you can put theta
as t and r will be r,

or you can use
whatever letters you

want that are roman letters.

So you have to
integrate smartly, here,

switching to r and
theta, and think

about the meaning of that.

So first of all, if
I put dr d theta,

I'm not worried that you won't
be able to get r and theta,

because I know you can do it.

You can prove it to me
right now. r between 0 and

STUDENT: 1.

MAGDALENA TODA: Excellent.

And theta, pay
attention, between 0 and

STUDENT: pi over 2

MAGDALENA TODA: Excellent.

I'm proud.

Yes, sir?

STUDENT: Is it supposed
to be r dr 2 theta,

or are you going
to add that later?

MAGDALENA TODA: I
will add it here.

So the integrand
will contain the r.

Now what do I put
in terms of this?

I left enough room.

STUDENT: Is it pi over 2,
or is it negative pi over 2?

MAGDALENA TODA:
It doesn't matter,

because I'll have to take that--
we assume always theta to go

counterclockwise, and go
between 0 and pi over 2,

so that when you start--
let me make this motion.

You are here at theta equals 0.

STUDENT: Oh, OK.

Sorry.

I got my coordinates
mixed around--

MAGDALENA TODA: --and
counterclockwise to pi over 2.

[INTERPOSING VOICES]

MAGDALENA TODA: Yeah.

So you go in the trigonometric--
Here, you have 8 minus,

and who tells me what
I'm supposed to type?

STUDENT: r over x.

MAGDALENA TODA: r
cosine theta minus

STUDENT: Sine.

MAGDALENA TODA: r sine theta.

And let mister
whatever his name is,

the computer, find the answer.

Can I do it by hand?

Actually, I can.

I can, but again, it's not worth
it, because it drives me crazy.

How would I do it by hand?

I would split the
integral into three,

and I would easily
compute 8 times r,

integrand is going to be easy.

Right?

Agree with me?

Then what am I going to do?

I'm going to say, an r out
times an r, out comes r squared.

And I have integral of r
squared times a function

of theta only,
which is going to be

sine theta plus cosine theta.

We are going to say, yes,
with a minus, with a minus.

Now, when I compute
r and theta thingy,

theta will be between
0 and pi over 2.

r will be between 0 and 1.

But I don't care, because
Matthew reminded me,

if you have a product
of separate variables,

life becomes all of the
sudden easier for you.

STUDENT: You've also got to
add your integral of [? 8r ?]

[? dr. ?]

MAGDALENA TODA: Yeah.

At the end, I'm going to
add the integral of 8r.

So I take them separately.

I just look at one chunk.

And this chunk will be what?

Can you even see how easy it's
going to be with the naked eye?

Firs of all,
integral from 0 to 1,

r squared dr is a piece of cake.

How much is that--
piece of salami.

STUDENT: 1/3.

MAGDALENA TODA: 1/3.

Right?

Because it's r cubed over 3.

Then you have 1/3.

That's easy.

With a minus in front, but I
don't care about it in the end.

What is the integral of
sine theta cosine theta?

STUDENT: Negative [INAUDIBLE].

MAGDALENA TODA: Minus
cosine theta plus sine theta

taken between 0 and pi over 2.

Will this be hard?

Who's going to tell me what,
or how I'm going to get what--

we don't compute it now,
but I just give you.

Cosine of pi over 3 is?

STUDENT: 0.

MAGDALENA TODA: 0.

Sine of pi over 2 is?

STUDENT: Oh yeah.

1.

MAGDALENA TODA: 1.

So this is going to
be 1 minus, what's

the whole thingy computed at 0?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: It's going to
be minus 1, but minus minus 1

is plus 1.

So I have 2.

So only this chunk of the
integral would be easy.

Minus 2/3.

OK?

So it can be done by hand,
but why waste the time when

you can do it with Maple?

Yes, sir?

STUDENT: Where did
you get rid of 8?

On the second, after the 8--

MAGDALENA TODA: No, I didn't.

That's exactly what
we were talking.

Alex says, but you just
talked about integral of 8r,

but you didn't want to do it.

I said, I didn't want to do it.

This is just the second
chunk of this integral.

So I know that I can do integral
of integral of 8r in no time.

Then I would need to
take this and add that,

and get the number.

I don't care about the number.

I just care about the method.

Yes, sir?

STUDENT: Why are the limits
from 0 to 1 instead of like 0

to r squared?

Because didn't we say
earlier the domain

is x squared plus y squared?

Wouldn't that be r squared?

MAGDALENA TODA: No.

No, wait.

This is r squared.

STUDENT: Right.

Why didn't we plug r
squared into the 1 again.

MAGDALENA TODA: And that means
r is between 0 and 1, right?

STUDENT: Oh, OK.

MAGDALENA TODA: r squared
being less than 1.

That means r is between 0 and 1.

OK?

And one last problem-- no.

No last problem.

We have barely 10 minutes.

So you read from the book some.

I will come back to this
section, and I'll do review.

Have a wonderful
spring break, and I'm

going to see you after
spring break on Tuesday.

[INTERPOSING VOICES]