TTU Math2450 Calculus3 Secs 12.6 - 12.7

Integration in Rectangular, Cylindrical and Spherical coordinates

MAGDALENA TODA: Midterm
covers the following sections.

Chapter 9, no-- that
was practically--

that is review, mostly review.

10.1, 10.2, 10.4, and
what I'm writing down

will be the same section
covered in the final.

11.1 through 11.8, all of them.

11.8 is LaGrange multipliers.

That may or may not be
on the actual midterm,

but it's included in
the [? midterm. ?]

12.1 through 12.5, all of them.

Skip 12.6 as you know.

And 12.7 will be covered
today-- will be covered today--

in the classroom
after the review

because today is the
long day with all three

hours [INAUDIBLE].

We have almost
three hours today.

[? Just do that on ?] the
midterm, and then I'll

move onto 12.7.

12.7 is the last required
section in chapter 12.

And then there is nothing else
for Chapter 13 and that's it.

12.7, you will see it's called
cylindrical and spherical

coordinates.

And you use these
type of coordinates

for triple integrals--
for triple integrals.

Everything is in 3D, three
dimensional Euclidean stuff.

All right.

Is it clear?

So I'm going to start
with the specification.

[INAUDIBLE] and portable
electronic devices of all sorts

are not allowed during
the examination.

No calculator, no pagers,
no cellphones open.

Taking this exam is subjected to
the existing rules [INAUDIBLE]

regarding academic honesty.

So don't attempt cheating,
looking at the notes,

talking to other
people during the exam.

That's not going to be allowed.

Write in the spaces provided
under every problem.

No additional papers
have to be turned in.

When you'll actually
get the exam,

it's going to be typed a little
bit larger so you will have

spaces between the problems.

In those spaces, you
will be able to write.

The answers are really short.

You'll be surprised
how short they will be.

On top of those two
pages, there will

be a third page which is blank.

In that third blank page, you
can scribble your computations.

You will not have many
computations to scribble,

but the space provided
should be practically enough

for you to write your answers.

Two or three of
those 10 problems

will have multiple
choice answers.

So you're going to
ask me, what do I

expect about that
multiple choice answer?

I will tell you in
a second and I'm

going to also give
you an example.

So I'm just expecting some
circling some numerical answer

that you have [INAUDIBLE].

And I'll tell you in the context
of this sample what those are.

Who's missing the midterm?

Come and pick it up, please.

And would you distribute
this to the others?

[INAUDIBLE]

All right.

Number 1, write the
total differential

of the following function.

You will have something like
that for a different function.

But you will have to do
the total differential.

And remember the
total differential

is not a sum or
partial derivative.

So if you want to
kill me, say that.

You're going to get a
big 0 on that, right,

but you're also going to
kill me in the process.

So what was the
total differential

in the function
of two variables?

Now you know it.

You can help me write it down.

STUDENT: Partial derivative
in the x direction

dx [? plus 1. ?]

MAGDALENA TODA: And that's
an infinitesimal element.

If you skip that, everything
you write is nonsense.

So pay attention that this
part, the whole combination

of scalar real value functions
and infinitesimal elements

would be good.

So what is the answer
for my problem here

if f is-- what's your problem?

Let's see, what's your problem?

Cosine of xy.

And I'm going to
do this in lime.

I hope it's visible
enough for the video.

What's the derivative with
respect to x of cosine of xy?

STUDENT: Negative y sine xy.

MAGDALENA TODA:
Negative y sine xy.

So pay attention
to the chain rule.

You'll have to remember
that the derivative

of this function with respect to
x assumes that y is a constant.

So when you differentiate
xy with respect to xy,

it remains [INAUDIBLE]
of the product.

All right.

dx, [INAUDIBLE] and say dx
plus the infinitesimal element

[? dx ?] plus-- what next?

The same thing.

It's a symmetric polynomial.

But exactly like
Aaron said, now I'm

going to have minus x sine xy.

So the same kind of policy.

You've seen this
before on homework.

Guys, I don't like
surprises on the exam,

so what I picked for the
midterm-- and you will see also

the final will be
in the same screen--

it is out of those 200
problems, whatever you

have covered in the homework.

We don't give you new
stuff, stuff that you

haven't seen in the homework.

All right.

So with this, your
answer is ready.

You have one or two rows are
enough for you to write down

the answer for that.

Actually, one line
would be enough.

Do I expect more than this line?

Take one from the
desk, [INAUDIBLE].

So do I expect more than
this answer under problem 1?

No.

You either know it
or you don't know it.

If you know the definition,
you don't have to repeat it.

If you know how to
differentiate, you're fine.

You're in business.

You've got a perfect score on
this first problem out of 10.

They are not time consuming.

I don't like exams that
are time consuming.

They are not testing anything.

So I want to walk you
through the material

through many questions so
that I see what you understood

and what concepts
are well understood

and which ones are not.

OK.

All right.

Let's move on to number two.

That was number 1.

Number 2.

Number 2 says find the
directional derivative

of a function f of xy, which
is equal to cosine of xy.

What do you see?

I have already prepared
you for this problem

because it's the same
function as before.

You have computed f sub
x and f sub y already.

That should be
helpful when you write

the directional derivative.

In the direction--
and by direction,

we mean unit vector-- direction
of something means unit vector.

What direction do they say
the highest [? ascent? ?]

So this problem is walking you
through the whole chapter 11

by testing what you understood.

So what is the direction
of the highest [? ascent ?]

for a function like that?

STUDENT: The gradient
of the [INAUDIBLE].

MAGDALENA TODA: The gradient.

So if you know what the
direction of the gradient

is [INAUDIBLE], at the
point p of coordinate 0, 0--

so p is the point at 0, 0--
if such a direction exists,

can you say-- well
of course it exists.

It's a gradient.

Justify your answer.

All right.

So when would it not exists?

STUDENT: If there's no
direction of highest decent.

MAGDALENA TODA: Right.

STUDENT: All of
them are the same.

MAGDALENA TODA: Right.

All right.

So what would you
have to write in terms

of directional derivative
for an arbitrary function?

What do you know about the
[? directional ?] derivative

of a function at the point in
the direction of a vector u?

You can write the
definition with limit.

That's going to be an
expression like that eating up

all your space, but
that's not what I mean.

I mean that you'll have
to know the theorem that

says this is partial
derivative with respect

to x measure that
the point p times u1

plus partial derivative
with respect to y

measured at the point
p times u2 where

u1 and u2 are the components
of your unit vector

that we talked about before.

Now, when I compute
f sub x and p,

I have to say oh, I
know where sub x is.

But if I compute it at
p and p is 0, 0, my god,

this is going to be 0, right?

And if I do f sub
y at p and p is 0,

well, you also know that the
second partial derivative is 0.

So at 0, this is
going to be 0 as well.

So the directional
derivative is 0.

Did I have to find the
direction of the gradient?

No, because in the
end, you got 0.

So it's always
important to write down

the final thing you
want, the definition

of the directional derivative.

In the actual exam, you may
have a different situation.

So let's work on alternative
one, alternative problem

like that.

I'll pick an easy f of x and y.

All of the examples would
be [INAUDIBLE] numerically,

computationally.

The [? point ?] components
[? 1,1 ?] in the [INAUDIBLE].

Do the same problem for
this different data.

Who will be u?

u will be the direction
of the gradient.

If I'm talking about the
direction of the highest

asset for-- or highest
asset direction,

Sometimes I may give
you that direction,

and it doesn't have to be
the highest asset direction.

But in this case, it has to
be because the problems are

solved.

So you have to say, what
is the gradient in my case?

Well, piece of cake.

That's going to be 1 times
i plus 1 times j divided

by square root of 2.

So u1 is 1 over root
2, u2 is 1 over root 2.

And can I write [? this view ?]
in such an alternative problem,

ah, it should be easy.

Well, it is easy.

The first number will
be 1 because that's

the derivative with respect
to x times u1, 1 over root 2

plus the derivative with
respect to y, 1, times u2 1

over root 2.

Only now do you
leave it like this.

Or you leave it like
that and you get 100%.

I really don't care.

But what if the
problem is formulated

as a multiple choice?

It will be formulated as
multiple choice with A, B,

C choices like that, we'll
never write it-- well,

if we are normal people, we'll
never write it like that.

So what do you think the
answer, the correct answer,

out of those possible
answers will be in this case?

One answer will be 0, one
answer will be 2, one answer--

STUDENT: Root 2.2

MAGDALENA TODA: Root 2.

So what are you supposed to do?

Just go to the answer
[INAUDIBLE] circle it.

Do you have to show anyone?

No.

But you only have, like, two
such multiple choice problems.

Everything else is show work.

All right.

Yes, sir.

STUDENT: How can a
problem change because

of the steepest descent?

MAGDALENA TODA: The
steepest descent

will happen in the opposite
direction of the gradient.

So it's going to happen exactly
in the direction minus i,

minus j over root 2.

And if you compute the
directional derivative

in that direction, you're going
to have exactly the opposite

of that minus root 2.

All right.

STUDENT: It says
justify your answer.

How would you write that?

MAGDALENA TODA: Oh.

Justify your answer.

Let me justify the
answer [INAUDIBLE] here.

The steepest ascent takes
place in the direction u

corresponding to the gradient.

Delta f formula [? au ?]
bar f of p formula

above means the theorem
that was from chapter 11.

You don't have to
know which theorem,

but you just write down
this [? was a ?] theorem.

You plugged in, you
got 0 for your example.

For the actual midterm,
it may be different.

The same kind of
problem, absolutely.

So you have to say--
justify your answer.

I may or may not say justify
your answer in the end,

but you have to
know this anyway.

You have to know
this part anyway.

You have to know
this formula anyway.

So justification means
showing the formula in itself

and explaining that the
steepest ascent take place

in the direction
of the gradient.

All right.

OK.

Number 2.

Number 3.

They're all related.

They're all the
same [? constant ?].

So I'm going to go ahead
and erase this part.

And it says find number three.

For the function f of
x equals cosine of xy,

as if we never saw that
before, find the gradient.

What are you supposed to
write on such a problem

to justify your answer?

And also, it says
write parenthesis

specify the equation of the
tangent plane to the graph z

equals f of xy at
the point-- I forgot

the word at-- at the point
a of coordinate 0, 0, 1.

All right.

So let's say-- this
problem has two parts,

but they shouldn't be long.

Number 3 says what?

Find me the gradient
of this function.

What are you supposed to write?

You have two alternatives here.

Either you write f sub xi plus
f sub yj equals minus y sine xy

i minus x sine xy j.

Or this is up to you.

This is how I would
write the gradient.

Some people prefer to
forget the x and y.

They don't write the
gradient in standard basis.

They're assuming it
is the standard basis

and they only write
the coordinates.

In angular brackets,
assume with respect

to the standard basis,
canonical basis.

Whichever you want is fine.

Are you guys with me?

Any questions?

All right.

Is this hard?

No.

Let's see how hard
is the next one.

What do you think I have to do?

I'm waiting for your input.

[? Write, ?]
parenthesis, specify

the equation of the tangent
plane to the graph z

equals f of xy.

That graph is going to be the
graph of a surface, right?

z equals f of x sub y.

We want the equation
of the tangent plane

to the graph at
the point 0, 0, 1,

which means I have to take
x, 0 to be 0, y, 0 to be 0.

And consequently, z, 0 will
be cosine of 0, which is 1.

So it makes sense that
point p is on the graph.

Or a is on the graph, a.

[INAUDIBLE]

Good.

At least I know I didn't
make a mistake about that.

How do we write
the tangent plane

when someone gives us the
explicit equation of the graph?

z equals f of xy.

Knowing that f is a function,
how is this function?

Well look at it.

It's z infinity.

It has differentiable
infinitely many times,

and all the derivatives
are continuous.

So we are really
lucky we don't have

any headaches in this case.

[INAUDIBLE] both
partial derivatives

existed that are
continuous, I'm going

to write down the equation
of the tangent [INAUDIBLE].

But not by myself.

With your help because
it's been a long time.

But you are not allowed
to forget the formula.

What was the general formula for
the tangent plane in this case?

[INAUDIBLE]

And I'll try to
keep my mouth shut.

z--

STUDENT: Minus z0.

MAGDALENA TODA: Minus z0.

Remember the linear
approximation

via the surface, the current
surface, with the plane.

That's the linear approximation.

STUDENT: x with
respect [INAUDIBLE].

MAGDALENA TODA: That's right.

I want to say with respect to
[INAUDIBLE] x and y at 0, 0--

good, we'll do that
later-- times--

STUDENT: x minus x0.

MAGDALENA TODA:
--x minus x0 plus--

STUDENT: f sub y.

MAGDALENA TODA: f
sub y at 0, 0 times--

STUDENT: y minus y0.

MAGDALENA TODA: y minus y0.

But again, x0 is 0.
y0 is 0. z0 is 1.

Now, we know more than
that because I gave you

a problem that's too easy.

[INAUDIBLE] trying to
do something impossible.

f sub x at 0, 0, we
already computed that.

And we noticed
that was 0 and this

was-- so we are exceedingly
lucky in this case.

Not all the problems
are so beautiful.

The problems that
we give in the exam

generally are more beautiful
than all the problems

that happen in
reality, real life.

z equals 1 will be your
answer in this case.

What is z equals 1?

It's the horizontal plane
that z equals [? 1. ?] So

that will be the [INAUDIBLE].

And you're done.

You have problem--
what problem was that?

Problem 3 finished.

They are short and they
are, again, [INAUDIBLE]

the ones on Thursday.

Next Thursday,
not this Thursday.

It will be very easy to solve.

I'm just testing
you on the concepts,

see what do you remember
from the concepts.

So now you know that
write a note to yourself.

So she expects me to know the
equation of the tangent plane

for both the midterm and
for the final by heart.

Yes, I do.

And I think compared to
other things that you

have to learn, memorize for
your engineering courses,

this is a piece of cake.

You only have a few things
that you have to memorize.

Very few things you
have to memorize.

All right.

Number 4, find the direction
in which the function--

I'm changing the
function, finally.

I say function x
plus y increases

most rapidly at the
point p of 1, 1.

Or in which direction is
this function increasing

most rapidly at 1, 1?

In which direction
is this function

increasing most rapidly at 1, 1.

That means that the 0
equals 1, 1, [? 0 ?].

We've done these kind
of things before.

We know the direction
will be the direction

corresponding to that gradient.

But the tension, when I say
direction, what do I mean?

Normalize it.

OK.

All right.

So gradient of f at 1, 1 will
be 1 times i plus 1 times j.

I'm not allowed to
leave it like this.

I would say the answer
is u equals gradient

of f over normal
gradient of f, which

is i plus j over square root of
2, which is the same as writing

1 over 2, 1 over root 2.

And that's it.

So if x is this axis
and y is this other one,

I'm modeling it with my hands.

The direction I'm talking
about is the first [INAUDIBLE]

of this quadrant.

[INAUDIBLE] If I'd done
steepest as a steepest descent,

it's going to be in
the opposite direction.

I'm going to go
ahead and erase-- I

don't have [INAUDIBLE] number.

All right.

Is it the same problem?

Yes.

So let's keep this on the
board and [? find ?] what is

the maximal rate of change of
the function f of xy equals x

plus y at the point
[? p, 1,1. ?] That is another

theorem.

So how do I justify my
answer in this case?

There is another theorem that
says the maximum rate of change

will be the
directional derivative.

So the maximum
rate of change will

be the directional
derivative of f

in the direction u of
the gradient lambda

f at the point p.

This is what I would
have done as a student

because I think I would
have remembered only

the theorem without
remembering that this

is the same time, the
same thing, as what?

The length of the
radian at that point.

I think I would've
forgotten that because I'm

trying to remember who I
was more than 20 years ago

and see how I would have
answered this question.

Now, I think I would've
gotten to the same answer.

Nevertheless, I would have
struggled a little bit

and I would've thought just
in the middle of the exam,

why in the world is that?

Because I always find myself
not remembering things

in the exams, and I would
try to remember why is that?

So by the way, why is that?

The directional derivative
in the direction

of u of f at the point
p also was returned.

I wrote it it down
before f sub x u1 plus f

sub y u2 where you guys
told me what that was.

That was a certain scalar
product, or dot product,

between the direction of the
gradient and the vector u.

And I was grateful you told me
that that was a long time ago,

or it was two weeks, three
weeks ago, something like that.

So we noted together
that it's the same thing

if you write [INAUDIBLE] f
gradient [INAUDIBLE] product

with u.

But if u itself is gradient
of f, well, over the length,

let's write it down.

If I'm talking
highest ascent, then

u itself is the gradient
of f over the form,

the lengths of the gradient.

So what is the answer?

That is-- [INAUDIBLE].

It would be exactly
the square of the norm,

the square of the length,
divided by the length.

So I proved again, because
I'm forgetful and I'm silly,

I put again that this
maximum rate of change

is exactly the length
of the gradient.

For somebody who
is better than me

when it comes to
memorizing things,

they wouldn't have
had to do that.

I'm just doing review.

So what was expected from
you if you studied and you're

a good student.

We expected only
that you would write.

The answer would
be-- which is what?

Who's telling me what this is?

STUDENT: The square root of 2.

MAGDALENA TODA: The
square root of 2.

Piece of cake.

So it depends on how
much you remember,

how much you understood
from the material,

and also a little bit of
memorization, yes, helps.

If you're not able
to memorize, you

can always come up to
the right conclusion,

but it takes forever
and you don't

have time, 1 hour and 20 minutes
or whatever that is, 1 hour 20

minutes to [? see ?] too much.

OK.

I'm going to erase--
what should I erase?

Number 4, maybe I should erase.

And--

STUDENT: [INAUDIBLE],
I was going

to ask you about number 4.

MAGDALENA TODA: Huh?

STUDENT: On problem 6, it
asks you about that one

specifically?

MAGDALENA TODA: And
number 5 is asking you--

STUDENT: We just did number 5.

MAGDALENA TODA: Yeah.

And the answer is root 2.

And number 4--

STUDENT: It pretty much just
asks you about that direction.

MAGDALENA TODA: It asks
you about that direction.

That's what you have to print
on the exam [INAUDIBLE].

So again, the actual
exam may ask you

to do this by printing it out.

Which of those following
numbers will represent

the maximum rate of change?

0, 1, the square root of
2, 792, stuff like that.

So you would have to
circle root 2 in this case.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: What?

STUDENT: I'm saying
you might not

want to erase number 4
because I was about to ask you

about number 4.

MAGDALENA TODA: Oh my God.

You are deeper than me.

OK.

Is the direction that you
found that point for unique

for the given point?

Yes or no?

What do you think?

Well, OK.

By direction, we mean what?

Not all the colinear vectors
that go in that direction,

but the actual uniquely
defined number as that, right?

So you can write an essay
based on the uniqueness

of that [INAUDIBLE].

Let's just say the
direction is this.

Some people use direction
in other courses,

like linear algebra.

In linear algebra, you will
see they say 1, 1, 1 direction.

2, 2, 2 is the same direction.

3, 3, 3 is the same direction.

But I thought you mean
that restrictive case

of this course, my
direction will mean that.

So it's 0.

All right.

What else do we have here?

STUDENT: I have a
question about that.

MAGDALENA TODA: Huh?

STUDENT: Why would
that be unique

if it really doesn't
matter what the point is

if you're going to get that?

MAGDALENA TODA:
For a given point.

So once somebody gave
you a given point in 0,

y0, you have only
one answer, which

is f sub x and x0,
y0, f sub y at x0,

y0 in angular brackets
divided by-- I'm

too lazy to write
it down, but I will.

This is the answer.

So somebody [? got ?],
gave you a point x0, y0.

And it asks you--
and a function.

A point and a function.

And the point is on the
graph of the function.

It asks you what is that
direction? [INAUDIBLE].

So again, in other
classes, you will

see people who say, well,
after changing orientation,

I can go back to Amarillo
if I was driving this way.

It's the same direction.

For us, no.

So for us, that is exactly
the opposite direction.

Compute the volume.

Now, you like this one and I
know you remember it very well.

If you don't like
this kind of question,

I can switch to another one.

But all the questions
should be easy.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Yes, ma'am.

STUDENT: I don't
understand why number 6 is

unique at that point.

Isn't the gradient the
same at every other point

since the derivatives of f of
x-- or since f of x and f of y

don't have x's and y's in them.

They're both 1.

So is it not the same
for the entire graph,

but at a different point?

MAGDALENA TODA: No.

If you take another point--
OK, let's take this one, right.

STUDENT: Isn't it
asking specifically

for the graph of x plus y?

MAGDALENA TODA: Specifically
for the graph of x plus y.

[INAUDIBLE]

What is that direction for here?

i plus j over square root of 2.

That's unique.

STUDENT: OK.

So it's not asking
for the point.

It's asking for the function.

MAGDALENA TODA: At
the given point.

So let me eliminate--
this is too philosophical.

But it says, unique
for the given point

because the point is
still given as 1, 1.

For that given point 1, 1
the direction is unique.

In the sense, this
is exactly what

I wanted to emphasize
that in other classes,

they will say the direction 1, 1
is the same as the direction 2,

2 is the same as
the direction 3,

3 is the same as the
direction 1, 1, 2, 1, 1, 1, 2.

All the colinear vectors
are the same direction.

They point in the
same direction.

But in our course,
this direction

is uniquely defined as the
gradient over its norm,

especially to make it unique.

We build that especially
to make it unique.

I'm pointing in your
direction, but every arrow

that I'm pointing
in your direction

could have any magnitude, and
[INAUDIBLE] in everyday life

speaking, in physics-- ask
anybody in physics, mechanics--

they will say it's the
same direction pointing

towards her whether you
call it this or that

or that, it's the
same direction.

You are pointing in
the same direction

because they consider direction
being all the colinear vectors.

I consider direction only this.

Then [INAUDIBLE].

It's just-- it's not even
a mathematical problem.

It's a philosophical concept
I'm trying to make [? out ?].

But now that is more
like the homework problem

you had before.

OK.

This one you had
in your homework.

[INAUDIBLE] the volume of the
tetrahedron, x plus y plus z

equals one, situating
the [INAUDIBLE].

Maybe in the actual
exam, I'm going

to say let's run up the integral
of the volume without solving

it, or something like that.

Let's do both because
I am afraid that you're

going to cheat.

And you're welcome to
cheat on this problem

as much as you want.

But I would like you to also
show me the actual integral.

So the volume will be how much?

This is 1, 0, 0;
0, 1, 0; 0, 0, 1.

The plane is x plus
y plus z equals 1.

Am I right?

Did I put 1 or I
put another number?

I put 1.

OK.

So you know how to
cheat on this problem.

So volume means area of the base
times the height divided by 3,

which is 1/2 times 1/3 is 1/6.

But what I'm going
to get, I think

I'm going to
rephrase the problem,

say set up the integral
as in triple integral

with or without solving it.

If you show me the answer,
you get extra credit.

You don't have to solve
it by hand in this case.

So you guys showed
me last time what

the answer was for dz, dy,
dx order of integration.

And how was it, guys?

Mr. x is moving freely.

He's the only free
man in the picture

because even he
has bound limits.

But he's free to move
between 0 and 1 [INAUDIBLE].

And then y will go
to x plus y plus 1.

So y will go all the way to
between 0 and 1 minus x, right?

So let me write it down.

0 to 1.

Next, 0 to--

STUDENT: [INAUDIBLE].

MAGDALENA TODA: 1 minus--

STUDENT: Oh, I'm sorry.

[INAUDIBLE]

MAGDALENA TODA: --x.

STUDENT: I thought you were
asking about [INAUDIBLE].

MAGDALENA TODA: You're fine.

I mean, this is--
you can change.

Actually one time, I think,
last year-- maybe two years ago,

last year-- we said
to the students,

without solving this,
show the volume integral

in three different
ways by changing

your order of integration.

So you can have dz, dy, dx.

Or you can have dy, dz, dx.

Or you can have dx, dy, dz.

So they have to turn
their heads in many ways,

and rotate, and see how it is.

But it's a nice problem because
it's very easy to figure out

what the endpoints are.

The last endpoint-- actually,
the first endpoint if you think

of integration-- 0 to--

STUDENT: 1 minus y.

MAGDALENA TODA: --1 minus x
minus y because you see Mr. z,

Mr. z goes up like a helium
balloon from the floor

and he hits his head here.

And he says, I hit my head there
because I'm between 0 and 1

minus x minus y.

And that's the
upper bound for z.

And you're done.

And that's what I
would-- the problem

would be the same problem.

Well, maybe a problem like that.

But not hard.

Number-- what was that?

STUDENT: 7.

MAGDALENA TODA: Thank you.

Yes.

STUDENT: Can you change the
order of dx, dy, and dz?

MAGDALENA TODA: Yes
OK, let's do one.

Of course I can, but
do you want to see?

STUDENT: Yeah.

I want to see, please.

MAGDALENA TODA: 1,
dx-- whatever I want.

I can do whatever.
dz, dy or dy, dz?

STUDENT: dy, dz.

MAGDALENA TODA: dy, dz.

So z is still between 0 and 1.

He is the guy to start with.

So what do you think?

If you want to do it like this--

STUDENT: Next one is a dy.

Then we'll do the 1 minus x.

MAGDALENA TODA: The
relationship that

counts in the first picture,
the first frame that you see

[INAUDIBLE], is the
relationship between y and z.

So Mr. z is free to move from
0 to 1, and Mr. y is here,

and he's conditioned by z.

And you have to write down the
equation of this line, which

is y plus z equals 1.

[INAUDIBLE]

So y goes all the way
from 0 to y minus z.

And x goes all the
way from 0 to what?

STUDENT: [INAUDIBLE].

MAGDALENA TODA: y minus z.

The same thing.

So how can you even [INAUDIBLE]?

Do you guys play cards?

You know what a-- OK.

It's like in cards at the end.

After king, the ace is coming.

And after the ace, a [? 2y ?].

So it's a circular
permutation in the sense

that x can be replaced by
y, can be replaced by z,

can be replaced by x, and so on.

So what do you
replace in this case

to make a circular permutation?

z becomes x.

y becomes y.

This is held in place.

And z becomes and
x becomes the z.

So what I can do is just replace
the x's by z's and I'm done.

And the x, the x by
z and the z by x.

This is circular permutation.

Have you heard the
presentation before?

You can write them
even in matrix form.

I don't know if
in linear algebra

they will do permutations,
but they should.

They cover permutations usually
in higher math, in 3310.

Or in graduate school, they
start with that in algebra.

All right.

I'm going to move on.

I'm thinking I'm not sure
what I want to do really.

But I'm also thinking of
proposing an extra credit

problem.

I will say maybe
something like that.

y equals x, or maybe
any one of the problems

could be considered
extra credit problem.

I don't know.

The same idea of reversing
the order of the integral,

I would like you guys
to-- Where do they meet?

At 1?

Right.

So keep in mind that I may
not give you the points

and may only give
you the functions.

And you have to find
the points where

the two functions intersect,
the two lines intersect.

And it's a piece of cake
because you set x as x squared.

x squared minus x is 0.

x times x minus 1 is 0.

So you have solution
x1 equals 0.

That gives you the origin.

And x2 equals 1.

That gives you 1 and 1 and
then becomes a piece of cake.

I may ask you to
say the following.

Find me the area of this
beautiful leaf by both methods,

or reverse the order
of integration.

Set up the integrals in both
ways and get me the answer.

I have no idea
what the answer is.

I used to know these things
by heart, [INAUDIBLE].

So you have integral
from 0 to 1, integral

from x squared to x.

Am I right?

1 dy dx.

Or-- and now I'll shut up
Magdalena, and let people

talk-- reverse the
order of integration.

What do you have?

STUDENT: Still at 0, 1.

MAGDALENA TODA: Still 0 to 1,
but Mr. y is happy about that.

[INAUDIBLE].

We have to [INAUDIBLE] in
our head and [INAUDIBLE].

STUDENT: Square root of y.

MAGDALENA TODA: y
squared is smaller.

Yes.

He wants to be on the
bottom and this is why.

So I'm too lazy
to write it down.

If I have three people
who will tell me

what the answer is [? so I'll ?]
write the answer down.

The area.

The area of this
beautiful shape.

STUDENT: [INAUDIBLE].

MAGDALENA TODA: Is it?

Yeah.

I need three people to tell
me what it is because it's not

that I don't believe you.

I know you're a good
student, but I just--

I need witnesses so I
can sign the document.

Integrate.

STUDENT: [INAUDIBLE] integration
[INAUDIBLE] y squared.

MAGDALENA TODA: x is
between y squared and y.

y [? is usually ?] [INAUDIBLE].

STUDENT: It would
be square root of y.

Square root of y, yeah.

Square root of y.

MAGDALENA TODA: Oh.

Thank you.

So x is square root of y,
and x-- thank you very much.

I'm very happy.

You told me and I believed you.

See why you should
never believe me.

Square root of y.

And this is x
equals y, of course.

Good.

So very good.

So which one are
you going to do?

This one or this one?

Hmm.

It's easier to do this
one numerically quickly

in your mind.

Come on. [INAUDIBLE].

STUDENT: We have 1/6.

MAGDALENA TODA: OK.

So square root over
2, x cubed, 1/6.

So I'm sorry for not believing,
but I should [INAUDIBLE].

STUDENT: That's still
only two people.

MAGDALENA TODA: Huh?

STUDENT: That's still
only two people.

MAGDALENA TODA: Yes.

All right.

Don't expect something hard,
but of course, always check it.

Check your work even
if you trust yourself

because any of us
can make a mistake.

This was [INAUDIBLE].

So you see just
think of everything

you write down and think,
is this what [INAUDIBLE]?

Even if you know the
material very well,

you can make an algebra
mistake any time.

All right.

[INAUDIBLE] number 8.

Compute the area of the
part of the plane which

lies about the xy lane
in the first [INAUDIBLE].

So practically they say
line about the xy plane,

there is this part of
the x plus y plus z.

You want the area of that guy.

Now, two ways to do it.

I would not feel like repeating
it because it's in your notes.

How did you do this-- OK.

How can you do this by
elementary mathematics?

STUDENT: It's an
equilateral triangle.

MAGDALENA TODA: It's an
equilateral triangle of side l,

and then you said
area is l squared.

Square root of 3 over 4.

How di we do that with a
height-- this is l over 2.

This is l.

This is square
root of 3l over 2.

We did the area.

We came up with that.

So the answer would
be l is root 2, right.

This is 1 and this is 1.

So root 2, root 3 over
4 is root 3 over 2.

But that's not how I
wanted to solve it.

So I may say using the
formula of the surface area,

find me the area
of the triangular

surface in the picture.

And in that case,
what do I need to do?

We need to know a
formula, which is what?

Which says double integral over
the domain d square root of 1

plus f sub x squared plus
x sub y squared dx dy

where f is easy to pull out
from here-- 1 minus x minus y.

Who is d?

Who remembers who d is?

He must be the
[? projection ?] on the floor.

So for d, I say x
is between 0 and 1.

Am I right or am I wrong?

y is between 0 and 1 minus x.

This is x, y, and z.

What is this number?

Formulas we give on the exams,
usually a piece of cake.

This problem was
picked so that z

would be so beautiful in order
for you to integrate quickly

to the square root of--
what is this number?

Minus 1 squared is 1.

1 plus 1 plus 1.

Root 3, square root of 3.

And square root of
3 will multiply.

OK, what is the meaning
of double integral 1, 1,

dx dy over the domain?

It's actually the
area of that domain.

Say what, Magdalena?

Remember that.

When you had double
integral, you

don't have to do the
work here because double

integral over domain will be
easy to compute [INAUDIBLE].

This is [INAUDIBLE].

So you cheat only in part.

You pretend that you've
already done the triangle area.

Otherwise, you must
say area of the d

will be my hard worker, 0 to
1, 0 to 1 minus x 1 dy dx.

And [INAUDIBLE].

STUDENT: Squared.

MAGDALENA TODA: Yeah.

So it's x over x between 0 and 1
minus x squared [? over 2, ?] 0

and 1, 1 minus 1/2, 1/2.

I actually proved
what in this part,

that the area of this
triangle on the floor

is 1/2 as if we didn't
know that from school.

But I have to
pretend that I don't

know the area, the
pink area, is 1/2.

All right.

OK.

In this [INAUDIBLE], where
else do I want to go?

I want to do this
with your help.

It's sort of like this problem
before but a little bit

different.

Can I go ahead and erase or--

STUDENT: [INAUDIBLE].

MAGDALENA TODA: OK.

So I'll write it down.

Yes.

STUDENT: In this
case, [INAUDIBLE]

x plus y [INAUDIBLE].

MAGDALENA TODA: Mm-hmm.

STUDENT: How would
we go from there?

MAGDALENA TODA: So you
know that that's above

because the plane
intersects the axes

and coordinates by this line,
by this line, and by this line.

If you think that line's
above in this first octent

because it is said-- although
maybe lies above the plane

but everywhere.

Then you will be in
trouble because you'll

get an infinite area.

But in the first octent,
you have a finite volume

because your triangle coming
from here all the way down,

it cuts the planes and you
get that the only thing

you have above the floor is this
triangle, the triangle that you

draw in your imagination because
if I start drawing they will

fire me, even if I have tenure.

If I start drawing with
the marker on the board,

I'll get in trouble.

Or maybe they make me
pay like $10,000 fine.

But you have to
imagine this corner

as being exactly that one and
those three being your 1, 2,

and 3 lines.

That's the only thing that's
above in the first octet.

All right.

[INAUDIBLE] you
have no alternative,

you don't have something like
that. [INAUDIBLE] yes, sir.

STUDENT: I was wondering,
where'd the square root of 1

and then the partial derivative
of x, partial derivative of 1

come from just that--

MAGDALENA TODA: This?

STUDENT: Yes, that part.

Like, I understand the
area and everything,

but I don't understand--

MAGDALENA TODA: We've
done that before.

I'll tell you in a second.

So when we computed this kind
of problem, there was a surface

and this was parameterized.

And we proved that there
is a position vector,

and for that, you have--
imagine like a car

with coordinate lines.

And we've said we want to
approximate the curvilinear

domain between some mesh,
or a rectangular mesh.

And then you have a
curvilinear square.

And the best approximation for
the area of that curvilinear

square was the parallelogram,
the parallelogram

that had the velocities-- so
you have this kind of stuff.

You have [INAUDIBLE], and
you have a delta y here,

and here you have r sub x,
and here you have r sub y.

And you say I don't want
the big parallelogram.

I want the one that is
approximately infinitesimally

small.

So you make

make pixels, and
pixels, and pixels

that are smaller and smaller.

And you can say, if
this curvilinear pixel

is small enough, then
its area will be what?

r sub x, r sub y,
and cross product

in length, which is the area of
the parallelogram times dx dy.

Right?

It's like delta x and
delta y displacement.

When we did the kind of--
this is infinitesimal area.

When we did the integral,
what we said, write rx,

ry cross product dx dy.

And I'm going to say, oh
my God, this is such a pan

to write the surface area.

That's going to be the
surface area of the graph.

This is the graph
s, from surface.

Area of s.

We came up with this.

If you remember, it
was a long time ago.

It was before the spring break.

OK.

We took r as the
parametrization,

and we said, wait a minute,
if r is my parametrization,

it better be easy,
otherwise my life

is going to be too complicated.

So I g x is x, y is y, and
z will be instead of z,

we put the graph f of x, y.

And then we did what?

Then we said, r sub x was 1, and
this is where the 1 comes from.

1, 0, f sub x, r sub y
equals 0, 1, f sub y.

And when we computed
the cross product,

we got-- what is the cross
product, guys, I forget.

ijk, the determinant that
has ijk and the coordinates

on the first row and
on the second row.

So I'm practically
proving the theorem.

This is the theorem.

Most people don't prove
anything in the classroom.

They just say, swallow
these formulas.

They come from God.

You have to believe them.

Right?

But the whole
artificial argument

was brought up by people later.

The way Gauss and Euler,
independently anyway,

came up with formulas
like that was

by finding this
curvilinear little red area

of this curvilinear square
written as an integral.

And this becomes
much more beautiful

when you have z equals f
of x, y as parametrization.

What do you do in that case?

Well, you take the
lengths of this vector.

What is the length
of this vector?

What is this vector, actually?

This vector should
be-- somebody help me,

please-- minus f sub x, i,
minus f sub y, j plus k.

OK, so what is the
length of this vector?

The length of this vector
will be exactly that.

So this is exactly the
length of the-- the area

of the little parallelogram,
curvilinear parallelogram.

That area element will be this.

And that's why you
have this area element.

Let's call this area element.

In general, it doesn't
have to be so beautiful.

In general, it's really ugly.

It's an integral that you have
to involve Matlab in, or Maple,

or something really hard.

You cannot solve this by hand.

Imagine even if you have a
simple example like-- guys,

think of the paraboloid, or
the paraboloid that you liked,

that looks like a vase.

What would be the
parametrization?

You have to do square root,
double integral 1 plus

STUDENT: 4x squared.

MAGDALENA TODA: 4x
squared plus 4y squared.

If you were to do that in a
domain that's rectangular,

you'd just kill yourself.

STUDENT: Could you
say change to polar?

MAGDALENA TODA: With polar, it's
easy, but writing the domain

is ugly.

If it's a round domain, like
a disc on top of a disc,

you switch to polar,
and you say, thank

God I'm above a disc,
but otherwise, it

would be really hard.

So what do you do when
you cannot integrate

because the integration
is too hard to do?

You go to Matlab, which
is written for engineers,

but mathematicians
use Matlab a lot.

Or Mathematica,
which was primarily

written for mathematicians,
I think, by Wolfram

in Urbana-Champaign, Illinois.

And then, what other softwares?

Mainly Matlab.

Matlab would be the
one that helps you.

And Maple.

You can use Maple.

It's very user friendly.

You just plug in the endpoints
for the integral domain,

and then you have the
integrand to that.

But Matlab has much
higher capabilities

from computational viewpoints.

You can also write
you own programs.

If you are a computer
science major,

you would be able to
program in C or C++,

write a program on how
to compute the area

of a curvilinear graph about
some-- and that's neat.

That's just for fun.

Of course, you would neglect
your other responsibilities,

and you say, I'm not
going to a Saturday party,

I am writing this program.

And that's how we become nerds,
by saying no to the parties.

All right.

So let's do this one.

I'm not afraid.

I know that you guys
would do the problem

with no problem
over the interval,

but I've never done
exactly this one,

so I want to see
how you respond.

Set up a double
integral, it says,

to compute the area of the
domain between the two lines,

so that you integrate first
with respect to y and then

with respect to x.

Reverse the order
of integration.

What is the area?

Remark that no written
computation is necessary.

Why?

I don't know why-- why
am I asking you that?

Oh, because it's an
elementary problem.

I didn't think of a graph.

So the answer would
be, draw the picture.

You will see how funny it is.

y equals x.

y equals x plus 2.

It looks horrible.

Guys, you have to forgive me.

These are parallel
lines, for God's sake.

Do they look parallel to you?

Let's say we are in
hyperbolic space,

and then they should--
could be anything.

Now, we [? split them up ?]
[? separate. ?] So you have x

between-- did they say-- x
has to be between 0 and 2.

So this is the point 0, 0.

This is the point 0, 2.

This is the point 2 and 2,
and this is the point 2 and 4.

OK.

Set up a double integral
to compute the area.

Then, reverse the
order of integration,

and tell me the
area, and tell me

why I shouldn't worry too
much about the computation.

So integral from what to what?

STUDENT: 0 to 2.

MAGDALENA TODA: 0.

To 2.

STUDENT: And x to x plus 2.

MAGDALENA TODA:
And x to x plus 2.

Very good.

Of 1, because it's
in the area, right?

dy dx.

Or reverse the order
of integration,

but do it carefully, right?

It's that easy.

And then, it's not easy at all.

Why is this not easy?

First of all, it's not easy
because my picture sucks.

Let's see what it is.

I'll do it again.

Oh.

Will you forgive me for
this horrible picture?

I'm going to draw a better one.

First of all, this is a square.

2, 2, and this is 2, 4.

And I'll draw the lines.

OK, now it looks better.

Because before, your domains
would intersect like crazy.

So right now I can
see that-- if I

want to do the strip test,
vertical strip test, that's

already done, and I'm glad.

But if I want to do
the horizontal test,

I'm in trouble, at
first, a little bit.

Why?

STUDENT: [INAUDIBLE] two
separate [INAUDIBLE].

MAGDALENA TODA: Right.

So I have to divide, split
this integral into two.

So it's saying,
area 1 plus area 2.

And so when you actually
decide which method to use,

you're never going to
compute it with this one.

Yes, ma'am?

STUDENT: Aren't they equal?

MAGDALENA TODA: They are
equal, but I'm saying,

when you compute, you don't
take the worst-- well,

you mean the two?

STUDENT: Yeah.

So can't you just multiply--

MAGDALENA TODA: So you
can say bisymmetry.

In this case, you're lucky.

In this case, you're lucky.

But [? life, ?] in general--
attention in the actual exam.

And you can have
something like that.

So in general, you add them.

You add them, and
you do like this.

In this case, they added them.

So I'll write twice,
but you have to say,

you observe that
bisymmetry, we have

to-- the diagonal in
this parallelogram,

we'll divide the parallelogram
into two congruent triangles.

Yes, sir?

STUDENT: I don't
want to sound dumb,

but if I take integration
between 0 and 2,

and choose as the top function
my biggest function, then--

MAGDALENA TODA: 0 to 2 is,
this is the top function,

this is the bottom function.

STUDENT: No, no.

X plus--

MAGDALENA TODA:
For the 2, 2, 4--

STUDENT: You have x plus 2--

MAGDALENA TODA: --this
is the top function,

this is the bottom function.

STUDENT: I take the
integration between 0 and 2,

minus integration--

MAGDALENA TODA: Plus, plus.

No?

STUDENT: No, he's doing
something different--

STUDENT: Different.

Integration of 0 to
2 that is only in x.

That will be the
same area that--

MAGDALENA TODA: You
mean with respect

to y, or with respect to x?

STUDENT: I'm taking it
with respect to-- I'm just

looking at the graph.

MAGDALENA TODA: Which graph?

Are you looking like this,
or are you looking like that?

STUDENT: No, this way.

Whatever the graph is.

MAGDALENA TODA: This way?

STUDENT: Yes.

MAGDALENA TODA: No.

The x is between 0 and 2, or
y is between 0 and 4 for you.

Are you doing horizontal
strips or vertical strips?

STUDENT: No.

I'm just using the Calc
I, base, then finding

the area of up to two lines.

MAGDALENA TODA: Ah.

Then you are using
vertical strips.

Because this is the same as
Calc I. f of x minus g of x

equals x plus 2.

We haven't solved
the integral yet, OK?

So we didn't get to solving.

We are saying, how do I set up
the reverse integration thingy.

Which I can say observe
congruent triangles,

and then I say twice the
integral from, this is 0 to 2.

And then this guy is
y equals x on top.

This guy is-- well,
[? who ?] is first?

STUDENT: There
should be x equals y.

MAGDALENA TODA: Should
be, this is dx dy.

OK?

This is dx dy.

STUDENT: x should go--

MAGDALENA TODA: So x is between
0 on the bottom, and y on top.

And you have a 1.

And after you do this,
you would double it.

Right?

What I wrote is just-- What
I wrote is just this part.

So in the end,
which one do you do?

Which one is easier
for you to do?

For me, what you
said is the easiest.

STUDENT: Yeah.

MAGDALENA TODA: So
you get-- Right?

So how would my son solve it?

Or your nephew, by the way?

He is third grade, fifth grade?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: He would
say, take this triangle,

cut it, and glue it here.

He doesn't know
parallelogram or anything.

So he says, cut this
triangle, and glue it here.

And he knows the
area of 2 times 2,

and he would have gotten it in
like 5 seconds, what we just

discussed for 20 seconds.

Now, unfortunately,
life is not linear.

Life is very non-linear,
and as you saw,

you get the most surprising
fight with a parent,

or breakup.

That's all examples of
life being non-linear.

We have two
non-linear functions,

or we have some
curvilinear functions

that intersect,
elementary geometry is not

going to help you here.

Only calculus for everything
that is non-linear.

STUDENT: That's why
we have calculus.

MAGDALENA TODA: That's
why we have calculus.

STUDENT: Now, question is--

MAGDALENA TODA: Yes.

I have a question,
as well, so go ahead.

STUDENT: Would you accept
if I use the Calc-based,

even though--

MAGDALENA TODA: This is
what my question was.

Would I accept?

What do you guys think?

So I would like you
to write this down,

because I'm saying it.

I'm saying actually write
down, reverse the integration,

and stuff.

But if you don't,
you just say, I

see from the
beginning I have a 4,

that the area is 4, that's
fine with me, as well.

I don't care how
you get the answer.

The simplest possible
way to get an answer

is the best thin, as long
as the answer is correct.

STUDENT: So, will there be
partial credit on the exam?

MAGDALENA TODA: And,
yeah, by the way.

So like, when I say set
up a double integral,

I would say at 50% to do that.

And for somebody
who doesn't know

how to do that, I would
say-- and the person says,

I know it's 4, I saw it's 4.

It's elementary.

50%.

Partial credit.

Is that fair?

I think so.

So for setting up
the two things, half,

and for doing the numerical
part by any method, half.

And I do partial credit.

All right.

The last problem.

The Last of the Mohicans.

Write and set up only a
triple integral in spherical

coordinates.

Now, you cannot do that, because
I haven't taught spherical

coordinates, and that's
I'm going to do today.

In order to compute the volume
inside the unit sphere, which

is a sphere radius 1
centered at the origin,

we have to do that using
spherical coordinates.

And that's what I'm
going to do today.

So good, that was
the time to stop.

I'm moving to 12.7, Cylindrical
and Spherical Coordinates.

And you take a
five minute break.

STUDENT: Starting now?

MAGDALENA TODA: Yeah.

Because it's also time to stop.

It's between the review
and the new thing to teach.

How was your break?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Did you go
to Florida, or any nice place

with lots of palm trees?

You went home.

That is the best.

Home.

STUDENT: [INAUDIBLE]

[SIDE CONVERSATIONS]

MAGDALENA TODA: All right.

[SIDE CONVERSATIONS]

MAGDALENA TODA: Now, if you
remember polar coordinates,

polar coordinates are
cylindrical coordinates,

so they are the
easier guys to learn.

Practically, I'm talking about
switching from x, y, z in R3

to-- back and forth--
to r, theta, and z.

In which way?

Well, x, y, r, your old friends.

r cosine theta, and r
sine theta, and z is z.

So this is the change
between Cartesian coordinates

and polar coordinates.

And it's a
differentiable function.

So you can go back and forth.

So it's going to be r
is a function of x, y,

square root of x
squared plus y squared,

and then if you look at tangent
of theta, tangent of theta

is always y over x.

So theta will be
r tan of y over x.

So you have a bunch
of inverse functions

that are differentiable.

OK?

Both functions are c1.

The one that goes from r
theta x to x, y, z and back.

Circles, c1 functions.

Differentiable, and the
derivatives are continuous.

So you have no problem
setting up a Jacobian.

What will the Jacobian be?

That's going to be dx
dr, dx d theta, dx dz.

So you take the x, y,
z prime with respect

to r, theta, and z.

Then you have dy dr,
dy d theta, dy dz.

So you can take this function
of these three variables

prime with respect to each
of the three variables,

and then you take this
z with respect to r,

with respect to theta, and
with respect to z itself.

And then you say, oh my
God, that should be easy,

because they are independent.

Exactly.

So you have independence here.

Declaration of independence
between x and z.

So this is 0.

This is 0.

This is 0.

This is 0.

This is 1.

He's a lucky guy.

And these parts, you've
seen these guys before.

Let's see how well
you remember them.

You don't have to remember them.

You compute them again.

So if you differentiate
x with respect to r,

you get cosine theta.

Taking x prime with
respect to theta,

you get minus r sine theta.

Here, you get y with respect
to r will be sine theta,

and y with respect to theta
will be r equals sine theta.

And again, 0, 0, 0, and 1.

So who can compute
this determinant

and tell me quickly, well, if
you are taking Linear Algebra,

and two of you are, you
can tell me immediately.

STUDENT: So it's r squared
[? minus ?] 2r squared?

STUDENT: It should just be r.

MAGDALENA TODA: --just r.

STUDENT: r.

MAGDALENA TODA: It's just r.

You can expand
along the last row,

and you have 1 times the
determinant over here.

The determinant
over here is just

r cosine squared plus r
sine squared, which is r.

So it's r times 1, which is r.

Your old friend r, which was the
Jacobian for polar coordinates,

is still the Jacobian for
cylindrical coordinates.

So that's why I'm saying,
cylindrical coordinates

are your friends.

They want to be your friends.

I will give you an example.

Yes?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: Sorry, sorry.

No, no.

We take that eggshell
from last time.

We put it upside down.

Only one eggshell, OK?

We start with that easy problem.

And this is z equals y minus
x squared minus y squared.

It's half of the egg that we
built from plastic last time.

And I want to express the
triple integral of the volume--

express triple integral
of the volume in terms

of first, Cartesian coordinates,
and second, cylindrical

coordinates.

And you're going to say, well,
I have a circular domain, right?

The disc is a domain
inside the circle.

So it's a lot more natural to
use cylindrical coordinates

and not Cartesian coordinates.

Why do you want us to use
Cartesian coordinates at all?

I just beg you to bear with
me, and not actually compute

the integral in
Cartesian coordinates,

because you're going
to kill each other

in the process about the
square roots you have inside.

So you can go ahead and
write it, integral over

d, of the function on top,
which is going to be-- well,

let's put it in triple form.

The egg, e, you're
going to laugh at me.

It's the triple
integral over the egg.

But actually, it's
triple integral

over the half of the
egg, but I don't know

how to denote half of an egg.

It's [? all the ?] volume
of the body inside the egg.

1 dV.

I can write it as a
double integral over D--

is that clear?

of the z function on top.

1 minus x squared minus y
squared minus the function

below, which is 0.

Which is going to tell me dx dy.

Which is also denoted
[? da. ?] If you're

going to do this in
Cartesian coordinates,

well, good luck to you.

Because you're going to
have something like that.

Mister x says I want to
go between minus 1 and 1.

And that's x.

He wants to go all the
way between minus 1 and 1.

And meanwhile, y says,
but I'm restricted by you,

because I'm in a marriage.

x squared plus y
squared equals 1,

so y goes between square
root 1 minus x squared

and minus square root
1 minus x squared.

So he goes, look how ugly I am.

Minus square root
1 minus x squared

plus square root
1 minus x squared.

And then you have to put
the integrand, 1 minus x

squared minus y squared.

And then you go dx dy.

So it's not the best thing.

It's not the most
beautiful thing.

Now, you say, but
I'm going to write it

in cylindrical coordinates, and
you will see how beautiful it's

going to become, and I agree.

The same thing can be written
as triple integral over the egg,

over the half of the egg,
1 dV, will be 1, 2, 3.

What will be the volume element?

Because the volume
element is not

going to be-- when I switch
to polar-- not going to be dr,

d theta, dz.

It's going to be r
rimes dz dr d theta.

Remember that, because
I've seen students

forget this r, the
Jacobian, all the time.

This Jacobian is very
[? useful. ?] All right.

So I'll put here
1r, dz, dr, d theta.

Mister z says, I know I'm going
from the bottom to the top,

from the bottom to the top.

So z goes, I'm
between 0 and whoever

you gave me, 1 minus x
squared minus y squared.

And he's happy about that.

It doesn't look so ugly.

And now r and theta say, well,
we are sort of independent.

r is always between 0
and 1 in my picture,

because I have the unit disc
on the bottom, on the floor.

And theta is between
0 and 2 pi, because I

have a complete rotation.

If this were chocolate
cake, or just something

made of-- something
good, some mayo tower,

or something, cheese tower.

And I wanted to cut it in
half or cut it in a quarter,

I would be very careful
about the theta.

For example, this is chocolate,
and I cut it as a quarter.

I have to say, just
the first octant.

So you see the piece of the
cake that would be in the first

octant-- now,
[? our coordinate ?]

[? here ?].

So for that kind of problem,
you would have to say,

r is between 0 and 1, but
theta would be between 0 and pi

over 2.

Are you guys with me?

So you were lucky, in
this case, that you have

a full rotation all around.

You eat the entire cake.

But in other problems,
you may have a half

of a cake, a quarter of a cake.

Yes, sir?

STUDENT: Why is it dx dy
down there on the area?

MAGDALENA TODA: dy dx.

You are very smart,
and I appreciate that.

Thank you.

In a regular class--
oh, I'm not even saying.

I made the same mistake one
time, just right in class,

and nobody notices.

I noticed later, but
it was a little late.

Thanks [INAUDIBLE].

So this is y integration
first, x integration.

Good.

Let's see about that.

What am I going to do with that?

That worries me, because
this looks like it

has nothing to do there.

Everything should be
in z, r, and theta,

and this guy is 1
minus r squared.

Otherwise, he doesn't
belong in there.

So I have to write it down.

0 to 2 pi, 0 to 1, 0 to 1 minus
r squared, r, dz, dr, d theta.

Is it hard to do?

Is it hard to solve?

Shouldn't be.

Why?

Well, let's go from the
inside to the outside.

First of all, what is the
beautiful thing about this?

As soon as I'm done
with this integration,

I have a product of
functions in theta and r

that are independent.

So let me write it down.

Integral from 0 to 2 pi,
integral from 0 to 1,

and I decide, I'm all for it.

I'll have integral of r dz.

But z has nothing to do with
r, so he says, well, I'm z,

you're r.

We are independent.

r is like a constant
for the time being.

He can even go here and wait.

And then the integral
you have, it goes out.

He gets out.

You see, guys?

He gets out.

And then, you have 1,
integral of 1, this is z,

z between-- I'm taking
this alone-- between 0 down

and 1 minus r squared up.

I could have
written this faster,

but that's not the point.

So this will be integrated
with respect to r and theta

in the end.

So, again, what did I do?

I said that was the r.

But the r, when I integrate
with z, they're independent.

So r gets out.

Integral of 1 is z.

z between these two guys
is 1 minus r squared.

I'm going to write it out.

Integral from 0 to 2 pi.

Integral from 0 to 1.

Your old friend,
y minus r squared,

times your old
friend r, dr d theta.

Is this hard?

Come on, this is not hard.

This is a piece of cake,
a piece of chocolate cake.

Right?

So, what is this?

It's the integral
of r minus r cubed.

What is your luck here?

Theta is not in the picture.

So theta says, I'm going
to go out for a walk.

You guys finish integration
in r, I'm by myself.

0 to 2 pi d theta.

Say bye, bye, because theta is
not contained in the integrand.

And then you go integral from
0 to 1, r minus r cubed dr.

And now, we have two beautiful,
simple integrals from Calc I

that we should be able
to solve in no time.

I'm too lazy to write,
again, the anti-derivative,

which would be, r squared over
2 minus r to the 4 over 4.

What do I get her in the end?

STUDENT: [INAUDIBLE]

MAGDALENA TODA: 1/2
minus 1 over 4, right?

So I have 1/4.

Did you say the same thing?

STUDENT: I said 1/4.

MAGDALENA TODA: It's
the same thing, 1/4.

And this is?

STUDENT: 2 pi.

MAGDALENA TODA: 2 pi.

So in the end, I allow
him to write it like that.

Of course-- but you'll
say, well, it's silly.

I should be able to simplify.

Well, go ahead and simplify.

How much do we get?

STUDENT: It's 5/2.

MAGDALENA TODA: 5/2.

So this says that the
volume of the chocolate

cake-- now I'm dreaming, now
it's not an eggshell that's

empty in the middle
anymore, it's just

a chocolate cake all
round up and made

in the shape of a paraboloid,
or some hill made of chocolate

mousse that I made.

So 5 over 2 will be
the volume of that.

Is this a hard problem?

It shouldn't be.

If you think about
it, it's nothing new.

It's all polar coordinates
and nothing else.

But you may want to
see one more like that.

Do you want to see another
one like that, or not?

Yes?

STUDENT: I did the same, but I
couldn't get the right answer

in the homework.

Because the first
level is in Cartesian,

and then you have to change
to spherical, and then--

MAGDALENA TODA: Algebra is
a problem for all of us.

Did you remember to include the
r, and do everything slowly?

STUDENT: Yes.

My issue was the [? borders. ?]

MAGDALENA TODA: The
boundaries, the midpoints?

STUDENT: --it's not-- if I
put the 0, it's not right.

So--

MAGDALENA TODA:
Well, send it to me.

Send it to me from WeBWorK,
and we will see what I can do.

It's still unsolved?

OK, I'll help you on it.

All right.

So, let's move on.

STUDENT: Yeah, I
tried [? 5 and 4. ?]

MAGDALENA TODA: --to something.

Let's move on to
something trivial.

You are going to say, come on,
Magdalena, this is trivial.

I don't want to do this.

Of course I know we'll
find it really easy,

the volume of the can of
Pepsi Cola or whatever.

It's a cylinder of
radius r and height h,

and I want to set
up the integral

to compute the volume
of the cylinder

in cylindrical coordinates.

So problem 2 for
cylindrical coordinates

to compute the
volume of a cylinder.

Did we know that from before?

I have my helper, here.

He has [INAUDIBLE].

You all know him.

Area of the--

STUDENT: Volume [INAUDIBLE]

MAGDALENA TODA: --base.

STUDENT: The area of
the base, I'm sorry.

MAGDALENA TODA:
--times the height,

which will be pi r
squared h, right?

This is what we should get
when we compute the integral.

If we don't, that means
we messed up somewhere.

STUDENT: Wow.

Well get pi r squared z.

MAGDALENA TODA: Let's put big
R, because I reserved little r

for the variable
between 0 and big R

for the polar coordinates.

So this is big R,
and this is little h.

And this is your domain.

So how do you write cylinder?

Well, let's say the
body inside is called c.

I have triple
integral over c of 1.

dz, dy, dx.

Of course, we won't do this
in cylindrical coordinates,

so this dV becomes
r, dr, d theta, dz.

So I will have to do
it in proper order.

r dz first, dz, dr, d
theta, because-- well, it

doesn't matter in
this case, z is still

between some fixed values.

z is between 0 and h.

Again, if this were a
plane or another surface,

you would write it as a variable
function of two variables.

Then you have r between 0 and r.

You have theta
between 0 and 2 pi.

I could give you on
the final something,

a quarter of a salami.

Just cut the salami
on the first octant,

and get the same problem
as this, but instead

of theta from 0 to 2 pi, you
would have theta from 0 to pi

over 2.

So, is this hard?

It shouldn't be hard.

r, again, gets out.

He says, I'm out of her.

I have nothing to do with z.

So z is by himself.

Z by himself gets integral of
1 dz, z. z between 0 and h.

So h is by himself.

So I have, again, integral,
integral 0 to 2 pi, 0 to big R.

And here, I have h.

Are you guys with me?

Are you following?

dr d theta.

STUDENT: h should [INAUDIBLE]

MAGDALENA TODA: And
the r-- I forgot.

You see how easy it is
to forget. r is here,

but I forgot it here.

r is here.

He was hiding behind the
bush, so I pulled him back

in the picture, because I need
to integrate with respect to r.

Now, fortunately for
us, this is easy.

It's a piece of cake.

What is a constant
for everybody?

Little h.

He goes out in a hurry.

He says, I'm hurrying out.

I have h out.

But then you have
a double integral

of a product of two
functions in one variable.

So you have one integral in
r only, which is this one.

Are you guys with me?

And one integral in theta only,
which is integral of 1 d theta.

Can you tell me the limit
points for the integrals?

STUDENT: 0, r.

STUDENT: 0 and r.

MAGDALENA TODA: 0
and r for this guy,

STUDENT: And 0 and 2 pi.

MAGDALENA TODA: And 0
to 1 pi for that guy.

And now, it should be easy.

Am I getting the right answer?

Was this the [? guy ?] and why?

Because I have h r
squared over 2 times 2 pi.

2 goes away, so I've got exactly
h times r squared times pi.

Which was the volume
of the cylinder

that our teachers gave to us
in K-12 with no explanation.

They said, well, you
have to understand,

it's basically, if you
believe that the disc has

area pi r squared, you multiply
that, it'll make 8 sheets, 8

tiny sheets of that.

Then you multiply,
and that's how

we understood the
notion of a volume,

by repeating those
cross-sections.

You have an altitude
of h, and you

have cross-section
after cross-section

after cross-section.

But still you have to
believe the area of the disc.

How did we believe the area of
the disc when we were little?

I remember I didn't believe it.

My teacher in fourth grade came
up with some graphing paper.

And we went ahead
and compared what

I got from the little tiny
squares and the actual formula,

and it was hard
for me to believe.

But I was surprised
that I couldn't

find an explanation for my
pi r squared formula, even

in middle school.

And then I started high school,
and still no explanation

for that.

So I was very
confused, I was like,

is math more like history, where
we have to memorize everything

and believe everything?

But thank God for
calculus, as you said.

In college, I
understood, finally,

why the area of the
disc is pi r squared.

OK, let's move on to our
friend that I announced today

with lots of advertisement.

Yes, sir?

STUDENT: What happens if
you ask the area [INAUDIBLE]

between x and y plane?

MAGDALENA TODA: Between?

STUDENT: x and y plane.

What happens in that?

[INAUDIBLE]

MAGDALENA TODA:
Between which plane?

STUDENT: x and y.

MAGDALENA TODA: x and y plane?

STUDENT: Let's
say z and y plane.

MAGDALENA TODA: z
and y-- which planes?

STUDENT: This--

MAGDALENA TODA: z equals y?

You mean x equals y is a plane.

STUDENT: Yeah.

MAGDALENA TODA:
You can ask, well,

yeah-- let's say I'm asking the
volume of the slice of the cake

that is between
x equals y plane,

and this plane, x equals
0, and z equals something.

Flat.

Then I have to think of,
again-- so theta, right?

I have to take out a 4.

0 to 5 or 4, and
measure differently

when I measure the angle, right?

We will see some
sorts of measurements

for angles that
will surprise you.

When we look at
geography on an atlas,

we all those latitude and
longitude coordinates,

but nobody ever told us
about the angles very well.

How do we actually define
longitude and latitude?

Well, between mathematicians and
people who work in geography,

there is a consensus.

Longitude is this thing that
you measure around the equator.

That is the angle.

For people in
geography, it starts

at the Greenwich Village.

I mean, the meridian passes
through Greenwich Village.

I went there out of curiosity.

And there is nothing there.

It's just like a suburb
of London, or something.

It goes through-- so
meridian 0 goes through that.

That's where you
measure theta equal 0.

And you go around up to
say that equals 2 pi.

And how many meridians
do we actually have?

STUDENT: There are 24.

MAGDALENA TODA: So when we
them 1 degree apart, there

should be 360 things like that.

STUDENT: [INAUDIBLE]

MAGDALENA TODA: What?

STUDENT: Yeah.

MAGDALENA TODA: Huh?

STUDENT: Yes, but how
many do we define as--

MAGDALENA TODA: Yeah.

In geography, we say 0 degree,
1 degree, the discretely many 0,

1 degree, and so on.

So the one that's
at 180-- OK, it

goes 0, 180, is the meridian
that, if we were to continue,

it would go to meridian
0, past the north pole.

All right?

So we have this
notion of longitude

that we are going to
introduce from 0 to 2 pi.

And that's fine.

But how do you measure latitude?

Longitude is up.

Latitude, in geography
and in mathematics,

they may differ a little bit.

We can even change our notation
for latitude sometimes.

In geography, always from
0, the equator, going up.

At the North Pole,
what is your latitude?

STUDENT: It's going
to be 90 degrees.

MAGDALENA TODA: 90 degrees.

And how do we say that?

STUDENT: [INAUDIBLE] Well, in
math we say-- I would, too,

but-- Now we say North, 90
degrees North in geography.

For a mathematician,
it's not like that.

Mathematicians are crazy people.

The go minus pi over
2, which is minus 90.

The geography guy
says 90 degrees South.

OK?

What latitude are
we here in Lubbock?

Do we know?

Come on, you know.

STUDENT: 32.

33.

MAGDALENA TODA: Something.

33, I thought, but
maybe I'm wrong.

Something like that.

STUDENT: 32.8.

MAGDALENA TODA: 32.8.

I don't know.

In any case, when
we measure latitude,

we should always start-- if we
are normal, we should always

start at the equator,
because that's

how we did it in school, right?

To go this way and this way.

Mathematicians don't.

They say, no, I'm
going like this.

I'm going from minus pi
over 2 to 0 and 2 pi over 2.

That's one way to do it.

One way.

Minus pi over 2 to
0 and pi over 2.

But here in calculus, we
are absolutely not normal.

So we have another way
of defining latitude.

We say latitude
from the North Pole.

Latitude from the North
Pole I know how to spell it.

So we measure like this.

Look at me.

You are going to be amused,
because you will see that we

are crazy, which is true.

But this is already
accepted as the notation.

So we go like that
all the way to 180.

So this-- the North
Pole would be-- so what

is the angle for the point p?

The angle for the
point p of latitude--

not theta, theta was the
larger-- let's call it phi.

Phi is between 0
degrees and 180 degrees,

which is the South Pole.

So we measure-- I think
because mathematicians still

believe in Santa, I don't know--
they go from the North Pole, 0,

90, 180.

So this is the angle phi
that we are defining.

Let me show you on a picture
what the spherical coordinates

are, the way we are
going to use them.

So I take a big
sphere-- a big sphere.

And these are the axes
the way you know them.

This is x-axis, this is y-axis,
and z-axis, and the origin.

And you pick your favorite
point somewhere on the sphere.

And you say, I want to see
what spherical coordinates are.

Those spherical coordinates
are three coordinates, r theta,

and phi.

r is the radius from p to o.

Theta is the longitude, and phi
is the latitude from the North

Pole.

OK.

So let me draw.

Let's take this red,
and let's draw op.

This is r.

But then we say, well, I could
still be inside the sphere.

Yes, you can still
be inside the sphere.

It could be anywhere
you want in space.

Then, how do you measure the
latitude and the longitude?

First of all, to measure,
let's go from p down,

and that's going to
be p prime somewhere.

Then I'm going to
say that's op prime.

And then I will project on
the x-axis and the y-axis.

So I'm putting 90-degree
angle, 90-degree angle.

This is the x coordinate,
this is the y coordinate,

and this is-- how do
I do the z coordinate?

Through the point
p, I should draw

a plane that's horizontal,
until that horizontal plane is

going to cut the axis z.

I'm going to do that.

And then what I get is something
that finishes my rectangle.

I have a 90-degree, 90-degree,
90-degree, 90-degree.

And this is z.

This is z.

So we have to write down
everybody, x, y, and z,

in terms of r theta and phi.

This is going to be
just trigonometry.

We are going to get a
little bit of a headache,

but I know that you
are going to help me.

So the latitude measured
from the North Pole

is this angle, this
angle phi, which

is the same thing
as this angle phi.

And where is the angle theta?

I need a line-- this is
line color, and I have here.

That's theta.

So somebody needs to help me.

The easiest is z.

Why is z the easiest
to write down?

Because-- It's going to come
from the Pythagorean theorem,

for whatever
trigonometric angle phi.

I have a 90-degree.

This is the triangle.

What do you think
z is going to be?

I'm going to go 90.

In this triangle, I
have an angle phi.

And z is the adjacent.

Phi is the hypotenuse.

So let me write down.

z is r cosine phi.

Yes, in this case.

Why is that?

Because I'll take z over
r, and that's cosine phi.

z over r is cosine phi.

It's the adjacent
over the hypotenuse.

All right, that was easy.

But the problem is, how
do I get to x and y?

Now, unless I find
the protection,

I need you to give me the
line op prime measure.

This is the hard--
not hard, but-- It's

from this other triangle.

I'm going to fill it
in red dots everywhere.

STUDENT: That lined
line is r the sine of 5.

MAGDALENA TODA: r sine 5.

Why is that?

Because sine of phi will
be exactly the-- where

is the 90-degree angle here?

So the sine phi will be the
opposite over hypotenuse.

Say it again.

Opposite over hypotenuse.

And that's the 90-degree angle.

So good.

So it's op prime over r.

Good.

But now, I have to go on,
because the [? lined ?]

guy is not the end of the game.

The end of the game is x and y.

So x is op prime cosine theta,
and y op prime sine theta.

Who can tell me why?

STUDENT: Because x
is your adjacent,

and y is your opposite.

MAGDALENA TODA: Right.

Because this is also
a 90-degree angle.

So I have a right triangle,
and if I take the angle theta,

this is the adjacent
for the angle theta.

So cosine theta is x over r, x
over op prime, the hypotenuse.

So again, let's
write, x over op prime

is going to equal sine theta.

y over op prime is
going to be sine theta.

That's why I have those.

Good.

So can I write x and y?

X will be-- now comes a
little bit of headache.

op prime is r sine
phi, cosine theta.

r sine phi, cosine theta.

Y equals r sine phi sine theta,
and z equals r cosine phi.

Good.

These are the famous
cylindrical-- spherical

coordinates.

Spherical coordinate
transformation.

So what does this
transformation do?

This transformation takes
the r, phi, and theta,

and moves them into
the image x, y, z.

So this is the transformation
I was talking about.

STUDENT: Doesn't it
go from x, y, z to r--

MAGDALENA TODA: It goes back.

So you have a
one-to-one function.

You have an inverse, and
the inverse is also c1.

They are both c1 functions.

The inverse.

Who is the inverse?

I don't know, guys.

OK, I'll only give you one piece
of information that's crucial,

and you'll have to find
phi and theta, which

are not hard to find.

Who is r?

STUDENT: r is still x
squared plus y squared.

MAGDALENA TODA: r is x squared,
plus y squared, plus z squared,

under square root.

Let's see why that is.

You may not believe
me, but I'll do it.

So assume that you square x
squared plus y squared, right?

Then you have r squared
sine phi squared.

But the square of this plus
the square of that collapse

and will become a 1
when you add them up.

So When you add up x
squared plus y squared,

this goes away and
becomes 1, and you have

r squared sine squared phi.

Now, if you make z
squared, who is the square?

R squared cosine squared phi.

Add them all up.

x squared plus y
squared plus z squared

equals little r squared.

So now we are ready to
prove the volume of a ball.

We have this ball of radius r.

Find the volume of this ball
with any kind of coordinates

you think.

But now, you say, OK, if you do
cylindrical coordinates, good

luck to you.

It's going to be, we'll
stay here until midnight

computing the volume.

Let's try both,
and let's see how

they differ from one another.

How hard they are--

STUDENT: Can you do Cartesian?

MAGDALENA TODA: Hmm?

STUDENT: Are you
going to do Cartesian?

MAGDALENA TODA: I'll
write Cartesian,

but I'll write Cartesian,
I'll write cylindrical.

I'm not going to
use them, though.

I'm going to say goodbye
to them, and in the end,

I'm going to use spherical.

Now, the beautiful thing about
the Jacobian for spherical

is that it has to be memorized.

Because for you to write
it down as a determinant,

dx dr, dx d phi, dx d theta, and
there are six more like that,

it takes forever.

So it's better that I
tell you what it is.

I compute it a few times, and
we get r squared sine phi.

So what is the volume of a
sphere, the volume of the ball?

I can write a triple integral
of 1 dV over the ball

B. Let's call this ball
B. Script, beautiful B.

And that will take care of
the-- this is dx, dy, dz, right?

That's the first formula.

I cannot use it.

It's going to be a mess.

Now, let's try this written
at least in cylindrical

coordinates.

You will still see
that is not easy.

When we write it in
cylindrical coordinates,

I'm going to have cylindrical
coordinates, [? still. ?]

It's going to be what?

It's going to be an r,
and then dz, dr, d theta.

And z, well, we can
do what [INAUDIBLE]

said, mapping several of them.

When you have a symmetric
object, you split it in half,

and you say double the area.

We can always do that.

Or you could go from
the Southern Hemisphere

to the Northern Hemisphere,
and you practically

get the same thing.

Now, for the ball
we have here, you

have x squared,
plus y squared, plus

z squared equals-- what was the
radius of the ball? r squared.

OK.

I don't like it, but that's OK.

z will have to go
between what and what?

Between two nasty square
roots and a minus.

r squared minus x
squared, minus y squared,

and r squared minus x squared
minus y squared with minus.

Do I like it?

No.

But at least--
that's in Cartesian.

At least in polar
coordinates, I can

say, mister z decides to go from
minus square root r squared,

minus r squared,
because x squared

plus y squared is cylindrical.

This is r squared
in cylindrical.

OK?

So I would do it like that.

Square root r squared minus
r squared, r from 0 to r,

and theta from 0 to 2 pi.

And I don't want the headache
of these square roots,

because as soon as I do
that-- well, I can do it,

bit it's obnoxious.

I take the z out.

z between the 2, I'll take
z between this and that.

So I'll have twice the
square root times little r.

I have to perform some u
substitution, and to be smart,

and I think I can do it.

This in cylindrical.

But I want to show you how
beautiful it is in spherical.

In spherical-- what are you
going to have in spherical?

In spherical, you have
this, the ball in spherical.

1, 2, 3 of 1.

1 times the Jacobian.

Who is the Jacobian?
r squared sine phi.

Good.

What else?

Anything else you want,
dr, d phi, d theta,

they are all between
fixed endpoints.

So you are really lucky.

Why are you lucky?

Because all these guys,
r, phi, and theta,

are between fixed
endpoints for a ball.

Do we know those endpoints?

Well, we better.

If we don't, then
we are not with it.

So r for the ball goes little
r from 0 all the way to r.

Good.

Now, the latitude
measured from Santa Claus.

That is the question.

It's phi, right?

It's from 0 to the
South Pole, phi.

And theta is from

STUDENT: 0 to 2 pi.

MAGDALENA TODA: 0 to 2 pi.

From the Greenwich
meridian all the way back.

Good.

So now we are ready.

Now it's a blessing, because
we apply the Fubini theorem,

and all the endpoints
being fixed,

I can separate this
integral into a product

of three integrals that
are all independent.

And I'm going to ask you
to tell me what those are.

Yes, sir?

STUDENT: Why don't we
just go from 0 to pi?

MAGDALENA TODA: Because--
you mean for phi?

OK, phi is the latitude
measured from the North Pole.

That's why I said Santa Claus.

Because mathematicians,
many of them,

are split into two categories.

Some say, I measure
latitude between 0 and pi,

and 0 minus pi over 2, so I can
go all the way between minus pi

over 2 to pi.

But some say, no,
I don't like is.

I measure the phi
from 0, because it's

easier for the picture.

And I measure phi from North
Pole all the way to the South

Pole, and the latitude goes
all the way to pi, to 180.

Now, for me, it was
easier, because I

was able to draw
this picture, and I

was able to write those
formulas very easily.

Now, for those integrals, can
you tell me, the first one

in r, can you tell?

Who can tell me the r?

STUDENT: 0, r, r squared u.

MAGDALENA TODA: Very good.

So that's the only thing
that's a little bit hard.

Everything is the same.

The integral from--
sine phi d phi from

STUDENT: 0 to pi.

MAGDALENA TODA: 0 to pi.

And finally, integral from 0 to

STUDENT: 2 pi.

MAGDALENA TODA: 2 pi.

Of 1 d theta, and you
say that's just 2 pi.

I know.

That is the beauty of it.

It's 2 pi.

So that simplifies a
little bit my life.

How about the first fellow?

I know you love
the picture, right?

You love this picture.

But I need to say goodbye to it.

OK.

So I'll erase it.

And I have r cubed over 3. r
cubed over 3 starts showing

the meaning of
the volume formula

that you knew from high school.

So this guy becomes
r cubed over 3

means big R cubed over 3 here.

The first guy.

This first guy is
this first guy.

This guy, not a problem.

Integral of sine phi
is minus cosine phi.

And that, you have to measure
that between 0 and phi.

So you have to do it a
little bit carefully.

One time, I had a student
who got 0 on that.

After he plugged in
here, he got a 0,

so he came up with
this new theorem

that the volume
of the ball was 0,

no matter what the
radius was for him.

But he didn't realize
what was going on.

So you realize what
his mistake was.

He didn't get the signs right.

So minus cosine of pi.

What's cosine of pi?

STUDENT: Negative 1.

MAGDALENA TODA: Negative 1.

With a minus, it's minus
minus 1, which is plus 1.

Right?

But then you have
minus, and he wrote 1,

and this is where
the mistake was.

Minus the other 1 is minus
cosine of 0, which is minus 1.

So you get 2.

All right, and
then, finally, 2 pi.

So what did we get here?

We got something beautiful.

We got--

STUDENT: 4/3 pi r cubed over 3.

MAGDALENA TODA: Huh?

STUDENT: 4/3 pi r cubed.

MAGDALENA TODA: Yeah.

We got 4 pi r cubed over 3.

This is the thing we knew
from when we were little.

How old were we when our
teachers told us that they

cannot tell us why, but this
is the volume of a sphere?

Were we in ninth grade?

Younger?

Eight grade?

But I don't think I
was able to memorize.

That was one of my problems.

I think in eight grade, my
teacher gave me the formula.

But I wasn't able to memorize
that in eighth grade.

And I think in high school,
I memorized it finally,

like when I was 16 or something.

So this is 4 pi r cubed over
3, and it closes that question.

Next time, I want to do one
or two more applications

for the spherical coordinates,
because you're not

yet used to them.

Maybe I'm going to do a
mixture of the problems,

cylindrical, spherical.

You know what to expect.

Don't expect
anything really hard.

But it walks you slowly
through the whole [? deal ?]

if you go over the set.

OK?

enjoy the day, and ask
me questions if you have.

I'll be in my office hours.