TTU Math2450 Calculus3 Secs 12.7 - 12.8

Review of Integration in 3 dimensions and Jacobian of a transformation

PROFESSOR: I know you have
encountered difficulties

on the last few problems,
maybe four, maybe five, maybe

the last 10, I don't know.

But today, I want to--
we have plenty of time.

We still have time
for chapter 13,

and plenty of time
for the final review.

I can afford to spend two or
three hours just reviewing

chapter 12, if I wanted to.

All right, so I have this
question from one of you saying

what part of the problem is
that in terms of a two point

integral.

We have a solid
bounded by z equals 3x,

and z equals x squared,
and is a plane.

And can anybody tell
me what this is?

Just out of curiosity,
you don't have to know.

STUDENT: It's a parabola.

PROFESSOR: It would
be a parabola,

if we were in
[INTERPOSING VOICES]

if we were in 2D.

So the parabola is missing
the y, and y could be anybody.

So it's a parabola
that's shifted along y.

It's going to give you
a cylindrical surface.

It's like something used
for drainage, I don't know.

Water, like a valve.

So this is what it is,
a cylindrical surface.

And you know that z must be
between x squared and 3x.

How do you know which one is
bigger, which one is smaller?

You should think about it.

When you draw, you
draw like that.

Do these guys intersect?

We are in the xz plane.

Do these guys intersect?

x squared equals
3x intersect where?

They intersect at 0, and at 3.

x1 is 0, and x2 is 3.

So when I want to draw this,
I would say that indeed, it's

a bounded domain.

If it where unbounded, it
wouldn't ask for the volume,

because the volume
would be nothing.

So this thing must
be a bounded domain.

x cannot go on, this
is the infinite part.

So we are thinking of just this
striped piece of a domain x

here, where this piece is
between z equals x squared,

then z equals 3x.

This is 0 origin, and this is 3.

So at 3, they meet again.

Are you guys with me?

At 3 o'clock they meet again.

I'm just kidding, x
doesn't have to be time.

It's a special coordinate.

And y is looking at you,
and is going towards you.

Well, if it's toward like
that, it's probably not

positively oriented.

So y should come from you,
and go into the board,

and then keep going in that
direction for the frame

to be positive oriented.

Positively oriented means x like
that, y like this, z like that.

So k must be the
crossproduct between i and j.

i cross j must be k.

If I use the right hand
rule, and I go y like that,

that means I changed
the orientation.

So the y, you have to imagine
the y coming from you, going

perpendicular to
the board, then keep

going inside the
board infinitely much.

Now, if we were to
play with Play-Doh,

and we were on the other side of
that, like Alice in the mirror,

we would have the y in the
mirror world, going between 0

and 2, inside the board.

If I were to draw
this piece of cake,

I start dreaming
again, I apologize.

But I'm dreaming of very
nice bounded pieces of solids

that would be made of cheese.

This is a perfect example where
you have something like curve

or linear shape, and you
kind of slice the cheese,

and that's a piece
of the Parmesan.

OK.

So the y here is
going from 0 to 2.

It's sort of the altitude.

And this is the piece
of cake that you

were looking-- or the cheese,
or whatever you're looking at.

So what do you put here?

You put z between
x squared and 3x.

You put y between--
y is between 0 and 2.

And you Mr. x as the last
of them, he can go from-- he

goes from 0 to 3.

So x has the freedom
to go from 0 to 3.

y has the freedom to go
from flat line to flat line,

from between to flat planes--
from two horizontal planes.

But Mr. z is married to
x, he cannot escape this

relationship.

So we can only take this z
with respect between these two

values that depend on x.

That's all.

Now why would we have 1?

Because by definition,
if you remember

the volume was the
triple integral

of 1 dv over any
solid value domain.

Right?

So whenever you see
a problem like that,

you know how to start it.

One triple integrate,
and that's going to work.

Something else that
gave you a big headache

was the ice cream cone.

The ice cream cone problem
gave a big headache

to most of my students
over the past 14 years

that I've been teaching Cal 3.

It's a beautiful problem.

It's one of my
favorites problems,

because it makes me
think of food again.

And not just any food, but
some nice ice cream cone

that's original ice cream, not
the one you find in a box like

Blue Bell or Ben and Jerry's.

All right, so how is the
ice cream cone problem

that-- he showed it to me, but
I forgot the problem number.

It was--

STUDENT: It's number 20.

PROFESSOR: Number?

STUDENT: 20.

PROFESSOR: Number 20, thanks.

And I want the data [INAUDIBLE].

I want to test my memory,
see how many neurons

died since last time.

Don't tell me.

So I think the sphere was
a radius 2, and the cone

that we picked for you,
we picked it on purpose.

So that the results that
come up for the ice cream

cone boundaries will
be nice and workable.

So we can propose some data
where the ice cream cone will

give you really nasty radii.

Can I draw it?

Hopefully.

This ice cream cone is
based off the waffle cone.

I don't like the waffle
cone, because I'm dreaming.

But the problem is, the waffle
cone is not a finite surface.

It's infinite,
it's a double cone.

It's the dream of every binger.

So it goes to infinity,
and to negative infinity,

and that's not my problem.

My problem is to intersect
this cone with the sphere,

and make it finite.

So to make it the
true waffle cone,

I would have to draw a
sphere of what radius?

Root 2.

I'll try to draw a
sphere of root 2,

but I cannot
predict the results.

Now I'm going to only
look at this v1 cone.

I don't know what the problem
wanted, but I'm looking at,

do they say in what domain?

Above the plane?

STUDENT: It just says
lies above the cone.

PROFESSOR: That
lies above the cone.

But look, if I turn my head like
this, depending on my reference

frame, I have
cervical spondylosis,

this is also lying
above the cone.

So the problem is
a little bit silly,

that whoever wrote
it should have

said the sphere lies above a
cone, for z greater than 0.

In the basement,
it can continue--

that's for z-- for z
less than 0 can continue

upside down, and then between
the sphere and the cone,

you'll have another ice
cream cone outside that.

But practically what
they mean is just do v1

and forget about this one.

It's not very nicely phrased.

Above, beyond.

Are we above Australia?

That's stupid, right?

Because they may say, oh no,
depends on where you are.

We are above you guys.

You think you're better
than us because you

are closer to the North Pole.

But who made that rule that
t the North Pole is superior?

If you look at the universe,
who is above, who is beyond?

There is no direction.

So they would be very offended.

I have a friend who
works in Sydney.

She is a brilliant geometer.

And I bet if I asked
her, she would say

who says you guys are above?

Because it depends on where your
head is and how you look at it.

The planet is the
same, so would you

say that the people
who are walking closer

to the North Pole have
their body upside down?

So it really matters how
you look, what's above.

So assume that above means--
the word above means positive.

And this is the ice cream cone.

Now how do I find out
where to cut the waffle?

Because this is the question.

I need to know where the
boundary of the waffle is.

I'm not allowed to eat
anything above that.

So that's going
to be the waffle.

And for that, any ideas-- how do
I get to see what the circle--

where the circle will be?

STUDENT: Do they
meet each other?

PROFESSOR: They intersect.

Excellent, Matthew.

Thank you.

So intersect the two surfaces
by setting up a system to solve.

Solve the system.

And the intersection of the two
is Mr. z, which is Mr. circle.

All right.

So what do I do?

I'm going to say I have to
be smart about that one.

So if z squared from
here plugging in,

substitute, is the same as
x squared plus y squared.

So that means this is if
and only if 2x squared

plus y squared.

2 times x squared
plus y squared.

We have an x squared
plus y squared,

and another x squared
plus y squared equals 2.

You see how nicely the
problem was picked?

It was picked it's going
to give you some nice data.

z equals-- z
squared equals that.

Keep going with if and only if.

If you're a math major,
you will understand

why x squared plus
y squared equals 1.

And z squared must be 1.

Well so, we really
get two solutions,

the one close to the South
Pole, and the one close

to the North Pole,
because I'm going

to have z equals plus minus 1.

Where's the North Pole?

The North Pole would be
0, 0, square root of 2.

The South Pole would be 0,
0, minus square root of 2.

And my plain here is
cut at which altitude?

z equals 1.

And I have another plane here,
and an imaginary intersection

that I'm not going to talk
about today. z equals minus y.

I don't care about the mirror
image of the-- with respect

to the equator of the cone.

All right, good.

So we know who this
guy is, we know

that he is-- I have a
red marker, and a green,

and a blue.

I cannot live without colors.

Life is ugly enough.

Let's try to make it colorful.

x squared plus y
squared, equals 1.

We are happy, because
that's a nice, simple circle

of radius 1.

Now you have to think
in which coordinate

you can write this problem.

And I'm going to
beg of you to help

me review the material for
the final and for the midterm.

Then again, on the midterm, I'm
not going to put this problem.

So for the final, do
expect something like that.

We may have, instead of
the cone in the book,

you'll have a paraboloid
and a sphere intersecting.

It's sort of the same thing, but
instead of the ice cream cone,

you have the valley
full of cream.

I cannot stop, right?

So if you do the volume
like I told you before,

you simply have
[INTERPOSING VOICES]

No, I'll do the volume first.

I know you have
the surface there,

but what-- I'm doing the
volume because I have plans,

and I didn't want
to say what plans.

You forgot what I said, right?

So suppose somebody's
asking you for the volume.

The volume-- how much
ice cream you have inside

depends on that very much.

And I'd like you to
remember that the v is

piratically against
the ybc, right?

Right until the Cartesian
coordinates would be a killer,

we try to write it
in either cylindrical

or spherical to make
our life easier.

If I want to make my life
easier, first in cylindrical

coordinates, and then in
spherical coordinates.

Could you help me
find the limit points?

And then we'll do the surface.

Just remind me, OK?

[INTERPOSING VOICES]
The volume of the-- OK.

The volume occupied
by the ice cream.

The ice cream is between
this plastic cap,

that is the sphere.

We cover it for
hygiene purposes.

So for cylindrical coordinates,
rho and theta are really nice.

We don't worry about them yet.

But z should be between?

OK.

Really, 0?

So because it's
between the ice cream

the cone-- do you
think the waffle

cone-- what's the equation
of the waffle cone?

And how do you get to the
equation of the waffle cone?

The waffle cone meant--
oh guys, help me.

z equals-- now you
take the square root

of x squared plus y squared,
because the other one would

be here.

The imaginary one, z equals
minus the square root

of x squared plus y squared.

Forget about the
world in the basement.

So you take z to
the square root of,

with a plus, plus y squared.

In cylindrical coordinates,
what does this mean?

In cylindrical coordinates?

STUDENT: It equals r.

PROFESSOR: This
equals r, very good.

He's thinking faster.

Do you guys
understand why he said

x squared plus y
squared, if we work

with polar coordinates--
which is cylindrical

coordinates it's the same
thing-- polar coordinates

and cylindrical coordinates.

x squared plus y squared
would be little r squared.

Under the square
root would be r.

So you'll see, we're between r.

And now, another hard part.

What is the z equals plus square
root of-- What was it, guys?

STUDENT: So are we taking
the volume of all of the ice

cream inside the cone?

PROFESSOR: So between the
cone, ice cream lies here.

Ice cream chips,
chocolate chips,

between the cone and the sphere.

The sphere is the bottom
function, the lower function.

Is that good?

STUDENT: 2 minus r squared.

PROFESSOR: So I have the square
root of 2 minus r squared.

I don't like the
square root, I hate it.

Can I do it with it?

Yes, I can.

Maybe I can apply the
use of solution later.

Don't worry about
me, I'll make it.

I will live better if I didn't
have any ugly things like that.

So let's see what we have.

Theta.

Your cone is not just sliced
cone, it's all the cone.

So you have 0 to 2 pi.

One revolution,
complete revolution.

How about rho?

Rho is limited.

Rho is what lies in the
plane in terms of radius.

0, 2?

How much is from here to here?

How much is from here to here?

Didn't we do it x squared
plus y squared equals 1?

So what is this radius
from here to here?

1.

And what is this radius
from here to here?

1.

So the projection
of this ice cream

cone-- if you had the
eye of god is here,

sun, you have a
shadow on the ground,

coming from your ice cream
cone, and this is the shadow.

Your shadow is simply
a disk of radius 1.

Good, that's the
projection you have.

So Mr. rho is between 0 and 1.

For polar coordinates,
it's not so ugly actually.

0 to 2 pi.

0 to 1.

r to square root of
2 minus r squared.

Instead of r, it's
OK to react with rho.

And here's the
big j coordinates.

I'm going to erase
j, don't write j.

Jacobian in general is Jacobian.

The one that does the
transformation between

coordinates to
other coordinates.

Let me finish on that.

How about this one?

This is simply r, very good.

Your old friend.

So you [INAUDIBLE]
the dz d, theta,

dr, d is your [INAUDIBLE]
It's not easy there.

If I were to continue--
maybe on the final-- OK

I'm talking too much, as usual.

Maybe on whatever test
you're going to have,

this kind of stuff, with
a formulation saying you

do not have to compute it.

But if you wanted to compute it,
would it be hard from 0 to 1,

from 0 to 2 pi, and say
forgot the stinky pi?

Take the 2 pi out to
make your life easier.

Because the theta isn't
depending from-- there

is no theta inside.

So take the 2 pi
out, and then you

have an integral from 0 to
1, and integral from-- now.

R got out for a walk.

This is r going out for a walk.

Integral of 1 with
respect to dz.

So z is taken between r and
root of 2 minus r squared.

Right?

So I would have to write
here the 1 on top minus the 1

on the bottom, which is a
little bit of a headache for me.

I'm looking at it,
I'm getting angry.

Now, times the r that
went out for a walk.

So practically, the
0 is solved, and I

have the dr. And from 0 to 1.

So I took care or who?

I took care of the
integral with respect to z.

This is r here.

This was done first.

And you gave me that
between those two.

So I got that.

Times the r, between 0 and 1,
with respect to r, and then

the 2 pi gets outside.

Now, if I split this
into two integrals,

it's going to be easy, right?

Because I go--
the first integral

will be r times square
root of 2 minus r squared.

How can you do such an integral?

We do substitution.

For example, your u would
b 2 minus r squared.

Can you keep going?

The second integral
is a piece of cake.

A piece of ice cream.

The integral of r squared.

r cubed over 3,
between 0 and 1, 1/3.

So we can still solve the ice
cream cone volume like that.

Do I like it?

No.

Can you suspect why
I don't like it?

Oh, by the way.

Suppose you got to
this on the final,

how much do you get
for-- you mess up

the algebra, how
much do you get?

You say, I can do
that in my sleep,

u equals 2 minus r squared, u
equals minus 2r, I can go on.

Even if you mess up the
algebra, you get most of it.

Why don't I like it?

Because it involves
work, and I'm lazy.

So can I find a
better way to do it?

Can I get use
spherical coordinates?

And how do I use
spherical coordinates?

So let me see how I do that.

In spherical coordinates,
it should be easier.

Remember that for
mathematicians, they

include this course Cal
3 multivariable calculus.

We are not studying geography.

So for us, a lot can happen
between minus 90 and plus 90

degrees, but it measures
from the North Pole,

because we believe
in Santa Clause.

Always remember that.

So we go all the way from
0 degrees to 180 degrees.

So your-- in principle--
your latitude

will go from 0 to
all the way to pi.

But it doesn't, because
it gets stuck here.

What is the latitude
of the ice cream cone?

So what is the pi angle
for this ice cream cone?

It's a 45 degree angle.

That is true.

For anything like that-- I'm
looking again at this cone.

z squared equals x
squared plus y squared.

I just want to talk a
little bit about that.

So if you have x and
y, this is the x.

This is x, you have to use
your imagination on me.

And the hypotenuse would be
x squared plus y squared.

And this is the z.

And then, I draw
what is in between.

This has to be 45 degrees.

Can you see what's going on?

So theta has to be
between 0 and what?

STUDENT: 2 pi.

PROFESSOR: Yes, you
are smarter than me.

That was the longitude.

Thank you.

I'm sorry, I meant to
write the latitude.

Phi is between 0 and pi/4.

How about the radius?

Are you afraid of the radius?

No.

Why?

The radius is your friend.

It was not your friend before.

Look how wobbly it is.

But in this case, the radius
goes all the way from 0

to a finite value,
which is exactly

the radius of the sphere.

Because you have rays of light
coming from the source origin,

and they bounce
against this profile,

which is the profile
of the sphere, which

has radius square root of 2.

So life is good for
you in this case.

Are you guys with me?

Should it be easy?

Yes, it should be easy to
write that in the integral,

if you know how to write it.

So you have.

OK.

What do you want to do first?

It doesn't matter that you
apply Fubini's theorem.

You have fixed limits.

You have 0 to 2 pi, 0 to
pi/4, 0 to square root 2.

Inside, there must be a
Jacobian that you know by heart,

and I'm asking you to learn
it by heart before the final,

if not for now, but
maybe before the final.

But by now, you should
know it by heart.

Thank you so much, Matthew.

Yes.

You don't have much to memorize,
but this is one of the things

that I told you I did not
memorize it, I was a freshman,

I was stubborn and silly.

So I have to compute what?

I have to compute the Jacobian.

Imagine what work you have
when you're limited in time.

dx, dr. dx, d theta.

dx, d phi.

I thought I was
about to kill myself.

dy, dr. dy, d theta. dy, d
phi, and finally, dz, dr. dz,

d theta.

dz d phi.

And I did this.

And I thought I was about to
just collapse and not finish

my exam.

I finished my exam, but since
then, I didn't remember that.

I had to compute it.

It took me 10 minutes
to compute the Jacobian.

So this is r squared, psi, phi.

If you have nothing better
to do, you can do that.

Do you remember what the
spherical coordinates were, out

of curiosity?

Who remembers that?

There are some
pre-med majors here,

who probably remember that.

So when you have a phi here,
you have r-- sine or cosine?

Cosine.

r cosine phi.

And then r sine phi
for both times what?

The first one comes
from theta, like that.

It's going to be cosine
theta, and sine theta.

Well imagine me taking these
functions and differentiating,

partial derivatives.

And after I differentiated down,
compute the 3 by 3 determining.

It's an error, no matter how
good you are at computing.

So don't do that,
just memorize it.

Don't do like I did.

And then you have d what?

dr, d phi, d theta.

Now what is the volume
of the ice cream cone?

Let me erase.

This shouldn't be hard.

This is the type
of problem where

you have a product of
functions of several variables.

You can separate as a product
of three independent integrals

as a consequence of
Fubini's theorem.

So you have integral from,
integral from, integral from.

Who's your friend?

Who do you like the most?

STUDENT: Theta.

PROFESSOR: You like
theta the most?

Because it comes
from Santa Clause?

No, the theta doesn't.

This is the easiest step.

So that's why you
like it, because it's

the easiest to deal with.

How about phi?

Sine phi, d phi.

I agree with you,
it's not so easy,

but it's going to be a
piece of cake anyway.

How about this one, 0
to root 2 r squared dr.

Is this guy hard to do?

r cubed over 3 will give
me root 2 cubed over 3.

How much is that, by the way?

2 root 2 over 3.

Oh bless your heart,
that's not so hard.

This is not a problem.

How about that?

What do you have?

What is the integral of sine?

Negative cosine.

So you have minus cosine phi
between pi/4 up and 0 down.

Good luck to you.

Well, the first guy.

Good, minus root 2 over 2.

Minus, second guy?

Minus, minus 1.

Don't fall into the trap.

Pay attention to the signs.

Don't mess up,
because that's where

you can hurt your grade by
messing up with minus signs.

So this is 1 plus 1,
1 minus root 2 over 2.

And finally, let's see what
that is, the whole thing being.

Can we write it nicely?

What's 2 times-- 4.

4, root 2 over 3 pi.

The first and the last.

4 times 2, pi/3 times this
nasty guy 1 minus root 2.

I don't like it, let's
make it look better.

Well OK, you can
give me this answer,

of course you'll get 100%.

But am I happy with it?

If you were to publish
this in a journal,

how would you simplify?

This is dry.

OK so what do you have?

1 is 2/2.

2 minus root 2 pi,
4 and 2 simplify.

Are you guys with me?

There is a 2 down, and a 4
up, so I'm going to have a 2

and another 2, all over 3.

So I have 2 and 2 pi,
times 2 minus root 2.

Do you like it like that?

I don't.

So what do you do next?

I can even pull
the root 2 inside.

So I go 4 root 2, minus what?

Minus 4, because this is
2 times 3 is 4, pi over--

Do you like it?

Still I don't,
because I'm stubborn.

4 root 2 minus 1 over 3 is
the most beautiful form.

So I'll try to brush
it up, and put it

in the most elegant form.

It doesn't matter.

If you want to give
me a correct answer,

any form it would be OK.

Yes?

STUDENT: If it was
slightly different,

how would we find phi for the
limits in the second part?

PROFESSOR: If you have a what?

STUDENT: How would we find
phi if it wasn't obvious,

if it wasn't x
squared, or c squared?

PROFESSOR: If it wasn't
a 45 degree angle?

[INAUDIBLE]

It's not so bad, you
need a calculator.

Assume that I would have given
you the sphere of radius 7,

or square root of 7,
intersecting with this cone.

Then to compute
that phi, you would

have needed to intersect
the two surfaces

and then compute it, maybe
look at tangent inverse.

Compute phi with
tangent inverse.

And you will have tangent
inverse of a number.

Well, you cannot put tangent
inverse of a number everywhere,

it's not nice.

So what you would
do is in the end,

you would do it with
a calculator, come up

with a nice truncated
result with 5 decimals,

or 10 decimals, whatever the
calculator will give you.

OK?

Or, you can do it with MathLab.

You can do it with scientific
software, for sure.

Let's do what I-- Ryan
you said this was a what?

STUDENT: Number 20
is for surface area.

PROFESSOR: OK.

So, it's-- read it to me again.

What does it say?

I'm coming to you.

It says, find the surface area
of the part of the sphere that

lies where you have 64.

This is all because
of [INAUDIBLE]

But yours is not
very even, right?

You shouldn't have bad results.

And guess what?

If you do, you use
your calculator

to find out the upper limit
of the angle for the volume.

OK.

So now, you say oh
my god, this is ugly.

I agree with you, it's not nice.

You have square root of 2 minus
x squared minus y squared.

And when you compute the
surface area of the cap-- cup,

cap means spherical cap.

A little hat that
looks like this?

That's why it's called cap.

That will integrate
over the disk d.

Square root of 1
plus f of x squared

plus f of y squared, dx da.

Is that the only
way you can do this?

No.

You can actually do it with
parametrization of a sphere,

and you have the
element limit over here.

So that might be easier.

Yeah.

You can also do it in homework.

But what if you went up there--
let's see, how hard is life?

How hard would it be
to do it like this?

That's good.

First of all, let's
think everything

that's under the square root.

And write it down.

1 plus.

Now, computing this problem
with respect to x and you say,

oh my god, that's hard.

No, it's not.

If you want to do the
hard one, and most of you

were, and you have
that professors who

gave you enough practice,
what did you have done?

Chain rule.

On the bottom, you have
this nasty guy twice.

But on the top, your minus 2x.

So when you simplify
your life becomes easier.

And you will square it.

Are you guys with me,
have I lost you yet?

And then the same thing in y.

Minus 2y, over 2 square root 2
minus x square minus y squared,

square it.

Some things cancel out.

So let's be patient
and see what we have.

First of all, 1 is not going to
give you trouble, because let

write 1 as this over itself.

Plus, minus squared
is plus, thank god.

x squared over 2 minus x squared
minus y squared plus y squared

over 2 minus x squared
minus y squared.

And these guys go for a walk.

Minus x squared, minus y
squared, plus y squared.

They disappear
together in the dark.

So you have 2 over 2 minus
x squared minus y squared.

OK let's try to do that.

Guys, I have to erase.

I will erase.

So what you see
here, some people

call it ds, and use the
element of area on the surface.

It's like the area of
a small surface patch.

So the curve linear squared.

Alright.

So area of the cup will be--
now you say, well over the d,

let me think.

d represents those
xy's with a property

that x squared plus y squared
was between what and what?

0 and 1, because
that was our, the

predicted domain on the
shadow on the ground.

OK, that was this.

And as you look at it, I have
to put it on the square root.

Don't be afraid of
it, because it's not

much up here than you thought.

And let's solve this together.

What is your luck that this is
a symmetric polynomial index,

and why x squared plus y squared
that you can rewrite as r

squared, polar coordinates?

And Ryan asked, can I
do polar coordinates?

That's exactly what
you're going to do.

You didn't know, unless
your intuition is strong.

Yes?

Alex tell me.

STUDENT: I was going to ask,
if you could have done that

by taking the r plane and
multiplying that by 2 pi r?

PROFESSOR: Yeah,
you can do that.

Well, that is a way to do that.

So practically, he's
asking-- I don't

know if you guys
remember, in Cal 2,

you have the surface
of revolution, right?

And if you knew the
length of an arc,

you would be able
to revolve that arc.

This is the cap.

And you take one of the
meridians of the hat,

and revolve it, can
redo with a form,

like you did the washer
and dryer method.

It always amuses me.

Yes, you could have
done that from Cal 2.

Computing the area of
the cap as a surface

of revolution, chapter-- c'mon,
I'm a co-author of this book.

Chapter 7?

What chapter?

Chapter 6?

No.

The washer and dryer?

Chapter 6, right?

OK.

But now we already
have three, and we

don't want to remember Cal 2
because it was a nightmare.

Several of you told
me that this is

easier, these things are
generally easier than Cal 2,

because Cal 2 was headache.

And what seemed to be giving
you most of the headache

was a salad of
different ingredients

that seemed to be unrelated.

Which I agree.

You have arcing, washer,
slices, then Greek substitution,

the partial fractions.

All sorts of things and
series and sequences.

And they are little
things that don't quite

follow one from another.

They are a little bit unrelated.

OK, how do you do that?

You have to help me because
that was the idea, that now you

can help me, right?

Square root of

STUDENT: 2 over 2 minus r
squared, times r, dr d theta.

PROFESSOR: And do we like it?

No, but we have to continue.

0 to 1, this is 0 to 2 pi.

I can get rid of the 2 pi,
and put it here and say, OK.

I should be as good as taking
out square root of 2 from here.

He goes out for a walk.

And then I have integral 1 over
this long line of fraction.

STUDENT: And that would be 2r
so that the r will cancel out.

PROFESSOR: So r dr, if
u is 2 minus r squared,

the u is minus.

I have to pay attention, so
I don't mess up the signs.

So rdr is a block.

And this block is
simply minus du/2.

So I write it here,
minus 1/2, du/2.

Don't be nervous
about this minus,

because it's not going to
give me a minus result,

a negative result.
If it did, that means

that I was drunk when I
did it, because I will

get the area of the cap as
a negative number, which

is impossible.

But it's going to happen
when I change the limits.

Yes?

STUDENT: Where did that
last one come from?

PROFESSOR: From this one.

[INTERPOSING VOICES]
Oh, I put too many.

So this guy is
this guy, which is

this guy minus the [INAUDIBLE].

Now, do you want me to go
ahead and cancel this out?

Right?

OK.

I have squared 2 pi.

I did not get the
endpoints, you have

to help me put the endpoints.

From 2 down.

To 1.

Which is crazy, right?

Because 2 is bigger than 1.

That's exactly where the next
minus is going to come from.

So integral from 2 to 1 is
minus integral from 1 to 2.

So I shouldn't be
worried, because I already

have the minus out, with the
minus that's going to come out,

I'm going to have a
positive variable.

Square root 2 pi.

Somebody was smarter than
me and said Magdalene,

I think Alex-- was it you?

You said, why don't you
take advantage of the fact

that you already have 1 over
2 square root u and integrate?

And that is going
to be squared u.

Can you understand?

Who said that?

I heard a voice, it
was not in my head.

I'm Innocent.

I heard a voice that told
me, if you are smart,

you would understand
to pull out the minus.

You would understand
that this exactly

is the derivative
of square root of u.

And I will be faster than
you, because you have just

computed made between 1 and 2.

Some people aren't too smart.

I didn't think of that.

Now I've been thing about that.

Why cancel out the 2
when you can [INAUDIBLE].

So you have a minus out.

STUDENT: There still needs to
be a 2 on the outside, right?

PROFESSOR: Yes, I have
to put 2 together.

Minus 2 and 2.

2 pi, that's a collective thing.

Squared and cubed
between 1 and 2.

Do I like this?

No, but you tell me
what is in the bracket.

How much u minus 12, 13.

1 minus square root of
2 is a negative number.

But with the minus outside,
I"m going to fix it.

And I'm going to get
something really ugly.

Yeah.

So when I'm multiplying sides,
this by that, I get 4, right?

Guys?

I get a 4.

This guy and this guy, minus,
minus, plus 2, 2 times 2 is 4.

And then minus, to make
it look better, 2 root 2.

And multiply out.

Minus root 2 and the pi.

Now, I don't care
where you stop.

I swear that if you stop
here, you'll still get 100%.

Because what I care about is
not to see a nice simplified

result, so much.

I won't go over your work.

But to see that
you understood how

you solve this kind of problem.

It's not the sign
of intelligence

being able to simplify
answers very much.

But the method in itself, why
and how, what the steps are,

that shows knowledge
and intelligence.

Have I mess up?

I don't think so.

STUDENT: Does the order matter,
of the dr d theta, or dx/dy,

does that matter which
order you put them in?

PROFESSOR: In this case, no.

STUDENT: Or over here?

PROFESSOR: In this
case, no again.

But if you were
to swap them, you

would have to swap
the values as well.

Why is that?

That's a very good question.

He's right, but why is that?

It doesn't matter, why?

In general, it matters.

They have to be from a given
number to a given number.

It's not like reversing--
when you reverse

the order of integrals,
it's usually harder,

because you have to draw
the vertical strips.

And you have it
between two functions.

And then from vertical strips,
you go to horizontal strips,

and you have other
two functions.

So you always have to think
how to change the function.

Here, you don't have
to think at all.

You have a function
that depends on r only.

There is no theta
in the picture.

Plus these two
are fixed numbers.

You can reverse the
integration in your sleep.

OK, you get the same thing.

All you have to do is
swap these two guys,

and swap-- the 0 is the same.

So I swap these two guys.

STUDENT: How did you
take the 2 pi out?

PROFESSOR: What did I do?

How did I take this out?

STUDENT: No, the 2 pi.

PROFESSOR: Oh, the 2 pi?

OK.

Let me show you it better
here, because we've

discussed about this before.

When you have
integral from a to b,

or integral from c to
d of a function or r

and a function of
theta, what do you go?

There is a theorem
that says that--

and thanks for this
theorem and the fact

that they're separable.

The variables are
separated in this product.

This is the product between
integral from a to b,

here of theta to
theta, and integral

from c to d, f of r/dr.
They are nothing to one

another but a product.

So what do you do?

You say this is integral
of 1, from 0 to pi d theta,

times the other guy.

So this is 2 pi.

When theta doesn't appear
inside, it's a blessing.

But if and if there
is, I have a question.

What if theta appeared inside?

Theta doesn't appear
inside by himself.

He appears inside
of a trig function.

So assume you have cosine
theta here times r.

You would have pulled
cosine theta out,

and integrated cosine
theta, that would be easy.

And if you have a problem like
that, you would have gotten 0.

Because integral of
course of cosine theta

would be sine of theta, and
then theta between 0 and 2

pi is sine of theta between,
which would give you 0.

It happened to me,
many times in the exam.

It was a blessing.

I was 19, and I was so happy.

Professors wanted to
see only the answer.

Because in Romania,
it's different.

You come take a written
exam, and the professor

has five hours to grade it.

The same day, two hours later,
you have the oral examination.

You pick up a ticket,
on the ticket,

you see three things to
solve, four things to solve.

You go take a seat.

And while the professor and
the assistant grade the exam,

you actually think
of your oral exam.

When you come and present
your results on the board,

they tell you,
you messed up, you

got a 60% on this sticking exam.

This is how it goes.

Or, on the contrary, hey,
listen, you got a 95%

on the written part, OK?

I don't want to see
what you have there,

it really looks good.

I don't want to see
it, it's clear to me

that you know what you're doing.

So it's a different
kind of examination.

I hear that Princeton
does that, I wonder

how are all the exams here.

I don't think people
are ready for them yet.

But at Princeton they
do a lot, all the same.

They make a hat, and take a
[INAUDIBLE], put tickets in it.

And the teacher comes,
and closes his eyes

and picks a ticket,
and says oh my god, I

got proof of Fubini's theorem.

And do these three
triple integrals.

This is a type of oral
exam that you would have.

But if you know that,
because you studied,

you're not afraid
to present them.

But you have to present them,
and you have a limited time.

Because there are other 30
students in the classroom.

You only have five minutes.

And I only pick-- I want to
see your work on all of them.

And I'll teach
you how to present

on this problem on the board.

And then you have five
minutes to present.

If you are really
embarrassed and you

don't want to speak in public,
then you have a problem.

I've had many fears--
and in other countries--

I heard that in England,
they have the same system.

There are people who are too
shy to show their results,

or too shy to talk.

And then they start stuttering.

But they have to do it.

There is no excuse,
they don't care if you

have problems with your speech.

So I asked the people I
knew and I went to London,

and they said most people
will stutter in there.

I was so scared.

Most people who stutter
in our oral exams

are people who
spend too much time

in the pub the previous day.

Pubs were everywhere and I saw
lots of students in the pubs.

I went to University of Durham--
this is where Harry Potter was

filmed, by the way.

I saw the castle, which
is a student dorm.

You pay something
like 500 pounds.

Which would be like $100?

$1,000?

Less, because I think it's
7.50, something like $800.

It used to be that the
pound was double the dollar.

[INTERPOSING VOICES]

So you could stay in that
dorm for $800 per month.

And you've got the same table
where they ate in the movie.

It was really nice.

But the University of
Durham is a isolated castle,

the cathedral,
everything is very old,

from the 11th
century, 12th century.

But if you go into the
city, it's full of pubs.

Who is in the pubs?

The calculus students.

This is where they
do their homework.

And it amazes me how
they don't get drunk.

I'm not used to alcohol,
because I don't drink.

Well they are used to it.

So they may nicely can do
their homework, beautifully,

next to a big draft
of Guinness like that.

And still makes sense when
they write the solution.

They don't miss a minus
sign, they're amazing.

Alright, is this hard?

If you are interested, you
can ask about study abroad.

We don't have big
business with England,

but you could go to Seville.

There are some programs
in the summer where

one of our professors teaches
differential equations like I

told you about.

He teaches differential
equations this summer

in Seville.

I think you can still
add in the next two days.

Some of you did,
some of you didn't.

All right.

Any questions about
other problems?

I have to apologize,
I played the game

without telling you the truth.

[INAUDIBLE] he came to
me last time and said,

you never showed
us this notation.

So what if one gives you x
of u, v equals 2x minus y.

y of u, v equals 3x plus y.

What the heck is that?

He didn't say heck, because
he's a gentlemen, right?

But he said this is the
notation used in web work,

and the book is actually not
emphasizing it, which is true.

The book is emphasizing the
Jacobian in section 12.8

only, which is not covered,
it's not part of the menu.

But the definition, you
should at least know it.

So what would be the
definition of this animal?

You see that we have to take
the partial derivative of x

with respect to u, the partial
derivative of x with respect

to v, the partial derivative
of y with respect to u,

and the partial derivative
of y with respect to v.

And that's exactly what
it is, indeterminate.

Not matrix, but indeterminate.

So do I bother to write it down?

If I wanted to write
down what it is,

of course I would write
it down like that.

I don't want to spend
all my time doing that,

because it's such
an easy problem.

What do you have to do?

Just compute for such a simple
transformation in plane.

Actually, if you
took linear-- again,

who is enrolled
in linear algebra?

Only 1, 2, 3?

Thought there were only 2.

OK.

In linear algebra, you
wrote this differently.

You wrote it like this. x and
y equals matrix multiplication.

You have 2, minus
1, 3, 1, by the way

it's obvious the
determinate of this matrix

is different from 0.

This is the linear map that you
are applying to the vector xy.

And in your algebra book,
you're using Larson, am I right?

Larson's book?

It's a good book.

So you have a of the vector x.

a of the vector x
is the vector v.

When you all get to
see linear algebra,

you'll like it more than Cal
3, because it's more fun.

So how do you do this
matrix multiplication?

It's very easy.

This time that, minus
this times this.

So can computers do that?

Yes, computers can, if you
have the right program.

And this is the first
program I learned in C++.

No, it was the second program.

How to write a little program
for multiplication of two

matrices.

The first program I
had, I learned in C++.

It was to build an ATM machine.

I hated that, because every time
I went under 0 with my balance,

I would have new word under 0.

So I would have to prepare
for all the possible cases

and save.

If you don't have
enough money, whatever.

So that was the first
program we wrote.

OK so, what do we have?

2 minus 1, 3 and 1.

What is the Jacobian
in this case?

It's 2 plus 3, 5.

Different from 0.

You have one or two
problems like that.

Three problems.

I was really mean.

I apologize.

But you still have time
to do those problems

in case of the review.

STUDENT: So we just take
the determinate of it?

PROFESSOR: And you take the
determinate of the matrix.

And that's you Jacobian.

STUDENT: What number is that?

PROFESSOR: I don't remember.

STUDENT: What if
it's the u and the v

is at the top and x and
the y at the bottom?

PROFESSOR: So the
determinate will be the same.

This is a very good question.

Are you guys with me?

So he said, what if you have
your first equation's name

would be this one.

And you have your equations
written like that.

Right?

And so, when you
look at this, you

will go-- it depends how you--
in which order you do that.

I wrote u, v. Sorry.

u and v, but you
understood what I meant.

Right?

u and v.

STUDENT: Can you
do number three?

It was a hard one.

PROFESSOR: I will,
just a second.

So d y, x with respect to u,
v. What would happen, I just I

would flip the x and y.

What will happen?

I get 3, 1, it's still
the same function.

2, and minus 1.

Why do I get minus 5?

So imagine guys, what happens
when you have x and y?

If you rotate, you don't
change the sign of your matrix,

or notation.

Matrix notation will
always have [INAUDIBLE].

But if you flip it,
if you swap x and y,

you are actually changing
the sign of the Jacobian,

the sign of the matrix.

You are changing
your orientation.

That would be a
hypothetical situation.

You are changing
your orientation.

Do you have a number, Ryan?

Is it hard?

Why is it hard?

Yeah, let me do that.

It's hard enough.

It's computation.

Were you able to do it?

Not yet, right?

So this is x, not-- OK.

So you can write this also,
differently, except the y sub

u, y sub v. Who
can tell me-- there

are ways to do it
in a simpler way.

But I don't want to tell
you yet what that way is.

And I'll show you next time.

What is x sub u?

It shouldn't be so hard
because it's the quotient rule.

You have 4 times u squared
plus v squared minus

[INAUDIBLE] minus 2u the
derivative of this times 4u

divided by the square of that.

Did I go too fast?

So what you have is 4u
squared minus 8u squared

equals minus 4u squared plus
4v squared divided by that.

v squared.

Squared, sorry.

x of v, that should be easier.

Why is it easy?

The first guy prime
minus the second guy

So the first primes,
second not prime.

Minus second prime, straight to
v, times the first not prime.

Divided by u squared.

Which is minus 8uv over that.

Is this one of those that you
said you couldn't do it yet?

You?

Both?

You did this one?

You got the right answer, good.

y sub v. Y sub u, it's OK
to have a minus 0 times

the second one.

Minus this prime with
respect to u, times

6v over the square of that.

And finally, y sub v
equals minus the derivative

of the top, with respect to
6v times u squared plus v

squared minus the derivative
of the bottom with respect

to v. v times 6v divided
by the whole shebang.

Now is it simplified?

No, I will simplify in a second.

You get minus 12uv.

I'm not going to finish
it, but we are almost done.

Why are we almost done?

This is very easy.

I mean, not very
easy, but doable.

How about this guy?

What do you get?

A 6u squared, a 6v squared,
a minus 12v squared.

It's not that bad.

So you have 6u squared
minus 6v squared,

over u squared plus v squared.

What did I do?

Add a minus in front.

I didn't copy.

Let me make room
for that, thank you.

STUDENT: It's also the 12.

It's 12uv, because there's
a negative in front of it.

It's minus times minus.

PROFESSOR: Here?

STUDENT: No, y sub u.

The third one.

PROFESSOR: Minus, minus,
plus, that's good.

Thanks for observing things.

Anything else that's fishy?

Minus, minus, plus.

OK that's better.

Change the signs.

When I move onto this one,
remind me to change the signs.

So what is the Jacobian?

I'm too lazy to
write this thing.

I'm going to have-- so, x sub u.

4 times v squared
minus u squared.

Let's me count the OK.

Let's do it over a.

I'll show you what happens.

Maybe you don't know
yet what happens,

but I'll show you what happens.

Then the next one is going to
be x sub v minus 8, uv over a.

y sub u, 12.

uv over a, and last,
with your help.

It's plus this was
my-- so 6 times v

squared minus u squared over a.

OK OK, let me erase.

So you guys know
what happens when

you have something like that?

A determinate has one
line multiplied or column

multiplied by a number.

If you have alpha a,
alpha b, alpha c and d.

The determinate of that
is alpha aut, a, b, c, d.

I assume you know
this from high school,

but I know very well
that many of you don't.

How do you prove this?

Very easily.

This times that would be
an alpha out, minus this,

and alpha out.

It's very easy to prove.

So when you have one line or one
column multiplied by an alpha,

that alpha gets out.

So if you have two lines
multiplied by an alpha,

or two rows, alpha
squared, excellent.

So who gets out?

1 over a squared, which
means this guy to the fourth.

Sorry that this is so long.

I don't like this problem,
because of this computation

you have to go through here.

So I would simplify
it as much as I could.

Let's see, before I
missed my a group.

So you have 24 times v squared
minus u squared, plus 96,

am I right?

v squared divided by all
this ugly guy which I hate,

to the fourth.

Fortunately, everybody's
a multiple of 24.

So we can pull a 24 out.

and get it out of our life,
because it drives us crazy.

And then you have v to the
fourth plus u to the 4,

minus twice.

Was that the binomial format?

Minus 2 us squared, v squared.

What was left when
I pull this out?

I pulled 24 out, 96 is what?

4.

So I have a 4 left.

So I would put that down.

4u squared, v squared over
the-- it looks symmetric

but-- that's OK.

It's not so bad.

So can you write this better?

Look at it.

Do you like it?

There is a 3.

The 4 the 2, 4
minus 2 is a plus 2.

Just like when we did those
tricky things in high school.

That would be, again,
the binomial formula.

u squared plus v squared.

Are you guys with me?

Because minus 2
plus 4 is plus 2.

This is exactly the
same thing as that.

Over u squared plus v
squared to the fourth.

If you have problems
computing that,

send me some emails from
WebWork, because I'm

going to help you do that, OK.

24 divided by what?

Yes?

u squared plus v
squared squared.

Oh my god.

All right.

So, I'm not going to
think lesser of you

if you don't put
all of this here.

Therefore, if you
get in trouble,

click from the expression from
the whatever you got, and say,

this horrible problem gives
me a headache, help me.

And I'm going to help you with
that simple computation that

is just algebra.

That's not going to teach you
anything more about Cal 3.

That's why I'm
going to help you.

I'll help you with
the answers on those.

Just send me an email.

I'm planning on still
reviewing even on Tuesday.

I don't want to teach anything
new, because I'm tired

and-- I'm just kidding.

I don't want to teach
anything new on Tuesday,

because I want you to be very
well prepared for the midterms.

So I'll do a general
review again,

and I'll go over some
homework like problems,

but mostly over
exam like problems.

So I want everybody to succeed,
to get very high scores.

But we need to practice,
practice, practice.

It's like you did
before your SATs.

It's not that much, I mean what
happens if you don't do great

on the midterm?

Well the midterms is
a portion the final.

But what I am trying
to do by reviewing

so much for the
midterm is also trying

to help you for the final.

Because on the final,
half of the problems

will be just like the ones
on the midterm Emphasizing

the same type of concepts.

It's good practice
for the final as well.

All right, good luck.

I'll see you Tuesday.

Let me know by email how
it goes with the problems.