TTU Math2450 Calculus3 Sec 12.4

Surface Area

DR. MAGDALENA TODA: Sorry.

I really don't mind if you
walk in a little bit late.

I know that you guys come
from other buildings,

and some professors
keep you overtime.

So as long as you
quietly enter the room,

I have no problem with
walking in a little bit late.

Would anybody want to
start an attendance sheet?

Who wants to be the one?

Roberto, please.

Thank you so much.

All right.

We went through chapter
12 on Monday fast.

And I would like to start with
a review of 12.1, 12.2, 12.3.

So two thing we will do today.

Part one will be review of
chapter 12, sections to 12.1,

12.3 from the book and
starting chapter 12,

section 12.4 today later.

What is that about?

This is about the surface
integrals, surface area,

and [INAUDIBLE].

All right.

What have you seen
in 12.1, 12.3?

Let's review quickly
what you've learned.

You've learned about
how to interpret

an integral with a positive
function that is smooth.

Well, we said
continuous-- that would

be enough-- over a
rectangular region.

And the geometric meaning
of such a problem,

integrate f of x, y positive
over a domain was what?

The volume of a body under the
graph and above that domain,

so projected down, protecting
down on the domain.

Evaluate that body.

How did we do it?

Double integral of f
of x, y, dxdy or dA.

But then we said, OK, if you
have a rectangular region

on the ground, then it's easy.

You apply the Fubini theorem.

And then you'll have
integral from A to B,

integral from C to
D, fixed end points.

When you didn't have
a rectangular region

to integrate over, you
would have such a type one,

type two regions, who
are easy to deal with,

which were the case of regions
like the ones between two

straight lines
and two functions.

And then you had the type
two, two straight lines

and two functions,
where the functions

were assumed differentiable
actually in our examples.

Type one, type two.

What did we do after that?

After that, we said, well,
what if you're not so lucky

and have such nice domains?

Or maybe you have
something with a corner.

What do you do if
you have a corner?

Well, you'd still be able to
divide the surface into two,

where you have two
separate areas.

And then you integrate on them
separately at the same time.

And you have an
additive integral.

The integral would be additive.

Those are easy to deal with.

Well, what if you had something
that is more sophisticated,

like a disk or an annulus?

And in that case, it's
really a big headache,

considering how to do this
using one of the previous steps.

So we had to introduce
polar coordinates.

And we have to
think, what change

do I have from x, y to r,
theta, polar coordinates

back and forth?

And when we did the double
integral over a domain f of x,

y function positive dA in
the Cartesian coordinates.

When we switched to
polar coordinates,

we had a magic thing
happen, which was what?

Some f of x of r,
theta, y of r, theta.

I say theta.

I put phi.

It doesn't matter.

Let me put theta if
you prefer theta.

A change of
coordinates, a Jacobian.

That was what?

Do you guys remember that?

r.

Very good.

I'm proud of you, r.

And then drd theta.

So you're ready do that
kind of homework, integrals,

double integrals in
polar coordinates.

dr will be between
certain values,

hopefully fixed
values because that

will make the Fubini-Tonelli
a piece of cake.

Theta, also fixed values.

But not always will you have
fixed values, especially

in the first part.

You may have some function
of r, function of r.

And here, theta 1 and theta 2.

So I want to see a
few more examples

before I move on to section 12.4
because, as the Romans said,

review is the
mother of studying,

which is [LATIN], which means
go ahead and do a lot of review

if you really want to
master the concepts.

OK.

I'm going to take the plunge
and go ahead and help you

with your homework.

I've been pondering
about this a lot.

We've done problems that I made
up, like the ones in the book.

And I also took problems
straight out of the book.

But I would like to go over
some homework type problems

in order to assist you in more
easily doing your homework.

In chapter 12, homework
four-- am I right,

homework number four?

You have a big array of
problems, all sorts of problems

because mathematicians
have all sorts of problems.

For example, an easy
one that you're not

going to have a problem with--
and I'm using my own end

points.

Your end points may be
different in the homework.

It would be homework four,
chapter 12, number four.

And you say-- most of
you should say, oh,

that's a piece of cake.

I don't know why she even talks
about such a trivial problem,

right?

Many of you have said that.

Well, I am willing
to review everything

so that you have a better
grasp of the material.

On this one, since it's so
easy, I want you to help me.

What kind of problem is that?

As I said, mathematicians have
all sorts of problems, right?

So a problem where you
have a product inside

as an integrand, where the
variables are completely

separated-- what does it mean?

The function
underneath is a product

of two functions, one
function of x only,

the other function
of y only, which

is a blessing in disguise.

Why is that a blessing?

I told you last time that you
can go ahead and write this

as product of integrals.

Is there anybody seeing already
what those integrals will be?

Let's see how much you
mastered the material.

STUDENT: x over 2y times--

DR. MAGDALENA TODA:
From 1 to 2, you said?

STUDENT: Yeah, from 1 to 2.

I'm sorry.

DR. MAGDALENA TODA: Of what?

X, dx times the integral
from 0 to pi of what?

STUDENT: Cosine y.

DR. MAGDALENA TODA: Cosine y.

Do we need to
re-prove this result?

No, we proved it last time.

But practically, if
you forget, the idea

is a very simple thing.

When you integrate
with respect to y,

Mr. X said, I'm
not married to y.

I'm out of here.

I'm out of the picture.

I'm going for a walk.

So the integral of cosine is
in itself to be treated first,

independently.

And it's inside,
and it's a constant.

And it pulls out in the end.

And since it pulls
out, what you're

going to be left with afterwards
will be that integral of 1

to 2x dx.

So we've done that
last time as well.

Yes, sir?

STUDENT: So you
would-- would you

not be able to do that
if it was cosine x, y?

DR. MAGDALENA TODA: Absolutely.

If you had cosine
x, y, it's bye, bye.

STUDENT: So it's only when
they're completely separate--

DR. MAGDALENA TODA: When
you are lucky enough

to have a functional of only
that's a function of y only.

And if you had another
example, sine of x plus y,

anything that mixes them up--
that would be a bad thing.

Do I have to compute this?

Not if I'm smart.

At the blink of an eye,
I can sense that maybe I

should do this one first.

Why?

Integral of cosine is sine.

And sine is 0 at both 0 and pi.

So it's a piece of pie.

So if I have 0, and
the answer is 0.

So you say, OK,
give us something

like that on the midterm
because this problem is

a piece of cake.

Uh, yeah.

I can do that.

Probably you will have something
like that on the midterm,

on the April 2 midterm.

So since Alex just
entered, I'm not

going to erase this for a while
until you are able to copy it.

I announced starting the
surface area integral today.

Section 12.4, we'll
do that later on.

And I will move on to
another example right now.

Oh, now they learn.

Look, they learned about me.

They learned about me,
that I have lots of needs.

And I don't complain.

But they noticed that these
were disappearing really fast.

Everybody else told me that
I write a lot on the board,

compared to other professors.

So I don't know if that is true.

But I really need
this big bottle.

OK.

So you can actually
solve this by yourself.

You just don't realize it.

I'm not going to take
any credit for that.

And I'm going to go ahead
and give you something

more challenging, see if
you are ready for the review

and for the midterm.

OK.

That's number nine
on your homework

that may have again
the data changed.

But it's the same
type of problem.

Now you cannot ask me about
number nine anymore directly

from WeBWork, because I'll
say, I did that in class.

And if you have
difficulty with it,

that means you did
not cover the notes.

This is pretty.

You've seen that one before.

And I would suspect
that you're not

going to even let me
talk, because look at it.

Evaluate the following integral.

And it doesn't
matter what numbers

we are going to put on that
and what funny polynomial I'm

going to put here.

You are going to have
all sorts of numbers.

Maybe these are not
the most inspired ones,

but this is WeBWork.

It creates problems at
random, and every student

may have a different
problem, that is,

in order to minimize cheating.

And that's OK.

The type of the problem
is what matters.

So if we were in
Calc 1 right now,

and somebody would say, go
ahead and take an integral of e

to the x squared dx and compute
it by hand, see what you get,

you already know.

They don't know, poor people.

They don't know.

But you know because I told
you that this is a headache.

You need another way out.

You cannot do that in Calc 2.

And you cannot do that in
an elementary way by hand.

This is something that MATLAB
would solve numerically for you

in no time if you gave
certain values and so on.

But to find an explicit
form of that anti-derivative

would be a hassle.

The same thing would happen
if I had the minus here.

In that case, I wouldn't be able
to express the anti-derivative

as an elementary
function at all.

OK.

So this is giving
me a big headache.

I'm going to make a face.

And I'll say, oh, my god.

I get a headache.

Unless you help me get out of
trouble, I cannot solve that.

MATLAB can do that for me.

On Maple, I can go in and
plug in the endpoints and hope

and pray that I'm
going to get the best

numerical approximation
for the answer.

But what if I want a precise
answer, not a numerical answer?

Then I better put my
mind, my own mind,

my own processor to work and
not rely on MATLAB or Maple.

OK.

Hmm.

Understandable, precise answer.

And I leave it unsimplified
hopefully, yes.

We need to think of what
technique in this case?

STUDENT: Changing the order.

DR. MAGDALENA TODA: Change
the order of integration.

OK.

All right.

And in that case, the
integrand stays the same.

These two guys are
swapped, and the end points

are changing
completely because I

will have to switch from one
domain to the other domain.

The domain that's given here by
this problem is the following.

x is between 7y 7, and
y is between 0 and 1.

So do they give you horizontal
strip or vertical strip domain?

Horizontal.

Very good.

I wasn't sure if
I heard it right.

But anyway, what is this
function and that function?

So x equals 7 would be what?

x equals 7 will be far away.

I have to do one, two, three,
four-- well, five, six, seven.

Then is a vertical
line. x equals 7.

That's the x-axis.

That's the y-axis.

I'm trying to draw the domain.

And what is x equals 7y?

X equals 7y is the
same as y equals 1/7x.

Uh-huh.

That should be a
friendlier function

to draw because I'm smart
enough to even imagine

what it looks like.

y equals mx is a line that
passes through the origin.

It's part of a pencil of planes.

A pencil of planes is infinitely
many-- pencil of lines,

I'm sorry.

Infinitely many lines that all
pass through the same point.

So they all pass
through the origin.

For 7, x equals 7 is going
to give me y, 1, y equals 1.

So I'm going to erase this
dotted line and draw the line.

This is y equals x/7,
and we look at it,

and we think how nice
it is and how ugly it

is because it's [? fat ?].

It's not a straight line.

Now it looks straighter.

So simply, I get to 1, y equals
1 here, which is good for me

because that's
exactly what I wanted.

I wanted to draw the horizontal
strips for y between 0 and 1.

I know I'm going
very slow, but that's

kind of the idea
because-- do you

mind that I'm going so slow?

OK.

This is review for the
midterm slowly, a little bit.

So y between 0 and 1.

I'm drawing the
horizontal strips,

and this is exactly
what you guys have.

This is the red domain.

Let's call it d.

It's the same domain but
with horizontal strips.

And I'm going to
draw the same domain.

What color do you like?

I like green because it's
in contrast with red.

I'm going to use green to
draw the vertical strip domain

and say, all right.

Now I know what I'm
supposed to say,

that d with vertical strips is
going to be x between-- what?

Yes.

First the fixed
numbers, 0 and 7.

And y between--

STUDENT: 0 and x plus 7.

STUDENT: And 1.

DR. MAGDALENA TODA: This one.

x/7, 1/7x.

Right?

Is it x/7?

x/7y equals x is the same
thing as y equals x/7.

So y equals x/7 is
this problem, which was

the same as x equals 7y before.

OK.

So how do I set up
the new integral?

I'm going to say dydx, and then
y will be between 0 and x/7.

And x will be between 0 and 7.

Is it solved?

No.

But I promise from my heart that
if you do that on the midterm,

you'll get 75% on this
problem, even if doesn't say

don't compute it.

If it says, don't compute
it or anything like that,

you got 100%.

OK?

So this is the most
important step.

From this on, I know
you can do it with what

you've learned in Calc 1 and 2.

It's a piece of cake, and you
should do it with no problem.

Now how are we going
to handle this fellow?

This fellow says, I have nothing
to do with you, Mr. Y. I'm out,

and you're alone.

I don't need you as my friend.

I'm out.

I'm independent.

So Mr. Y starts sulking.

And say I have an integral
of 1dy between 0 and x/7.

I'm x/7.

So you are reduced to
a very simple integral.

That is the integral that you
learned in-- was it Calc 1?

Calc 1, yes, the end of Calc 1.

All right.

So you don't need
the picture anymore.

You've done most of
the work, and you say,

I have an integral from 0 to
7, x over-- so this guy-- which

one shall I put first?

It doesn't matter.

e to the x squared
got out first.

He said, I'm out.

And then the integral of
1dy was y between these two,

so it's x/7 dx.

And this is a 7.

All right.

We are happy.

So what happens?

1/7 also goes for a walk.

And xdx says, OK, I need
to think about who I am.

I have to find my own
identity because I

don't know who I am anymore.

So he says, I need
a u substitution.

u substitution is
u equals x squared.

du equals 2xdx.

So xdx says, I know at least
that I am a differential

form, a 1 form, which is du/2.

And that's exactly what you
guys need to change the inputs.

1/7 was a [? custom ?].

He got out of here.

But you have to think,
when x is 0, what is u?

0.

When x is 7, what is u?

49.

Even my son would know this one.

He would know more.

He would know
fractions and stuff.

OK.

So e to the u.

And the 1/7 was out.

But what is xdx?

du/2.

So I'll say 1/2 du.

Are you guys with me?

Could you follow everything?

Yes.

It shouldn't be a problem.

Now 1/7 got out.

1/2 gets out.

Everybody gets out.

And the guy in the middle who is
left alone, the integral from e

to the u du-- what is he?

e to the u.

Between what values?

Between 49 and 0.

So I'm going to--
shall I write it again?

I'm too lazy for that.

e to the u-- OK, I'll write it.

e to the u between 49 and 0.

So I have 1/14, parentheses,
e to the 49 minus e to the 0.

That's a piece of cake.

1.

OK, so presumably if you
answered that in WeBWork,

this the precise answer.

Finding it correctly, you
would get the right answer.

Of course, you could do
that with the calculator.

MATLAB could do it for you.

Maple could do it for you.

Mathematica could do it for you.

But they will come up
with a numerical answer,

an approximation.

And you haven't learned
anything in the process.

Somebody just served you
the answer on a plate,

and that's not the idea.

Is this hard?

I'm saying on the midterm that's
based on the double integral

with switching order integrals,
this is as hard as it can get.

It cannot get worse than that.

So that will tell you about
the level of the midterm that's

coming up on the 2nd of April,
not something to be worried

about.

Do you need to learn a little
bit during the spring break?

Maybe a few hours.

But I would not worry my
family about it and say,

there is this witch.

And I'm going back
to [? Lubbock ?],

and I have to take
her stinking midterm.

And that stresses me out, so I
cannot enjoy my spring break.

By all means, enjoy
your spring break.

And just devote a few
hours to your homework.

But don't fret.

Don't be worried
about the coming exam,

because you will be prepared.

And I'm going to do
more review so that you

can be confident about it.

Another one.

Well, they're all easy.

But I just want to help you
to the best of my extent.

One more.

Here also is-- I don't-- OK.

Let's take this one because
it's not computational.

And I love it.

It's number 14.

Number 14 and number 15
are so much the same type.

And 16.

It's a theoretical problem.

It practically tests if
you understood the idea.

That's why I love this problem.

And it appears
obsessively, this problem.

I saw it in-- I've
been here for 14 years.

I've seen it at least on 10
different finals, the same type

of theoretical problem.

So it's number 14
over homework four.

Find an equivalent integral
with the order of integration

reversed.

So you need to
reverse some integral.

And since you are so savvy about
reversing the ordered integral,

you should not have
a problem with it.

And WeBWork is
asking you to fill

in the following expressions.

You know the type.

f of y, you have to
type in your answer.

And g of y, to type
in your answer.

OK.

So you're thinking, I know
how to do this problem.

It must be the idea as before.

This integral should
be-- according

to the order of
integration, it should

be a vertical strip thing
switching to a horizontal strip

thing.

And once I draw the domain,
I'm going to know everything.

And the answer is, yes,
you can do this problem

in about 25 seconds.

The moment you've learned
it and understood it,

it's going to go very smoothly.

And to convince
you, I'm just going

to go ahead and say, 0 and 1.

And draw, Magdalena.

You know how to draw.

Come on.

OK.

From 1-- 1, 1, right?

Is this the corner-- does
it look like a square?

Yes.

So the parabola y equals x
squared is the bottom one.

Am I right?

That is the bottom one, guys?

But when you see-- when
you are between 0 and 1,

x squared is a lot less
than the square root of x.

The square root of x is the
top, is the function on top.

And then you say, OK, I
got-- somebody gave me

the vertical strips.

I'll put the [INAUDIBLE],
but I don't need them.

I'll just go ahead
and take the purple,

and I'll draw the
horizontal strips.

And you are already
there because I

see the light in your eyes.

So tell me what you
have. y between n--

STUDENT: 0 and 1.

DR. MAGDALENA TODA: 0 and 1.

Excellent.

And y between what and x?

Oh, sorry, guys.

I need to protect my hand.

That's the secret recipe.

x is between a function of y.

Now what's the
highest function of y?

STUDENT: Square root of y.

DR. MAGDALENA TODA:
Square root of y.

And who is that fellow?

This one.

x equals square root
of y, the green fellow.

I should have written in
green, but I was too lazy.

And this one is going to
be just x equals y squared.

So between y square
down, down, down, down.

Who is down? f is down.

Right, guys?

The bottom one is f.

The bottom one is y squared.

The upper one is the
square root of y.

You cannot type that
in WeBWork, right?

You type sqrt, what?

y, caret, 2.

And here, what do you have?

0 and 1.

So I talk too much.

But if you were on your
own doing this in WeBWork,

it would take you no
more than-- I don't

know-- 60 seconds to type in.

Remember this problem
for the midterm.

It's an important idea.

And you've seen it emphasized.

You will see it emphasized
in problems 14, 15, 16.

It's embedded in this
type of exchange,

change the order of
integration type problem.

OK?

Anything else I would like
to show you from-- there

are many things I
would like to show you.

But I better let you
do things on your own.

How about 17, which is a
similar type of problem,

theoretical, just like this one?

But it's testing
if you know the--

if you understood the idea
behind polar integration,

integration in
polar coordinates.

Can I erase?

OK.

So let's switch to number
17 from your homework.

Write down the problems
we are going over,

so when you do
your homework, you

refer to your lecture notes.

This is not a lecture.

What is this, what
you're doing now?

It's like-- what is this?

An application session,
a problem session.

OK.

Number 17, homework four.

On this one, unfortunately
I'm doing just your homework

because there is no data.

So when-- it's the unique
problem you're going to get.

You have a picture, and that
picture looks like that.

From here, [INAUDIBLE]
a half of an annulus.

You have half of a ring.

And it says, suppose that
r is the shaded region

in the figure.

As an iterated integral
in polar coordinates,

the double integral
over R f of x, y dA

is the integral from A to
B of the integral from C

to B of f of r, theta times r
drd theta with the following

limits of integration.

A. And WeBWork says, you say it.

You say.

It's playing games with you.

B, you say.

It's a guessing game.

C, you say.

Then D, you say it.

And let's see what you say.

Well, we say, well,
how am I going to go?

I have to disclose
the graphing paper.

They are so mean.

They don't show you
the actual numbers.

They only give you
graphing paper.

I'm not good at graphing, OK?

So you will have to
guess what this says.

That should be good enough.

Perfect.

So the unit supposedly
is this much.

1 inch, whatever.

I don't care.

So is it hard?

It's a piece of cake.

It's a 10 second problem.

It's a good problem for the
midterm because it's fast.

Theta is a wonderful angle.

It is nice to look at.

And they really don't
put numbers here?

They do.

They do on the margin
of the graphing paper.

They have a scale.

OK.

So come on.

This is easy.

You guys are too smart
for this problem.

From what to what?

STUDENT: [INAUDIBLE].

DR. MAGDALENA TODA: Nope.

No, that's a problem.

So when we measure
the angle theta,

where do we start measuring?

STUDENT: 0.

DR. MAGDALENA TODA: Over here.

So we go down there
clockwise because that's

how we mix in the bowl,
counter-clockwise.

So 0-- so this is
going to be pi.

Pi.

And what is the end?

STUDENT: 2 pi.

DR. MAGDALENA TODA: 2 pi.

2 pi.

Don't type-- oh, I mean, you
cannot type the symbol part,

right?

And then what do you
type, in terms of C and D?

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA: Nope.

No, no.

The radius is positive only.

STUDENT: 0 to 1.

DR. MAGDALENA TODA: 1 to 2.

Why 1 to 2?

Excellent.

Because the shaded area
represents the half of a donut.

You have nothing inside.

There is a whole in
here, in the donut.

So between 0 and 1,
you have nothing.

And the radius-- take
a point in your domain.

It's here.

The radius you
have, the red radius

you guys see on the picture is
a value that's between 1 and 2,

between 1 and 2.

And that's it.

That was a 10 second problem.

So promise me.

You are going to do
the homework and stuff.

You have two or three like that.

If you see this on
the midterm, are you

going to remember the procedure,
the idea of the problem?

OK.

I'm going to also think
of writing a sample.

I promised Stacy I'm
going to do that.

And I did not forget.

It's going to happen.

After spring break, you're
going to get a review

sheet for the midterm.

I promised you a sample, right?

OK.

Shall I do more or not?

Yes?

You know what I'm
afraid of, really?

I think you will be able to do
fine with most of the problems

you have here.

I'm more worried about
geometric representations

in 3D of quadrics that
you guys became familiar

with only now,
only this semester.

And you have a grasp of them.

You've seen them.

But you're still not
very friendly with them,

and you don't
quite like to draw.

So let's see if we can learn
how to draw one of them together

and see if it's a big
deal or not because it's

pretty as a picture.

And when we set it
up as an integral,

it should be done wisely.

It shouldn't be hard.

We have to do a good job
from the moment we draw.

And if we don't do that,
we don't have much chance.

The problem is going to
change the data a little bit

to numbers that I like.

29.

You have a solid.

And I say solid gold, 24 k.

I don't know what.

That is between two paraboloids.

And those paraboloids
are given, and I'd

like you to tell me
what they look like.

One paraboloid is y-- no.

Yeah.

One paraboloid is
y-- I'll change it.

z.

So I can change your
problem, and then you

will figure it out by yourself.

z equals x squared
plus y squared.

They give you y equals x
squared plus d squared.

So you have to change
completely the configuration

of your frame.

And then z equals 8 minus
x squared minus y squared.

I'm I'm changing problem 29,
but it's practically the same.

Find the volume of the solid
enclosed by the two paraboloids

and write down the answer.

Find the volume of
the solid enclosed

by the two paraboloids.

You go, oh, my god.

How am I going to do that?

STUDENT: Draw the pictures.

DR. MAGDALENA TODA:
Draw the pictures.

Very good.

So he's teaching
my sensing to me

and says, OK, go ahead
and draw the picture.

Don't be lazy, because
if you don't, it's

never going to happen.

You're never going
to see the domain

if you don't draw the pictures.

So the first one will
be the shell of the egg.

Easter is coming.

So that's something
like the shell.

It's a terrible shell,
a paraboloid, circular

paraboloid.

And that is called z equals
x squared plus y squared.

OK.

This guy keeps going.

But there will be
another paraboloid

that has the shape of exactly
the same thing upside down.

STUDENT: Where's 8?

DR. MAGDALENA TODA: Where is 8?

The 8 is far away.

STUDENT: It's on--

DR. MAGDALENA TODA: I'll try.

STUDENT: Did they tell you
that a had to be positive?

DR. MAGDALENA TODA: Huh?

STUDENT: Did they tell
you a had to be positive?

DR. MAGDALENA TODA: Which a?

STUDENT: That a or whatever.

DR. MAGDALENA TODA: 8.

STUDENT: Oh, 8.

DR. MAGDALENA TODA: 8.

STUDENT: Oh, that's
why I'm confused.

DR. MAGDALENA TODA:
How do I know it's 8?

Because when I put
x equals 7 equals 0,

I get z equals 8
for this paraboloid.

This is the red paraboloid.

The problem-- my
question is, OK, it's

like it is two
eggshells that are

connecting, exactly this egg.

But the bound-- the--
how do you call that?

Boundary, the thing where
they glue it together.

What is the equation
of this circle?

This is the question.

Where do they intersect?

How do you find out where
two surfaces intersect?

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA:
Solve a system.

Make a system of two equations
and solve the system.

You have to intersect them.

So whoever x, y, z will be, they
have to satisfy both equations.

Oh, my god.

So we have to look for the
solutions of both equations

at the same time, which
means that I'm going

to say these are equal, right?

Let's write that down.

x squared plus y
squared equals 8 minus x

squared minus y squared.

Then z is whatever.

What is this equation?

We'll find out who
z is in a second.

z has to be x squared
plus y squared.

If we find out who the sum
of the squares will be,

we'll find out the altitude z.

z equals what number?

This is the whole idea.

So x squared.

I move everything to
the left hand side.

So I have 2x squared
plus 2y squared equals 8.

And then I have z equals
x squared plus y squared.

And then that's if and only
if x squared plus y squared

equals 4.

STUDENT: Then z equals 4.

DR. MAGDALENA TODA:
So z equals 4.

So z equals 4 is exactly what
I guessed because come on.

The two eggshells
have to be equal.

So this should be in the
middle between 0 and 8.

So I knew it was z equals 4.

But I had to check
it mathematically.

So z equals 4, and x
squared plus y squared

equals 4 is the boundary.

Let's make it purple
because it's the same

as the purple equation there.

So the domain has to be the
projection of this purple--

it looks like a sci-fi thing.

You have some hologram.

I don't know what it is.

It's all in your imagination.

You want to know the domain
D. Could somebody tell me

what the domain D will be?

It will be those x's and y's
on the floor with the quality

that x squared plus y squared
will be between 0 and--

STUDENT: 4

DR. MAGDALENA TODA: --4.

So I can do everything
in polar coordinates.

This is the same thing as
saying rho, theta-- r, theta.

Not rho.

Rho is Greek.

It's all Greek to me.

So rho is sometimes
used by people

for the polar coordinates,
rho and theta.

But we use r.

r squared between 0 and 4.

You'll say, Magdalena, come on.

That's silly.

Why didn't you write
r between 0 and 2?

I will.

I will.

I will.

This is 2, right?

So r between 0 and 2.

I erase this.

And theta is between 0 and 2 pi.

And I'm done.

Why 0 and 2 pi?

Because we have the whole egg.

I mean, I could
cut the egg in half

and say 0 to pi or something,
invent a different problem.

But for the time
being, I'm rotating

a full rotation of 2 pi to
create the egg all around.

So finally, what is
the volume of-- suppose

this is like in the story
with the golden eggs.

They are solid gold eggs.

Wouldn't that be wonderful?

We want to know the
volume of this golden egg.

What's inside the solid egg, not
the shell, not just the shell

made of gold.

The whole thing is made of gold.

And who's coming tomorrow
to the-- sorry, guys.

Mathematician talking.

Switching from
another-- who's coming

tomorrow to the honors society?

Do you-- did you decide?

You have.

And Rachel comes.

Are you coming?

No, no, no, no.

Tomorrow night.

Tomorrow night.

Tomorrow.

What time does that--

STUDENT: 3:00.

DR. MAGDALENA
TODA: At 3 o'clock.

At 3 o'clock.

OK.

So if you want, I can
pay your membership.

And then you'll be members.

I saw one of the certificates.

It was really beautiful.

That one [INAUDIBLE].

It was really-- some
parents frame these things.

My parents don't care.

But I wish they cared.

So the more certificates you
get, and the older you get,

the nicer it is to put them,
frame them and put them

on the wall of
fame of the family.

This certificate, the KME
one, looks so much better

than my own diplomas, the PhD
diplomas, the math diplomas.

And it's huge, and it
has a golden silver seal

will all the stuff.

And it's really nice.

OK.

Now coming back to this thing.

STUDENT: Can we multiply by 2?

Just find the--

DR. MAGDALENA TODA: Exactly.

That's what we will do.

We could set up the integral
from whatever it is.

My one function to
another function.

But the simplest way
to compute the volume

would be to say
there are two types.

And set up the
integral for this one,

for example or the other one.

It doesn't matter which one.

It doesn't really
matter which one.

Which one we would prefer?

I don't know.

Maybe you like the bottom
part of the [INAUDIBLE].

I don't know.

Do you guys understand
what I'm talking about?

STUDENT: If we
just-- I don't know

where to find B. Find the
area left, like indented?

Because if you did it at the
bottom, the domain is zero.

Then you have-- wouldn't
it find the stuff that

was not cupped out, the edges?

DR. MAGDALENA TODA:
Isn't it exactly

the same volume up and down?

STUDENT: Yes.

DR. MAGDALENA TODA: It's
the same volume up and down.

So it's enough for
me to take the volume

of the lower part and w.

Can you help me set
up the lower part?

So I'm going to have two types.

Can I do that directly
in polar coordinates?

That's the thing.

1 is 1.

r is r.

r is going to be-- this is
the Jacobian r drd theta.

OK?

But now let me ask you, how
do we compute-- I'm sorry.

This is the function f of x, y.

Yeah, it's a little
bit more complicated.

So you have to subtract
from one the other one.

So I'm referring to the domain
as being only the planar

domain.

And I have first a graph
and then another graph.

So when I want to compute,
forget about this part.

I want to compute the
volume of this, the volume

of this egg, the inside.

I have to say, OK, integral over
the d of the function that's

on top.

The function that's on top
is the z equals f of x, y.

And the function that's
on the bottom for this egg

is just this.

So this is just
a flat altitude g

of x, y equals-- what is that?

4.

So I have to subtract the two
because I have first this body.

If this would not exist, how
would I get the purple part?

I would say for the function f,
the protection on the ground,

I have this whole body
that looks like a crayon.

A whole body that
looks like crayon.

This is the first integral.

I minus the cylinder
that's dotted with floating

points, which is this part.

So it's V1 minus V2.

V1 is the volume of the whole
body that looks like a crayon.

V2 is just the volume of the
cylinder under the crayon.

We want-- minus
V2 is exactly half

of the egg, the volume of the
half of the egg, give or take.

So is this hard?

It shouldn't be hard.

f of x, y-- can you guys
tell me who that is?

A minus x squared
minus y squared.

And who is g?

4.

Just the altitude, 4.

OK.

So I'm going to go
ahead and say, OK,

I have to integrate 2 double
integral over D. 8 minus 4

is 4 minus x squared
minus y squared dA.

dA is the area element dxdy.

Now switch to polar.

How do you switch to polar?

You can also set this
up as a triple integral.

And that's what I
wanted to do at first.

But then I realized that you
don't know triple integrals,

so I set it up as
a double integral.

For a triple integral,
you have three snakes.

And you integrate the
element 1, and that's

going to be the volume.

And I'll teach you in
the next two sessions.

2 times the double integral.

Who is this nice fellow?

Look how nice and sassy he is.

4 minus r squared times--
never forget the r drd theta.

Theta goes between 0
and 2 pi and r between--

STUDENT: 0 and 2.

DR. MAGDALENA TODA: 0 and--

STUDENT: 2.

DR. MAGDALENA TODA: 2.

Excellent.

Because when I had 4 here,
that's the radius squared.

So r is 2.

Look at this integral.

Is it hard?

Not so hard.

Not so hard at all.

So what would you
do if you were me?

Would you do a u substitution?

Do you need a u
substitution necessarily?

You don't need it.

So just say 4r minus r cubed.

Now what do you see again?

Theta is missing
from the picture.

Theta says, I'm out of here.

I don't care.

So you get 2 times the integral
from 0 to 2 pi of nothing--

well, of 1d theta, not of
nothing-- times the integral

from 0 to 2 of 4r
minus r cubed dr.

4r minus r cubed dr, the
integral from 0 to 2.

Good.

Who's going to help me?

I give you how
much money-- money.

Time shall I give
you to do this one?

And I need three people
to respond and get

the same answer.

So [INAUDIBLE] 2r squared
minus r to the 4 over 4

between 0 and 2.

Can you do it please?

STUDENT: 16 [INAUDIBLE].

DR. MAGDALENA TODA:
How much did you get?

STUDENT: 4.

DR. MAGDALENA TODA:
How much did you get?

STUDENT: For just this one?

DR. MAGDALENA
TODA: For all this.

STUDENT: 4.

DR. MAGDALENA TODA:
Yes, it is, right?

Are you with me?

You have 2r squared
when you integrate

minus r to the 4 over
4 between 0 and 2.

That means 2 times 4 minus 16/4.

8 minus 4 is 4.

So with 4 for this guy, 2 pi
for this guy, and one 2 outside,

you have 16 pi.

And that was-- I remember
it as if it was yesterday.

That was on a final
two or three years ago.

OK.

So you've seen many
of these problems now.

It shouldn't be complicated
to start your homework.

Go ahead.

If you want, go ahead and
start with the problems

that we did today.

And when you see numbers
changed or something,

go ahead and work the
problem the same way.

Make sure you understood it.

I'm going to do more.

Is this useful for you?

I mean-- OK.

So you agree that
every now and then,

we do homework in the classroom?

Homework like problems
in the classroom.

In the homework, you
may have different data,

but it's the same
type of problem.

OK.

I'm going to remind you
of some Calc 2 notions

because today I will
cover the surface area.

STUDENT: Dr. Toda?

DR. MAGDALENA TODA: Yes, sir?

STUDENT: I have a question
on the last problem.

DR. MAGDALENA TODA: Yes, sir?

STUDENT: If we had seen
something like that on the exam

and had done it using the fact
that it's a solid revolution--

DR. MAGDALENA TODA:
Yeah, you can do that.

There are at least four
methods to do this problem.

One would be with
triple integral.

One would be with
a double integral

of a function on top
minus the function below.

One would be with solid of
revolution like in Calc 2,

where your axis is the z axis.

I don't care how you
solve the problem.

Again, if I were
the CEO of a company

or the boss of a
firm or something,

I would care for my employees to
be solving problems the fastest

possible way.

As long as the
answer is correct,

I don't care how you do it.

STUDENT: Thank you, Doctor.

DR. MAGDALENA TODA: So go ahead.

All right.

Oh, and by the way, I want to
give you another example where

the students were able
to very beautifully cheat

and get the right answer.

That was funny.

But that is again
a Calc 3 problem

in an elementary way
that can be solved

with the notions you have from
K-12, if you mastered them

[INAUDIBLE].

So you are given x
plus y plus z equals 1.

Before I do the
surface integral--

I could do the surface
integral for such a problem.

This is a plane that intersects
the coordinate planes

and forms a
tetrahedron with them.

Find the volume of
that tetrahedron.

Now I say, with Calc 3,
because the course coordinator

several years ago did not
specify with what you learned.

Set up a double integral
or set up-- he simply

said, find the volume.

So the students-- what's
the simplest way to do it?

STUDENT: That's
just half a cube.

DR. MAGDALENA TODA:
Just draw the thingy.

And they were smart.

They knew how to draw it.

The knew what the vertices were.

The plane looks like this.

If you shade it, you see
that it's x plus y plus z.

And I'm going to try
and write with my hands.

It's very hard.

But it comes from 0, 0, 1 point.

This is the 0, 0, 1.

And it comes like that.

And it hits the floor over here.

And these points are 1,
0, 0; 0, 1, 0; and 0, 0,

1 on the vertices
of a tetrahedron,

including the origin.

How do I know those are
exactly the vertices

of the tetrahedron?

Because they verify x
plus y plus z equals 1.

As long as the point
verifies the equation,

it is in the plane.

For example, another point
that's not in the picture

would be 1/3 plus 1/3 plus 1/3.

1/3 and 1/3 is 1/3.

Anything that verifies the
equation is in the plane.

So the tetrahedron has a name.

It's called-- let's call
this A, B, C, and O. OABC.

It's a tetrahedron.

It's a pyramid.

So how does the smart
student who was not given

a specific method solve that?

They did that on the final.

I'm so proud of them.

I said, come on now.

The final is two
hours and a half.

You don't know what to do first.

So they said-- they
did the base multiplied

by the height divided by 3.

So you get 1 times 1.

So practically, divided by 2.

1/2.

You don't even have
to do the-- even

my son would know that this
is half of a square, a 1

by 1 square.

So it's half the area of the
base times the height, which

is 1, divided by 3 is 1/6.

And goodbye and see you later.

But if you wanted-- if
the author of the problem

would indicate, do
that with Calculus 3,

then that's another
story because you

have to realize what the domain
would be, the planar domain.

You practically have a surface.

The green-shaded
equilateral triangle

is your surface, which-- let's
call it c of f from surface.

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

How do you get to that?

You get it from here.

The explicit equation
is-- [INAUDIBLE].

1 minus x minus y.

That is the surface,
the green surface.

And the domain--
let's draw that in.

Do you prefer red or purple?

You don't care?

OK, I'll take red.

Red.

Red.

That's the domain D. So you'll
have to set up I, integral.

I for an I. And volume, double
integral over D of f of x, y,

whatever that is, dA.

That's going to be-- who is D?

Somebody help me, OK?

That's not easy.

So to draw the domain D, I have
to have a little bit of skill,

if I don't have any skill, I
don't belong in this class.

What do I have to draw?

Guys, tell me what to do.

0, x, and y.

To draw z, 0, z equals 0 gives
me x plus y equals 1, right?

So this is the floor.

Guys, this is the floor.

So why don't I shade it?

Because I'm not sure
which one I want.

Do I want vertical strips
or horizontal strips?

You're the boss.

You tell me what I want.

So do you want vertical strips?

Let's draw vertical strips.

So how do I
represent this domain

from the vertical strips?

x is between 0 and 1.

These are fixed
variable values of x

between fixed values 0 and 1.

For any such blue choice
of a point, I have a strip,

a vertical strip that goes
from y equals 0 down to--

STUDENT: 1 minus x.

DR. MAGDALENA TODA:
--1 minus x up.

Excellent, excellent.

This is exactly-- Roberto, you
were the one who said that?

OK.

So this is the domain.

So how do we write it down?

0 to 1.

0 to 1 minus x.

That is what I want to write.

No polar coordinates.

Goodbye.

There is no problem.

This is all a typical
Cartesian problem.

f-- f.

f is 1 minus x minus
y, thank you very much.

This is f dydx.

Homework, get 1/6.

So trying to do
that and get a 1/6.

And of course in the
exam-- oh, in the exam,

you will cheat big time, right?

What would you do?

You would set it up and forget
about computing it, integrating

one at a time, doing this.

And you would put equals 1/6.

Thank you very much.

Right?

Can I check that you
didn't do the work?

No.

You trapped me.

You got me.

I have no-- I mean, I need
to say, is this correct?

Yes.

Is the answer correct?

Yes.

Do they get full credit?

Yes.

So it's a sneaky problem.

OK.

Now finally, let's plunge
into 12.4, which is-- can you

remember this problem for 12.4?

I want to draw this again.

So I'll try not to
erase the picture.

I'll erase all the
data I have here.

And I'll keep the future because
I don't want to draw it again.

When we were small-- I
mean small in Calc 1 and 2,

they gave us a function
y equals f of x.

That is smooth, at least C1.

C1 means differentiable, and
the derivative is continuous.

And we said, OK, between
x equals a and x equals b,

I want you-- you, any
student-- to compute

the length of the arch.

Length of the arch.

And how did we do
that in Calc 2?

I have colleagues who
drive me crazy by refusing

to teach that in Calc 2.

Well, I disagree.

Of course, I can teach
it only in Calc 3,

and I can do it with
parametrization, which we did,

and then come back to the case
you have, y equals f of x,

and get the formula.

But it should be taught at
both levels, in both courses.

So when you have a
general parametrization

rt equals x of ty
of t [INAUDIBLE],

this is a parametrized
curve that's in C1, in time.

What is the length of the
arch between time t0 and time

t1 [INAUDIBLE]?

The integral from t0 to t1 or
the speed because the space

is the integral
of speed in time.

So in terms of speed,
remember that we

put square root of x prime of
t squared plus y prime of t

squared dt.

Why is that?

Somebody tell me.

That was the speed.

That was the magnitude
of the velocity vector.

And we've done that, and we
discovered that in Calculus 3.

In Calculus 2, they
taught this for what case?

The case when x is t--
say it again, I will now.

The case when x is
t, and y is f of t,

which is f of x--
and in that case,

as I told you before,
the length will

be the integral from A to B.
Whatever, it's the same thing.

A to B.

Square root-- since x
is t, x prime of t is 1.

So you get 1 plus-- y is just f.

y is f.

So you have f prime
of x squared dx.

So the length of this arch--
let me draw the arch in green,

so it's beautiful.

The length of this green
arch will be the length of r.

The integral from A to B square
root of 1 plus f prime 1x

squared dx.

Now there is a beautiful,
beautiful generalization

of that for-- generalization
for extension gives you

the surface area of a graph z
equals f of x, y over domain D.

Say what?

OK, it's exactly the
same formula generalized.

And I would like you to guess.

So I'd like you to
experimentally get

to the formula.

It can be proved.

It can be proved by taking
the equivalence of some sort

of Riemann summation
and passing to the limit

and get the formula.

But I would like you
to imagine you have--

so you have z equals f of x, y.

That projects exactly over
D. The area of the surface--

let's call it A of s.

The area of the
surface will be--

what do you think it will be?

You are smart people.

It will be double integral
instead of one integral

over-- what do you think?

Over the domain.

What's the simplest
way to generalize this

through probably [INAUDIBLE]?

Another square root.

We don't have just one
derivative, f prime of x.

We are going to have two
derivatives, f sub x and f sub

y.

So what do you think the
simplest generalization

will look like?

STUDENT: 1 plus [INAUDIBLE].

DR. MAGDALENA TODA: 1 plus
f sub x squared plus f sub y

squared dx.

That's it.

So you say, oh, I'm a genius.

I discovered it.

Yes, you are.

I mean, in a sense that-- no.

In the sense that we
all have that kind

of mathematical intuition,
creativity that you come up

with something.

And you say, OK, can I verify?

Can I prove it?

Yes.

Can you discover
things on your own?

Yes, you can.

Actually, that's how all
the mathematical minds came.

They came up to it
with a conjecture based

on some prior experiences, some
prior observations and said,

I think it's going
to look like that.

I bet you like 90% that it's
going to look like that.

But then it took them
time to prove it.

But they were convinced that
this is what it's going to be.

OK.

So if you want the area
of the patch of a surface,

that's going to be 4.1, and
that's page-- god knows.

Wait a second.

Wait.

Bare with me.

It starts at page 951,
and it ends at page 957.

It's only seven pages, OK?

So it's not hard.

You have several examples.

I'm going to work
on an example that

is straight out of the book.

And guess what?

You see, that's why I like
this problem, because it's

in-- example one is exactly the
one that I came up with today

and says, find the
same tetrahedron thing.

Find the surface area of
the portion of the plane

x plus y plus z equals
1, which was that, which

lies in the first octant
where-- what does it mean,

first octant?

It means that x is
positive. y is positive.

z is positive for z.

OK.

Is this hard?

I don't know.

Let's find out.

So if we were to apply this
formula, how would we do that?

Is it hard?

I don't know.

We have to recollect
who everybody

is from scratch, one at a time.

Hmm?

STUDENT: Could we just
use our K-12 knowledge?

DR. MAGDALENA TODA: Well,
you can do that very well.

But let's do it
first-- you're sneaky.

Let's do it first as Calc 3.

And then let's see who comes
up with the fastest solution

in terms of surface area.

By the way, this individual--
this whole fat, sausage

kind of thing is ds.

This expression is called
the surface element.

Make a distinction
between dA, which is

called area element in plane.

ds is the surface element on the
surface, on the surface on top.

So practically, guys, you have
some [? healy ?] part, which

projects on a domain in plane.

The dA is the infinite
decimal area of this thingy.

And ds is the infinite
decimal area of that.

What do you mean by that?

OK.

Imagine this grid of pixels
that becomes smaller and smaller

and smaller.

OK?

Take one pixel already and
make it infinitesimally small.

That's going to be
da dxdy, dx times dy.

What is the corresponding
pixel on the round surface?

I don't know.

It's still going to
be given by two lines,

and two lines form a
curvilinear domain.

And that curvilinear tiny-- do
you see how small it is that?

I bet the video cannot see it.

But you can see it.

So this tiny infinitesimally
small element

on the surface-- this is ds.

This is ds.

OK?

So if it were between a plane
and a tiny square, dxdy dA

and the ds here, it would
be easy between a plane

and a floor because
you can do some trick,

like a projection
with cosine and stuff.

But in general,
it's not so easy,

because you can have
a round patch that's

sitting above a domain.

And it's just-- you
have to do integration.

You have no other choice.

Let's compute it both ways.

Let's see.

A of s will be integral over
domain D. What in the world

was the domain D?

The domain D was the
domain on the floor.

And you told me what that
is, but I forgot, guys.

x is between 0 and 1.

Did you say so?

And y was between what and what?

Can you remind me?

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA:
Between 0 and--

STUDENT: 1 minus x.

DR. MAGDALENA TODA:
1 minus x, excellent.

So this is the
meaning of domain D.

And the square root
of-- who is f of x, y?

It's 1 minus x minus what?

Oh, right.

So you guys have to help
me compute this animal.

f sub x is negative 1.

Attention, please.

f sub y is negative 1.

OK.

So I have to say 1 plus negative
1 squared plus negative 1

squared dA.

Gosh, I'm lucky.

Look.

I mean, I'm not
just lucky, but that

has to be-- it has to
be something beautiful

because otherwise the
elementary formula will not

be so beautiful.

This is root 3, and
it brings it back.

Root 3 pulls out
of the whole thing.

So you have root 3.

What is double integral--
OK, let's compute.

So first you go dy and dx.

x, again, you gave
it to me, guys.

0 to 1.

y between 0 and 1 minus x.

Great.

We are almost there.

We are almost there.

I just need your
help a little bit.

The square root of 3 goes out.

The integral from 0 to 1.

What is the integral of 1dy?

It's y, y between 1 minus x
on top and 0 on the bottom.

That means 1 minus x.

If I make a mistake, just shout.

dx.

The square root of 3 times
the integral of 1 minus x.

STUDENT: x minus
the square root.

DR. MAGDALENA TODA: x
minus the square root of 2.

Let me write it separately
because I should do that fast,

right?

Between 0 and 1.

What is that?

1/2.

That's a piece of cake.

This is 1/2.

So 1/2 is this thing.

And root 3 over 2.

And now Alex says,
congratulations

on your root 3
over 2, but I could

have told you that much faster.

So the question
is, how could Alex

have shown us this root
3 over 2 much faster?

STUDENT: Well, it's
just a triangle.

DR. MAGDALENA TODA:
It's just a triangle.

It's not just a triangle.

It's a beautiful triangle
that's an equilateral triangle.

And in school, they used
to teach more trigonometry.

And they don't.

And if I had the
choice-- I'm not

involved in the K-12 curriculum
standards for any state.

But if I had the choice, I
would say, teach the kids

a little bit more
geometry in school

because they don't know
anything in terms of geometry.

So there were
triumphs in the past,

and your parents may
know those better.

But If somebody gave you
an equilateral triangle

of a certain side, you
would be able to tell them,

I know your area.

I know the area.

I know the area being l squared,
the square root of 3 over 4.

Your parents may know that.

Aaron, ask your dad.

He will know.

But we don't teach
that in school anymore.

The smart kids do this
by themselves how?

Can you show me how?

They draw the
perpendicular bisector.

And there is a
theorem actually--

but we never prove
that in school--

that if we draw that
perpendicular bisector,

then the two triangles
are congruent.

And as a consequence,
well, that is l/2, l/2.

OK?

So if you draw just
the perpendicular,

you can prove it using some
congruence of triangles

that what you get here
is also the median.

So it's going to keep
right in the middle

of the opposite side.

So you l/2, l/2.

OK.

That's what I wanted to say.

And then using the
Pythagorean theorem,

you're going to get the height.

So the height will be the square
root of l squared minus l/2

squared, which is the
square root of l squared

minus l squared over 4, which
is the square root of 3l squared

over 4, which simplified
will be l root 3 over 2.

l root 3 over 2 is
exactly the height.

And then the area will be
height times the base over 2

for any triangle.

So I have the height times
the base over 2, which

is root 3l squared over 4.

So many people when
they were young

had to learn it in
seventh grade by heart.

Now in our case, it
should be a piece of cake.

Why?

Because we know who l is.

l is going to be the hypotenuse.

We have here a 1 and
a 1, a 1 and a 1.

So this is going to be the
hypotenuse, square root of 2.

So if I apply this
formula, which

is the area of the
equilateral triangle,

that says root 2 squared root
3 over 4 equals 2 root 3 over 4

equals root 3 over 2.

So can you do that?

Are you allowed to do that?

Well, we never formulated
it actually saying

compute the surface of
this patch of a plane using

the surface integral.

We didn't say that.

We said, compute it,
period We didn't care how.

If you can do it
by another method,

whether to stick to that
method, elementary method

or to just check
your work and say,

is it really a square
root of 3 over 2?

You are allowed to do that.

Questions?

STUDENT: So would the
length be square root

of 2 squared, which is 2.

2 divided by 4 is [INAUDIBLE]
square root of 3 over 2.

I'm just talking-- oh, yeah.

DR. MAGDALENA TODA: You are
just repeating what I have.

So the answer i got
like this is elementary.

And the answer I got
like this is with Calc 3.

It's the same answer, which
gives me the reassurance

I wasn't speaking nonsense.

I did it in two different ways,
and I got the same answer.

Let's do one or
two more examples

of surface integrals, surface
areas and surface integrals.

It's not hard.

It's actually quite fun.

Some of them are
harder than others.

Let's see what we can do.

Oh, yeah.

I like this one very much.

I remember we gave it several
times on the final exams.

So let's go ahead
and do one like that

because you've
seen-- why don't we

pick the one I picked before
with the eggshells for Easter,

like Easter eggs?

What was the paraboloid I
had on top, the one on top?

STUDENT: 8 minus x--

DR. MAGDALENA TODA: 8 minus
x squared [INAUDIBLE].

That's what I'm going to pick.

And I'll say, as Easter
is coming, a word problem.

We want to compute the
surface of an egg that

is created by intersecting
the two paraboloids 8 minus x

squared minus y squared and
x squared plus y squared.

So let's see.

No.

Not y, Magdalena.

Intersect, make
the egg intersect.

Create the eggshells.

Shells.

Compute the area.

And you say, wait a minute.

The two eggshells were equal.

Yes, I know.

I know that the two
eggshells were equal.

But they don't look
equal in my picture.

I'll try better.

Assume they are parabolas.

Assume this was z equals 4.

This was 8 minus x
squared minus y squared.

This was x squared
plus y squared.

How do we compute-- just
like Matthew observed, 8

for the volume.

I only need half of the
8 multiplied by the 2.

The same thing.

I'm going to take one of
the two shells, this one.

And the surface of the egg will
be twice times the surface S1.

All I have to
compute is S1, right?

It shouldn't be a big problem.

I mean, what do I
need for that S1?

I only need the shadow of
it and the expression of it.

The shadow of it and the--
the shadow of it is this.

The shadow of this is this.

And the expression-- hmm.

It shouldn't be hard.

So I'm going to-- I'm
going to start asking

you to tell me what to write.

What?

STUDENT: Square root of 1 plus--

DR. MAGDALENA TODA: No.

First I will write
the definition.

Double integral over D,
square root of, as you said--

say it again.

STUDENT: 1 plus f
of x squared plus f

of y squared [INAUDIBLE] dA.

DR. MAGDALENA TODA: All right.

So this is ds, and I'm
integrating over the domain.

Should this be hard?

No, it shouldn't be hard.

But I'm going to get
something a little bit ugly.

And it doesn't matter, because
we will do it with no problem.

I'm going to say, integral
over-- now the domain

D-- I know what it is
because the domain D will

be given by x squared
plus y squared less than

or equal to 4.

So I would know how to
deal with that later on.

Now what scares me off
a little bit-- and look

what's going to happen.

When I compute f sub x and f sub
y, those will be really easy.

But when I plug
everything in here,

it's going to be
a little bit hard.

Never mind, I'm
going to have to have

to battle the problem
with polar coordinates.

That's why polar coordinates
exist, to help us.

So f sub x is minus 2x, right?

f sub y is minus 2y.

OK.

So what am I going
to write over here?

Minus 2x squared plus
minus 2y squared dx.

I don't have much room.

But that would mean dxdy.

Am I happy with that?

No.

I'm not happy with
it, because here it's

going to be x squared plus
y squared between 0 and 4.

And I'm not happy with it,
because it looks like a mess.

And I have to find this area
integral with a simple method,

something nicer.

Now the question is,
does my elementary math

help me find the
area of the egg?

Unfortunately, no.

So from this point on, it's
goodbye elementary geometry.

STUDENT: Unless you
know the radius.

DR. MAGDALENA TODA: But they
are not spheres or anything.

I can approximate the
eggs with spheres,

but I cannot do anything with
those paraboloids [INAUDIBLE].

STUDENT: I know the
function of the top.

DR. MAGDALENA TODA:
Yeah, yeah, yeah.

You can.

STUDENT: [INAUDIBLE]
the integration f prime.

DR. MAGDALENA TODA: But
it's still integration.

So can I pretend like I'm a
smart sixth grader, and I can--

how can I measure that if I'm
in sixth grade or seventh grade?

With some sort of graphic paper,
do some sort of approximation

of the area of the egg.

It's a school project that's
not worth anything because I

think not even at a science
fair, could I do it.

STUDENT: Unless--
in the same radius,

I can draw the sphere in.

Then if I apply the
distance between the sphere

and the [INAUDIBLE] the
distance between [INAUDIBLE]

and take it all from there.

But then the function
actually will look easier

because it will go from the
y axis up to the A axis,

and they meet each other.

So I took up the
area and took up

the other area to [INAUDIBLE].

DR. MAGDALENA TODA: Yeah.

Well, wouldn't that
surface of the egg still

be an approximation
of the actual answer?

Anyway, let's come
back to the egg.

The egg, the egg.

The egg is [INAUDIBLE].

It's nice.

1 plus 4 x squared
plus y squared.

Look at the beauty of the
symmetry of polynomials.

x squared plus y squared says,
I'm a symmetric polynomial.

You're my friend
because I'm r squared,

and I know what I'm going to do.

So how do we compute?

What kind of integral
do we need to compute?

So S1 will be the integral of
integral of the square root

of 1 plus 4r squared.

Don't forget the dA
contains the Jacobian.

So don't write drd theta.

I had a student who wrote that.

That is worth
exactly zero points.

So say, times r.

r between-- oh, my god,
the poor egg-- 0 to 2.

And theta between 0 to 2 pi.

And come on.

We've done that in Calc 2.

I mean, it's not so hard.

So u substitution.

u is 4r squared plus 1.

That's our only hope.

We have no other hope.

du is going to be 8rdr.

And rdr is a married couple.

They stick together.

Where is the purple?

The purple is here.

rdr, rdr.

rdr is du/8.

This fellow's name is u.

He is u.

He is not u, but he's like u.

OK, not necessary.

OK.

So you go 2 pi-- because
there is no theta.

So no theta means-- let me
write it one more time for you.

The integral from
0 to 2 pi 1d theta.

And he goes out and has fun.

This is 2 pi.

But then all you have left
inside is the integral of u.

Square root of u times
1/8 du, close the bracket,

where u is between 1 and 17.

Isn't that beautiful?

That's 17.

So you have 2 squared
times 416 plus 117.

But believe me that from this
viewpoint, from this point on,

it's not really hard.

It just looks like the
surface of that egg

is-- whenever it was produced,
in what factory, in whatever

country is the toy factory, they
must have done this area stage.

So you have 2 pi.

1/8 comes out, whether
he wants out or not.

Integral of square root of u.

Do you like that?

I don't.

You have--

STUDENT: 2/3.

DR. MAGDALENA TODA:
2/3 u to the 3/2

between-- down is u equals 1.

Up is u equals 17.

So I was asked,
because we've done

this in the past
reviews for the finals--

and several finals
are like that.

My students asked me, what
do I do in such a case?

Nothing.

I mean, you do nothing.

You just plug it in
and leave it as is.

So you have-- to simplify
your life a little bit, what

you can do is 2, 2, and 8.

What is 2 times 2?

4.

Divided by 8-- so you
have pi/6 overall.

pi/6 times 17 to
the 3/2 and minus 1.

One of my students, after he
got such an answer last time

we did the review, he
said, I don't like it.

I want to write this as
square root of 17 cubed.

You can write it
whatever you want.

It can be-- it
has to be correct.

I don't care how you write it.

What if you mess up?

You say, well, this
woman is killing me

with her algebra over here.

OK.

If you understood--
suppose that you

are taking the final right now.

You drew the
picture beautifully.

You remember the problem.

You remember the formula.

You write it down.

You wrote it down.

You got to this point.

At this point, you already
have 50% of the problem.

Yup.

And then from this point on,
you do the polar coordinates,

and you still get another 25%.

You messed it up.

You lose some partial credit.

But everything you
write correctly

earns and earns
and earns points.

OK?

So don't freak out
thinking, I'm going

to mess up my algebra for sure.

If you do, it doesn't matter,
because even if this would

be a multiple choice--
some problems will

be show work completely,
and some problems

may be multiple
choice questions.

Even if this is going
to be a multiple choice,

I will still go over the entire
computation for everybody

and give partial credit.

This is my policy.

We are allowed to choose
our policies as instructors.

So you earn partial credit
for everything you write down.

OK.

Was this hard?

It's one of the harder
problems in the book.

It is he similar to
example number-- well,

this is exactly like
example 2 in the section.

So we did these two
examples from the section.

And I want to give you one
more piece of information

that I saw, that unfortunately
my colleagues don't teach that.

And it sort of bothers me.

I wish they did.

Once upon a time,
a long time ago,

I taught you a little bit
more about the parametrization

of a surface.

And I want to give you yet
another formula, not just

this one but one more.

So what if you have a
generalized surface that

is parametrized, meaning
that your surface is not

given as explicitly
z equals f of x, y?

That's the lucky case.

That's a graph.

We call that a graph,
z equals f of x and y.

And we call ourselves lucky.

But life is not always so easy.

Sometimes all you can get
is a parametrization r

of v, v for a surface.

And from that, you
have to deal with that.

So suppose somebody says,
I don't give you f of x, y,

although locally every
surface looks like the graph.

But a surface doesn't have
to be a graph in general.

Locally, it does look like
a graph on a small length.

But in general, it's
given by r, v, v equals--

and that was what?

I gave you something like x of
u, v I plus y of u, v J plus z

of u, v-- let's not put
things in alphabetical order.

z of u, v J and K.

And we said that
we have a point.

P is our coordinate u0, v0.

And we said we
look at that point,

and we try to draw the partials.

What are the partials from
a geometric viewpoint?

Well, if I want to
write the partials,

they would be various.

It's going to be the vector
x sub u, y sub u, z sub

u, and the vector x sub v, y
sub v, z sub v, two vectors.

Do you remember when I
drew them, what they were?

We said the following.

We said, let's assume
v will be a constant.

So we say, v is a constant.

And then v equals v0.

And then you have P of u0, v0.

And then we have another,
and we have u equals u0.

This guy is going to
be who of the two guys?

r sub u.

When we measure out
the point P, r sub u

is this guy, who is tangent
to the line r of u, v zero.

Does it look tangent?

I hope it looks tangent.

And this guy will be r of
u-- because u0 means what?

u0 and v. So who
is free to move?

v. So this guy, this r sub
v-- they are both tangents.

So do you have a surface?

This is the surface.

This is the surface.

And these two horns or
whatever they are-- those

are the tangents r sub u, r
sub v, the two tangent vectors,

the partial velocities.

And I told you before, they
form the tangent plane.

They are partial velocities.

They are both tangent to
the surface at that point.

They form a basis.

They are linearly independent.

Always?

No.

But we assume that r sub u
and r sub v are non-zero,

and they are not co-linear.

How do I write that?

They are not parallel.

So guys, what does it mean?

It means-- we talked
about this before.

The velocities cannot be 0.

And r sub u, r sub v
cannot be parallel,

because if they are parallel,
there is no area element.

There is no tangent
plane between them.

What they form is
the area element.

So what do you think the
area element will look like?

It's a magic thing.

The surface element
actually will

be exactly the area between ru
and rv times the u derivative.

Say it again, Magdalena.

What the heck is the area
between the vectors r sub

u, r sub v?

You know it better than
me because you're younger,

and your memory is better.

And you just covered
this in chapter nine.

When you have a vector A
and a vector B that are not

co-linear, what was the
area of the parallelogram

that they form?

STUDENT: The
magnitude [INAUDIBLE].

DR. MAGDALENA TODA:
Magnitude of--

STUDENT: The cross product.

DR. MAGDALENA TODA:
The cross product.

Excellent.

This is exactly what
I was hoping for.

The magnitude of the
cross product is the area.

So you have ds, infinitesimal
of an answer plus area, surface

element will be
exactly the magnitude

of the cross product of the
two velocity vectors, dudv.

dudv can also be
written dA in this case

because it's a flat
area on the floor.

It's the area of a tiny
square on the floor,

infinitesimally small square.

So remember that.

And you say, well, Magdalena,
you are just feeding us

formula after formula.

But we don't even know.

OK, this makes sense.

This looks like I have some
sort of tiny parallelogram,

and I approximate the
actual curvilinear

patch, curvilinear
patch on-- I'm

going to draw it on my hand.

So this is-- oh, my god.

My son would make fun of me.

So this curvilinear patch
between two curves on my hand

will be actually
approximated by this.

What is this rectangle?

No, it's a--

STUDENT: Parallelogram.

DR. MAGDALENA TODA:
Parallelogram.

Thank you so much.

So this is an
approximation, again.

So this is the area
of the parallelogram.

And that's what we defined
as being the surface element.

It has to do with
the tangent plane.

But now you're asking,
but shouldn't this

be the same as the formula
root of 1 plus f sub x squared

plus f sub y squared dxdy?

Yes.

Let's prove it.

Let's finally prove that
the meaning of this area

will provide you
with the surface

element the terms of x
and y, just the way you--

you did not prove it.

You discovered it.

Remember, guys?

You came up with a
formula as a conjecture.

You said, if we generalize
the arch length,

it should look like that.

You sort of smelled it.

You said, I think.

I feel.

I'm almost sure.

But did you prove it?

No.

So starting from the
idea of the area element

that I gave before,
do you remember

that we also had that signed
area between the dx and dy,

and we used the area of
the parallelogram before?

We also allowed it to
go oriented plus, minus.

OK.

All right.

So this makes more sense
than what you gave me.

Can I prove what you
gave me based on this

and show it's one
and the same thing?

So hopefully, yes.

If I have my explicit form
z equals f of x and y,

I should be able to
parametrize this surface.

How do I parametrize
this surface

in the simplest possible way?

x is u.

y is v. z is f of
u, v. And that's it.

Then it's r of u, v as a vector
will be angular bracket, u, v,

f of u, v. Now you have to
help me compute r sub u and r

sub v. They have to
be these blue vectors,

the partial velocities.

r sub u, r sub v.

Is it hard?

Come on.

It shouldn't be hard.

I need to change colors.

So can you tell
me what they are?

What's the first-- 1?

Good.

What's next?

0.

Thank you.

STUDENT: F sub u.

DR. MAGDALENA TODA: f sub u.

Very good.

f sub u or f sub x is the same
because x and u are the same.

So let me rewrite
it 1, 0, f sub x.

Now the next batch.

0, 1, f sub v, which
is 0, 1, f sub y.

0, 1, f sub y.

Now I need to cross them.

And I need to cross them, and
I'm too lazy because it's 2:20.

But I'll do it.

I'll do it.

I'll cross.

So with your help
and everything,

I'm going to get to
where I need to get.

You can start.

I mean, don't wait for me.

Try it yourselves
and see what you get.

And how hard do you think
it is to compute the thing?

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA: I
will do the normality

at the magnitude later.

r sub u, r sub v's cross product
will be I, J, K. 1, 0, f sub x,

0, 1, f sub y.

Is this hard?

It shouldn't be hard.

So I have minus f sub x, what?

I. I'm sorry.

I for an I. OK.

J, again minus because
I need to change.

When I expand along the row, I
have plus, minus, plus, minus,

plus, alternating.

So I need to have minus.

The determinant is f sub y times
J plus K times this fellow.

But that fellow is 1, is the
minor 1, for god's sakes.

So this is so easy.

I got the vector,
but I need the norm.

But so what?

Do you have it?

I'm there, guys.

I'm really there.

It's a piece of cake.

I take the components.

I squeeze them a little bit.

No.

I square them,
and I sum them up.

And I get the square
root of 1 plus-- exactly.

1 plus f sub x squared
plus f sub y squared d.

This is u and v, and this
is dxdy, which is dA.

This is the tiny floor square
of an infinitesimally square

on the floor.

OK?

And what is this again?

This is the area of
a tiny curvilinear

patch on the surface that's
projected on that tiny square

on the floor.

All right?

OK.

So Now you know why
you get what you get.

One the last problem
because time is up.

No, I'm just kidding.

We still have plenty of time.

This is a beautiful,
beautiful problem.

But I don't want to finish it.

I want to give you
the problem at home.

It's like the one
in the book, but I

don't want to give you
exactly the one in the book.

I want to cover
something special today.

We are all familiar with their
notion of a spiral staircase.

But spiral staircases
are everywhere,

in elegant buildings, official
buildings, palaces, theaters,

houses of multimillionaires
in California.

And even people who
are not millionaires

have some spiral
staircases in their houses,

maybe made of wood
or even marble.

Did you ever wonder why
the spiral staircases

were invented?

If you go to most of the
castles on the Loire Valley,

or many European castles
have spiral staircases.

Many mosques, many churches
have these spiral staircases.

I think it was about a few
thousand years ago that it

was documented
for the first time

that the spiral staircases
consumed the least

amount of materials to build.

Also what's good about them
is that for confined spaces--

you have something like a
cylinder tower like that--

that's the only shape you
can build that minimizes

the area of the staircase
because if you start building

a staircase like
ours here, it's not

efficient at all in
terms of construction,

in terms of materials.

So you get a struggle
at actually making

these stairs that are not even.

You know, they're not even even.

Each of them will have a
triangle, or what is this?

Not a triangle, but
more like a trapezoid.

And it keeps going up.

This comes from a
helix, obviously.

And we have to understand
why this happens.

And I will introduce the
surface called helicoid.

And the helicoid will have
the following parametrization

by definition.

u cosine v, u sine v, and v.

Assume u is between 0 and 1 and
assume v is between 0 and 2 pi.

Draw the surface.

Also find the surface area of
the patch u between 0 and 1,

v between 0 and pi/2.

So I go, uh oh.

I'm in trouble.

Now how in the world am I
going to do this problem?

It looks horrible.

And it looks hard.

And it even looks hard to draw.

It's not that hard.

It's not hard at
all, because you

have to think of these
extreme points, the limit

points of u and v and see what
they really represent for you.

Put your imagination
to work and say,

this is the frame
I'm starting with.

This is the x and y and
z frame with origin 0.

And I better draw this helicoid
because it shouldn't be hard.

So for u equals
0, what do I have?

I don't know.

It looks weird.

But Alex said it.

0, 0, v. v is my
parameter in real life.

So I have the whole z-axis.

So one edge is going to
be the z-axis itself.

Does it have to be
only the positive one?

No.

Who said so?

My problem said so,
that v only takes

values between 0 and 2 pi.

Unfortunately, I'm limiting
v between 0 and 2 pi.

But in general, v could
be any real number.

So I'll take it from
0 to 2 pi, and this

is going to be what I'm thinking
of, one edge of the staircase.

It's the interior age, the axis.

Let's see what happens
when u equals 1.

That's another curve
of the surface.

Let's see what I get.

Cosine v, sine v, and v. And it
looks like a friend of yours.

And you have to
tell me who this is.

If v were bt, what
is sine, cosine tt?

STUDENT: A sphere.

DR. MAGDALENA TODA: It's your--

STUDENT: Helicoid.

DR. MAGDALENA TODA: --helix
that you loved in chapter 10

and you made friends with.

And it was a curve that had
constant curvature and constant

portion, and it
had constant speed.

And the speed of
this, for example,

would be square root of 2,
if you have the curiosity

to compute it.

It will have square root of 2.

And can I draw it?

I better draw it,
but I don't know how.

So I have to think of drawing
this for v between 0 and 2 pi.

When I'm at 0, when I
have time v equals 0,

I have the point 1, 0, and 0.

And where am I?

Here.

1, 0, 0.

And from here, I start moving
on the helix and going up.

And see, my hand should be
on-- this is the stairs.

It's obviously a smooth surface.

This is a smooth
surface, but the stairs

that I was talking about
are a discretization

of the smooth surface.

I have a step, another step,
another step, another step.

So it's like a smooth helicoid
but discretized step functions.

Forget about the step functions.

Assume that instead of the
staircase in the church--

you don't want to go to church.

You want to go to
the water park.

You want to go to Six Flags.

You want to go to whatever,
Disney World, San Antonio,

somewhere.

This is a slide.

You let yourself go.

This is you going
down, swimming--

I don't know-- upside down.

I don't know how.

So this is a smooth
slide in a water park.

That's how you should
be imagining it.

And it keeps going.

If I start here-- if I
start here, again, look,

this is what I'm describing.

A helicoid.

My arm moved on this.

Again, I draw the same motion.

My elbow should not
do something crazy.

It should keep
moving on the z-axis.

And I perform the pi/2 motion
when the stair-- not the stair.

I don't know what to call it.

This line becomes horizontal
when v equals pi/2.

So for v equals pi/2, I moved
from here straight to here.

STUDENT: Doesn't it go around?

DR. MAGDALENA TODA:
It goes around.

But see, what I asked-- I
only asked for the patch.

First of all, I
said it goes around.

So I'll try to go around.

But it's hard.

Oh, wish me luck.

One, two, three.

I cannot go higher.

It goes to pi.

STUDENT: If it went to
2 pi, it would actually

wrap completely around?

DR. MAGDALENA TODA: It
would wrap like that.

STUDENT: Right above
where it started?

DR. MAGDALENA TODA: Exactly.

So if I started here, I end
up parallel to that, up.

But 2 pi is too high for me.

So I should go slowly.

I'm up after 2 pi in
the same position.

STUDENT: But since it's pi/2,
it's just kind of like--

DR. MAGDALENA TODA:
When I'm pi/2,

I just performed
from here to here.

STUDENT: So that's it's
asking you for the patch?

DR. MAGDALENA TODA:
And it's asking for,

what area does my arm from here
to here sweep at this point?

From this point to this point.

It's a smooth surface.

So it's generated by my motion.

And stop.

It's a root surface.

It's a root patch of a surface.

Somebody tell me how I'm going
to do this because this is not

with square root of 1 plus f sub
x squared plus f sub y squared.

That is for normal people.

You are not normal people.

They never teach this in honors.

In honors, we don't
cover this formula.

But you're honors.

So do I want you to finish
it at home with a calculator?

All I want is for you to be
able to set up the integral.

And I think, knowing you
better and working with you--

I think you have the potential
to do that without my help,

with all the elements
I gave you until now.

So the area of s-- it
will be the blue slide.

These are all--
when you slide down,

you slide down along helices.

You and your friends-- you're
going down along helices.

OK?

So that's what you have.

[INAUDIBLE] double integral
for a certain domain D.

Which is that domain D?

That domain D is u between 0
and 1 and v between 0 and pi/2

because that's what
I said I want here.

Of what?

Of magnitude of r sub u times
r sub v cross product dudv.

You need to help me
though because I don't

know what I'm going to do.

So who starts?

r sub u is-- I'm not doing
anything without you, I swear.

OK?

STUDENT: Cosine.

DR. MAGDALENA TODA: Cosine v.

STUDENT: Sine v.

DR. MAGDALENA TODA: Sine v.

STUDENT: And 0.

DR. MAGDALENA TODA: 0.

r sub v equals--

STUDENT: Negative u sine v.

DR. MAGDALENA TODA: Very good.

u cosine v. And 1, and
this goes on my nerves.

But what can I do?

Nothing.

OK?

All right.

So I have to compute the what?

STUDENT: The cross product.

DR. MAGDALENA TODA:
The cross product.

I, J, K, of course, right?

I, J, K, cosine v, sine v,
0, minus u sine-- oh, my god.

Where was it?

It's there.

You gave it to me.

u cosine v and 1.

Minus u sine v, u
cosine v, and 1.

OK.

I times that.

Sine v, I. sine v,
I. J. J has a friend.

Is this mine or--
cosine v times 1.

But I have to change the sine.

Are you guys with me?

It's really serious that I have
to think of changing the sine.

Minus cosine v times
J. Are you with me?

This times that
with a sign change.

And then plus-- what is--
well, this is not so obvious.

But you have so much practice.

Make me proud.

What is the minor
that multiplies K?

This red fellow.

You need to compute
it and simplify it.

So I don't talk too much.

u, excellent.

How did you do it, [INAUDIBLE]?

STUDENT: [INAUDIBLE].

DR. MAGDALENA TODA: So
you group together cosine

squared plus sine squared.

Minus, minus is a plus.

And you said u, it's u.

Good.

Plus u times K. Good.

It doesn't look so bad.

Now that you look at it,
it doesn't look so bad.

You have to set up the integral.

And that's going to be what?

The square root of
this fellow squared

plus that fellow squared
plus this fellow squared.

Let's take our time.

You take these three fellows,
square them, add them up,

and put them under
a square root.

Is it hard?

STUDENT: 1 plus u squared.

DR. MAGDALENA TODA:
1 plus u squared.

Now the thing is I
don't have a Jacobian.

This is dudv.

The Jacobian is what?

So this is what I have.

Now between the end
points, I have to think.

v has to be between 0 and pi/2.

And u has to be between 0 and 1.

Do you notice anything?

And that's exactly what
I wanted to tell you.

v is not inside.

v says, I'm independent.

Please leave me alone.

I'll go off, take a break. pi/2.

But then you have
integral from 0

to 1 square root of
1 plus u squared du.

I want to say a remark.

Happy or not happy, shall I be?

Here, you need either
the calculator,

which is the simplest
way to do it, just

compute the simple integral.

Integral from 0 to 1 square
root of 1 plus u squared du.

Or what do you have in this
book that can still help you

if you don't have a calculator?

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA: A table of
integration, integration table.

Please compute this.

I mean, you cannot
give me an exact value.

But give me an approximate
value by Thursday.

Is today Tuesday?

Yes.

By Thursday.

So please let me know how much
you got from the calculator

or from integration tables.

So we have this
result. And then I

would like to interpret
this result geometrically.

What we can say about the
helicoid that I didn't tell you

but I'm going to tell you
just to finish-- have you

been to the OMNIMAX
Science Spectrum, the one

in Lubbock or any other?

I think they are
everywhere, right?

I mean-- everywhere.

We are a large city.

Only in the big cities can
you come to a Science Spectrum

museum kind of like that.

Have you played
with the soap films?

OK.

Do you remember what kind
wire frames they had?

They had the big tub with soapy
water, with soap solution.

And then there were all sorts
of [INAUDIBLE] in there.

They had the wire with pig
rods that looked like a prism.

They had a cube that they wanted
you to dip into the soap tub.

They had a heart.

And there comes the
beautiful thing.

They had this, a
spring that they took

from your grandfather's bed.

No.

I don't think it
was flexible at all.

It was a helix made with
a rod inside, a metal rod

inside and attached
to the frame of that.

There was the metal rod and
this helix made of hard iron,

and they were sticking together.

Have you dipped this
into the soap solution?

And what did you get?

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA: That's
exactly what you get.

You can get several surfaces.

You can even get the one on
the outside that's unstable.

It broke in my case.

It's almost like a cylinder.

The one that was pretty
stable was your helicoid.

The helicoid is a so-called
soap film, soap film

or minimal surface.

Minimal surface.

So what is a minimal surface?

A minimal surface
is a soap film.

A minimal surface
is a surface that

tends to minimize the area
enclosed in a certain frame.

You take a wire that looks like
a loop, its own skew curve.

You dip that into
the soap solution.

You pull it out.

You get a soap film.

That a minimal surface.

So all the things
that you created

by taking wires and dipping
them into the soap tub

and pulling them out-- they
are not just called soap films.

They are called
minimal surfaces.

Somewhere on the wall
of the Science Spectrum,

they wrote that.

They didn't write much about
the theory of minimal surfaces.

But there are people-- there
are famous mathematicians who

all their life studied
just minimal surfaces, just

soap films.

They came up with the results.

Some of them got very
prestigious awards

for that kind of thing
theory for minimal surfaces.

And these have been known
since approximately the middle

of the 19th century.

There were several
mathematicians

who discovered the most
important minimal surfaces.

There are several that
you may be familiar with

and several you may
not be familiar with.

But another one that you may
have known is the catenoid.

And that's the last thing I'm
going to talk about today.

Have you heard of a catenary?

Have you ever been to St. Louis?

St. Louis, St. Louis.

The city St. Louis
with the arch.

OK.

Did you go to the arch?

No?

You should go to the arch.

It looks like that.

It has a big base.

So it looks so beautiful.

It's thicker at the base.

This was based on a
mathematical equation.

The mathematical equation
it was based on was cosh x.

What is cosh as a function?

Now I'm testing you,
but I'm not judging you.

If you forgot--

STUDENT: e to the x.

DR. MAGDALENA TODA:
e to the x plus e

to the negative x over 3.

If it's minus, it's called sinh.

So the one with [? parts. ?]
You can have x/a,

and you can have 1/a in front.

It's still called a catenary.

Now what is-- this
is a catenary.

The shape of the arch of
St. Louis is a catenary.

But you are more used to the
catenary upside down, which

is any necklace.

Do you have a necklace?

If you take any
necklace-- that is,

it has to be homogeneous,
not one of those,

like you have a pearl hanging,
or you have several beads.

No.

It has to be a
homogeneous metal.

Think gold, solid gold, but
that kind of liquid gold.

Do you know what
I'm talking about?

Those beautiful bracelets
or necklaces that are fluid,

and you cannot even see
the different links.

So you hang it at
the same height.

What you get-- Galileo
proved it was not

a parabola because people at
that time were really stupid.

So they thought, hang a
chain from a woman's neck

or some sort of
beautiful jewelry,

it must be a parabola because
it looks like a parabola.

And Galileo Galilei says,
these guys are nuts.

They don't know any mathematics,
any physics, any astronomy.

So he proved in no time
that thing, the chain,

cannot be a parabola.

And he actually
came up with that.

If a is 1, you just
have the cosh x.

So this is the chain.

If you take the chain--
if you take the chain--

that's a chain upside down.

If you take a chain--
let's say y equals cosh

x to make it easier--
and revolve that chain,

you get a surface of revolution.

And this surface of
revolution is called catenoid.

How can you get the catenoid
as a minimal surface, expressed

as the minimum surface?

There are people
who can do that.

They have the
ability to do that.

And they tried to have
an experimental thing

at the Science Spectrum as well.

And they did a beautiful job.

I was there.

So they took two
circles made of plastic.

You can have them be
circles made of wood,

made of iron or
steel or anything.

But they have to be
equal, equal circles.

And you touch them,
and you dip them

both at the same time
into the soap solution.

And then you pull
it out very gently

because you have
to be very smart

and very-- be like a surgeon.

If your hands start shaking,
it's goodbye minimal surfaces

because they break.

They collapse.

So you have to pull those
circles with the same force,

gently, one away from the other.

What you're going
to get is going

to be a film that looks
exactly like that.

These are the circles.

After a certain distance
of moving them apart,

the soap film will
collapse and will burst.

There's no more surface inside.

But up to that moment,
you have a catenoid.

And this catenoid is
a minimal surface.

It's trying to-- of
the frame you gave it,

which is the wire frame--
minimize the area.

It's not going to be a cylinder.

It's way too much area.

It's going to be something
smaller than that,

so something that says,
I'm an elastic surface.

I'm occupying as
little area as I

can because I live in
a world of scarcity,

and I try to occupy
as little as I can.

So it's based on a
principle of physics.

The surface tension
of the soap films

will create this
minimization of the area.

So all you need to do is
remember you know the helicoid

and catenoid only because
you're honors students.

So thank [INAUDIBLE] college
for giving you this opportunity.

All right?

I'm not kidding.

It may sound like a joke, but
it's half joke, half truth.

We learn learn a little
bit more interesting stuff

than other kids.

Enjoy your week.

Good luck with homework.

Come bug me abut any kind
of homework [INAUDIBLE].

STUDENT: Do you
know if I can talk

to people about [INAUDIBLE]?

DR. MAGDALENA TODA:
Actually, my [INAUDIBLE].

She is the one who
does [INAUDIBLE].

But I can take you to her so
you can start [INAUDIBLE].

I think it would be a very
interesting [INAUDIBLE].

Maybe.

[INAUDIBLE]

STUDENT: OK.

DR. MAGDALENA TODA:
Second floor, [INAUDIBLE].

This is the [INAUDIBLE].

You have to go all
the way behind.

STUDENT: OK.

OK.

STUDENT: [INAUDIBLE]

DR. MAGDALENA TODA:
[INAUDIBLE] because I don't

want to give you anything new.

I don't want to get you
in any kind of trouble.

The problems that I solved
on the board are primarily,

I would say, 60% of what
you'll see on the midterm.

It's something that
we covered in class.

And the other 40% will be
something not too hard,

but something standard out of
your WeBWork homework, the one

that you studied.

It shouldn't be hard.

STUDENT: Thank you.

DR. MAGDALENA TODA:
You're welcome.

STUDENT: We're trying to
join the Honors Society.

But we can't make
that thing tomorrow.

Can we still join?

DR. MAGDALENA TODA:
You can still join.

Remind me to give you the
golden pin, the brochures, all

the information.

And then when you get those,
you'll give me the $35.

It's a lifetime thing.

STUDENT: Awesome.

Thank you.

DR. MAGDALENA TODA:
You're welcome.

Both of you want to--

STUDENT: Yeah.

DR. MAGDALENA TODA: And
you cannot come tomorrow?

STUDENT: Yeah.

I have my--

DR. MAGDALENA TODA: I wanted to
bring something, some snacks.

But I don't know.

I need to count and see
how many people can come.

And it's going to
be in my office.

STUDENT: OK.

DR. MAGDALENA TODA: All right.

STUDENT: Were you in
the tennis tournament?

STUDENT: Yeah.

STUDENT: You're the guy who won?

STUDENT: Yeah.

STUDENT: Congratulations.

I was like, I know that name.

He's in my [INAUDIBLE].

DR. MAGDALENA TODA: You won it?

STUDENT: Yeah.

DR. MAGDALENA TODA:
The tennis tournament?

STUDENT: Yeah.

It was [INAUDIBLE].

DR. MAGDALENA TODA:
Congratulations.

Why don't you blab a
little bit about yourself?

You're so modest.

You never say anything.

STUDENT: It's all right.

STUDENT: I'm sorry.

STUDENT: No.

STUDENT: It's not a big deal.

STUDENT: Were people good?

Yeah?

DR. MAGDALENA TODA: All right.

STUDENT: I have my extra credit.