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.
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.
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.
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.
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.
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.