PROFESSOR: I know you have encountered difficulties
on the last few problems, maybe four, maybe five, maybe
the last 10, I don't know.
But today, I want to-- we have plenty of time.
We still have time for chapter 13,
and plenty of time for the final review.
I can afford to spend two or three hours just reviewing
chapter 12, if I wanted to.
All right, so I have this question from one of you saying
what part of the problem is that in terms of a two point
integral.
We have a solid bounded by z equals 3x,
and z equals x squared, and is a plane.
And can anybody tell me what this is?
Just out of curiosity, you don't have to know.
STUDENT: It's a parabola.
PROFESSOR: It would be a parabola,
if we were in [INTERPOSING VOICES]
if we were in 2D.
So the parabola is missing the y, and y could be anybody.
So it's a parabola that's shifted along y.
It's going to give you a cylindrical surface.
It's like something used for drainage, I don't know.
Water, like a valve.
So this is what it is, a cylindrical surface.
And you know that z must be between x squared and 3x.
How do you know which one is bigger, which one is smaller?
You should think about it.
When you draw, you draw like that.
Do these guys intersect?
We are in the xz plane.
Do these guys intersect?
x squared equals 3x intersect where?
They intersect at 0, and at 3.
x1 is 0, and x2 is 3.
So when I want to draw this, I would say that indeed, it's
a bounded domain.
If it where unbounded, it wouldn't ask for the volume,
because the volume would be nothing.
So this thing must be a bounded domain.
x cannot go on, this is the infinite part.
So we are thinking of just this striped piece of a domain x
here, where this piece is between z equals x squared,
then z equals 3x.
This is 0 origin, and this is 3.
So at 3, they meet again.
Are you guys with me?
At 3 o'clock they meet again.
I'm just kidding, x doesn't have to be time.
It's a special coordinate.
And y is looking at you, and is going towards you.
Well, if it's toward like that, it's probably not
positively oriented.
So y should come from you, and go into the board,
and then keep going in that direction for the frame
to be positive oriented.
Positively oriented means x like that, y like this, z like that.
So k must be the crossproduct between i and j.
i cross j must be k.
If I use the right hand rule, and I go y like that,
that means I changed the orientation.
So the y, you have to imagine the y coming from you, going
perpendicular to the board, then keep
going inside the board infinitely much.
Now, if we were to play with Play-Doh,
and we were on the other side of that, like Alice in the mirror,
we would have the y in the mirror world, going between 0
and 2, inside the board.
If I were to draw this piece of cake,
I start dreaming again, I apologize.
But I'm dreaming of very nice bounded pieces of solids
that would be made of cheese.
This is a perfect example where you have something like curve
or linear shape, and you kind of slice the cheese,
and that's a piece of the Parmesan.
OK.
So the y here is going from 0 to 2.
It's sort of the altitude.
And this is the piece of cake that you
were looking-- or the cheese, or whatever you're looking at.
So what do you put here?
You put z between x squared and 3x.
You put y between-- y is between 0 and 2.
And you Mr. x as the last of them, he can go from-- he
goes from 0 to 3.
So x has the freedom to go from 0 to 3.
y has the freedom to go from flat line to flat line,
from between to flat planes-- from two horizontal planes.
But Mr. z is married to x, he cannot escape this
relationship.
So we can only take this z with respect between these two
values that depend on x.
That's all.
Now why would we have 1?
Because by definition, if you remember
the volume was the triple integral
of 1 dv over any solid value domain.
Right?
So whenever you see a problem like that,
you know how to start it.
One triple integrate, and that's going to work.
Something else that gave you a big headache
was the ice cream cone.
The ice cream cone problem gave a big headache
to most of my students over the past 14 years
that I've been teaching Cal 3.
It's a beautiful problem.
It's one of my favorites problems,
because it makes me think of food again.
And not just any food, but some nice ice cream cone
that's original ice cream, not the one you find in a box like
Blue Bell or Ben and Jerry's.
All right, so how is the ice cream cone problem
that-- he showed it to me, but I forgot the problem number.
It was--
STUDENT: It's number 20.
PROFESSOR: Number?
STUDENT: 20.
PROFESSOR: Number 20, thanks.
And I want the data [INAUDIBLE].
I want to test my memory, see how many neurons
died since last time.
Don't tell me.
So I think the sphere was a radius 2, and the cone
that we picked for you, we picked it on purpose.
So that the results that come up for the ice cream
cone boundaries will be nice and workable.
So we can propose some data where the ice cream cone will
give you really nasty radii.
Can I draw it?
Hopefully.
This ice cream cone is based off the waffle cone.
I don't like the waffle cone, because I'm dreaming.
But the problem is, the waffle cone is not a finite surface.
It's infinite, it's a double cone.
It's the dream of every binger.
So it goes to infinity, and to negative infinity,
and that's not my problem.
My problem is to intersect this cone with the sphere,
and make it finite.
So to make it the true waffle cone,
I would have to draw a sphere of what radius?
Root 2.
I'll try to draw a sphere of root 2,
but I cannot predict the results.
Now I'm going to only look at this v1 cone.
I don't know what the problem wanted, but I'm looking at,
do they say in what domain?
Above the plane?
STUDENT: It just says lies above the cone.
PROFESSOR: That lies above the cone.
But look, if I turn my head like this, depending on my reference
frame, I have cervical spondylosis,
this is also lying above the cone.
So the problem is a little bit silly,
that whoever wrote it should have
said the sphere lies above a cone, for z greater than 0.
In the basement, it can continue--
that's for z-- for z less than 0 can continue
upside down, and then between the sphere and the cone,
you'll have another ice cream cone outside that.
But practically what they mean is just do v1
and forget about this one.
It's not very nicely phrased.
Above, beyond.
Are we above Australia?
That's stupid, right?
Because they may say, oh no, depends on where you are.
We are above you guys.
You think you're better than us because you
are closer to the North Pole.
But who made that rule that t the North Pole is superior?
If you look at the universe, who is above, who is beyond?
There is no direction.
So they would be very offended.
I have a friend who works in Sydney.
She is a brilliant geometer.
And I bet if I asked her, she would say
who says you guys are above?
Because it depends on where your head is and how you look at it.
The planet is the same, so would you
say that the people who are walking closer
to the North Pole have their body upside down?
So it really matters how you look, what's above.
So assume that above means-- the word above means positive.
And this is the ice cream cone.
Now how do I find out where to cut the waffle?
Because this is the question.
I need to know where the boundary of the waffle is.
I'm not allowed to eat anything above that.
So that's going to be the waffle.
And for that, any ideas-- how do I get to see what the circle--
where the circle will be?
STUDENT: Do they meet each other?
PROFESSOR: They intersect.
Excellent, Matthew.
Thank you.
So intersect the two surfaces by setting up a system to solve.
Solve the system.
And the intersection of the two is Mr. z, which is Mr. circle.
All right.
So what do I do?
I'm going to say I have to be smart about that one.
So if z squared from here plugging in,
substitute, is the same as x squared plus y squared.
So that means this is if and only if 2x squared
plus y squared.
2 times x squared plus y squared.
We have an x squared plus y squared,
and another x squared plus y squared equals 2.
You see how nicely the problem was picked?
It was picked it's going to give you some nice data.
z equals-- z squared equals that.
Keep going with if and only if.
If you're a math major, you will understand
why x squared plus y squared equals 1.
And z squared must be 1.
Well so, we really get two solutions,
the one close to the South Pole, and the one close
to the North Pole, because I'm going
to have z equals plus minus 1.
Where's the North Pole?
The North Pole would be 0, 0, square root of 2.
The South Pole would be 0, 0, minus square root of 2.
And my plain here is cut at which altitude?
z equals 1.
And I have another plane here, and an imaginary intersection
that I'm not going to talk about today. z equals minus y.
I don't care about the mirror image of the-- with respect
to the equator of the cone.
All right, good.
So we know who this guy is, we know
that he is-- I have a red marker, and a green,
and a blue.
I cannot live without colors.
Life is ugly enough.
Let's try to make it colorful.
x squared plus y squared, equals 1.
We are happy, because that's a nice, simple circle
of radius 1.
Now you have to think in which coordinate
you can write this problem.
And I'm going to beg of you to help
me review the material for the final and for the midterm.
Then again, on the midterm, I'm not going to put this problem.
So for the final, do expect something like that.
We may have, instead of the cone in the book,
you'll have a paraboloid and a sphere intersecting.
It's sort of the same thing, but instead of the ice cream cone,
you have the valley full of cream.
I cannot stop, right?
So if you do the volume like I told you before,
you simply have [INTERPOSING VOICES]
No, I'll do the volume first.
I know you have the surface there,
but what-- I'm doing the volume because I have plans,
and I didn't want to say what plans.
You forgot what I said, right?
So suppose somebody's asking you for the volume.
The volume-- how much ice cream you have inside
depends on that very much.
And I'd like you to remember that the v is
piratically against the ybc, right?
Right until the Cartesian coordinates would be a killer,
we try to write it in either cylindrical
or spherical to make our life easier.
If I want to make my life easier, first in cylindrical
coordinates, and then in spherical coordinates.
Could you help me find the limit points?
And then we'll do the surface.
Just remind me, OK?
[INTERPOSING VOICES] The volume of the-- OK.
The volume occupied by the ice cream.
The ice cream is between this plastic cap,
that is the sphere.
We cover it for hygiene purposes.
So for cylindrical coordinates, rho and theta are really nice.
We don't worry about them yet.
But z should be between?
OK.
Really, 0?
So because it's between the ice cream
the cone-- do you think the waffle
cone-- what's the equation of the waffle cone?
And how do you get to the equation of the waffle cone?
The waffle cone meant-- oh guys, help me.
z equals-- now you take the square root
of x squared plus y squared, because the other one would
be here.
The imaginary one, z equals minus the square root
of x squared plus y squared.
Forget about the world in the basement.
So you take z to the square root of,
with a plus, plus y squared.
In cylindrical coordinates, what does this mean?
In cylindrical coordinates?
STUDENT: It equals r.
PROFESSOR: This equals r, very good.
He's thinking faster.
Do you guys understand why he said
x squared plus y squared, if we work
with polar coordinates-- which is cylindrical
coordinates it's the same thing-- polar coordinates
and cylindrical coordinates.
x squared plus y squared would be little r squared.
Under the square root would be r.
So you'll see, we're between r.
And now, another hard part.
What is the z equals plus square root of-- What was it, guys?
STUDENT: So are we taking the volume of all of the ice
cream inside the cone?
PROFESSOR: So between the cone, ice cream lies here.
Ice cream chips, chocolate chips,
between the cone and the sphere.
The sphere is the bottom function, the lower function.
Is that good?
STUDENT: 2 minus r squared.
PROFESSOR: So I have the square root of 2 minus r squared.
I don't like the square root, I hate it.
Can I do it with it?
Yes, I can.
Maybe I can apply the use of solution later.
Don't worry about me, I'll make it.
I will live better if I didn't have any ugly things like that.
So let's see what we have.
Theta.
Your cone is not just sliced cone, it's all the cone.
So you have 0 to 2 pi.
One revolution, complete revolution.
How about rho?
Rho is limited.
Rho is what lies in the plane in terms of radius.
0, 2?
How much is from here to here?
Didn't we do it x squared plus y squared equals 1?
So what is this radius from here to here?
1.
And what is this radius from here to here?
1.
So the projection of this ice cream
cone-- if you had the eye of god is here,
sun, you have a shadow on the ground,
coming from your ice cream cone, and this is the shadow.
Your shadow is simply a disk of radius 1.
Good, that's the projection you have.
So Mr. rho is between 0 and 1.
For polar coordinates, it's not so ugly actually.
0 to 2 pi.
0 to 1.
r to square root of 2 minus r squared.
Instead of r, it's OK to react with rho.
And here's the big j coordinates.
I'm going to erase j, don't write j.
Jacobian in general is Jacobian.
The one that does the transformation between
coordinates to other coordinates.
Let me finish on that.
How about this one?
This is simply r, very good.
Your old friend.
So you [INAUDIBLE] the dz d, theta,
dr, d is your [INAUDIBLE] It's not easy there.
If I were to continue-- maybe on the final-- OK
I'm talking too much, as usual.
Maybe on whatever test you're going to have,
this kind of stuff, with a formulation saying you
do not have to compute it.
But if you wanted to compute it, would it be hard from 0 to 1,
from 0 to 2 pi, and say forgot the stinky pi?
Take the 2 pi out to make your life easier.
Because the theta isn't depending from-- there
is no theta inside.
So take the 2 pi out, and then you
have an integral from 0 to 1, and integral from-- now.
R got out for a walk.
This is r going out for a walk.
Integral of 1 with respect to dz.
So z is taken between r and root of 2 minus r squared.
Right?
So I would have to write here the 1 on top minus the 1
on the bottom, which is a little bit of a headache for me.
I'm looking at it, I'm getting angry.
Now, times the r that went out for a walk.
So practically, the 0 is solved, and I
have the dr. And from 0 to 1.
So I took care or who?
I took care of the integral with respect to z.
This is r here.
This was done first.
And you gave me that between those two.
So I got that.
Times the r, between 0 and 1, with respect to r, and then
the 2 pi gets outside.
Now, if I split this into two integrals,
it's going to be easy, right?
Because I go-- the first integral
will be r times square root of 2 minus r squared.
How can you do such an integral?
We do substitution.
For example, your u would b 2 minus r squared.
Can you keep going?
The second integral is a piece of cake.
A piece of ice cream.
The integral of r squared.
r cubed over 3, between 0 and 1, 1/3.
So we can still solve the ice cream cone volume like that.
Do I like it?
No.
Can you suspect why I don't like it?
Oh, by the way.
Suppose you got to this on the final,
how much do you get for-- you mess up
the algebra, how much do you get?
You say, I can do that in my sleep,
u equals 2 minus r squared, u equals minus 2r, I can go on.
Even if you mess up the algebra, you get most of it.
Why don't I like it?
Because it involves work, and I'm lazy.
So can I find a better way to do it?
Can I get use spherical coordinates?
And how do I use spherical coordinates?
So let me see how I do that.
In spherical coordinates, it should be easier.
Remember that for mathematicians, they
include this course Cal 3 multivariable calculus.
We are not studying geography.
So for us, a lot can happen between minus 90 and plus 90
degrees, but it measures from the North Pole,
because we believe in Santa Clause.
Always remember that.
So we go all the way from 0 degrees to 180 degrees.
So your-- in principle-- your latitude
will go from 0 to all the way to pi.
But it doesn't, because it gets stuck here.
What is the latitude of the ice cream cone?
So what is the pi angle for this ice cream cone?
It's a 45 degree angle.
That is true.
For anything like that-- I'm looking again at this cone.
z squared equals x squared plus y squared.
I just want to talk a little bit about that.
So if you have x and y, this is the x.
This is x, you have to use your imagination on me.
And the hypotenuse would be x squared plus y squared.
And this is the z.
And then, I draw what is in between.
This has to be 45 degrees.
Can you see what's going on?
So theta has to be between 0 and what?
STUDENT: 2 pi.
PROFESSOR: Yes, you are smarter than me.
That was the longitude.
Thank you.
I'm sorry, I meant to write the latitude.
Phi is between 0 and pi/4.
How about the radius?
Are you afraid of the radius?
No.
Why?
The radius is your friend.
It was not your friend before.
Look how wobbly it is.
But in this case, the radius goes all the way from 0
to a finite value, which is exactly
the radius of the sphere.
Because you have rays of light coming from the source origin,
and they bounce against this profile,
which is the profile of the sphere, which
has radius square root of 2.
So life is good for you in this case.
Are you guys with me?
Should it be easy?
Yes, it should be easy to write that in the integral,
if you know how to write it.
So you have.
OK.
What do you want to do first?
It doesn't matter that you apply Fubini's theorem.
You have fixed limits.
You have 0 to 2 pi, 0 to pi/4, 0 to square root 2.
Inside, there must be a Jacobian that you know by heart,
and I'm asking you to learn it by heart before the final,
if not for now, but maybe before the final.
But by now, you should know it by heart.
Thank you so much, Matthew.
Yes.
You don't have much to memorize, but this is one of the things
that I told you I did not memorize it, I was a freshman,
I was stubborn and silly.
So I have to compute what?
I have to compute the Jacobian.
Imagine what work you have when you're limited in time.
dx, dr. dx, d theta.
dx, d phi.
I thought I was about to kill myself.
dy, dr. dy, d theta. dy, d phi, and finally, dz, dr. dz,
d theta.
dz d phi.
And I did this.
And I thought I was about to just collapse and not finish
my exam.
I finished my exam, but since then, I didn't remember that.
I had to compute it.
It took me 10 minutes to compute the Jacobian.
So this is r squared, psi, phi.
If you have nothing better to do, you can do that.
Do you remember what the spherical coordinates were, out
of curiosity?
Who remembers that?
There are some pre-med majors here,
who probably remember that.
So when you have a phi here, you have r-- sine or cosine?
Cosine.
r cosine phi.
And then r sine phi for both times what?
The first one comes from theta, like that.
It's going to be cosine theta, and sine theta.
Well imagine me taking these functions and differentiating,
partial derivatives.
And after I differentiated down, compute the 3 by 3 determining.
It's an error, no matter how good you are at computing.
So don't do that, just memorize it.
Don't do like I did.
And then you have d what?
dr, d phi, d theta.
Now what is the volume of the ice cream cone?
Let me erase.
This shouldn't be hard.
This is the type of problem where
you have a product of functions of several variables.
You can separate as a product of three independent integrals
as a consequence of Fubini's theorem.
So you have integral from, integral from, integral from.
Who's your friend?
Who do you like the most?
STUDENT: Theta.
PROFESSOR: You like theta the most?
Because it comes from Santa Clause?
No, the theta doesn't.
This is the easiest step.
So that's why you like it, because it's
the easiest to deal with.
How about phi?
Sine phi, d phi.
I agree with you, it's not so easy,
but it's going to be a piece of cake anyway.
How about this one, 0 to root 2 r squared dr.
Is this guy hard to do?
r cubed over 3 will give me root 2 cubed over 3.
How much is that, by the way?
2 root 2 over 3.
Oh bless your heart, that's not so hard.
This is not a problem.
How about that?
What do you have?
What is the integral of sine?
Negative cosine.
So you have minus cosine phi between pi/4 up and 0 down.
Good luck to you.
Well, the first guy.
Good, minus root 2 over 2.
Minus, second guy?
Minus, minus 1.
Don't fall into the trap.
Pay attention to the signs.
Don't mess up, because that's where
you can hurt your grade by messing up with minus signs.
So this is 1 plus 1, 1 minus root 2 over 2.
And finally, let's see what that is, the whole thing being.
Can we write it nicely?
What's 2 times-- 4.
4, root 2 over 3 pi.
The first and the last.
4 times 2, pi/3 times this nasty guy 1 minus root 2.
I don't like it, let's make it look better.
Well OK, you can give me this answer,
of course you'll get 100%.
But am I happy with it?
If you were to publish this in a journal,
how would you simplify?
This is dry.
OK so what do you have?
1 is 2/2.
2 minus root 2 pi, 4 and 2 simplify.
Are you guys with me?
There is a 2 down, and a 4 up, so I'm going to have a 2
and another 2, all over 3.
So I have 2 and 2 pi, times 2 minus root 2.
Do you like it like that?
I don't.
So what do you do next?
I can even pull the root 2 inside.
So I go 4 root 2, minus what?
Minus 4, because this is 2 times 3 is 4, pi over--
Do you like it?
Still I don't, because I'm stubborn.
4 root 2 minus 1 over 3 is the most beautiful form.
So I'll try to brush it up, and put it
in the most elegant form.
It doesn't matter.
If you want to give me a correct answer,
any form it would be OK.
Yes?
STUDENT: If it was slightly different,
how would we find phi for the limits in the second part?
PROFESSOR: If you have a what?
STUDENT: How would we find phi if it wasn't obvious,
if it wasn't x squared, or c squared?
PROFESSOR: If it wasn't a 45 degree angle?
[INAUDIBLE]
It's not so bad, you need a calculator.
Assume that I would have given you the sphere of radius 7,
or square root of 7, intersecting with this cone.
Then to compute that phi, you would
have needed to intersect the two surfaces
and then compute it, maybe look at tangent inverse.
Compute phi with tangent inverse.
And you will have tangent inverse of a number.
Well, you cannot put tangent inverse of a number everywhere,
it's not nice.
So what you would do is in the end,
you would do it with a calculator, come up
with a nice truncated result with 5 decimals,
or 10 decimals, whatever the calculator will give you.
OK?
Or, you can do it with MathLab.
You can do it with scientific software, for sure.
Let's do what I-- Ryan you said this was a what?
STUDENT: Number 20 is for surface area.
PROFESSOR: OK.
So, it's-- read it to me again.
What does it say?
I'm coming to you.
It says, find the surface area of the part of the sphere that
lies where you have 64.
This is all because of [INAUDIBLE]
But yours is not very even, right?
You shouldn't have bad results.
And guess what?
If you do, you use your calculator
to find out the upper limit of the angle for the volume.
OK.
So now, you say oh my god, this is ugly.
I agree with you, it's not nice.
You have square root of 2 minus x squared minus y squared.
And when you compute the surface area of the cap-- cup,
cap means spherical cap.
A little hat that looks like this?
That's why it's called cap.
That will integrate over the disk d.
Square root of 1 plus f of x squared
plus f of y squared, dx da.
Is that the only way you can do this?
No.
You can actually do it with parametrization of a sphere,
and you have the element limit over here.
So that might be easier.
Yeah.
You can also do it in homework.
But what if you went up there-- let's see, how hard is life?
How hard would it be to do it like this?
That's good.
First of all, let's think everything
that's under the square root.
And write it down.
1 plus.
Now, computing this problem with respect to x and you say,
oh my god, that's hard.
No, it's not.
If you want to do the hard one, and most of you
were, and you have that professors who
gave you enough practice, what did you have done?
Chain rule.
On the bottom, you have this nasty guy twice.
But on the top, your minus 2x.
So when you simplify your life becomes easier.
And you will square it.
Are you guys with me, have I lost you yet?
And then the same thing in y.
Minus 2y, over 2 square root 2 minus x square minus y squared,
square it.
Some things cancel out.
So let's be patient and see what we have.
First of all, 1 is not going to give you trouble, because let
write 1 as this over itself.
Plus, minus squared is plus, thank god.
x squared over 2 minus x squared minus y squared plus y squared
over 2 minus x squared minus y squared.
And these guys go for a walk.
Minus x squared, minus y squared, plus y squared.
They disappear together in the dark.
So you have 2 over 2 minus x squared minus y squared.
OK let's try to do that.
Guys, I have to erase.
I will erase.
So what you see here, some people
call it ds, and use the element of area on the surface.
It's like the area of a small surface patch.
So the curve linear squared.
Alright.
So area of the cup will be-- now you say, well over the d,
let me think.
d represents those xy's with a property
that x squared plus y squared was between what and what?
0 and 1, because that was our, the
predicted domain on the shadow on the ground.
OK, that was this.
And as you look at it, I have to put it on the square root.
Don't be afraid of it, because it's not
much up here than you thought.
And let's solve this together.
What is your luck that this is a symmetric polynomial index,
and why x squared plus y squared that you can rewrite as r
squared, polar coordinates?
And Ryan asked, can I do polar coordinates?
That's exactly what you're going to do.
You didn't know, unless your intuition is strong.
Yes?
Alex tell me.
STUDENT: I was going to ask, if you could have done that
by taking the r plane and multiplying that by 2 pi r?
PROFESSOR: Yeah, you can do that.
Well, that is a way to do that.
So practically, he's asking-- I don't
know if you guys remember, in Cal 2,
you have the surface of revolution, right?
And if you knew the length of an arc,
you would be able to revolve that arc.
This is the cap.
And you take one of the meridians of the hat,
and revolve it, can redo with a form,
like you did the washer and dryer method.
It always amuses me.
Yes, you could have done that from Cal 2.
Computing the area of the cap as a surface
of revolution, chapter-- c'mon, I'm a co-author of this book.
Chapter 7?
What chapter?
Chapter 6?
No.
The washer and dryer?
Chapter 6, right?
OK.
But now we already have three, and we
don't want to remember Cal 2 because it was a nightmare.
Several of you told me that this is
easier, these things are generally easier than Cal 2,
because Cal 2 was headache.
And what seemed to be giving you most of the headache
was a salad of different ingredients
that seemed to be unrelated.
Which I agree.
You have arcing, washer, slices, then Greek substitution,
the partial fractions.
All sorts of things and series and sequences.
And they are little things that don't quite
follow one from another.
They are a little bit unrelated.
OK, how do you do that?
You have to help me because that was the idea, that now you
can help me, right?
Square root of
STUDENT: 2 over 2 minus r squared, times r, dr d theta.
PROFESSOR: And do we like it?
No, but we have to continue.
0 to 1, this is 0 to 2 pi.
I can get rid of the 2 pi, and put it here and say, OK.
I should be as good as taking out square root of 2 from here.
He goes out for a walk.
And then I have integral 1 over this long line of fraction.
STUDENT: And that would be 2r so that the r will cancel out.
PROFESSOR: So r dr, if u is 2 minus r squared,
the u is minus.
I have to pay attention, so I don't mess up the signs.
So rdr is a block.
And this block is simply minus du/2.
So I write it here, minus 1/2, du/2.
Don't be nervous about this minus,
because it's not going to give me a minus result,
a negative result. If it did, that means
that I was drunk when I did it, because I will
get the area of the cap as a negative number, which
is impossible.
But it's going to happen when I change the limits.
Yes?
STUDENT: Where did that last one come from?
PROFESSOR: From this one.
[INTERPOSING VOICES] Oh, I put too many.
So this guy is this guy, which is
this guy minus the [INAUDIBLE].
Now, do you want me to go ahead and cancel this out?
Right?
OK.
I have squared 2 pi.
I did not get the endpoints, you have
to help me put the endpoints.
From 2 down.
To 1.
Which is crazy, right?
Because 2 is bigger than 1.
That's exactly where the next minus is going to come from.
So integral from 2 to 1 is minus integral from 1 to 2.
So I shouldn't be worried, because I already
have the minus out, with the minus that's going to come out,
I'm going to have a positive variable.
Square root 2 pi.
Somebody was smarter than me and said Magdalene,
I think Alex-- was it you?
You said, why don't you take advantage of the fact
that you already have 1 over 2 square root u and integrate?
And that is going to be squared u.
Can you understand?
Who said that?
I heard a voice, it was not in my head.
I'm Innocent.
I heard a voice that told me, if you are smart,
you would understand to pull out the minus.
You would understand that this exactly
is the derivative of square root of u.
And I will be faster than you, because you have just
computed made between 1 and 2.
Some people aren't too smart.
I didn't think of that.
Now I've been thing about that.
Why cancel out the 2 when you can [INAUDIBLE].
So you have a minus out.
STUDENT: There still needs to be a 2 on the outside, right?
PROFESSOR: Yes, I have to put 2 together.
Minus 2 and 2.
2 pi, that's a collective thing.
Squared and cubed between 1 and 2.
Do I like this?
No, but you tell me what is in the bracket.
How much u minus 12, 13.
1 minus square root of 2 is a negative number.
But with the minus outside, I"m going to fix it.
And I'm going to get something really ugly.
Yeah.
So when I'm multiplying sides, this by that, I get 4, right?
Guys?
I get a 4.
This guy and this guy, minus, minus, plus 2, 2 times 2 is 4.
And then minus, to make it look better, 2 root 2.
And multiply out.
Minus root 2 and the pi.
Now, I don't care where you stop.
I swear that if you stop here, you'll still get 100%.
Because what I care about is not to see a nice simplified
result, so much.
I won't go over your work.
But to see that you understood how
you solve this kind of problem.
It's not the sign of intelligence
being able to simplify answers very much.
But the method in itself, why and how, what the steps are,
that shows knowledge and intelligence.
Have I mess up?
I don't think so.
STUDENT: Does the order matter, of the dr d theta, or dx/dy,
does that matter which order you put them in?
PROFESSOR: In this case, no.
STUDENT: Or over here?
PROFESSOR: In this case, no again.
But if you were to swap them, you
would have to swap the values as well.
Why is that?
That's a very good question.
He's right, but why is that?
It doesn't matter, why?
In general, it matters.
They have to be from a given number to a given number.
It's not like reversing-- when you reverse
the order of integrals, it's usually harder,
because you have to draw the vertical strips.
And you have it between two functions.
And then from vertical strips, you go to horizontal strips,
and you have other two functions.
So you always have to think how to change the function.
Here, you don't have to think at all.
You have a function that depends on r only.
There is no theta in the picture.
Plus these two are fixed numbers.
You can reverse the integration in your sleep.
OK, you get the same thing.
All you have to do is swap these two guys,
and swap-- the 0 is the same.
So I swap these two guys.
STUDENT: How did you take the 2 pi out?
PROFESSOR: What did I do?
How did I take this out?
STUDENT: No, the 2 pi.
PROFESSOR: Oh, the 2 pi?
OK.
Let me show you it better here, because we've
discussed about this before.
When you have integral from a to b,
or integral from c to d of a function or r
and a function of theta, what do you go?
There is a theorem that says that--
and thanks for this theorem and the fact
that they're separable.
The variables are separated in this product.
This is the product between integral from a to b,
here of theta to theta, and integral
from c to d, f of r/dr. They are nothing to one
another but a product.
So what do you do?
You say this is integral of 1, from 0 to pi d theta,
times the other guy.
So this is 2 pi.
When theta doesn't appear inside, it's a blessing.
But if and if there is, I have a question.
What if theta appeared inside?
Theta doesn't appear inside by himself.
He appears inside of a trig function.
So assume you have cosine theta here times r.
You would have pulled cosine theta out,
and integrated cosine theta, that would be easy.
And if you have a problem like that, you would have gotten 0.
Because integral of course of cosine theta
would be sine of theta, and then theta between 0 and 2
pi is sine of theta between, which would give you 0.
It happened to me, many times in the exam.
It was a blessing.
I was 19, and I was so happy.
Professors wanted to see only the answer.
Because in Romania, it's different.
You come take a written exam, and the professor
has five hours to grade it.
The same day, two hours later, you have the oral examination.
You pick up a ticket, on the ticket,
you see three things to solve, four things to solve.
You go take a seat.
And while the professor and the assistant grade the exam,
you actually think of your oral exam.
When you come and present your results on the board,
they tell you, you messed up, you
got a 60% on this sticking exam.
This is how it goes.
Or, on the contrary, hey, listen, you got a 95%
on the written part, OK?
I don't want to see what you have there,
it really looks good.
I don't want to see it, it's clear to me
that you know what you're doing.
So it's a different kind of examination.
I hear that Princeton does that, I wonder
how are all the exams here.
I don't think people are ready for them yet.
But at Princeton they do a lot, all the same.
They make a hat, and take a [INAUDIBLE], put tickets in it.
And the teacher comes, and closes his eyes
and picks a ticket, and says oh my god, I
got proof of Fubini's theorem.
And do these three triple integrals.
This is a type of oral exam that you would have.
But if you know that, because you studied,
you're not afraid to present them.
But you have to present them, and you have a limited time.
Because there are other 30 students in the classroom.
You only have five minutes.
And I only pick-- I want to see your work on all of them.
And I'll teach you how to present
on this problem on the board.
And then you have five minutes to present.
If you are really embarrassed and you
don't want to speak in public, then you have a problem.
I've had many fears-- and in other countries--
I heard that in England, they have the same system.
There are people who are too shy to show their results,
or too shy to talk.
And then they start stuttering.
But they have to do it.
There is no excuse, they don't care if you
have problems with your speech.
So I asked the people I knew and I went to London,
and they said most people will stutter in there.
I was so scared.
Most people who stutter in our oral exams
are people who spend too much time
in the pub the previous day.
Pubs were everywhere and I saw lots of students in the pubs.
I went to University of Durham-- this is where Harry Potter was
filmed, by the way.
I saw the castle, which is a student dorm.
You pay something like 500 pounds.
Which would be like $100?
$1,000?
Less, because I think it's 7.50, something like $800.
It used to be that the pound was double the dollar.
[INTERPOSING VOICES]
So you could stay in that dorm for $800 per month.
And you've got the same table where they ate in the movie.
It was really nice.
But the University of Durham is a isolated castle,
the cathedral, everything is very old,
from the 11th century, 12th century.
But if you go into the city, it's full of pubs.
Who is in the pubs?
The calculus students.
This is where they do their homework.
And it amazes me how they don't get drunk.
I'm not used to alcohol, because I don't drink.
Well they are used to it.
So they may nicely can do their homework, beautifully,
next to a big draft of Guinness like that.
And still makes sense when they write the solution.
They don't miss a minus sign, they're amazing.
Alright, is this hard?
If you are interested, you can ask about study abroad.
We don't have big business with England,
but you could go to Seville.
There are some programs in the summer where
one of our professors teaches differential equations like I
told you about.
He teaches differential equations this summer
in Seville.
I think you can still add in the next two days.
Some of you did, some of you didn't.
All right.
Any questions about other problems?
I have to apologize, I played the game
without telling you the truth.
[INAUDIBLE] he came to me last time and said,
you never showed us this notation.
So what if one gives you x of u, v equals 2x minus y.
y of u, v equals 3x plus y.
What the heck is that?
He didn't say heck, because he's a gentlemen, right?
But he said this is the notation used in web work,
and the book is actually not emphasizing it, which is true.
The book is emphasizing the Jacobian in section 12.8
only, which is not covered, it's not part of the menu.
But the definition, you should at least know it.
So what would be the definition of this animal?
You see that we have to take the partial derivative of x
with respect to u, the partial derivative of x with respect
to v, the partial derivative of y with respect to u,
and the partial derivative of y with respect to v.
And that's exactly what it is, indeterminate.
Not matrix, but indeterminate.
So do I bother to write it down?
If I wanted to write down what it is,
of course I would write it down like that.
I don't want to spend all my time doing that,
because it's such an easy problem.
What do you have to do?
Just compute for such a simple transformation in plane.
Actually, if you took linear-- again,
who is enrolled in linear algebra?
Only 1, 2, 3?
Thought there were only 2.
OK.
In linear algebra, you wrote this differently.
You wrote it like this. x and y equals matrix multiplication.
You have 2, minus 1, 3, 1, by the way
it's obvious the determinate of this matrix
is different from 0.
This is the linear map that you are applying to the vector xy.
And in your algebra book, you're using Larson, am I right?
Larson's book?
It's a good book.
So you have a of the vector x.
a of the vector x is the vector v.
When you all get to see linear algebra,
you'll like it more than Cal 3, because it's more fun.
So how do you do this matrix multiplication?
It's very easy.
This time that, minus this times this.
So can computers do that?
Yes, computers can, if you have the right program.
And this is the first program I learned in C++.
No, it was the second program.
How to write a little program for multiplication of two
matrices.
The first program I had, I learned in C++.
It was to build an ATM machine.
I hated that, because every time I went under 0 with my balance,
I would have new word under 0.
So I would have to prepare for all the possible cases
and save.
If you don't have enough money, whatever.
So that was the first program we wrote.
OK so, what do we have?
2 minus 1, 3 and 1.
What is the Jacobian in this case?
It's 2 plus 3, 5.
Different from 0.
You have one or two problems like that.
Three problems.
I was really mean.
I apologize.
But you still have time to do those problems
in case of the review.
STUDENT: So we just take the determinate of it?
PROFESSOR: And you take the determinate of the matrix.
And that's you Jacobian.
STUDENT: What number is that?
PROFESSOR: I don't remember.
STUDENT: What if it's the u and the v
is at the top and x and the y at the bottom?
PROFESSOR: So the determinate will be the same.
This is a very good question.
Are you guys with me?
So he said, what if you have your first equation's name
would be this one.
And you have your equations written like that.
Right?
And so, when you look at this, you
will go-- it depends how you-- in which order you do that.
I wrote u, v. Sorry.
u and v, but you understood what I meant.
Right?
u and v.
STUDENT: Can you do number three?
It was a hard one.
PROFESSOR: I will, just a second.
So d y, x with respect to u, v. What would happen, I just I
would flip the x and y.
What will happen?
I get 3, 1, it's still the same function.
2, and minus 1.
Why do I get minus 5?
So imagine guys, what happens when you have x and y?
If you rotate, you don't change the sign of your matrix,
or notation.
Matrix notation will always have [INAUDIBLE].
But if you flip it, if you swap x and y,
you are actually changing the sign of the Jacobian,
the sign of the matrix.
You are changing your orientation.
That would be a hypothetical situation.
You are changing your orientation.
Do you have a number, Ryan?
Is it hard?
Why is it hard?
Yeah, let me do that.
It's hard enough.
It's computation.
Were you able to do it?
Not yet, right?
So this is x, not-- OK.
So you can write this also, differently, except the y sub
u, y sub v. Who can tell me-- there
are ways to do it in a simpler way.
But I don't want to tell you yet what that way is.
And I'll show you next time.
What is x sub u?
It shouldn't be so hard because it's the quotient rule.
You have 4 times u squared plus v squared minus
[INAUDIBLE] minus 2u the derivative of this times 4u
divided by the square of that.
Did I go too fast?
So what you have is 4u squared minus 8u squared
equals minus 4u squared plus 4v squared divided by that.
v squared.
Squared, sorry.
x of v, that should be easier.
Why is it easy?
The first guy prime minus the second guy
So the first primes, second not prime.
Minus second prime, straight to v, times the first not prime.
Divided by u squared.
Which is minus 8uv over that.
Is this one of those that you said you couldn't do it yet?
You?
Both?
You did this one?
You got the right answer, good.
y sub v. Y sub u, it's OK to have a minus 0 times
the second one.
Minus this prime with respect to u, times
6v over the square of that.
And finally, y sub v equals minus the derivative
of the top, with respect to 6v times u squared plus v
squared minus the derivative of the bottom with respect
to v. v times 6v divided by the whole shebang.
Now is it simplified?
No, I will simplify in a second.
You get minus 12uv.
I'm not going to finish it, but we are almost done.
Why are we almost done?
This is very easy.
I mean, not very easy, but doable.
How about this guy?
What do you get?
A 6u squared, a 6v squared, a minus 12v squared.
It's not that bad.
So you have 6u squared minus 6v squared,
over u squared plus v squared.
What did I do?
Add a minus in front.
I didn't copy.
Let me make room for that, thank you.
STUDENT: It's also the 12.
It's 12uv, because there's a negative in front of it.
It's minus times minus.
PROFESSOR: Here?
STUDENT: No, y sub u.
The third one.
PROFESSOR: Minus, minus, plus, that's good.
Thanks for observing things.
Anything else that's fishy?
Minus, minus, plus.
OK that's better.
Change the signs.
When I move onto this one, remind me to change the signs.
So what is the Jacobian?
I'm too lazy to write this thing.
I'm going to have-- so, x sub u.
4 times v squared minus u squared.
Let's me count the OK.
Let's do it over a.
I'll show you what happens.
Maybe you don't know yet what happens,
but I'll show you what happens.
Then the next one is going to be x sub v minus 8, uv over a.
y sub u, 12.
uv over a, and last, with your help.
It's plus this was my-- so 6 times v
squared minus u squared over a.
OK OK, let me erase.
So you guys know what happens when
you have something like that?
A determinate has one line multiplied or column
multiplied by a number.
If you have alpha a, alpha b, alpha c and d.
The determinate of that is alpha aut, a, b, c, d.
I assume you know this from high school,
but I know very well that many of you don't.
How do you prove this?
Very easily.
This times that would be an alpha out, minus this,
and alpha out.
It's very easy to prove.
So when you have one line or one column multiplied by an alpha,
that alpha gets out.
So if you have two lines multiplied by an alpha,
or two rows, alpha squared, excellent.
So who gets out?
1 over a squared, which means this guy to the fourth.
Sorry that this is so long.
I don't like this problem, because of this computation
you have to go through here.
So I would simplify it as much as I could.
Let's see, before I missed my a group.
So you have 24 times v squared minus u squared, plus 96,
am I right?
v squared divided by all this ugly guy which I hate,
to the fourth.
Fortunately, everybody's a multiple of 24.
So we can pull a 24 out.
and get it out of our life, because it drives us crazy.
And then you have v to the fourth plus u to the 4,
minus twice.
Was that the binomial format?
Minus 2 us squared, v squared.
What was left when I pull this out?
I pulled 24 out, 96 is what?
4.
So I have a 4 left.
So I would put that down.
4u squared, v squared over the-- it looks symmetric
but-- that's OK.
It's not so bad.
So can you write this better?
Look at it.
Do you like it?
There is a 3.
The 4 the 2, 4 minus 2 is a plus 2.
Just like when we did those tricky things in high school.
That would be, again, the binomial formula.
u squared plus v squared.
Are you guys with me?
Because minus 2 plus 4 is plus 2.
This is exactly the same thing as that.
Over u squared plus v squared to the fourth.
If you have problems computing that,
send me some emails from WebWork, because I'm
going to help you do that, OK.
24 divided by what?
Yes?
u squared plus v squared squared.
Oh my god.
All right.
So, I'm not going to think lesser of you
if you don't put all of this here.
Therefore, if you get in trouble,
click from the expression from the whatever you got, and say,
this horrible problem gives me a headache, help me.
And I'm going to help you with that simple computation that
is just algebra.
That's not going to teach you anything more about Cal 3.
That's why I'm going to help you.
I'll help you with the answers on those.
Just send me an email.
I'm planning on still reviewing even on Tuesday.
I don't want to teach anything new, because I'm tired
and-- I'm just kidding.
I don't want to teach anything new on Tuesday,
because I want you to be very well prepared for the midterms.
So I'll do a general review again,
and I'll go over some homework like problems,
but mostly over exam like problems.
So I want everybody to succeed, to get very high scores.
But we need to practice, practice, practice.
It's like you did before your SATs.
It's not that much, I mean what happens if you don't do great
on the midterm?
Well the midterms is a portion the final.
But what I am trying to do by reviewing
so much for the midterm is also trying
to help you for the final.
Because on the final, half of the problems
will be just like the ones on the midterm Emphasizing
the same type of concepts.
It's good practice for the final as well.
All right, good luck.
I'll see you Tuesday.
Let me know by email how it goes with the problems.