PROFESSOR: I have some assignments
that I want to give you back.
And I'm just going to put them here,
and I'll ask you to pick them up as soon as we take a break.
There are explanations there how they were computed in red.
If you have questions, you can as me
so I can ask my grader about it.
Now, I promised you that I would move on today,
and that's what I'm going to do.
I'm moving on to something that you're gong to love.
[? Practically ?] chapter 12 is integration of functions
of several variables.
And to warn you we're going to see
how we introduce introduction to the double integral.
But you will say, wait a minute.
I don't even know if I remember the simple integral.
And that's why I'm here.
I want to remind you what the definite integral was both
as a formal definition-- let's do it as a formal definition
first, then come up with a geometric interpretation based
on that.
And finally write down the definition
and the fundamental theorem of calculus.
So assume you have a function that's continuous.
Continuous over a certain integral of a, b interval in R.
And you know that in that case, you
can "define the definite integral of f
of x from or between a and b."
And as the notation is denoted, by integral from a to b f of x
dx.
Well, how do we define this?
This is just the notation.
How do we define it?
We have to have a set up, and we are thinking of a x, y frame.
You have a function, f, that's continuous.
And you are thinking, oh, wait a minute.
I would like to be able to evaluate
the area under the integral.
And if you ask your teacher when you are in fourth grade,
your teacher will say, well, I can give you some graphing
paper.
And with that graphing paper, you
can eventually approximate your area like that.
Sort of what you get here is like you draw a horizontal
so that the little part above the horizontal
cancels out with the little part below the horizontal.
So more or less, the pink rectangle
is a good approximation of the first slice.
But you say yeah, but the first slice is a curvilinear slice.
Yes, but we make it like a stop function.
So then you say, OK, how about this fellow?
I'm going to approximate it in a similar way,
and I'm going to have a bunch of rectangles on this graphing
paper.
And I'm going to compute their areas,
and I'm going to come up with an approximation,
and I'll give it to my fourth grade teacher.
And that's what we did in fourth grade,
but this is not fourth grade.
And actually, it's very relevant to us
that this has applications to our life,
to our digital world, that people did not
understand when Riemann introduced the Riemann sum.
They thought, OK, the idea makes sense that practically we
have a huge picture here, and I'm
taking a and b and a function that's continuous over a and b.
And then I say I'm going to split this
into a equidistant intervals.
I don't know how many I want, but let
me make them eight of them.
I don't know.
They have to have the same length.
And I'll call this delta x.
It has to be the same.
And, you guys, please forgive me for the horrible picture.
They don't look like the same step, delta x,
but it should be the same.
In each of them I arbitrarily, say it again,
Magdalena, arbitrarily pick x1 star, and another point,
x2 star wherever I want inside.
I'm just getting [INAUDIBLE].
X4 star, and this is x8 star.
But let's say that in general I don't know they are 8.
They could be n.
xn star.
And passing to the limit with respect
to n going to infinity, what am I going to get?
Well, in the first cam I'm going up,
and I'm hitting at what altitude?
I'm hitting at the altitude called f of x1 star.
And that's going to be the height of this-- what is this?
Strip?
Right?
Or rectangle.
OK.
And I'm going to do the same with green
for the second rectangle.
I'll pick x2 star, and then that doesn't work.
And I'll take this.
Let's see if I can do the light green one,
because spring is here.
Let's see.
That's beautiful.
I go up.
I hit here at x2 star.
I get f of x2 star.
And so on and so forth.
Until I get to, let's say, the last of the Mohicans.
This will be xn minus 1, and this is going
to be xn star, the purple guy.
And this is going to be the height
of that last of the Mohicans.
So when I compute the sum, I call that approximating sum
or Riemann approximating sum, because Riemann had nothing
better to do than invent it.
He didn't even know that we are going
to get pixels that are in larger and larger quantities.
Like, we get 3,000 by 900.
He didn't know we are going to have all those digital gadgets.
But passing to the limit practically should
be easier to understand for teenagers now
age, because it's like making the number of pixels
larger and larger, and the pixels practically invisible.
Remember, I mean, I don't know, those old TVs,
color TVs where you could still see the squares?
STUDENT: Mm-hm.
PROFESSOR: Well, yeah.
When you were little.
But I remember them much better than you.
And, yes, as the number of pixels will increase,
that means I'm taking the limit and going larger and larger.
That means practically limitless.
Infinity will give me an ideal image.
My eye will be as if I could see the image that's a curvilinear
image as a real person.
And, of course, the quality of our movies
really increased a lot.
And this is what I'm trying to emphasize here.
So you have f of x1 star delta x plus the last rectangle
area, f of xn star delta x.
Well, as a mathematician, I don't write it like that.
How do I write it as a mathematician?
Well, we are funny people.
We like Greek.
It's all Greek to me.
So we go sum and from-- no. k from 1 to n, f of x sub k star.
So I have k from 1 to n exactly an rectangles area to add.
And this is going to be [INAUDIBLE], which
is the same everywhere.
In that case, I made the partition is equal.
So practically I have the same distance.
And what is this limit? [? Lim ?]
is going to be exactly integral from a to b of f of x dx.
And I make a smile here, and I say I'm very happy.
This is as a meaning is the area under the graph.
If-- well, I didn't say something.
If I want it to be positive, otherwise it's
getting not to be the area under the graph.
The integral will still be defined like that.
But what's going to happen if I have, for example, half of it
above and half of it below?
I'm going to get this, and I'm going to get that.
And when I add them, I'm going to get a negative answer,
because this is a negative area, and that's a positive area
and they try to annihilate each other.
But this guy under the water is stronger,
like an iceberg that's 20% on tip of the water,
80% of the iceberg is under the water.
So the same thing.
I'm going to get a negative answer in volume [INAUDIBLE].
OK.
Now, we remember that very well, but now we
have to generalize this thingy to something else.
And I will give you a curvilinear domain.
Where shall I erase?
I don't know.
Here.
What if somebody gives you the image of a potatoe-- well,
I don't know.
Something.
A blob.
Some nice curvilinear domain-- and says, you know what?
I want to approximate the area of this image, curvilinear
image, to the best of my abilities.
And compute it, and eventually I have some weighted sum of that.
So if one would have to compute the area,
it wouldn't be so hard, because we would say,
OK, I have to "partition this domain
into small sections using a rectangular partition
or square partition."
And how?
Well, I'm going to-- you have to imagine
that I have a bunch of a grid, and I'm
partitioning the whole thing.
And you say, wait a minute.
Wait a minute.
It's not so easy.
I mean, they are not all the same area, Magdalena.
Even if you tried to make these equidistant in both directions,
look at this guy.
Look at that guy.
He's much bigger than that.
Look at this small guy, and so on.
So we have to imagine that we look at the so-called normal
the partition.
And let's say in the normal, or the length of the partition,
is denoted like that.
We have to give that a meaning.
Well, let's say "this is the highest diameter
for all subdomains in the picture."
And you say, wait a minute.
But these subdomains should have names.
Well, they don't have names, but assume they have areas.
This would be-- I have to find a way to denote them
and be orderly.
A1, A2, A3, A4, A5, AN, AM, AN, stuff like that.
So practically I'm looking at the highest diameter.
When I have a domain, I look at the largest instance
inside that domain.
So what would be the diameter?
The largest distance between two points in that domain.
I'll call that the diameter.
OK.
I want that diameter to go got 0 in the limit.
So I want this partition to go to 0 in the limit.
And that means I'm "shrinking" the pixels.
"Shrinking" in quotes, the pixels.
How would I mimic what I did here?
Well, it would be easier to get the area.
In this case, I would have some sort of A sum limit.
I'm sorry.
The curvilinear area of the domain.
Let's call it-- what do you want to call it?
D for domain-- inside the domain.
OK?
This whole thing would be what?
Would be limit of summation of, let's say, limit of what kind?
k from 1 to n.
Limit n goes to infinity.
K from 1 to n of these tiny A sub k's,
areas of the subdomain.
Wait a minute.
But you say, but what if I want something else?
Like, I'm going to build some geography.
This is the domain.
That's something like on a map, and I'm
going to build a mountain on top of it.
I'll take some Play-Do, I'll take some Play-Do,
and I'm going to model some geography.
And you say, wait a minute.
Do you make mountains?
I'm afraid to make Rocky Mountains, because they
may have points where the function is not smooth.
If I don't have derivative at the peak,
them I'm in trouble, in general.
Although you say, well, but the function
has to be only continuous.
I know.
But I don't want any kind of really nasty singularity
where I can have a crack in the mountain
or a well or something like that.
So I assume the geography to be smooth,
the function of [INAUDIBLE] is continuous,
and the picture should look something
like-- let's see if I can do that.
The projection, the shadow of this geography,
would be the domain, [? D. ?] And this is equal, f of x what?
You say, what?
Magdalena, I don't understand.
The exact shadow of this fellow where I have the sun on top
here-- that's the sun.
Spring is coming-- the shade is the plain, or domain, x, y.
I take all my points in x, y.
I mean, I take really all my points in x, y,
and the value of the altitude on this geography at the point
x, y would be z equals f of x, y.
And somebody's asking me, OK, if this would be a can of Coke,
it would be easy to compute the volume, right?
Practically you have a constant altitude everywhere,
and you have the area of the base times the height,
and that's your volume.
But what if somebody asks you to find the volume under the hat?
"Find the volume undo this graph."
STUDENT: I would take it more as two functions.
So the top line would be the one function,
and the bottom line would be another function.
So if you take the volume of the top function minus the volume
of the bottom function, it'd give you
the total volume of the object.
PROFESSOR: And actually, I want the total volume
above the sea level.
So I'm going to-- sometimes I can take it up
to a certain level where-- let's say the mountain is up to here,
and I want it only up to here.
So I want everything, including the-- the walls
would be cylindrical.
STUDENT: Yeah.
PROFESSOR: If I want all the volume,
that's going to be a little bit easier.
Let's see why.
I will have limit.
The idea is, as you said very well, limit.
n goes to infinity.
A sum k from 1 to n.
And what kind of partition can I build?
I'll take the line, and I'll say,
I'll build myself a partition with a,
let's say, the typical domain, AK.
I have A1, A2 A3, A4, AK, AN.
How may of those little domains?
AN.
That will be all the little subdomains
inside the green curve.
The green loop.
In that case, what do I do?
For each of these guys, I go up, and I go, oh, my god, this
looks like a skyscraper, but the corners,
when I go through this surface, are
in the different dimensions.
What am I going to do?
That forces me to build a skyscraper
by thinking I take a point in the domain,
I go up until that hits the surface, pinches the surface,
and this is the altitude that I'm going
to select for my skyscraper.
And here I'm going to have another skyscraper, and here
another one and another one, so practically it's dense.
I have a skyscraper next to the other or a less like
[INAUDIBLE].
Not so many gaps in certain areas.
So I'm going to say f of x kappa star.
Now those would be the altitudes of the buildings.
Magdalena, you don't know how to spell.
Altitudes of the buildings.
What are they?
Parallel [INAUDIBLE] by P's.
Can you say parallel by P?
OK.
[INAUDIBLE] what.
Ak where Ak will be the basis of the area of the basis.
is of my building.
OK.
The green part will be the flat area
of the floor of the skyscraper.
Is this hard?
Gosh, yes.
If you want to do it by hand and take the limit
you would really kill yourself in the process.
This is how you introduce it.
You can prove this limit exists, and you can prove that limits
exist and will be the volume of the region under the geography
z equals f of x,y and above the sea level.
The seal level meaning z equals z.
STUDENT: What's under a of k?
PROFESSOR: Ak.
STUDENT: What is [INAUDIBLE]
PROFESSOR: Volume of the region.
STUDENT: Oh, I know, like what under it?
PROFESSOR: Here?
STUDENT: No, up.
PROFESSOR: Here?
STUDENT: Yes.
PROFESSOR: Area of the basis of a building.
STUDENT: Oh, the basis.
PROFESSOR: So practically this green thingy
is a basis like the base rate.
How large is the basement of that building.
Ak.
Now how am I going to write this?
This is something new.
We have to invent a notion for it, and since it's Ak,
looks more or less like a square or a rectangle.
You think, well, wouldn't-- OK, if it's a rectangle,
I know I'm going to get delta x and delta y right?
The width times the height, whatever those two dimensions.
It makes sense.
But what if I have this domain that's
curvilinear or that domain or that domain.
Of course, the diameter of such a domain
is less than the diameter of the partition, so I'm very happy.
The highest diameter, say I can get it here,
and this is shrinking to zero, and pixels
are shrinking to zero.
But what am I going to do about those guys?
Well, you can assume that I am still approximating
with some squares and as the pixels are getting
to be many, many, many more, it doesn't
matter that I'm doing this.
Let me show you what I'm doing.
So on the floor, on the-- this is the city floor, whatever.
What we do in practice, we approximate that
like on the graphing paper with tiny square domains,
and we call them delta Ak will be delta Sk times delta Yk,
and I tried to make it a uniform partition as much as I can.
Now as the number of pixels goes to infinity
and those pixels will become smaller and smaller,
it doesn't there that the actual contour of your Riemann sum
will look like graphing paper.
It will get refined, more refined, more refined, smoother
and smoother, and it's going to be
really close to the ideal image, which is a curve.
So as that end goes to infinity, you're
not going to see this-- what is this called-- zig zag thingy.
Not anymore.
The zig zag thingy will go into the limit to the green curve.
This is what the pixels are about.
This is how our life changed a lot.
OK?
All right.
Now good.
How am I going compute this thing?
Well, I don't know, but let me give it a name first.
It's going to be double integral over-- what
do want the floor to be called?
We called d domain before.
What should I call this?
Big D. Not round.
Over D. That's the floor, the foundation
of the whole city-- of the whole area of the city
that I'm looking at.
Then I have f of xy, da, and what is this?
This is exactly that.
It's the limit of sum of the-- what is the difference here?
You say, wait a minute, Magdalena,
but I think I don't understand what you did.
You tried to copy the concept from here,
but you forgot you have a function of two variables.
In that case, this mister, whoever it is that goes up
is not xk, it's XkYk.
So I have two variables-- doesn't change anything
for the couple.
This couple represents a point on the skyscraper
so that when I go up, I hit the roof with this exact altitude.
So what is the double integral of a continuous function
f of x and y, two variables, with respect to area level.
Well, it's going to be just the limit of this huge thing.
In fact, it's how do we compute it?
Let's see how we compute it in practice.
It shouldn't be a big deal.
What if I have a rectangular domain,
and that's going to make my life easier.
I'm going to have a rectangular domain in plane,
and which one is the x-axis?
This one.
From A to B, I have the x moving between a and Mr. y
says, I'm going to be between c and d.
C is here, and d is here.
So this is going to be the so-called rectangle
a, b cross c, d meaning the set of all the pairs--
or the couples xy-- inside it, what does it mean?
x, y you playing with the property there.
X is between a and b, thank god.
It's easy.
And y must be between c and d, also easy.
A, b, c, d are fixed real numbers in this order.
A is less than b, and c is less.
And we have this geography on top,
and I will tell you what it looks like.
I'm going to try and draw some beautiful geography.
And now I'm thinking of my son, who is 10.
He played with this kind of toy that was exactly this color,
lime, and it had needles.
Do you guys remember that toy?
I am sure you're young enough to remember that.
You have your palm like that, and you see this square thingy,
and it's all made of needles that
look like thin, tiny skyscrapers,
and you push through and all those needles go up and take
the shape of your hand.
And of course, he would put it on his face,
and you could see his face and so on.
But what is that?
That's exactly the Riemann sum, the Riemann approximation,
because if you think of all those needles or tiny--
what are they, like the tiny skyscrapers--
the sum of the them approximates the curvilinear shape.
If you put that over your face, your face is nice and smooth,
curvilinear except for a few single areas,
but if you actually look at that needle
thingy that is giving the figure,
you recognize the figure.
It's like a pattern recognition, but it's not your face.
I mean it is and it's not.
It's an approximation of your face, a very rough face.
You have to take that rough model of your face
and smooth it out.
How?
By passing to the limit, and this is what
animation is doing actually.
On top of that you want this to have some other properties--
illumination of some sort-- light coming from what angle.
That is all rendering techniques are actually
applied mathematics.
In animation, the people who programmed
Toy Story-- that was a long time ago,
but everything that came after Toy Story 2
was based on mathematical rendering techniques.
Everything based on the notion of length.
All right.
So the way we compute this in practice
is going to be very simple, because you're going to think,
how am I going to do the rectangle for the rectangles?
That'll be very easy.
I split the rectangle perfectly into other tiny rectangle.
Every rectangle will have the same dimension.
Delta x and delta y.
Does it makes sense?
So practically when I go to the limit,
I have summation f of xk star, yk star
inside the delta x delta y delta Magdalena,
the same kind of displacement when I take k from 1 to n,
and I pass to the limit according to the partition,
what's going to happen?
These guys, according to Mr. Linux,
will go to be infinitesimal elements, dx, dy.
This whole thing will go to double integral
of f of x and y, and Mr. y says, OK
it's like you want him to integrate him one at a time.
This is actually something that we are going to see in a second
and verify it.
X goes between a and b, and y goes between c and d,
and this is an application of a big theorem called
Fubini's Theorem that says, wait a minute,
if you do it like this over a rectangle a,b cross c,d,
you're double integral can be written as three things.
Double integral over your square domain f of x,y dA,
or you integral from c to d, integral from a to b,
f of x,y dx dy, or you can also swap the order,
because you say, well, you can do the integration with respect
to y first.
Nobody stops you from doing that,
and y has to be between what and what?
STUDENT: C and d.
PROFESSOR: C and d, thank you.
And then whatever you get, you get to integrate that
with respect to x from a to b.
So no matter in what order you do it,
you'll get the same thing.
Let's see an easy example, and you'll say, well,
start with some [INAUDIBLE] example, Magdalena,
because we are just starting, and that's
exactly what I'm going to.
I will just misbehave.
I'm not going to go by the book.
And I will say I'm going by whatever I want to go.
X is between 0, 2, and y is between 0 and 2
and 3-- this is 2, this is 3-- and my domain
will be the rectangle 0, 2 times 0, 3.
This is neat on the floor.
Compute the volume of the box of basis d and height 5.
Can I draw that?
It gets out of the picture.
I'm just kidding.
This is 5, and that's sort of the box.
And you say, wait a minute, I know that from third grade--
I mean, first grade, whenever.
How do we do that?
We go 2 units times 3 units that's
going to be 6 square inches on the bottom of the box,
and then times 5.
So the volume has to be 2 times 3 times 5, which
is 30 square inches.
I don't care what it is.
I'm a mathematician, right?
OK.
How does somebody who just learned Tonelli's-- Fubini
Tonelli's Theorem do the problem.
That person will say, wait a minute,
now I know that the function is going to be z
equals f of xy, which in this case happens to be cost.
According to what you told us, the theorem you claim Magdalena
proved to this theorem, but there is a sketch
of the proof in the book.
According to this, the double integral
that you have over the domain d, and this is dA.
DA will be called element of area, which is also dx dy.
This can be solved in two different ways.
You take integral from-- where is x going?
Do we want to do it first in x or in y?
If we put dy dx, that means we integrate with respect
to y first, and y goes between 0 and 3,
so I have to pay attention to the limits of integration.
And then x between 0 and 2 and again
I have to pay attention to the limits of integration
all the time and, here, who is my f?
Is the altitude 5 that's constant in my case?
I'm not worried about it.
Let me see if I get 30?
I'm just checking if this theorem was true or is just
something that you cannot apply.
How do you integrate 5 with respect to y?
STUDENT: 5y.
PROFESSOR: 5y, very good.
So it's going to be 5y between y equals 0 down and y equals 3
up, and how much is that 5y, we're
doing y equals 0 down and y equals 3 up,
what number is that?
STUDENT: 25.
PROFESSOR: What?
STUDENT: 25.
PROFESSOR: 25?
STUDENT: One [INAUDIBLE] 15.
PROFESSOR: No, you did-- you are thinking ahead.
So I go 5 times 3 minus 5 times 0 equals 15.
So when I compute this variation of 5y between y
equals 3 and y equals 0, I just block in
and make the difference.
Why do I do that?
It's the simplest application of that FT, fundamental theorem.
The one that I did not specify in [INAUDIBLE].
I should have specified when I have a g function that
is continuous between alpha and beta, how do we
integrate with respect to x?
I get the antiderivative of rule G. Let's call
that big G. Compute it at the end points,
and I make the difference.
So I compute the antiderivative at an endpoint--
at the other endpoint-- then I'm going to make the difference.
That's the same thing I do here, so 5 times 3 is 15,
5 times 0 is 0, 15 minus 0 is 15.
I can keep moving.
Everything in the parentheses is the number 15.
I copy and paste, and that should be a piece of cake.
What do I get?
STUDENT: 15.
PROFESSOR: I get 15 times x between 0 and 2.
Integral of 1 is x.
Integral of 1 is x with respect to x,
so I get 15 times 2, which is 30, and you go, duh,
[INAUDIBLE].
That was elementary mathematics.
Yes, you were lucky you knew that volume of the box,
but what if somebody gave you a curvilinear area?
What if somebody gave you something quite complicated?
What would you do?
You have know calculus.
That's your only chance.
If you don't calculus, you are dead meat.
So I'm saying, how about another problem.
That look like it's complicated, but calculus
is something [INAUDIBLE] with that.
Suppose that I have a square in the plane between-- this
is x and y-- do you want square 0,1 0,1
or you want minus 1 to 1 minus 1 to 1.
It doesn't matter.
Well, let's take minus 1 to 1 and minus 1 to 1,
and I'll try to draw as well as I can,
which I cannot but it's OK.
You will forgive me.
This is the floor.
If I were just a little tiny square
in this room plus the equivalent square in that room
and that room and that room.
This is the origin.
Are you guys with me?
So what you're looking at right now
is this square foot of carpet that I have,
but I have another one here and another one behind the wall,
and so do I everything in mind?
X is between minus 1 and 1, y is between minus 1 and 1.
And somebody gives you z to be a positive function,
continuous function, which is x squared plus y squared.
And you go, already.
Oh, my god.
I already have this kind of hard function.
It's not a hard thing to do.
Let's draw that.
What are we going to get?
Your favorite [INAUDIBLE] that goes like this.
And imagine what's going to happen
with this is like a vase.
Inside, it has this circular paraboloid.
But the walls of this vase are-- I cannot draw better than that.
So the walls of this vase are squares.
And what you have inside is the carved circular paraboloid.
Now I'm asking you, how do I find
volume of the body under and above D, which is minus 1,
1, minus 1, 1.
It's hard to draw that, right?
It's hard to draw.
So what do we do?
We start imagining things.
Actually, when you cut with a plane that is y equals 1,
you would get a parabola.
And so when you look at what the picture is going to look like,
you're going to have a parabola like this,
a parabola like that, exactly the same,
a parallel parabola like this and a parabola like that.
Now I started drawing better.
And you say, how did you start drawing better?
Well, with a little bit of practice.
Where are the maxima of this thing?
At the corners.
Why is that?
Because at the corners, you get 1, 1 for both.
Of course, to do the absolute extrema, minimum, maximum,
we would have to go back to section 11.7 and do the thing.
But practically, it's easy to see that at the corners,
you have the height 2 because this is the point 1, 1.
And the same height, 2 and 2 and 2, are at every corner.
That would be the maximum that you have.
So you have 1 minus 1 and so on-- minus 1, 1, and minus 1,
minus 1, who is behind me, minus 1, minus 1.
That goes all the way to 2.
So it's hard to do an approximation with a three
dimensional model.
Thank god there is calculus.
So you say integral of x squared plus y squared,
as simple as that, da over the domain, D, which is minus 1,
1, minus 1, 1.
How do you write it according to the theorem
that I told you about, Fubini-Tonelli?
Then you have integral integral x squared plus y squared dy dx.
Doesn't matter which one I'm taking.
I can do dy dx.
I can do dx dy.
I just have to pay attention to the endpoints.
Lucky for you the endpoints are the same.
y is between minus 1 and 1.
x is between minus 1 and 1.
I wouldn't known how to compute the volume of this vase made
of marble or made of whatever you
want to make it unless I knew to compute this integral.
Now you have to help me because it's not hard
but it's not easy either, so we need a little bit of attention.
We always start from the inside to the outside.
The outer person has to be just neglected for the time being
and I focus all my attention to this integration.
And when I integrate with respect to y,
y is the variable for me.
Nothing else exists for the time being,
but y being a variable, x being like a constant.
So when you integrate x squared plus y squared
with respect to y, you have to pay attention a little bit.
It's about the same if you had 7 squared plus y squared.
So this x squared is like a constant.
So what do you get inside?
Let's apply the fundamental theorem of calculus.
STUDENT: x squared y.
PROFESSOR: x squared y.
Excellent.
I'm very proud of you.
Plus?
STUDENT: y cubed over 3.
PROFESSOR: y cubed over three.
Again, I'm proud of you.
Evaluated between y equals minus 1 down, y equals 1 up.
And I will do the math later because I'm getting tired.
Now let's do the math.
I don't know what I'm going to get.
I get minus 1 to 1, a big bracket, and dx.
And in this big bracket, I have to do the difference
between two values.
So I put two parentheses.
When y equals 1, I get x squared 1--
I'm not going to write that down-- plus 1 cubed
over 3, 1/3.
I'm done with evaluating this sausage thingy at 1.
It's an expression that I evaluate.
It could be a lot longer.
I'm not planning to give you long expressions in the midterm
because you're going to make algebra mistakes,
and that's not what I want.
For minus 1, what do we have Minus x squared.
What is y equals minus 1 plugged in here?
Minus 1/3.
I have to pay attention.
You realize that if I mess up a sign, it's all done.
So in this case, I say, but this I have minus, minus.
A minus in front of a minus is a plus,
so I'm practically doubling the x squared plus 1/3
and taking it between minus 1 and 1
and just with respect to x.
So you say, wait a minute.
But that's easy.
I've done that when I was in Calc 1.
Of course.
This is the nice part that you get, a simple integral
from the ones in Calc 1.
Let's solve this one and find out what the area will be.
What do we get?
Is it hard?
No.
Kick Mr. 2 out.
He's just messing up with your life.
Kick him out.
2, out.
And then integral of x squared plus 1/3
is going to be x cubed over 3 plus--
STUDENT: x over 3.
PROFESSOR: x over 3, very good.
Evaluated between x equals minus 1 down, x equals 1 up.
Let's see what we get.
2 times bracket.
I'll put a parentheses for the first fractions,
and another minus, and another parentheses.
What's the first edition of fractions that I get?
1/3 plus 1/3.
I'll put 2/3 because I'm lazy.
Then minus what?
STUDENT: Minus 1/3.
PROFESSOR: Minus 1/3 minus 1/3, minus 2/3.
And now I should be able to not beat around the bush.
Tell me what the answer will be in the end.
STUDENT: 8/3.
PROFESSOR: 8/3.
Does that make sense?
When you do that in math, you should always
think-- one of the famous professors at Harvard
was saying one time she asked the students,
how many hours of life do we have have
in one day, blah, blah, blah?
And many students came up with 36, 37.
So always make sure that the answer you get makes sense.
This is part of a cube, right?
It's like carved in a cube or a rectangle.
Now, what's the height?
If this were to go up all the way to 2,
it would be 2, 2, and 2.
2 times 2 times 2 equals 8, and what we got is 8 over 3.
Now, using our imagination, it makes sense.
If I got a 16, I would say, oh my god.
No, no, no, no.
What is that?
So a little bit, I would think, does this make sense or not?
Let's do one more, a similar one.
Now I'm going to count on you a little bit more.
STUDENT: Professor, did you calculate that
by just doing a quarter, and then just multiplying it by 4?
Because then that would just leave us
with zeroes [INAUDIBLE].
PROFESSOR: You mean in that particular figure?
Yeah.
STUDENT: Yeah, because it was perfectly [INAUDIBLE].
PROFESSOR: Yeah.
It's nice.
It's a little bit related to some other problems that
come from pyramids.
By the way, how can you compute the volume of a square pyramid?
Suppose that you have the same problem.
Minus 1 to 1 for x and y.
Minus 1 to 1, minus 1 to 1.
Let's say the pyramid would have the something like that.
What would be the volume of such a pyramid?
STUDENT: [INAUDIBLE].
PROFESSOR: The height is h for extra credit.
Can you compute the volume of this pyramid
using double integrals?
Say the height is h and the bases is the square minus 1,
1, minus 1, 1.
I'm sure it can be done, but you know--
now I'm testing what you remember in terms of geometry
because we will deal with geometry a lot
in volumes and areas.
So how do you do that in general, guys?
STUDENT: 1/3 [INAUDIBLE].
PROFESSOR: 1/3 the height times the area
of the bases, which is what?
2 times 2.
2 times 2, 3, over 3, 4/3 h.
Can you prove that with calculus?
That's all I'm saying.
One point extra credit.
Can you prove that with calculus?
Actually, you would have to use what you learned.
You can use Calc 2 as well.
Do you guys remember that there were
some cross-sectional areas, like this would be made of cheese,
and you come with a vertical knife and cut cross sections.
They go like that.
But that's awfully hard.
Maybe you can do it differently with Calc 3 instead of Calc 2.
Let's pick one from the book as well.
OK.
So the same idea of using the Fubini-Tonelli argument
and have an iterative-- evaluate the following double integral
over the rectangle of vertices 0, 0--
write it down-- 3, 0, 3, 2, and 0, 2.
So on the bases, you have a rectangle of vertices 3, 0, 0,
0, 3, 2, and 0, 2.
And then somebody tells you, find us
the double integral of 2 minus y da
over r where r represents the rectangle that we talked about.
This is exactly [INAUDIBLE].
And the answer we should get is 6.
And I'm saying on top of what we said in the book,
can you give a geometric interpretation?
Does this have a geometric interpretation
you can think of or not?
Well, first of all, what is this animal?
According to the Fubini theorem, this animal
will have to be-- I have it over a rectangle,
so assume x will be between a and b, y
will be between c and d.
I have to figure out who those are.
2 minus y and dy dx.
Where is y between?
I should draw the picture for the rectangle
because otherwise, it's not so easy to see.
I have 0, 0 here, 3, 0 here, 3, 2 over here, shouldn't be hard.
So this is going to be 0, 2.
That's the y-axis and that's the x-axis.
Let's see if we can see it.
And what is the meaning of the 6, I'm asking you?
I don't know.
x should be between 0 and 3, right?
y should be between 0 and 2, right?
Now you are experts in this.
We've done this twice, and you already know how to do it.
Integral from 0 to 3.
Then I take that, and that's going
to be 2y minus y squared over 2 between y
equals 0 down and y equals 2 up dx.
That means integral from 0 to 3, bracket minus bracket
to make my life easier, dx.
Now, there is no x, thank god.
So that means I'm going to have a constant
minus another constant, which means I go 4 minus 4 over 2.
2, right?
The other one, for 0, I get 0.
I'm very happy I get 0 because in that case,
it's obvious that I get 2 times 3, which is 6.
So I got what the book said I'm going to get.
But do I have a geometric interpretation of that?
I would like to see if anybody can--
I'm going to give you a break in a few minues--
if anybody can think of a geometric interpretation.
What is this f of xy if I were to interpret this as a graph?
x equals f of x and y.
Is this--
STUDENT: 2 minus y.
PROFESSOR: So z equals 2 minus y is a plane, right?
STUDENT: Yes, but then you have the parabola is going down.
PROFESSOR: And how do I get to draw this plane the best?
Because there are many ways to do it.
I look at this wall.
The y-axis is this.
The z-axis is the vertical line.
So I'm looking at this plane.
y plus z must be equal to 2.
So when is y plus z equal to 2?
When I am on a line in the plane.
I'm going to draw that line with pink because I like pink.
This is y plus z equals 2.
And imagine this line will be shifted by parallelism as it
comes towards you on all these other parallel vertical planes
that are parallel to the board.
So I'm going to have an entire plane like that,
and I'm going to stop here.
When I'm in the plane that's called x equals 3--
this is the plane called x equals
3-- I have exactly this triangle, this [INAUDIBLE].
It's in the plane that faces me here.
I don't know if you realize that.
I'll help you make a house or something nice.
I think I'm getting hungry.
I imagine this again as being a piece of cheese,
or it looks even like a piece of cake would be with layers.
So our question is, if we didn't know calculus
but we knew how to draw this, and somebody gave you
this at the GRE or whatever exam,
how could you have done it without calculus?
Just by cheating and pretending, I know how to do it,
but you've never done a double integral in your life.
So I know it's a volume.
How do I get the volume?
What kind of geometric body is that?
STUDENT: A triangle.
STUDENT: It's a triangular prism.
PROFESSOR: It's a triangular prism.
Good.
And a triangular prism has what volume formula?
STUDENT: Base times height.
PROFESSOR: Base times the height.
And the height has what area?
Let's see.
The base would be that, right?
And the height would be 3.
Am I right or not?
The height would be 3.
This is not--
STUDENT: It's 2.
Yeah.
STUDENT: No, it's 3.
DR. MAGDALENA TODA: From here to here?
STUDENT: 3.
DR. MAGDALENA TODA: It's 3.
So how much is that?
How much-- OK.
From here to here is 2.
From here to here, it's how much?
STUDENT: The height is only-- I see--
STUDENT: It's also 2.
DR. MAGDALENA TODA: It's also 2 because look at that.
It's an isosceles triangle.
This is 45 to 45.
So this is also 2.
2 to-- that's 90 degrees, 45, 45.
OK.
So the area of the shaded purple triangle-- how much is that?
STUDENT: 2.
DR. MAGDALENA TODA: 2.
2 times 2 over 2.
2 times 3 equals 6.
I don't need calculus.
In this case, I don't need calculus.
But when I have those nasty curvilinear
z equals f of x, y, complicated expressions, I have no choice.
I have to do the double integral.
But in this case, even if I didn't know how to do it,
I would still get the 6.
Yes, sir?
STUDENT: What if we did that on the exam?
DR. MAGDALENA TODA: Well, that's good.
I will then keep it in mind.
Yes.
It doesn't matter to me.
I have other colleagues who really care about the method
and start complaining.
I don't care how you get to the answer
as long as you got the right answer.
Let me tell you my logic.
Suppose somebody hired you thinking you're a good worker,
and you're smart and so on.
Would they care how you got to the solution of the problem?
As long as the problem was solved correctly, no.
And actually, the elementary way is the fastest
because it's just 10 seconds.
You draw.
You imagine.
You know what it is.
So your boss will want you to find the fastest way to provide
the correct solution.
He's not going to care how you got that.
So no matter how you do it, as long
as you've got the right answer, I'm going to be happy.
I want to ask you to please go to page 927 in the book
and read.
It's only one page.
That whole end section, 12.1.
It's called an informal argument for Fubini's theorem.
Practically, it's a proof of Fubini's theorem, page 927.
And then I'm going to go ahead and start the homework
four, if you don't mind.
I'm going to go into WeBWork and give you homework four.
And the first few problems that you
are going to be expected to solve
will be out of 12.1, which is really easy.
I'll give you a few minutes back.
And we go on with 12.2, and it's very similar.
You're going to like that.
And then we'll go home or wherever we need to go.
So you have a few minutes of a break.
Pick up your extra credits.
I'll call the names.
Lily.
You got a lot of points.
And [INAUDIBLE].
And you have two separate ones.
Nathan.
Nathan?
Rachel Smith.
Austin.
Thank you.
Edgar.
[INAUDIBLE]
Aaron.
Andre.
Aaron.
Kasey.
Kasey came up with a very good idea
that I will write a review sample.
Did I promise that?
A review sample for the midterm.
And so I said yes.
Karen and Matthew.
Reagan.
Aaron.
When you submitted, you submitted.
Yeah.
And [INAUDIBLE].
here.
And I'm done.
STUDENT: Did we turn in [INAUDIBLE]?
DR. MAGDALENA TODA: Yes, absolutely.
Now once we go over 12.2, you will say, oh,
but I understand the Fubini theorem.
I didn't know whether there's room for Fubini,
because once I cover the more general case, which is in 12.2,
you are going to understand Why Fubini-Tonelli
works for rectangles.
So if I think of a domain that is of the following form,
in the x, y plane, I go x is between and and b, right?
That's my favorite x.
So I take the pink segment, and I
say, everything that happens-- it's going
to happen on top of this world.
I have, let's say, two functions.
To make my life easier, I'll assume both of them [INAUDIBLE]
one bigger than the other.
But in case they are not both positive,
I just need f to be bigger than g for every point.
And the same argument will function.
This is f, continuous positive.
Then g, continuous positive but smaller
in values than this one.
Yes, sir?
STUDENT: [INAUDIBLE] 12.2 that we're starting?
DR. MAGDALENA TODA: 12.2.
And you are more organized than I am, and I appreciate it.
So integration over a non-rectangular domain.
And we call this a type one because this
is what many books are using.
And this is that x is between two fixed end points.
But y is between two variable end points.
So what's going to happen to y?
y is going to take values between the lower,
the bottom one, which is g of x, and the upper one,
which is f of x.
So this is how we define the domain that's
shaded by me with black shades, vertical strips here.
This is the domain.
Now you really do not need to prove
that double integral over 1 dA over-- let's call
the domain D-- is what?
Integral between f of x minus g of x from a to b dx.
And you say, what?
Magdalena, what are you trying to say?
OK.
Let's go back and say, what if somebody
would have asked you the same question in calculus 2?
Saying, guys I have a question about the area
in the shaded strip, vertical strip thing.
How are we going to compute that?
And you would say, oh, I have an idea.
I take the area under the graph f, and I shade that in orange.
And I know what that is.
So you would say, I know what that is.
That's going to be what?
Integral from a to be f of x dx.
Let's call that A1, right?
A1.
Then you go, minus the area with-- I'm just going to shade
that, brown strips under g.
g of x dx.
And call that A2.
A1 minus A2.
We know both of these formulas from where?
Calc 1 because that's where you learned about the area
under the graph of a curve.
This is the area under the graph of a curve f.
This is the area under the graph of the curve g.
The black striped area is their difference.
All right.
And so how much is that?
I'm sorry I put the wrong thing.
a, b.
That's going to be integral from a to b.
Now you say, wait, wait, wait a minute.
Based on what?
Based on some sort of additivity property
of the integral of one variable, which says integral
from a to b of f plus g.
You can have f plus, minus g.
It doesn't matter.
dx.
You have integral from a to b f dx plus integral from a to b g
dx.
It doesn't matter what.
You can have a linear combination of f and g.
Yes, Matthew?
MATTHEW: So this is just for the domain?
So if you put it, that would be down.
So there might be another formula up here
that would be curved surface.
And this is the bottom, so you're using integral
to find the base, and then you're
going to plug that integral into the other integral.
DR. MAGDALENA TODA: So I'm just using the property that's
called linearity of the simple integral,
meaning that if I have even a linear combination like af
plus bg, then a-- I have not a.
Let me call it big A and big B. Big A Af integral
of f plus big B integral of g.
You've learned that in Calc 2.
I'm doing this to apply it for these areas that are subtracted
from one another.
If I were to add, as you said, I would put something
on top of that.
And then it would be like a superimposition onto it.
So I have integral from a to b of f of x minus g of x dx.
And I claim that this is the same
as double integral of the 1dA over the domain D.
How can you write that differently?
I'll tell you how you write that differently.
Integral from a to b of integral from-- what's
the bottom value of Mr. Y?
So Mr. X knows what he's doing.
He goes all the way from a to b.
The bottom value of y is g of x.
You go from the bottom value of y g of x to the upper value
f of x.
And then you here put 1 and dy.
Is this the same thing?
You say, OK, I know this one.
I know this one from calc 2.
But Magdalena, the one you gave us is new.
It's new and not new, guys.
This is Fubini's theorem but generalized
to something that depends on x.
So how do I do that?
Integral of 1dy.
That's what?
That's y measured between two values that don't depend on y.
They depend only on x, g of x on the bottom, f of x on the top.
So this is exactly the integral from a to b.
In terms of the round parentheses,
I put-- what is y between f of x and g of x?
f of x minus g of x dx.
So it is exactly the same thing from Calc 2
expressed as a double integral.
All right.
Now This is a type one region that we talked about.
A type two region is a similar region, practically.
What you have to keep in mind is they're both
given here as examples.
But the technique is absolutely the same.
If instead of taking this picture,
I would take y to move between fixed values,
like y has to be between c and d-- this is my y.
These are the fixed values.
And then give me some nice colors.
This curve and that curve-- OK, I
have to rotate my head because then this is going to be x.
This is going to be y.
And the blue thingy has to be a function of y.
x is a function of y.
So how do I call that?
I have x or whatever equals big F of y.
And here in the red one, I have x equals big G of y.
And how am I going to evaluate the striped area?
Of course striped because I have again y is between c and d.
And what's moving is Mr. X.
And Mr. X refuses to have fixed variables.
Now he goes, I move from the bottom, which is G of y,
to the top, which is F of y.
How am I going to write the double integral
over this domain of 1dA, where dA is dxdy.
Who's going to tell me?
Similarly, the same reasoning as for this one.
I'm going to have the integral from what to what
of integral from what to what?
Who comes first, dx or dy?
STUDENT: dx.
DR. MAGDALENA TODA: dx, very good.
And dy at the end.
So y will be between c and d, and x
is going to be between G of y and F of y.
And here is y.
How can I rewrite this integral?
Very easily.
The integral from c to d of the guy on top,
the blue guy, F of y, minus the guy on the bottom, G of y, dy.
Some people call the vertical stip method
compared to the horizontal strip method, where
in this kind of horizontal strip method,
you just have to view x as a function of y
and rotate your head and apply the same reasoning as before.
It's not a big deal.
You just need a little bit of imagination,
and the result is the same.
An example that's not too hard-- I
want to give you several examples.
We have plenty of time.
Now it says, we have a triangular region.
And that is enclosed by lines y equals 0, y equals 2x,
and x equals 1.
Let's see what that means and be able to draw it.
It's very important to be able to draw in this chapter.
If you're not, just learn how to draw,
and that will give you lots of ideas
on how to solve the problems.
Chapter 12 is included completely on the midterm.
So the midterm is on the 2nd of April.
For the midterm, we have chapter 10, those three sections.
Then we have chapter 11 completely,
and then we have chapter 12 not completely, up to 12.6.
All right.
So what did I say?
I have a triangular region that is obtained by intersecting
the following lines.
y equals 0, x equals 1, and y equals 2x.
Can I draw them and see how they intersect?
It shouldn't be a big problem.
This is a line that passes through the origin
and has slope 2.
So it should be very easy to draw.
At 1, x equals 1, the y will be 2 for this line of slope 2.
So I'll try to draw.
Does this look double to you?
So this is 2.
This is the point 1, 2.
And that's the line y equals 2x.
And that's the line y equals 0.
And that's the line x equals 1.
So can I shade this triangle?
Yeah, I can eventually, depending on what they ask me.
What do they ask me?
Find the double integral of x plus y dA
with respect to the area element over T, T being the triangle.
So now I'm going to ask, did they say by what method?
Unfortunately, they say, do it by both methods.
That means both by x intregration first and then
y integration and the other way around.
So they ask you to change the order of the integration
or do what?
Switch from vertical strip method
to horizontal strip method.
You should get the same answer.
That's a typical final exam problem.
When we test you, if you are able to do this
through the vertical strip or horizontal
strip and change the order of integration.
If I do it with the vertical strip method,
who comes first, the dy or the dx?
Think a little bit.
Where do I put d-- Fubini [INAUDIBLE]
comes dy dx or dx dy?
STUDENT: dy.
PROFESSOR: dy dx.
So VSM.
You're going to laugh.
It's not written in the book.
It's like a childish name, Vertical Strip Method,
meeting integration with respect to y
and then with respect to x.
It helped my students through the last decade
to remember about the vertical strips.
And that's why I say something that's not using the book, VSM.
Now, I have integral from-- so who is Mr. X going from 0 to 1?
He's stable.
He's happy.
He's going between two fixed values.
y goes between the bottom line, which is 0.
We are lucky.
It's a really nice problem.
Going to y equals 2x.
So it's not hard at all.
And we have to integrate the function x plus y.
It should be a piece of cake.
Let's do this together because you've accumulated seniority
in this type of problem.
What do I put inside?
What's integral of x plus y with respect to y?
Is it hard?
xy plus-- somebody tell me.
STUDENT: y squared.
PROFESSOR: y squared over 2, between y
equals 0 on the bottom, y equals 2x on top.
I have to be smart and plug in the values y.
Otherwise, I'll never make it.
STUDENT: Professor?
PROFESSOR: Yes, sir?
STUDENT: Why did you take 2x as the final value
because you have a specified triangle.
PROFESSOR: Because y equals 2x is the expression
of the upper function.
The upper function is the line y equals 2x.
They provided that.
So from the bottom function to the upper function,
the vertical strips go between two functions.
So when I plug in here y equals 2x,
I have to pay attention to my algebra.
If I forget the 2, it's all over for me, zero points.
Well, not zero points, but 10% credit.
I have no idea what I would get, so I have to pay attention.
2x times x is 2x squared plus 2x all squared-- guys,
keep an eye on me-- 4x squared over 2.
I put the first value in a pink parentheses,
and then I move on to the line parentheses.
Evaluate it at 0.
That line is very lucky.
I get a 0 because y equals 0 will give me 0.
What am I going to get here?
2x squared plus 2x squared.
Good.
What's 2x squared plus 2x squared?
4x squared.
So a 4 goes out.
Kick him out.
Integral from 0 to 1 x squared dx.
Integral of x squared is?
STUDENT: x cubed over 3.
PROFESSOR: x cubed over 3.
And if you take it between 1 and 0, you get?
STUDENT: 1.
PROFESSOR: 1/3.
1/3 times 4 is 4/3.
Suppose this is going to happen on the midterm,
and I'm asking you to do it reversing the integration
order.
Then you are going to check your own work very beautifully
in the sense that you say, well, now
I'm going to see if I made a mistake in this one.
What do I do?
I erase the whole thing, and instead of vertical strips,
I'm going to put horizontal strips.
And you say, well, life is a little bit harder in this case
because in this case, I have to look
at y between fixed values, y between 0 and 1.
So y is between 0 and 1-- 0 and 2, fixed values.
And Mr. X says, I'm going between two functions of y.
I don't know what those functions of y are.
I'm puzzled.
You have to help Mr. X know where
he's going because his life right now is a little bit hard.
So what is the function for the blue?
Now he's not blue anymore.
He's brown.
x equals 1.
So he knows what he's going to be.
What is the x function for the red line
that [INAUDIBLE] asked about?
STUDENT: y over 2.
PROFESSOR: x must be y over 2.
It's the same thing, but I have to express x in terms of y.
So I erase and I say x equals y over 2.
Same thing.
So x has to be between what and what, the bottom and the top?
Well, I turn my head.
The top must be x equals 1, and the bottom one is y over 2.
That's the bottom one, the bottom value for x.
Now wish me luck because I have to get the same thing.
So integral from 0 to 2 of integral from y over 2 to 1.
Changing the order of integration
doesn't change the integrand, which is exactly
the same function, f of xy.
This is the f function.
Then what changes?
The order of integration.
So I go dx first, dy next and stop.
I copy and paste the outer ones, and I focus my attention
to the red parentheses inside, which I'm
going to copy and paste here.
I'll have to do some math very carefully.
So what do I have?
I have x plus y integrated with respect to x.
If I rush, it's a bad thing.
STUDENT: So that would be x squared.
PROFESSOR: x squared.
STUDENT: Over 2.
PROFESSOR: Over 2.
STUDENT: Plus xy.
PROFESSOR: Plus xy taken between the following.
When x equals 1, I have it on top.
When x equals y over 2, I have it on the bottom.
OK.
This red thing, I'm a little bit too lazy.
I'll copy and paste it separately.
For the upper part, it's really easy to compute.
What do I get?
When x is 1, 1/2, 1/2 plus when x is 1, y.
Minus integral of-- when x is y over 2,
I get y squared over 4 up here over 2.
So I should get y squared over 8 plus--
I've got an x equals y over 2.
What do I get?
y squared over 2.
Is this hard?
It's very easy to make an algebra mistake on such
a problem, unfortunately.
I have y plus 1/2 plus what?
What is 1/2 plus 1/8?
STUDENT: 5/8.
PROFESSOR: 5 over 8 with a minus y squared.
So hopefully I did this right.
Now I'll go, OK, integral from 0 to 2 of all of this animal, y
plus 1/2 minus 5 over 8, y squared.
What happens if I don't get the right answer?
Then I go back and check my work because I
know I'm supposed to get 4/3.
That was easy.
So what is integral of this sausage, whatever it is?
y squared over 2 plus y over 2 minus 5 over 8--
oh my god-- 5 over 8, y cubed over 3, between 2 up
and 0 down.
When I have 0 down, I plug y equals 0.
It's a piece of cake.
It's 0.
So what matters is what I get when I plug
in the value 2 instead of y.
So what do I get?
4 over 2 is 2, plus 2 over 2 is 1, minus 2 cubed, thank god.
That's 8.
8 simplifies with 8 minus 5/3.
So I got 9/3 minus 5/3, and I did it carefully.
I did a good job.
I got the same thing, 4/3.
So no matter which method, the vertical strip
or the horizontal strip method, I get the same thing.
And of course, you'll always get the same answer
because this is what the Fubini theorem extended to this case
is telling you.
It doesn't matter the order of integration.
I would advise you to go through the theory in the book.
They teach you more about area and volume on page 934.
I'd like you to read that.
And let's see what I want to do.
Which one shall I do?
There are a few examples that are worth it.
I'll pick the one that gives people the most trouble.
How about that?
I take the few examples that give people the most trouble.
One example that popped up on almost each and every final
in the past 13 years that involves changing
the order of integration.
So example problem on changing the order of integration.
A very tricky, smart problem is the following.
Evaluate integral from 0 to 1, integral from x to 1,
e to the y squared dy dx.
I don't know if you've seen anything like that
in AP Calculus or Calc 2.
Maybe you have, in which case your professor probably told
you that this is nasty.
You say, in what sense is it nasty?
There is no expressible anti-derivative.
So this cannot be expressed in terms of elementary functions
explicitly.
It's not that there is no anti-derivative.
There is an anti-derivative-- a whole family, actually--
but you cannot express them in terms of elementary functions.
And actually, most functions are not so bad in real world,
in real life.
Now, could you compute, for example, integral from 1 to 3
of e to the t squared dt?
Yes.
How do you do that?
With a calculator.
And what if you don't have one?
You go to the lab over there.
There is MATLAB.
MATLAB will compute it for you.
How does MATLAB know how to compute it
if there is no way to express the anti-derivative
and take the value of the anti-derivative between b
and a, like in the fundamental theorem of calculus?
Well, the calculator or the computer program is smart.
He uses numerical analysis to approximate this type
of integral.
So he's fooling you.
He's just playing smarty pants.
He's smarter than you at this point.
OK.
So you cannot do this by hand, so this order of integration is
fruitless.
And there are people who tried to do this on the final.
Of course, they didn't get anywhere
because they couldn't integrate it.
The whole idea of this one is to-- some professors
are so mean they don't even tell you,
hint, change the order of integration
because it may work the other way around.
They just give it to you, and then people can spend an hour
and they don't get anywhere.
If you want to be mean to a student, that's what you do.
So I will tell you that one needs
to change the order of integration for this.
This is the function.
We keep the function, but let's see what happens if you draw.
The domain will be x between 0 and 1.
This is your x value.
y will be between x and 1.
So it's like you have a square.
y equals x is your diagonal of the square.
And you go from-- more colors, please.
You go from y equals x on the bottom and y equals 1 on top.
And so the domain is this beautiful triangle
that I make all in line with vertical strips.
This is what it means, vertical strips.
But if I do horizontal strips, I have to change the color, blue.
And for horizontal strips, I'm going
to have a different problem.
Integral, integral dx dy.
And I just hope to god that what I'm going to get
is doable because if not, then I'm in trouble.
So help me on this one.
If y is between what and what?
It's a square.
It's a square, so this will be the same, 0 to 1, right?
STUDENT: Yep.
PROFESSOR: But Mr. X?
How about Mr. X?
STUDENT: And then it will be between 1 and y.
PROFESSOR: Between-- Mr. X is this guy.
And he doesn't go between 1.
He goes between the sea level, which is
x equals 0, to x equals what?
STUDENT: [INAUDIBLE].
PROFESSOR: Right?
So from x equals 0 through x equals y.
And you have the same individual e to the y squared that before
went on your nerves.
Now he's not so bad, actually.
Why is he not so bad?
Look what happens in the first parentheses.
This is so beautiful that it's something
you didn't even hope for.
So we copy and paste it from 0 to 1 dy.
These guys stay outside and they wait.
Inside, it's our business what we do.
So Mr. X is independent from e to the y squared.
So e to the y squared pulls out.
He's a constant.
And you have integral of 1 dx between 0 and y.
How much is that?
1.
x between x equals 0 and x equals y.
So it's y.
So I'm being serious.
So I should have said y.
Now, if your professor would have given you, in Calc 2,
this, how would you have done it?
STUDENT: U-substitution.
PROFESSOR: U-substitution.
Excellent.
What kind of u-substitution [INAUDIBLE]?
STUDENT: y squared equals u.
PROFESSOR: y squared equals u, du equals 2y dy.
So y dy together.
They stick together.
They attract each other as magnets.
So y dy is going to be 1/2 du-- 1/2 pulls out--
integral e to the u du.
Attention.
When y is moving between 0 and 1,
u is moving also between 0 and 1.
So it really should be a piece of cake.
Are you guys with me?
Do you understand what I did?
Do you understand the words coming out of my mouth?
It's easy.
Good.
So what is integral of e to the u du?
e to the u between 1 up and 0 down.
So e to the u de to the 1 minus e to the 0 over 2.
That is e minus 1 over 2.
I could not have solved this if I tried it by integration
with y first and then x.
The only way I could have done this
is by changing the order of integration.
So how many times have I seen this in the past 12 years
on the final?
At least six times.
It's a problem that could be a little bit
hard if the student has never seen it before
and doesn't know what to do [? at that point. ?]
Let's do a few more in the same category.
STUDENT: Professor?
PROFESSOR: Yes?
STUDENT: Where did this shape-- where did this graph come from?
Were we just saying it was with the same--
PROFESSOR: OK.
I read it from here.
So this and that are the key.
This is telling me x is between 0 and 1, and at the same,
time y is between x and 1.
And when I read this information on the graph,
I say, well, x is between 0 and 1.
Mr. Y has the freedom to go between the first bisector,
which is that, and the cap, his cap, y equals 1.
So that's how I got to the line strips.
And from the line strips, I said that I need horizontal strips.
So I changed the color and I said
the blue strips go between x.
x will be x equals 0 and x equals y.
And then y between 0 and 1, just the same.
It's a little bit tricky.
That's why I want to do one or two more problems like that,
because I know that I remember 20-something years ago,
I myself needed a little bit of time understanding
the meaning of reversing the order of integration.
STUDENT: Does it matter which way you put it?
PROFESSOR: In this case, it's important that you do reverse.
But in general, it's doable both ways.
I mean, in the other problems I'm going to give you today,
you should be able to do either way.
So I'm looking for a problem that you could eventually
do another one.
We don't have so many.
I'm going to go ahead and look into the homework.
Yeah.
So it says, you have this integral,
the integral from 0 to 4 of the integral
from x squared to 4y dy dx.
Draw, compute, and also compute with reversing
the order of integration to check your work.
When I say that, it sounds horrible.
But in reality, the more you work on
that one, the more familiar you're going to feel.
So what did I just say?
Problem number 26.
You have integral from 0 to 4, integral
from x squared to 4x dy dx.
Interpret geometrically, whatever that means,
and then compute the integral in two ways,
with this given order integration, which
is what kind of strips, guys?
Vertical strips.
Or reversing the order of integration.
And check that the answer is the same just to check your work.
STUDENT: So first--
PROFESSOR: First you draw.
First you draw because if you don't draw,
you don't understand what the problem is about.
And you say, wait a minute.
But couldn't I go ahead and do it without drawing?
Yeah, but you're not going to get too far.
So let's see what kind of problem you have.
y and x.
y equals x squared is a what?
It's a pa--
STUDENT: Parabola.
PROFESSOR: Parabola.
And this parabola should be nice and sassy.
Is it fat enough?
I think it is.
And the other one will be 4x, y equals 4x.
What does that look like?
It looks like a line passing through the origin that
has slope 4, so the slope is really high.
STUDENT: Just straight.
PROFESSOR: y equals 4x versus y equals x squared.
Now, do they meet?
STUDENT: Yes.
PROFESSOR: Yes.
Exactly where do they meet?
Exactly here.
STUDENT: 4.
PROFESSOR: So 4x equals x squared, where do they meet?
They meet at-- it has two possible roots.
One is x equals 0, which is here,
and one is x equals 4, which is here.
So really, my graph looks just the way it should look,
only my parabola is a little bit too fat.
This is the point of coordinates 4 and 16.
Are you guys with me?
And Mr. X is moving between 0 and 4.
This is the maximum level x can get.
And where he stops here at 4, a miracle happens.
The two curves intersect each other exactly at that point.
So this looks like a leaf, a slice of orange.
Oh my god.
I don't know.
I'm already hungry so I cannot wait to get out of here.
I bet you're hungry as well.
Let's do this problem both ways and then go
home or to have something to eat.
How are you going to advise me to solve it first?
It's already set up to be solved.
So it's vertical strips.
And I will say integral from 0 to 4,
copy and paste the outer part.
Take the inner part, and do the inner part because it's easy.
And if it's easy, you tell me how I'm going to do it.
Integral of 1 dy is y.
y measured at 4x is 4x, and y measured at x squared
is x squared.
Oh thank god.
This is so beautiful and so easy.
Let's integrate again.
4 x squared over 2 times x cubed over 3 between x equals 0 down
and x equals 4 up.
What do I get?
I get 4 cubed over 2 minus 4 cubed over 3.
This 4 cubed is an obsession.
Kick him out.
1/2 minus 1/3.
How much is 1/2 minus 1/3?
My son knows that.
STUDENT: 1/6.
PROFESSOR: OK.
1/6, yes.
So we simply take it.
We can leave it like that.
If you leave it like that on the exam, I don't mind at all.
But you could always put 64 over 6 and simplify it.
Are you guys with me?
You can simplify it and get what?
32 over 3.
Don't give me decimals.
I'm not impressed.
You're not supposed to use the calculator.
You are supposed to leave this is exact fraction
form like that, irreducible.
Let's do it the other way around,
and that will be the last thing we do.
The other way around means I'll take another color.
I'll do the horizontal stripes.
And I will have to rewrite the meaning of these two
branches of functions with x expressed in terms of y.
That's the only thing I need to do, right?
So what is this?
If y is x squared, what is x?
STUDENT: Root y.
PROFESSOR: The inverse function. x will be root of y.
You said very well.
So I have to write.
In [INAUDIBLE], I have what I need
to have for the line horizontal strip method.
And then for the other one, x is going to be y over 4.
So what do I do?
So integral, integral, a 1 that was here hidden,
but I'll put it because that's the integral.
And then I go dx dy.
All I have to care about is the endpoints of the integration.
Now, pay attention a little bit because Mr. Y is not
between 0 and 4.
I had very good students under stress
in the final putting 0 and 4.
Don't do that.
So pay attention to the limits of integration.
What are the limits?
0 and--
STUDENT: 16.
PROFESSOR: 16.
Very good.
And x will be between root y-- well, which one is on top?
Which one is on the bottom?
Because if I move my head, I'll say that's on top
and that's on the bottom.
STUDENT: The right side is always on the top.
PROFESSOR: So the one that looks higher is this one.
This is more than that in this frame.
So square of y is on top and y over 4 is on the bottom.
I should get the same answer.
If I don't, then I'm in trouble.
So what do I get?
Integral from 0 to 16.
Tonight, when I go home, I'm going
to cook up the homework for 12.1 and 12.1 at least.
I'll put some problems similar to that
because I want to emphasize the same type of problem
in at least two or three applications for the homework
for the midterm.
And maybe one like that will be on the final as well.
It's very important for you to understand how,
with this kind of domain, you reverse
the order of integration.
Who's helping me here?
Root y.
What is root y when-- y to the 1/2.
I need to integrate.
So I need minus y over 4 and dy.
Can you help me integrate?
STUDENT: [INAUDIBLE].
PROFESSOR: 2/3 y to the 3/2 minus--
STUDENT: y squared.
PROFESSOR: y squared over 8, y equals 0
on the bottom, piece of cake.
That will give me 0.
I'm so happy.
And y equals 16 on top.
So for 16, I have 2/3.
And who's telling me what else?
STUDENT: 64.
PROFESSOR: 64.
4 cubed.
I can leave it 4 cubed if I want to minus another-- well here,
I have to pay attention.
So I have 16 here.
I got square root of 16, which is 4, cubed.
Here, I put minus 4 squared, which was there.
How do you want me to do this simplification?
STUDENT: [INAUDIBLE].
PROFESSOR: I can do 4 to the fourth.
Are you guys with me?
I can put, like you prefer, 16 squared over 8.
Is it the same answer?
I don't know.
Let's see.
This is really 4 to the 4, so I have 4 times 4 cubed.
4 cubed gets out and I have 2/3 minus 1/2.
And how much is that?
Again 1/6.
Are you guys with me?
1/6.
So again, I get 4 cubed over 6, so I'm done.
4 cubed over 6 equals 32 over 3.
I am happy that I checked my work
through two different methods.
I got the same answer.
Now, let me tell you something.
There were also times when on the midterm
or on the final, due to lack of time and everything,
we put the following kind of problem.
Without solving this integral-- without solving-- indicate
the corresponding integral with the order reversed.
So all you have to do-- don't do that.
Just from here, write this and stop.
Don't waste your time.
If you do the whole thing, it's going to take you
10 minutes, 15 minutes.
If you do just reversing the order of integration,
I don't know what it takes, a minute and a half, two minutes.
So in order to save time, at times,
we gave you just don't solve the problem. reverse
the order of integration.
One last one.
But I don't want to finish it.
I want to give you the answer at home,
or maybe you can finish it.
It should be shorter.
You have a circular parabola, but only the first quadrant.
So x is positive.
STUDENT: Question.
PROFESSOR: I don't know.
I have to find it.
Find the volume.
Example 4, page 934.
Find the volume of the solid bound
in the above-- this is a little tricky-- by the plane z
equals y and below in the xy plane
by the part of the disk in the first quadrant.
So z equals y means this is your f of x and y.
So they gave it to you.
But then they say, but also, in the xy plane,
you have to have the part of the disk in the first quadrant.
This is not so easy.
They draw it for you to make your life easier.
The first quadrant is that.
How do you write the unit circle, x squared equals 1,
x squared plus y squared less than or equal to 1,
and x and y are both positive.
This is the first quadrant.
How do you compute?
So they say compute the volume, and I say just
set up the volume.
Forget about computing it.
I could put it in the midterm just like that.
Set up an integral without solving it
that indicates the volume under z equals f of xy-- that's
the geography of z-- and above a certain domain in plane,
above D in plane.
So you have, OK, what this should teach you?
Should teach you that double integral over d f of xy da
can be solved.
Do I ask to be solved?
No.
Why?
Because you can finish it later, finish at home.
Or maybe, I don't even want you to compute on the final.
So how do we do that?
f is y.
Would I be able to choose whichever order integration I
want?
It shouldn't matter which order.
It should be more or less the same.
What if I do dy dx?
Then I have to do the Fubini.
But it's not a rectangular domain.
Aha.
So Magdalena, be a little bit careful
because this is going to be two finite numbers,
but these are functions.
STUDENT: It will be an x function.
PROFESSOR: So the x is between 0 and 1,
and that's going to be z.
You do vertical strips.
That's a piece of cake.
But if you do the vertical strips,
you have to pay attention to the endpoints for x and y,
and one is easy.
Which one is trivial?
STUDENT: Zero.
PROFESSOR: The bottom one, zero.
The one that's nontrivial is the upper one.
STUDENT: There will be 1 minus--
STUDENT: Square root of 1 minus y squared.
PROFESSOR: Very good.
Square root of 1 minus y squared.
So if I were to go one more step further without solving this,
I'm going to ask you, could this be solved by hand?
Well, so you have it in the book--
STUDENT: Professor, should be a [INAUDIBLE] minus x squared?
PROFESSOR: Oh yeah.
1 minus x squared.
Excuse me.
Didn't I write it?
Yeah, here I should have written y equals square root of 1
minus x squared.
So when you do it-- thank you so much-- you go integrate,
and you have y squared over 2.
And you evaluate between y equals 0
and y equals square root 1 minus x squared,
and then you do the [INAUDIBLE].
In the book, they do it differently.
They do it with respect to dx and dy and integrate.
But it doesn't matter how you do it.
You should get the same answer.
All right?
[INAUDIBLE]?
STUDENT: [INAUDIBLE] in that way,
doesn't the square root work out better because there's already
a y there?
PROFESSOR: In the other case--
STUDENT: Doing dy dx.
PROFESSOR: Yeah, in the other way,
it works a little bit differently.
You can do u-substitution, I think.
So if you do it the other way, it will be what?
Integral from 0 to 1, integral form 0
to square root of 1 minus y squared, y dx dy.
And what do you do in this case?
You have integral from 0 to 1.
Integral of y dx is going to be y is a constant.
x between the two values will be simply 1 minus y squared dy.
So you're right.
Matthew saw that, because he's a prophet,
and he could see two steps ahead.
This is very nice what you observed.
What do you do?
You take a u-substitution when you go home.
You get u equals 1 minus y squared.
du will be minus 2y dy, and you go on.
So in the book, we got 1/3.
If you continue with this method,
I think it's the same answer.
STUDENT: Yeah.
I got 1/3.
PROFESSOR: You got 1/3.
So sounds good.
We will stop here.
You will get homework.
How long should I leave that homework on?
Because I'm thinking maybe another month, but please
don't procrastinate.
So let's say until the end of March.
And keep in mind that we have included
one week of spring break here, which you
can do whatever you want with.
Some of you may be in Florida swimming and working on a tan,
and not working on homework.
So no matter how, plan ahead.
Plan ahead and you will do well.
31st of March for the whole chapter.