Mathematics - Multivariable Calculus - Lecture 16

So, next Thursday is the second midterm exam.
Who could have thought.
It feels like the first one was just yesterday.
Yeah.
It was much warmer, you know.
So, next Thursday, November 5, right here for most students.
As you know, last time we had an overflow.
And, again I apologize for this.
I apologize to those students who didn't
have comfortable seat.
So, we'll avoid this situation, this kind of
situation, this time.
We'll have some breathing room.
And, unfortunately, that means that two sections, students
from two sections, those which are led by Dario will meet at a
different, undisclosed, location. [LAUGHTER]
No, I just didn't want to write it.
You can find it, Dario will tell you.
If you are in Dario's section, Dario will tell you, and it's
also on the class home page.
It's in [? Barrows ?]
Hall.
And, if you're not in a section of Dario then you don't need
to worry about it, you
should just come here.
OK.
And, on the home page, you'll find more information.
But, it's basically the same, the same deal.
Someone asked me for the cheat sheet for the, you
know, page of formulas.
Can we use the old one.
The one from the first midterm, the answer is no.
You have to make a new one.
It will also be on one-sided, it has to one-sided on
one side of a standard size sheet of paper.
OK.
But, don't worry, we're not going to test you on the
material of the first midterm anyway, so it doesn't
make a difference.
The midterm will cover the material after
the first midterm.
OK.
Everything that we've done up to now, since the
first midterm up to
now, up to this week, including this week.
And, there are, on this webpage, you will find problems
for review, and you will also find a mock midterm.
It's not posted yet.
I will post it tonight, the mock
midterm.
Any questions about this, about the exam?
Next Tuesday, we will have a review lecture as before, in
which I will discuss all this material, and kind of try
to put it in perspective.
And, you will have a chance to ask me questions about it.
Yes.
[INAUDIBLE]
Oh, so there will be no quiz next week.
So, it's same deal as last time.
[INAUDIBLE]
Mock
midterm, the idea is to give you problems, which kind of
look similar to what you should expect.
Of course.
[INAUDIBLE]
Well difficulty is in the eye of the beholder, right.
So, for some people it will be more difficult, some people
it would be less difficult.
But, it should be in the same ballpark.
So, let's go back to the material of last week.
I mean last week, last Tuesday.
We talked about triple integrals, and we talked about
special coordinate systems for triple integrals -
cylindrical and spherical coordinate systems.
OK.
So, I want to say a few words, general words
about triple integrals.
When we see triple integrals, it means we
have something like this .
Where E is the region in three-dimensions space,
something we denote by R3 f is a function, and dv is what we
can think of as kind of a measure of integration, which
we would usually write as: dx, dy, dz.

In some order, but in fact we know that you can use
other coordinates as well.
And, today we'll talk more about the possibility of using
different coordinate systems.
So, the basic set up for calculating this integral is --
as I already explained last time, and has been explained
last week, but I want to repeat it one more time -- is that you
want to represent this triple integral as an
integrated integral.
In other words, you want to break it into steps.
We're at eat step you're just evaluating a single integral,
an integral in one variable.
Something which we learned before, in one
variable calculus.
And the way you do it, is first of all, you find, you find
the projection of your three-dimensional region.
So, you have some three-dimensional region here.
But, the first thing you do is you find its projection on
one of the planes, one of the coordinate planes.
Let's say xy plane, but it could be xz or yz plane.
So, that's the projection D.
Let's call it D.
So, this E is three-dimensional , this is three-dimensional.
Where as D is two-dimensional, two-dimensional region,
which is in the xy plane.
So, once you do that, you can re-write your integral as kind
of an integrated integral, where you have a double
integral over D.
And, here you put dA which is a measure of integration in a
two-dimensional space, which is essentially dx, dy.
And, then you have an integral of f with respect to the
remaining variable, where you have to put the limits, so
there'll be some limits, alpha and beta, which in general
will depend on xy.

So, here xy will be in D, and for each xy the limits for the
single integral will be from some alpha xy to beta of xy.
Now, the way we write it it's like a confusing.
And, you saw that confusion last time at the lecture
when we discussed it.
It is important to remember that you leave the integral
from right to left.
I mean that's the way, that's the way it's done.
This is a convention.
You read it from right to left, which means that
you first do this one.
You first, this is the first step.
You first evaluate this single integral, OK.
And, then as a second, you evaluate the resulting
double integrals.
So, onces you evaluate this, you will get something which
will depend on x and y.
This will be an expression depending on x and y.
Which, you then integrate over D in the xy plane.
So, we break it into two steps, the first step was a single
integral with respect to the remaining variable, along
which you have projected.
And, the second step is integrating over the image of
the projection, which is D.
This is the image of the projection.

So, we can actually write it further, further as a
combination of single integrals, because this
double integral in turn, can also be written as an
integrated integral.
OK.
So, the, that means that writing integral over d, dA as
an integrated integral, with respect to x and y, we actually
end up with three single integrals.
So, here we also have a choice.
There is a choice.
Which one goes first x or y.
OK.
Let's say, what do we mean by which goes first, which one
will appear first, or which one you will integrate first.
As I just said, you read it from right to left.
So, this is actually, the roles are reversed, right.
So, let's write it like this, let's say dx and then it would
be dy, and then finally it is dz, then you have
your function f.
Here , I wrote fdz and here I wrote dzf.
It doesn't really matter as we discussed last time.
It doesn't matter in which order you write, f and dz
or f and dy and so on.
I find this form a little bit more convenient to remember,
more intuitive, because the variable of integration is
written right next to the integral.
So you will not forget which one you're integrating.
But, the important thing to remember when you look at
this formula, is to read it from right to left.
You read it from right to left, which means that this is the
first step, integrate over z.
I'm skipping the limits.
Of course, there will be limits.
For example, here will be limits alpha over xy
and beta over xy like on that blackboard.
But, let me just keep that, and just giving you kind of
schematics of this process,
OK.
So, first step is integrating f over z.
The result will be a function which only depends on x and
y, because we have integrated out z.
Think of it as kind of removing the dependence on z by
averaging out over z, over the effect of z.
The result is a function of x and y.
That function you integrate over y.
This is step number two.
This is the second step.
And, finally you integrate the result, which will now be only
function of x over x, so that the end result is actually a
number, independent of all of the three variables.
So, at each of the steps, you get rid of one the variables.
You are kind of averaging out over that variable, so
that the result does not depend on that variable.
If you're doing an integral, and suddenly you noticed that
you have integrated over z, but the result somehow depends on
z, that you carry z somehow to the second, first or
third step, you made a mistake somewhere.
There should not be any depends on z on the second step.
There should not be any dependence on y and z at
the third step, you see.
So, this is a very easy test to see if you are doing
things in the right way.
Any questions about this?
So it's kind of a notational issue, but you just have to
remember it is actually, it makes sense, because you're
integrating this expression okay, but what is
this expression .
This expression is obtained by integrating this .
I mean integrating in respect to y, this expression.
So, unraveling it, you see that first you have to integrate
this and then you integrate this.
It makes perfect sense that you go from right to left, and
not from left to right, OK.
Now, there is a very special case of a triple integral.
There is a special case, and that's the case when the
function f, which in general is just any function
of xy and z, right.
So , f is really f of xyz.
In general, it could be anything.
It could be, you know, x squared times e to the y
times cosine z, or whatever, whatever you want.
But, there is a special case when this function is
actually just equal to 1.
Okay.
It's the simplest function you can imagine.
Well, except 0 perhaps.
But, if the function is 0, the integral is 0 so
there's nothing to discuss.
So, let's talk about the case when the function
is actually equal to 1.
And, also let's assume, in addition, that our region in
the three-dimensional space, E, is the region above this domain
D in a two-dimensional space, under the graph of a function.

Let's call this function F capital.
Note, that this has nothing to do with f small; f small is a
function of three variables.
It's a function on E, which is a region in 3-D, on R3, on
the three-dimensional space.
But, let's suppose that the region itself is -- there is
some functions, so let's draw a picture like this .
So, let's say D is kind of a rectangular region, and here we
have, we have some function, like this, so we have some.
This is graph of a function .
So, let's say that this is a graph of a function z
equals F capital of xy.
Yes.
So, this is what we discussed when we talked about
double integrals.
When we talked about double integrals, we discussed the
fact that a double integral represents the volume under the
graph, volume under the graph.
So, the volume under the graph, I put it in quotations.

What I mean is, first of all, that I'm assuming that this
functions is greater than or equal to 0 on D.
For otherwise, it would not be above the xy plane,
everywhere on this region D.
This is region D.
Which is now kind of rectangular region, I have
chosen to simplify things.
OK.
And, also, when I say under the graph, I really mean under the
graph, but above the xy plane.
And, also on the sides, it's confined by the
boundaries of D.
Which, in this case, are just vertical planes like this, OK.
So, when we talked about double integrals, we discussed the
fact that volume is equal to, this volume under the graph.
Is equal to the double integral over D of F
capital xy, dA, right.
So now, I want to show that you get exactly this answer if you,
also by calculating a triple integral, but with the
function 1, you see.
So, this is one of the points which many
people find confusing.
That, the volume under the graph is actually given by a
double integral, but it's also given by a triple integral.
But, in the special case, in a special case is that the
function which you integrate comes from three
variables, is 1.
In other words, what I'm integrating now is dv.
You see, this is a most general triple integral.
The integral of f dv over E.
But, now f is 1.
So, if you want, I can put 1 times dv, but it doesn't
matter; 1 times dv is dv.
OK.
So, what is it equal to.
Let's use this formula.
First of all, so I project unto D.
And, in this case I can surely do that, Because my region has
a very nice projection onto xy plane specifically.
Because, it's under the graph of the function.
And, so I will have here dA, then I will have to put the
limits here, alpha and beta and integrate my function.
But what are the limits in this situation.
In this situation, the limit, the lower limit is always 0.
I was counting z from 0, because my bottom, the bottom
lid of this region is just a rectangular xy plane.
so that means z is equal to 0 at the bottom.
Here z is equal to 0 across the bottom.
So, that's why here, I actually put 0, and then I'm integrating
z from 0 to something which lies on this graph.
So, that's sort of the top of this region, right.
And, so for the top point of that region, what is z
equal to; z is equal to F capital of xy right.
So, that means that my opera limit is F capital of xy.
And, then I have to put f dz, but f is 1, again.
So, it's 1 times dz, right.
So this is very easy to find, right.
This is just a very special case of this general set up.
I have to take the anti-derivative to
calculate this integral.
I simply to have to take the anti-derivative of 1 want.
The anti-derivative of 1 is z plus a constant, but anyway,
we're going to take the difference of the limit.
So, it's going to be z with the limits F of xy and 0.
So what is that.
That's just F of xy.
So, this inner integral, which we have agreed, should
represent the first step of the calculation.
This inner integral is simply equal to F of xy right.
Simply because the function is 1 this argument would not work
necessarily for a general function.
For general function we'll have some F here, which
depends on xy right.
So, xyz, in fact.
So, general function depends on xyz, so you wouldn't be able to
easily find the anti-derivative.
But, for the function 1, the derivative is z, and
this for sure we know.
So, that's the answer we get.
And, once we get this answer, let's substitute it here , so
we get double integral of F capital of xy, dA, so
that's the same answer.
Same answer as before, right.
The answer before was the integral of f xy, dA over D,
and that's exactly what we've got by integrating the
function 1 over this entire three-dimensional region, OK.
This makes sense?
Any questions about this?
So, the conclusion is that this integral is actually equal to
the triple integral over E.
Where E is this region inside.
This is E is this three-dimensional region, which
is drawn here under the graph.
So, one more time.
The volume of this region under the graph can be
represented in two ways.
One is triple integral, but a very special function,
namely function 1, OK.
And, can also be represented as a double integral of the
function F capital, of which this is the graph, over D.
The reason why the two things are the same, is just the
result of calculation of representing this
triple integral as an integrated integral.
Where, at the first step, we very easily evaluate what the
result is for this function, for this special function
1, that's what we get, OK.
So, now, of course, you can ask if the integral of the function
1 over this three-dimensional region is nothing but the
volume of that region, what about the general case, what
about in this integral of a general function f of xyz.
Is this also some sort of volume.
Well, there are two ways to interpret the triple integral
for a general function.
For a general function, function f of xyz not
necessarily 1, not 1 but some general function.
There are two ways to ways to interpret.
Two ways to interpet the integral.
Well, the first way, let's suppose you wanted to interpret
it as some sort of a volume, and let's argue by analogy.
Suppose we have, let's go down one dimension lower.
And, let's look at that integral of this, of
this function for the two-dimensional region,
double integral.
This is a double integral, but we interpreted it as
a volume of something three-dimensional, you see.
So, a double integral of a general function, actually
corresponds to something three-dimensional of
dimension greater by 1.
So, you start with a double integral, but you get a volume
of something three-dimensional.
That's not surprising, because likewise, if you have an
integral, a single integral of a function one variable, right,
what does this represent.
It represents the area of something two-dimensional,
namely, the area under the graph of this function f, let's
call it, let's choose a different notation for the
function, so that we don't confuse the two things, right.
So, a single integral corresponds to the volume,
but in this case we call volume area.
That's the terminology, proper terminology.
Area of something under the graph of this function.
And, this something, is actually two-dimensional,
even though, you start with a single integral.
If you start with a double integral, so 2 is a number
of integrations, you get volume of something
three-dimensional, right.
So, what will you get if you take triple integral.
You get this sort of hyper volume of something
four-dimensional, right, something that if insist on
interpreting the integral as a volume of sorts, you have to
introduce one extra dimension.
And, so you will get kind of a hyper volume.
I don't know, there is no good term for it.
Mathematicians would just call it a volume, because you don't
care about the dimension.
It's still called the volume.
But, let's call it hyper volume to emphasize that it is a
volume of something not quite, it's not that kind of an object
that we will need in real life.
Hyper volume over four-dimensional region, which
is the region under the graph -- again, continuation -- of f
of xyz.
So, you see their is a complete analogy.
A single integral over functions g will be the volume
or area under the graph of this function, and a doubling
integral will be the volume integral of a function and two
variables will be volume of something in three-dimensions
under the graph.
And, the triple-integral of a function f, is a hyper volume
of the four-dimenstional region under the graph
of this function.
So, you have to introduce an additional variable, u, and say
that the graph is given by, so you have a four-dimensional
space, let's call it f of xyz and u .
And, you write u is equal f of xyz.
Now, of course, we could not really visualize it, because
we can only visualize three-dimensions.
There is time, so can think of a region in space time if you
want, and the extra variable u, you can think of as time.
But, still it's kind of, so if you think about it,
you can imagine this.
Because, you can just count, you can look at a certain
region in three-dimensional space, and you can count time
from certain point 0 to some other point which
depends on x, y, z.
So, that would represent the kind of a region in space time.
And, what you're calculating is the hyper volume
of this region.
Now, still it's a little bit disconcerting, because we
really don't have a o good intuition for a
four-dimensional space.
So, it is natural to ask is it possible to give it a
three-dimensional interpretation.
Well, we can go back to the double integral and ask
the same question there.
We can ask what if f capital, the general function, we know
that this integral can be thought of as a volume
of this 3-D object.
But, can it also be interpreted in terms of just
two-dimensional geometry.
And, the answer is yes.
It is, if you think of this function as a density function,
mass density function, then this will give you
the total mass of
the lamina or some kind of a plate, which occupies
this region, right.
So, this is something that you discussed last week.
The total mass as a double integral, right.
So, in other words you are thinking of, you can think of
this calculation, where the function is 1, you can think of
calculating the volume, but you can also think of it as
calculating the mass.
But, in a special case when the density function is 1.
If the density function is 1, then calculating the mass is
the same as calculating the volume, that's why
we get the volume.
But, in general, the density function could
be more complicated.
Your object may not be homogeneous.
It could have some heavier parts and lighter parts,
and you want to kind of average out all of them.
You want to find the total mass.
The total mass would then be the integral of that function
of that density function.
So, from this perspective, you don't have to introduce an
extra dimension if your object , if you have a double
integral, you have region D.
You think of his function as a density function, and you
simply think that you're evaluating the mass of an
object occupying this region, where this function is a
density function, right.
And, likewise, if you have a three-dimensional object which
occupies some region E, you can think of the triple integral,
triple integral was a general function f, as the mass
of that object, you see.
So, that's the way to interpret it.
So, this is a total mass solid E, with mass
density function f.
Any questions about this?
Alright
So, our last topic in this chapter about integration is
about an general coordinate systems, and the general
change of variables.
We have already discussed some, new coordinate systems in two-
and three-dimensions spaces.
We have talked about the polar coordinate system and we
talked about spherical coordinate system.
Also cylindrical, but cylindrical in some sense is
not really a new coordinate syste, it's really more of a
cousin of the polar coordinate.
So, now we would like to talk more systematically about
different coordinates systems in two and
two-dimensional space.
You can also do it in three-dimensional space, but
we'll just focus on the two-dimensional space.
In 3-D, the conceptual is the same, but calculations are
much more complicated, so we basically, we'll mostly
confine ourselves to tow-dimensional cases.
And, as always, when we talk about this, conceptual matters,
it's always good to look at a simplified version, which
in this case would mean just single integrals.
Integrals in one variable, and see how things
work single integrals.
So, let's recall how this works.
Change of variable, variable in a single integral f of x dx.

So, we know how it works.
We simply write x as a function, let's say g of
some new variable, u.

And, then what we do is, we substitute this , g of u
instead of x in the integral.
But, in addition, in addition, we realize that dx is not du,
but rather dx is g prime of u, du.
So, what it means is that it would be wrong to just write
the integral as f times du.
But, instead there is this additional factor,
which is g prime.
So, we recognize that even in a one-dimensional case, there
is this additional factor.
And, this factor actually should be thought of as being
conceptually the same kind of object, same kind of
phenomenon, as the factors which we discussed, in the case
of polar coordinates, namely the factor r, and the factor
which we discussed in cases
of spherical coordinates last time, which is
rho squared times phi.
So, these are all brothers, in some sense.
They all appear for the same reason.
I mean, although, this is one-dimensional case, this is
two-dimensional case, this is three-dimensional case.
But, all of these factors are essentially, they are kind of a
distortion factors, they're area or volume distortion
factors, which happened, do to the fact that we introduced
a new coordinate system.
So, our task right now then is to understand what exactly this
factor is in the case of a most general coordinate change.
Not only in the one-dimensional case, but in the case of
two or three variables.
And like I said, mostly in two variables.
So, we would like to develop a formula [UNINTELLIGIBLE]
of change of variables, but for double integrals.
that's what
we would like to do, for which this r, this polar coordinate
case, will be just a special case.
So, let's try to figures this out, see how this works.

So, in a one-dimensionals case, we have you given variable x
and we have a new variable u.
And, we have a law, which allows us to express one of
them in terms of the other.
And, that's this formula x equals g of u.
But, now we have, now let's try to develop, let's try to think
of something similar in R2.
In R2, we have two given variables or coordinates, x
and y, and we would like to introduce two new coordinates.
Suppose, we introduced, new coordinates, let's
call them u and v.
So, here's an example, polar coordinates, something we have
learned already; uv as this new coordinate system is
our old friend, polar coordinate system.
So, u and v are r and theta.

So, when we do that, when we say that we have introduced
a new coordinate system, what do we mean by that.
We mean that we can express each of the old coordinates x
and y in terms of this new coordinate in a one-to-one way.
In a way that establishes a one-to-one correspondence
between points with xy coordinates, and points with
uv or r, theta coordinates.
In the case of polar coordinates, we know
those formulas.

And, we know that for every r and theta, where r is a
positive number and theta is between 0 and 2 pi, we have a
particular value of xy and conversely.
So, it's a one-to-one correspondence.
It's not, strictly speaking, correct, what I said.
Because of a special point, 0.
The origin, for which if r is equal to 0, then theta
is not well defined.
So, except for this one point, this flap, which I would like
to of these formulas as giving me a map, a correspondence from
between two coordinate systems.
This is our two physical coordinate systems xy.
This is a blackboard, and this is how we would present
points on this blackboard.
And, here is kind of a fake auxiliary, imaginary coordinate
system, with coordinates r and theta.
Let me write like this.
In fact, many you will recognize that we have used
this before when we tried to draw parameter curves
and polar coordinates.
We did use this on our coordinate system.
So, it doesn't, it has nothing to do immediately with the
plane, because the way a point will give us a point on the
plane is through this formulas.
But, I would like to think these formulas as giving us a
map from this auxiliary coordinate system, this
imaginary coordinate system, which we have in our head,
which really doesn't exist to the real coordinate system
under the physical coordinate system on this physical planse.
So, for example, let's say I take the region where theta
goes from 0 to pi, and r goes from 0 to 1.
This is a very simple region on the imaginary plane, right.
It's a rectangle, but what does it correspond to
in a physical world.
What do you think.
Circle.
OK.
Close.
Warm.
So, a circle will come, if we fix r, if we just look at this
integral, this interval will correspond to the part of the
circle, of radius 1, which lies above the x-axis, because
I said from 0 to pi.
If I said from 0 to 2 pi, it will be the
entire circle.
But, because I said from 0 to pi, it would be
half a circle, right.
Everyone agrees with that?
OK.
But, that's not the entire region, right.
Entire region is composed not just by this interval, but
also by all other intervals which are, which lie
beneath it, right.
And, so those intervals will correspond to circles
of a smaller radius.
So, altogether, here they give me this rectangle.
But, in a physical world, they give me this half
of disc, upper half of the disc, of radius 1.
So, you see, something very interesting happened
under this information.
We started out with a rectangle, and we ended up
with upper half of the disc.
And, I would like to say that this formulas give me a
transformation from this region to this region.
What do I mean by this.
I mean that each point here will correspond to a
particular point in here.
And, in fact, this correspondence is one-to-one.
In other words, no two points will go to the same point.
And, like I said, this is almost true.
Unfortunately, strictly speaking, it's not true
for this interval.
This entire interval where r is equal to 0 will
go to this one point.
Will go to this one point.
So, there is a subtly.
The point is that we are studying here
two-dimensional regional.
And, something bad happens on this interval, which
is one dimensional.
This is what mathematicians would call measure 0.
it's not measurable.
It's measure is negligibly small.
It's actually 0.
So, as far as computation of integrals is concerned, it's
not going to change anything.
But, if I wanted to be more precise, I would take say not
this rectangle, where r goes from 0 to 1, but I could
instead take a rectangle where r goes from 1/2 to 1, right.
Then, I would get instead of upper half of the disc, I would
get upper half of the annulus with a shorter radius 1/2, and
the longer, larger radius is 1.
So, if I do that, actually the mystery becomes, this
information becomes less mysterious.
You could certainly believe that I can transform
this strip into this.
What I do, I kind of twist it.
That's what this information, given by these two
formulas, does.
It kind of twists, and expands, and stretches in certain, very
controlled way, the fabric of this original domain, to create
a domain in the physical world.
That's what this information is.
And, then if you think about the big one, it's sort of, if
you get closer and closer to this red part, it
becomes closer.
The way I stretch it, is I stretch it on this side in
the same way, but I stretch it on the lower side.
Not stretch, but in fact, squeeze to make it shorter, and
shorter and short and as I approached here, actually
the whole thing opens up into a half of a disc.
You see what I mean?
You guys see what I mean?
Yeah?
OK.
So, this exactly is the kind of formulas I'm talking about.
This is a one-to-one transformation, one-to-one.
So, each point here corresponds to a point
here, and conversely.
Each point here correspond to one and only one point here.
When I'm in this situation, I would like to be able to write
a double integral of a given function f of xy, dA
as an integral with respect to u and v.
And, of course, so I'm kind of in the same situation as I was
when I talked about the one-dimensional case.
The first step would be to substitute the formula
for x and y in terms of the new coordinates.
So, this would be, let's say in general, here I have
x is r times theta.
So, it is some function of r and theta.
But, in general if I
have a coordinate system uv, not necessarily polar
but more complicated.
I'll give you an example in a of another change
variables like this.
But, let's say this will be some g of uv, and this
will be some h ov uv.
So, the first step would be to just substitute g you uv and h
of uv instead of x and y, right, and then just write
then just write dudv.
But, this is not enough.
The point is that there is always, there is always
an additional factor.
There is always a distortion factor here, which we are
now going to calculate.
An example of this factor is the factor r, which we have
found in the case of polar coordinates.
So for polar, we get r, and then we get dr, d theta.
What we want to di is to have a more conceptual way of deriving
this factor, not just for polar coordinate system, but for a
more general coordinate system, u,v.
This right here.
OK.
And, of course it's easier to explain having a particular
example in mind.
Well, here's one example, polar coordinates, but in polar
coordinates we have already calculate what happens.
You have a question.
[INAUDIBLE]
You're asking would this have that same area as this.
The answer is no.
I'm
sorry?
[INAUDIBLE]
Very good point.
So, that's a very good question.
So, when I say one-to-one doesn't it already assume that
this, we're going to have the same area, right.
And, the answer is no.
It's very easy to see that.
If you have, you know, if you
buy a garment which stretches, you can see immediately
the answer.
Unfortunately, none of what I'm wearing today can do the trick.
But, you can imagine that you have something
which is stretchable.
Maybe, no.
OK, let's just work with your imagination.
You stretch it right.
A stretch is a perfect.
A what, this?
Yeah, but then my hands are -- [LAUGHTER]
Can I use yours.
No. [LAUGHTER]
But, you see what I mean.
OK.
So, let's just use our imaginary clothes. [LAUGHTER]
So, OK.
So, when you stretch it, stretch is a perfect example
of a change of variable.
Because, there's a one-to-one.
Each particle, we don't create any new particles, right.
Clearly, it just, each point goes to a particular point,
and it's one-to-one.
It is like, okay it is even easier to do in
one dimension case.
You can stretch interval and make it, you know, just say a
map, x goes to 2x, you just, everything gets
stretch by factor 2.
Right.
So, it is one-to-one, but when you stretch the area
becomes bigger, right.
So, just the fact that it is one-to-one correspondents
doesn't mean that the areas stay the same.
And, that's exactly what we're fighting.
This factor is precisely the distortion factor for the are.
But the point, which is maybe not immediately obvious is that
it would not be a good idea to actually measure this area, and
then measure this area, and take the ratio of these
two areas, right.
Because, it would be okay if I were function, which we're
integrating function f.
If we chose the function 1, if we chose function 1,
then we would just be calculating the itself.
And, so if the area was something before, the two area
would be this, and the area which we would naively think is
correct in the imaginary r theta coordinate system,
would be something else.
And, there would be some distortion factor
between them, right.
But, that's not a good way to go.
Because, you see the point, when I make this
transformation, the distortion factor is that the
same everywhere.
Because, I explained, for example, that on this guy, this
guy, let's look at this guy at the top interval, the top
interval becomes part of the circle, segment of a
circle or radius 1.
Where as this guy, which has the same length on the
imaginary coordinate system, it becomes part of the
circle with radius of 1/2.
So, certainly this is shorter than this.
So, our distortion is not like just stretching, but it's kind
of stretching in different way, at different places.
So, what we need is a local extortion factor,
you see what I mean.
A local distortion factor, which, in other words, we want
to see not how this area gets distorted, this entire
area gets distorted.
In fact, we want to see how this very small area around
a particular point gets distorted.
And, that's the local distortion factor, and
that's exactly the factor which we'll put here.
Any other questions?
To see how this works, maybe it's better to.
Sorry, you have a question.
No?
Alright.
So, I want to give you another example of coordinate change
before we get to calculating this factor, just to, so that
we have a, so that we have a better variety of examples.
OK.
So, here is a second example, which is actually
straight from the book.
It's, and this I want to use as a motivation for everything
we are doing right now.
Because, perhaps for you it's not so clear yet why we
are discussing this issue.
But, well polar coordinates is already a good demonstration,
because we know that some of the integrals become much
simpler in polar coordinates.
But, here is another example where the coordinate system
of choice is not polar coordinate system.
So, this is 15.9 number 20.
So, let's say you need to calculate this integral
over some region D.
Where D is described as follows: Di is bounded
by lines x minus y
equals 0, x minus y equals 2, x x plus y equals 0,
and x plus y equals 3.

So, this is x plus y equals 0.
This is x minus y equals 0.
This is x minus y equals 2, this is negative 2.
Then, there is another one which goes like this.
This is 2 and this is 3.
So it is actually a rectangle.
Because, these two planes are parallel.
My picture is not so great, but they should be parallel.
And, this two should be parallel also.
Right, so this is actually a rectangle.
But it's a rectangle which is rotated.
So, first of all, if to do this integral in the usual way by
projecting onto the x line or projecting onto the y line,
it's going to be difficult.
Because, you see that first we have to break it into pieces,
and, you know, the bounds will be functions, will be some,
not very complicated linear functions.
But, then you look at the integrand, and you
see that it's really going to look awful.
So, what this picture immediately suggests is that we
should try to find another coordinate system in which this
integral will simplify.
And what is this coordinate system.
Well, what if I told you instead of calculating this
integral, you would have to calculate the integral of u
times e -- v times u times v, du dv, where over some region D
prime, where D prime would be simply u going from, so the, u
going from 0 to 2 -- let me do it a slightly different way --
those would be rectangle, but this rectangle would be turned
in the right direction.
So, this would be u v and this would be 2 and 3 You see.
This surely is much easier to calculate, right.
Because this immediately breaks into, I mean, integrated
integrals and you find the answer right away.
So, what have I done.
How can I get from this integral to this integral.
I simply introduced new variables, u and v, we're u is
x minus y, and v is x plus y.
If I do that, then my bounds which were x minus y equals 0
becomes just bound u equals 0.
And, u equals 2, right.
Which is what I did hear, from 0 to
2.
And, also v from 0 to 3.
Which is what I did here.
So, the region simplifies, it becomes, it kind of turns, we
turn it so that it really stands straight like this.
And, finally the function simplifies as well, because x
plus y becomes v and x squared minus y squared, I can write as
x minus y times x plus y, which is u times v, right.
But, it is not true that this integral is equal to this
integral, because, as I said, there has to be some factor,
which is introduced, some factor introduced here.
And, to understand why we should introduce this factor,
it's very easy to just look at the map, the transformation, so
we have here, we have here, this rectangle in uv
coordinate system.
See, this as an example also of this kind of transformation.
But, it's simpler, because it's linear.
And, those of you who have taken, or maybe are taking
Math 54 will recognize a linear transformation.
In fact, it is a rotation by 45 degrees, coupled with
an expansion by a square root of 2.
So, this is 2 and this is 3.
This formulas give me a map.
Because, actually here have expressed u and v in terms of x
and y, but I can also go back.
And, I can express x and y in terms of u and v.
And, I will find that x is u plus v over 2.
How to find this.
Well, this is u and this is v.
Take the sum.
When you take the sum, the y's will get canceled
and you'll get 2x.
So, if you divide by 2 you'll get x.
And, likewise, y is equal to negative u plus v divided by 2.
So, that's this map.
Each point here will correspond to a particular point here.
But now, why should there be a factor, a magnifying factor
or distortion factor.
It looks like, well, this is rectangle and this is a
rectangle, perhaps we could just get away
without any factor.
This is actually easy to find out, because in this case,
the stretching that occurs is actually homogeneous.
There is
a same distortion factor or stretching factor everywhere,
across the entire region, unlike this case.
So, in this case the stretching factor is actually r.
So, it does depend on the variables, on the polar
coordinates r and theta.
In this case is going to be a constant.
And, so to find out what it is, we could actually
look at this globally.
And, we can just, in other words assume that, suppose that
instead of this function, I would just take function 1.
Let's suppose I take function 1 instead of this.
If I take the function 1, what I'm calculating is just the
area of this guy, right.
And, this area is very to calcuate, in fact.
Because, well you just have to use the Pythagoras
Theoreom, right.
So, the point is that here you have 3, right.
And, so then you can find what this is.
This is a right rectangle, and this has 45 degrees This is 45
degrees and this is 45 degrees, or pi over 4.
So, this length is equal to this length divided
by square root of 2.
So, this is 3 divided by square roof of 2.
Let me use yellow.
This is 3 divided by square root of 2.
And this, comes from a triangle, same kind
of triangle, but now the long side is 2.
So, this is going to be square root of 2, right, 2 divided
by square root of 2.
So the area of this guy, the area here, and this a true area
in the physical world, is 3 over square root of 2 times 2
divided by square root of 2, which is 3.
But, the area of this guy, and that's the imaginary world,
because we're using coordinates which a priori have nothing
to do with the real world.
We are using them for convenience, in order to be
able to evaluate the integral.
The area here is 2 times 3 so it's 6.
So, you see what happened here is that the true area, if we
were calculating the integral, with the function f equals 1,
instead of this complicated function, we would
get the answer 3.
This is a true area.
I'm just simplifying the question.
The question is to calculate this function, but if the
function were actually 1, we wouldn't need to make any
integral whatsoever, we would just need to calculate the
area, which we can easily do by basic trigonometry,
which I just did.
You get 3.
But, if you calculate it the way I'm suggesting now, where
you put the u, dv over this imaginary region D
prime, you get 6.
Now, three is not equal to 6, right.
That's why, that's how you know that there is a factor.
And, now we know that this factor is actually 1/2.

So, it's actually a stretching factor, but it's really a
shrinking factor instead of a stretching factor.
And, we know that indeed that it has shrunk.
When we made this transformation, this guy has
shrunk by a factor of 2.
If you count the area.
I mean, it's linear sizes have shrunk by square root of 2.
So, that the area which is a product of the two sides has
shrunk by a factor of 2, you see.
So, in this particular case, because the transformation is
linear, the shrinking factor actually is a
constant, it's 1/2.
In general, it's going to depend on the point, where you
are making a transformation.
We will see now the general formula for
the shrinking factor.
But, I want to use this example to illustrate that even in a
very simple situation, supposedly, where you can
actually draw this quite easy and recognize it as a
rectangle, and so on, it is much easier to find it's this
new variable, u and v, which you can see -- they kind of ,
they leap in your eyes.
You see right away what the appropriate variables are,
because everything's expressed in terms of
x minus y, x plus y.
So, why not use them as a new variables.
When you do that, the only thing you need to remember is a
factor to put here, and the factor, actually in this
particular case, you can guess by measuring the areas.
In fact, it turns out to be 1/2.
So, the upshot of the discussion is that actually
this integral is equal to this integral.
Which you can easily evaluate.
And, that's a very good illustration of this power
of this, this method of changing variables.
But, now we have to address the question of how to calculate
this factor, this factor.
This question mark factor in the most general case.
So that, we will now obtain some universal formula which
will service both this example, this transformation, as well
as that example of polar coordinates.
And, it will give us, in this case, the distortion factor of
1/2 cop, and it will give us in this case, the distortion
factor of r, which you have already calculated before.
So, how to find the distortion factor in general.
Well for this, we simply have to compare the areas just the
way I have compared here, but locally.
Instead of globally, you just look at the areas on a very
small scale, in a small neighborhood of a point.
In other words, instead of this whole thing, you know,
comparing the area of the whole thing, and this whole thing,
and dividing, you want to take the area here where the sides
are some delta u and delta v.
So, this area would be delta u times delta v, and you want to
compare that to the area which you have in the real world.
Which, would be the are of the little piece of this region,
corresponding to this little rectangle.
So, let's draw this general picture.
So we will have a small rectangle here from some u,0 to
u,0 plus delta u, and here from v,0 to v,0 plus delta v.
Then we'll take delta u and delta v to 0, so,
as we did before.
And, let's look the image of this, of this guy under
this transformation.
And, it can become bigger or it can become smaller, depending
on what a transformation is.
So, I will draw it as if it becomes bigger, just
so that it is easier
to draw.
But, in principle, it doesn't have to become bigger,
it could become smaller.
So, in general, this rectangle is not going to
be a rectangle, right.
Because we know that for example, in a polar coordinate
system a rectangle goes to the annulus or part of the annulus.
So, there is no reason to expect that even a small
rectangle will go to rectangle.
It just distorts it in a sort of way.
And, so it will become, it will sort of be close to a
rectangle, but it will not be quite as same as a rectangle.
So, here's what it's going to look like in general.
Where, what do I mean by this picture.
I mean that say this side goes to this side, right, this
side goes to this side.
Kind of so distinguish between them.
And, then let's use a different color for the others sides,
for the verticalsides.
So let's say this one will go here, and this
one will go here.
So, that's what happens to this rectangle under
this transformation.
What we need to do, is to calculate the area of this guy.
We need to calculate this delta A, the area of this guy.
The area here we know.
It's delta u times delta v.
We'll have to calculate the this area as delta u times
delta v times the factor.

And, that's exactly what we need.
This will be the guy which will put the factor which we'll
put in the integral.
So, how to do this.
Well, here again we use one of the methods, which we've
developed for, namely cross product.
We know that areas can be calculated by
using cross product.
So, what we are going to do is we actually going
to approximate this.
I mean there's no way for us to calculate this area exactly.
What we'll do is we will shift it slightly , we'll deform it
slightly to make it into a parallelogram by
taking the tangent.
By taking the tangent vectors here.
So, the old idea again, is at work.
Which is that you're -- on a very small scale -- and we are
at a very small scale, Which is the point -- on a very small
scale, you can actually approximate smooth curves
by their tangent lines.
And, so the areas will be very well approximated by the areas
enclosed by those tangent lines.
So, in the case at hand, it simply means that instead of
this area of this curvy region, I will take the area
of this rectangle, OK.

So, what is the area of this rectangle.
So, delta A will approximate by the area of the rectangle.

And, I don't want to get into too much detail here, so I will
just give you the answer.
If you want to trace all the steps, you
can look in the book.
It's explained in great detail.
Basically, you need, once you talk about tangents, you
talk about derivatives.
So what you need to do, is you need to take derivatives
of this functions.
You have x is g of yv and y is h of uv, And, so what you'll do
is you would have r of uv which will be g of yv which will be
like x times i plus y times j, which is g of uv, i
plus h of uv, j.
And, this area of the parallelogram will be
approximately equal to following: will be cross
product, will be the absolute value, the norm, the length of
the cross product between r sub u of u0, u0 times delta u and r
sub v of u0, v0 times delta v.
So, why it's exactly this, I will skip, because I will
not have time to finish.
But, I think you can see clearly the basic idea,
something we've learned before.
And, how do you find tangent.
You find tangents by taking derivatives.
And, so that's, since I have replaced my rectangle, my curvy
kind of curved rectangle by this straight, sorry my curved
parallelogram by straight parallelogram, I used tangent
lines, because I used tangent lines, this vectors can be
expressed in terms of partial derivatives of r with
respect to u and v.
So, this is the first vector, this is the second vector, and
I take the cross product.
So, if you want to know more on how to get this formula,
just look in the book.
It's a very simple calculation.
But, once you get this, you can calculate very quickly the
cross product by using, you know, the methods,
which we've developed.
You have this ijk and then you're going to have t, dg, du,
dh, du, and then you'll have dg, dv, dh, dv and
you'll have 00.
And, you'll have to take the absolute value of this.

So, you see the only contribution will come
from k, and it will be this determinant.
So, it's dg, du, this guy times this guy, dh, dv
minus dh, du, dg, dv.
Right.
So, you see you take, you gave g and h, right.
You have this formulas.
So, you kind of know from the beginning, that this distortion
factor will be given by some sort of derivative or some
combinationof derivates.
So, it's sort of natural, because, to think in
the following way.
You've got two functions and two variables.
So, you have four possible derivatives you can take.
Right.
You can take two partial derivative of this function
two partial derivative of this function.
And, then you have to assemble them in some way into
a meaningful quantity.
And, that's how we assemble them.
We assemble them into this 2 by 2 matrix, and we
take its determinant.
So, it's really a combination of those four partial
derivative in this very special combination.
And what I claim, is that that's exactly the factor.
So, the factor is, so the distortion factor, which I
denoted by this question mark, equal to this.

Well let me write it, it's a little bit nicer to actually
write like this: dg, du, but g is like x, right, so I can
write dx, du, and then h is like y, right, so I have x is g
of uv and y is h of uv -- so it's dx, du, dy,
du dx, dv, dy dv.
So, you kind of assemble them in this way, where this is,
like x goes along the columns, the first column, y along the
second columns, u is in the first row and v is
in the second row.
You can actually transpose these two factors.
If you want you can switch rows and columns.
It's the same.
So in the book actually I think it's written in the other
way, but doesn't matter.
We will actually have a notation for this, it's like
taking d of xy over d of uv, and this is called a Jacobian
of this coordinate change, OK.
And, that's the factor which you need to introduce.
That's the distortion factor, because that's the area, you
see, the area of this parallelogram, which
approximates the physical area of the image of this guy, of
this rectangle in imaginary space is equal to this factor
times delta u, delta v.
Whereas, here you have delta u, delta v.
So, what's the difference, what's the ratio
between these two.
The ratio is this factor.
And, we find this factor by using cross product, because
that's how you find areas of parallelograms.
So, you can appreciate again, as a nice technical tool, and
we end up with this formula.
So, let's see if indeed we're going to get what is expected,
what we had expected on different grounds, like a
different grounds, in this two cases which we have discussed,
the polar coordinates and this problem which we talked about,
the integral we talked about.
So, in the case of, I am, made a slight mistake here, which
I forgot to write delta u, delta v.
Because, you had the factor of delta u and
the factor of delta v.
So, I just pull them out of the cross product, so
they will appear here.
OK.
So, let's check.
Suppose you have r and theta, and you have g is r cosine
theta, and h is r sine theta.
So, we have to take this derivative with
respect to r, right.
So we've got cosine theta, so then we take with respect to
theta so its minus r sine theta, and then we've got here,
derivatives of this function with respect r, that's sine
theta, and then derivative of this function with
respect to theta.
So that's r cosine theta.
So, we take this, this, minus this, this, so that's r times
cosine squared theta times sine squared theta,
which is r as expected.
That's the distortion factor that we had found earlier for
the polar coordinate system.
And, finally let's calculate other one.
So, here we have x and y are given by this formulas, so what
we need to do is we need to take the derivative of this,
the first function with respect to u.
So, it's 1/2, with respect to v, it's 1/2, and the second
function with respect to u is minus 1/2, with
respect to v is 1/2.

You calculate, 1/4 plus 1/4 is 1/2 as promised, right.
So, there is one important point which I want to
make, and we will stop.
Which is that, sometimes you will get a negative
number, so you always take the absolute value.
The point is that we are not keeping track of orientation
here, which we should have.
In other words, you could switch u and v for example, and
this will make the two rows switch, and if switch the rows,
you'll get a negative sign.
So, in fact, if you are careless, which we are kind of
careless, you might get a positive or negative answer.
So, the cure for this is, you always put absolute value.
That's the formula, the correct formula is always put up
absolute value at the end, so that the result is positive.
All right, so I'll see you on Tuesday.