Uploaded by UCBerkeley on 18.11.2009

Transcript:

OK.

How are you guys?

Good?

Did you have a good week?

I'm good, yeah.

So, but it looks like there are fewer of you.

I hope we haven't lost anyone, or maybe it's just the subject

is too easy for some people.

Well, we also have this wonderful resource.

So, for those watching online, welcome.

But do come to visit us sometimes.

All right, so we are talking about integrals, right.

And I know that last week we discussed double integrals and

triple integrals, and for double integrals you talked

about integration using polar coordinates.

And so this is actually a very -- it's a very useful tool for

integration that sometimes thing simplify when you use

-- some interesting stuff on the floor.

When you use a different coordinate system -- we saw

that when we talked about as early as like first or

second week of the class.

Are you guys lacking discipline after not having me for a week?

I hope not.

So, remember that I'm really strict, but fair.

All right.

So, already at the very beginning we saw that many --

our solutions simplify when we use polar coordinates, for

example, in describing curves on the plane.

And you saw last week that integration also simplifies,

double integrals simplify if you use polar coordinates.

So now the next question is what about triple integrals?

Is there an analog of polar coordinate system for triple

integrals and how can we use such a coordinate system to

effectively evaluate triple integrals.

And in fact, because now we are in three dimension, there is

more than one choice to make, there are more options, and

there are two particular coordinate systems, which are

very convenient in three dimension, which are called

cylindrical and spherical coordinate system.

So we will dicuss them in turn.

First we'll talk about cylindrical.

Cylindrical coordinate system.

And cylindrical coordinate system is really just an

offshot of the polar coordinate system in two dimensions.

In a way we don't introduce anything new in triple

integrals, we use the same tool, which we already use

effectively in two dimension.

So the way it works is like this.

The usual coordinate system, xyz is replaced by another

coordinate system where on the xy plane you use polar

coordinates on the xy plane.

And we just add the z variable , the way we

just added z to x and y.

So the result is that instead of the xyz coordinate system,

which we normally use, we pass through the coordinate

system, which

will have r, theta and z.

So x and y it get replaced by r and theta in the same way as

for polar coordinates and z is just thrown in as

an extra variable.

So the formula's expressing this coordinates in terms of

this new coordinates are exactly the same as for polars.

They are r cosine theta and r sine theta, and then we have

this z, and so z we have on both sides and it's

the same variable.

OK?

So what is this coordinate system good for?

This coordinate system is good for describing object

which are cylindrical.

Well, hence the name.

What do I mean by cylindrical?

A cylinder in general is something which comes from an

object on the xy plane by kind of sweeping a surface by using

that let's say a curve on the xy plane.

So a cylinder itself, a cylinder itself is like that.

We take a circle and if we just move it up and down vertically,

we sweep a surface and that is what we normally

call a cylinder.

So a circle, the equation of the circle on the plane

simplifies in polar coordinates, it's r equal r

capital where this is a number, and this is one of the

variables, one of the coordinates that's in the polar

coordinate system, and this number is just the

radius of the circle.

So in

other words it's a circle over radius r.

So when we move it up and down like this along the z-axis, we

create the surface in which the equation is the same in 3D, the

same equation means that we also have an arbitrary value of

z, and therefore, we get the cylinder, the cylinder

of radius r.

And so just like in integrating on the plane in double

integrals, it is beneficial to use polar coordinates in

describing things related to the circle or to the disk or

sectors of the disk or annuli and things like that.

It is beneficial to use the cylindrical coordinate system

for triple integrals in describing things -- in

integrating things related to cylinders and various things

that you can obtain from the cylinder.

So what does the triple integral look like in the

cylindrical coordinate?

There is an important point which we should not forget,

which is that when we pass to a new coordinate system, there is

some advantage which is that some of the formala simplify.

But there is a sort of a price to pay for it, which is that we

have to insert an additional factor in the integral.

So for polar coordinates, something that was discussed

last week: if you have an integral of a function f to the

a, you can write it as double integral in polar coordinates,

and have if of r and theta, but then instead of simply putting

the rd theta, which you would normally put, you insert an

additional factor, which is r.

So you get rdrd theta.

When instead of the usual dfdy, which we would have in the

polar coordinate, in the normal coordinate system.

What is the reason for this, just to go over it one more

time because we will now see how it works for cylindrical

and then for spherical coordinates.

Everything boils down to the area of an elementary object

with respect to this coordinate system.

An elementary object is a rectangle, which you obtain by

saying that x is -- in this case two coordinates, x and y,

x between some fixed value for x zero, and x zero plus delta

x, and y is between some fixed value and that value plus

some increment delta y.

For the xy coordinates for that rectangular coordinate system,

this inequalities describe a rectangle in which one of the

sides is delta x and the other side is delta y.

And so the area of this rectangle is simply

delta x times delta y.

And that is what gives us the expression for dadxdy.

And so it leads to a description of a double

integral as an iterated integral in x and then y

or first in y and then x, like Fubini's theorem.

But the key point is the area of the elementary rectangle,

which is given by this formula.

Now, when we pass to polar coordinates, the analog of

this elementary rectangle is the following.

We have to look at all points which are given

by this inequality.

Say r between r zero and r zero plus delta r, and theta is

between some fixed value theta zero, and theta zero

plus delta theta.

And the picture will be different, right because this

gives us a sector of angle delta theta, and then the

condition for r to be between r zero and r zero plus delta r,

let's assume that this is our zero, right, just like I am

assuming that this is theta zero.

Of course, in this calculation, eventually you would like delta

r and delta theta to become very small, whereas r zero and

theta zero are fixed, but I'm just drawing this picture.

I can magnify everything, so it looks like delta theta looks

like sort of the same magnitude as theta zero, but in fact

it should be much smaller.

And so then if this is delta r, and this is confined within

delta theta, and the elementary object instead of this one, we

get a kind of a -- it sort of looks like a rectangle but not

exactly because first of all there's a certain angle here

and these are not straight lines, but these are segments

of a circle, right.

So what's the area of this?

The area of this one -- actually we can not -- I mean

it will be -- we can but it will be different -- it's not

so simple to calculate, to give exact answer, but because if

I'm going to take the limit anyway, eventually, we only

need a sort of a good approximation to this.

And what's a good approximation to this?

We can think of this -- we can think that when delta r and

delta theta are very small, this will look like in a

rectangle with a side delta r, and what's the other side?

Well, this actually has length, r zero times delta theta.

The segment of a circle of radius r zero confined within

the angle delta theta is given by this formula, r zero

times delta theta.

And so we approximate this by the area of the rectangle with

the sides delta r and r zero times delta theta, and the

result of this is just the product of this and we get r

zero delta theta delta r.

And so that's where -- that's the r which shows up in the

formula for the integral.

Here I denote it by r zero, because I wanted to emphasize

that for calculating this particular area, I fix r zero,

I fix this length, but when we do a general calculation, this

will just become r, and delta theta will get replaced by d

theta and delta r by dr in the limit, right.

And so that's how we end up with this formula.

So this only sort of non trivial thing to remember

about polar coordinates.

Don't forget to put this factor r, which actually

could be a good thing.

I made it sound like it's a price to pay that we are being

taxed for sort of for the convenience of using these

coordinates, but actually sometimes, and oftentimes, this

could be a good thing, and the typical example is let's say

you have an integral like this, something like 1 minus x

squared minus y squared dfdy.

So you see this is complicated, right, this is actually

something we discussed, it's kind of similar to something

we discussed earlier.

This will be very difficult to take just as an integral in x

and y, but when you pass two polar coordinates this will

become square -- I'm not writing the limit, but

I'm just sort of giving a rough sketch here.

This becomes 1 minus r squared.

And now we get this extra factor r, and then

we get drd theta.

And so you see, this actually becomes a much better integral

because I could introduce a new variable, let's say t, which is

r squared, and if I do that then dt becomes two times rdr.

And then this actually gives me 1/2 of dt.

And this becomes a square root of 1 minus t instead of 1 minus

r squared, and this is much easier to calculate, because

the anti-derivative of this is just 1 minus t to the power

3/2 times 2/3, minus 2/3.

Whereas for this one it's much more complicated.

Well, imagine if it were like exponential function or

something, so then it becomes even worse, right.

So sometimes actually having this factor actually

works to our advantage.

But in any case we have to remember to insert it to

get the right answer.

Now, so this was all about the polar coordinate, but

cylindrical are not that far away, not that far apart,

because for cylindrical, the only thing that changes is that

we now throw in an extra variable.

Namely variable z, so we add the variable z, and now instead

of a double integral we have a triple integral over some

solid, some region in the three dimensional space, and so we

have some fda where this da, usually in the normal, in the

Cartesian coordinate system, we simply write as dxdydz.

But now if we want to use polar coordinate system, we'll have

to write this function as a function of r theta and z, and

then we'll have to write drd theta dz, and the question is

what should we put in front?

There has to be some factor, and that factor is the factor

responsible for the area of the elementary domain like this.

In the usual coordinate system in the three dimensional space

in the -- when I say usual I mean Cartesian

coordinate system.

The elementary object is a box of sides delta x,

delta y, and delta z.

So it's -- actually no one corrected me.

What is it mistake here? dv, that's right.

See, I'm a little rusty after break, but do correct me

when I make these mistakes.

So dv, because we emphasize that this is elementary

volume, its volume, it's not area anymore.

The a, of course, was for area and v is for volume.

So the elementary volume for this guy is just delta x, delta

y, delta z, which gives rise to dxdydz in the integral and

allows us to calculate this integral as it iterates

integral first x then y then z or any other order

if you choose.

But now we choose this coordinate system and so we

have to be careful, we have to now look at the elementary

object in this coordinate system, and what is this

elementary object?

Well, we just define it in the same way as before by simply --

I should have done it this way.

I much be jumping ahead.

Let's just do it like this.

It really should be cylindrical, so it

should be like this.

So it is, in a way, it's a kind of a -- it's what you get by

combining the two things.

In this phase, this side of this object looks like this.

It's--.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Why is it called cylindrical?

This is a cylinder, right.

He saved you, by the way, because I was kind of

approaching to count how many seconds it will take for you to

notice me, but he saved you so you should thank him.

So, cylindrical is called because you see the, you make a

change on the xy plane and then we just throw in

the z variable.

In other words, you change the coordinates just on x plane and

then you kind of let the third variable the same as before.

So it's the same idea as the idea for say taking the circle

and kind of just moving it up and down to get this object.

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: No, no, the circle is fixed, right,

it doesn't grow.

The radius of the circle is fixed, it's this r capital.

Just think of a ring which is made of metal or something and

you just move it up and down vertically along the --

parallel to the z-axis.

right.

So what's the result?

You sweep a surface, which is the surface of the cylinder.

You can only use it -- no, you don't use it only for cylinder,

but, well, what's in a name, you know.

It's a name, so it is derived from this, right, but it's not

supposed to explain everything we can do with this

coordinate system.

Any other questions?

So, what's the elementary volume of this?

Well, we already know the approximate area of this that

rdr, r delta r delta theta, and we know that this is delta z.

And so the volume is rdr delta -- sorry, r delta

r delta theta delta z.

So it's the same r which we had -- or maybe if you want r zero,

if you want this to be r zero.

The same r which we had in the calculation of the

polar coordinates.

So the bottom line is this factor is exactly the same as

in the polar coordinate system.

It's not more complicated.

So that's the formula we're

going to use.

And now let's do an example.

Let's do an example.

So, evaluate triple integral over e of the function e to the

z where e is enclosed by the paraboloid, z equals 1 plus

x squared plus y squared.

The cylinder x squared plus y squared equals

5, and the xy plane.

So, actually let me get -- let me draw a bigger picture.

So the paraboloid, what does a paraboloid look like?

It starts at the point 1 on the z-axis and it opens up like

this, and the cylinder is OK, I just erased it.

So that's a cylinder, let's assume that this is radius.

This is going to be radius square with a 5, right, because

we have 5 is the square of the radius, as always.

So the radius itself is square of the 5, and so that's a

circle at the base of the cylinder, and we want to figure

which is a solid, which is confined by this by the

paraboloid, by the cylinder and by the xy plane.

So it is the inside of this, so it's kind of concave, the

surface kind of concave, it goes inside.

It's like a fancy glass.

So what we need to do is to calculate the integral over

this -- this is our region, e, of the function e to the z.

So, when you do triple integrals, you have to --

so then you have to write as an iterated integral.

So you have to choose in which order you're

going to integrate.

So you're going to end up with an integral, so you have e to

the z, here actually z is one of the variables, so we just

write it like this if this is good, right, because these one

of the variables in the new cylindrical system.

And then you're going to have the r -- let me emphasize this

r again, that's the same r as before -- and then we

have rdrd theta and dz.

So now you want to write it as an iterated integral, so you

have to choose in which order to integrate.

So what's the -- what's the best way to do it?

For the cylindrical coordinate system it's always -- you see

the point is that you have to choose what the base of this --

actually this is a general approach to when you have

to choose the order.

You have to see whether, what's a good projection of your

object -- is there a good projection onto say xy plane or

yz plane or zx plane -- that's the first question, and once

you have a good projection, everything projects onto

something nicely, right, so in this particular case, it

certainly does projejct nicely onto the xy plane, it projects

just onto the disk, onto z which is r less than or

equal to square root of 5.

The disk of radius square root of 5.

So that's this, that's this yellow -- let's

draw it with yellow.

And then for each point in this disk, we'll have to integrate,

so first of all, we'll have to integrate over the disk, and

this will be the -- I'm talking about the outer integration,

I'm starting from outside.

When you project onto this, you are choosing the first two

variables, which normally if we were using x and y and z

coordinate, that would be integrating first x and y.

Then you decide, you decide between x and y later, but

first we choose the first two, and then you choose which

one goes first, right.

So that's a strategy which I propose that you're going

to end up with three different integrations.

So the last one will be the integration over the remaining

variable with respect to this projection.

So here I project onto the xy plane, so this will be dz.

But let's now see what happens here.

So this is going to be in the usual coordinate system, in the

Cartesian coordinates system, this would have been just dxdy

which you would have to decide, dxdy or dydx.

And once you decide, then in the last integration you have

the freedom to choose what x and y are, so that means that

you have a point here, xy inside, and then the limits

will be -- the limits you have to sort of draw the segment

starting from the bottom of this, of the object to the top

of this object, which in this case from this disk, which is

part of the xy plane to this paraboloid, and you would have

to integrate, in the last integration you would have to

integrate from the bottom value to the top value, which in

this particular case what does it mean?

The z goes from zero, right, to this value.

But what is this value?

This value's is going to be 1 plus this is going to be the

value of z on the paraboloid.

So that's -- in the Cartesian coordinate system, this

would have been 1 plus x squared plus y squared.

But now we're replacing this with r squared.

So, because we are doing the integral in the polar

coordinate system, we will actually write it as 1 plus r

squared where it is understood that r squared comprises x

squared and y squared -- the sum of x squared and y squared.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Well, up and down because I have chosen the

projection onto the xy plane.

Right, so the way I suggest to do the integral

is the following.

First let's choose the two outer integrations, the first

and the second, right.

So geometrically it means that you are projecting your three

dimensional region, three dimensional solid, onto one of

the planes, one of the coordinate planes.

I'm talking now about really the Cartesian coordinate

system, but cylindrical coordinate is not that far away

from Cartesian, so the same analysis sort of applies here.

So the projection here is going to be the disk, right, but in

general it could be square, it could be something else.

So that will be taken care of by these two integrations,

by the limits here.

STUDENT:

PROFESSOR: That's right.

The projection goes in those two.

And the inner integral is the remaining variable,

which in this case is z.

So to put the limit here, you have to pick one of the points

in the projection, in

the image of the projection, which is this point, say xy,

which actually because we are doing cylindrical coordinates,

this would have to be recorded as r theta.

So this point will be r theta.

And we'll have to take the segment along the third

variable, which is the z variable going from the bottom

to the top, from sort of the bottom lid to the top lid of

this figure, of this region.

Which, as I explained, the bottom is on the xy plane, so

the z is equal to zero at the bottom, and at the top, we have

to look at the equation of the paraboloid, it's 1 plus x

squared plus y squared, which because we are working with

cylindrical coordinates will be written as 1 plus r squared.

Let me finish the integral and then you ask me if

you have more questions.

So then what remains to be done?

What remains to be done is that we have to put also here r and

theta, so I have to choose some order in which we

will do r and theta.

So let's say we put here dr and put here d theta, because now

we are actually, see, once we are here in these two outer

integrals, we're actually doing a double integral.

We are deciding how to split the double integral

into iterated integral.

So, and what we are working with is already the image of

this three dimensional object onto the xy plane, which

is nothing but the disk.

So for the disk we know what the limits are -- first of all,

we know that it doesn't matter in which order to take r and

theta, and second of all we know that r goes from zero to

the radius, which is in the case square root of 5, and

theta goes from zero to 2 pi.

That's because the disk, this disk described by -- well,

I don't even have to write one more time.

I just have to add here theta from zero to 2 pi.

And finally we should not forget to put the function,

which is e to the z, and we should not forget to

put this factor r.

So I'm introducing a slightly different notation from before.

Normally we would write this as e to the z times r times dr d

theta and then we put sort of brackets and we think about

integrating first this with respect to r, and of this with

respect to theta -- sorry, integrating this with respect

to z then this with respect to theta and this with

respect to r.

But I'm writing it in this way, which I

think is a little bit more suggestive that you write the

dr, the d theta, dz, the d of the variable you're integrating

right next to the integral, so you remember which one it is.

You see because in the old way you would have to

put dr the last one.

So the last one will get paired with the first integral, the

next to last will get paired with the second

integral and so on.

This way it's a little bit more intuitive, so I prefer to write

in this way, but you can choose whichever way you like.

But I hope that it's clear what I mean.

Well, this certainly is unclear because it says onto and then

it says zero, so let me write it here.

I think now it should be clear.

OK, any questions about this?

No?

Yes?

[UNINTELLIGIBLE]

So, in fact, I could have put r all the way here.

[UNINTELLIGIBLE]

Oh, yeah, yeah, yeah, you're right.

[UNINTELLIGIBLE]

Right.

So I actually did it in the wrong way, in

the wrong quarter.

So--.

Did I do it in the wrong quarter?

Does the order matter?

No, the order, of course, matters, because see the

point is that -- no, no I did it correctly.

Who thinks that I did it correctly?

Who thinks that I didn't do it correctly?

OK, you guys, you fail the course, no.

We are here to find the truth.

But see here's the way -- I thought about it this

afternoon, actually, and I convinced myself this is right.

But see here's the point.

The point is--.

[UNINTELLIGIBLE]

Right.

So the point is that we are -- let me -- this is actually a

very -- it could be very confusing.

So let's actually do it slowly.

You should not worry so much about the fact that here

there is a dependence on r in the inner integral.

What you should worry about is the fact that there is a

dependence on r in the limit.

So you certainly don't want to put this outside.

Because you want to end up with a number.

In other words, we can disagree on many things, but there is

one thing we should certainly agree on, which is that

the integral is a number.

It should not depend on r, it should not depend on theta,

it should not depend on anything, it's a number.

You can not possibly get a number if the last integration

as the limit which depends on the variable.

That's not going the work.

But more conceptually, the way I'd like to think

about this is as follows.

And I would like to think of maybe in a slightly -- you

know, it could be some more complicated domain.

So what I'm doing is I'm projecting --

so there is some e.

Let's say I want to project it onto the xy plane.

But right now let's talk about just the Cartesian coordinate

system, let's not worry for now about cylindrical or spherical.

Even in the Cartesian coordinate system, the way I

would like to think about integration is as follows.

That I have a region in a three dimensional space, right, and

what I do is I project it onto the xy variable.

So now if I have a triple integral over e where I

have f dv, first of all I split it like this.

It's going to be a double integral over z, over the z.

Of the single integral with respect to the

remaining variable, dz.

You see where for each x and y in d, you will have some lower

limits, let's say l of xy, and you have some upper

limits, u of xy.

So that's what it's going to look like.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: So then I would just write, and here,

yeah, that's right.

So then here I would put da, like this.

Yes.

[UNINTELLIGIBLE]

That's right.

In the old way.

Right, so you want me to write this in the old way.

So this in the old way would be zero to square to 5, zero to 2

pi, zero to 1 plus r squared, e to the z, dz, d theta, dr.

Oh,

I'm sorry, rdr -- yes, very good, thank you.

See because it's red so I don't, I didn't see it. rdr --

well, usually we'd write rdr -- yeah, rdr, OK, fine, whatever.

We can put r inside here, maybe it's better, kind

of keep it up over here.

So then normally we would put brackets like this.

Is that OK?

Does it make sense?

So what I'm doing is just I'm trying to avoid using the

brackets and instead jut putting the differentials right

next to the integral, so it becomes a little bit more clear

to me anyway, but you are free to use whichever way you like.

So this is sort of a new way to write, and the old way to write

would be f dz -- well, it's almost the same, za over

-- maybe like this.

So but the main point which I would like to make is that the

first two integrations in the formula -- see, it's very

confusing because when I say the first integration, do I

mean the first integral or do I mean the actual

integration you perform.

This is opposite, right?

The first integral you write is the last one, it's the last

integral you write is the first you calculate.

That's right.

That's right, so this is the outer, the outer is

written first, so that's why it goes last.

So you have to figure this out on your own.

I guess at some point you should have to think about

it and have a clear picture because like I said

it's very confusing.

What do you call first, what do you call second.

Who's on first and, right.

So the last integration, is outer one, and the last

integration corresponds to the projection of your three

dimensional region onto the plane.

OK, That's the point.

So, having established that let's evaluate finally

this integral.

So what do we get?

We get zero square to 5, 2 pi, and then here we get r times e

to the 1 plus r squared minus 1, and you get dr -- this

integration just gives you 2 pi because the integrand does not

depend on theta, so you just get 2 pi times r e to the

1 plus r squared minus 1.

And finally you integrate over r, and so here you want to use

change variables, so I'll just write the answers so you've got

2 pi as an overall factor, and then you get 1/2 e to the 1

plus r squared minus 1/2 r squared between zero and square

root of 5, and so the answer is 2 pi times e to the 6 minus e

minus 5. 1/2 -- well, because 1/2 and 2 get canceled, so

you've got e to the 6 minus e and this guys gives you 5.

So that's correct.

OK?

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: It's pi that's right, so -- thanks.

OK, cool.

So we good with this?

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Is it universally accepted that--?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Are you talking about this formula

or this formula?

[UNINTELLIGIBLE]

You mean this integral?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: But I'm not sure I understand

what the question is.

Which line, first of all, this one or--?

This one, yes.

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Are you talking about the left hand side

or the right hand side?

The question is it OK to put dr before the function of r.

OK.

The answer is yes, it is OK.

I am suggesting this notation.

In the book it's not used, right, but as a practicing

mathematician I can assure you that a lot of people use that.

It is certainly interchangeable.

They do commute, in other words, it doesn't matter in

which order you write them.

It's a more interesting point whether this guy's commute, you

know, whether it is important how you write drd theta or d

theta dr, this is a much more subtle point, which I'm going

to get into too much, but certainly dr you can put on

the left or on the right of the function of r.

So no worries here.

Let's move on, so the next one is called spherical coordinates

and it's a little bit more interesting.

Sorry?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Spherical coordinates, and, of course,

the question is right away why are they called the spherical

coordinates, because they should remind you of a

sphere, as you will see.

So, let me actually keep this picture.

So, spherical coordinates work in the following way.

Suppose you want to invent a coordinate system in which

the sphere -- it is a sphere which gets the simplest

possible equation.

So, in the cylindrical coordinate system, it is a

circle that gets the -- say in a polar coordinate system, is a

circle which gets the simplest equation, r equals a constant.

That's the circle of a given radius.

In a cylindrical coordinate system, the simplest equation

is the equation for the cylinder, OK? r equals -- r is

a circle in polar coordinates because r is a cylinder in the

cylindrical coordinate system, in 3D, this is in 2D.

And now I want to have some variable rho such that this

equation will give me a sphere, and that's called the spherical

coordinate system and that's in 3D.

So that means that this rho should really be the distance

from the origin of my coordinate system to my point,

because a sphere -- well, sphere with the center

at the origin.

The sphere with a center at the origin of radius r capital,

this one, is going to be the set of all points such that the

distance from the origin to my point is given by this number

r, let's say square root of 5 like in the previous example.

So, note that this is not the same as r in the cylindrical

coordinate system, because r in the cylindrical coordinate

system is not the distance from the origin to this, but rather

it's the distance from the origin to the projection of my

point onto the xy plane, so that's not the same thing.

So you see right away that this is not the same coordinate

system, and sure enough, this simplest equation rho equals a

constant describes not a cylinder but a sphere.

But now, so now I got the first variable of my spherical

coordinate system, but it's not enough, I need three variables,

right, because I'm in three dimensional space.

So I need to complete this rho by two additional variables,

two additional degrees of freedom, if you will, so that I

could give each point a unique address by using those

three coordinates.

What are they?

So the standard -- there are different ways actually to do

it, but the standard convention which we are using in this

class and in this book is the following, that we

measure two angles.

In addition to rho we measure two angles, and the first

angle is exactly the same as before, theta.

It's the theta of the cylindrical or polar

coordinate systems.

So we need one more, and we find this one more by

measuring this angle, and we call this phi.

So the spherical coordinates in a three coordinates rho, phi

and theta where only one of them is part of the cylindrical

coordinate system, which theta is part of the cylindrical

coordinate system.

But the others are new.

So, by the way, notationally phi sometimes is also,

is written like this, it's the same thing.

So I find it easier to write phi like this, but when you

type, for example, and in the book it's written like this.

So you can use whichever notation you like.

So a row of phi and theta.

The first step, of course, is to express the usual

coordinates, the Cartesian coordinates in terms of a row

of phi and theta and to see that indeed these coordinates

are -- determine the points in space.

And also to see what the ranges are.

So what is the formula?

Well, what we can do is we can use the old formula, which is r

cosine theta, and then note that r is equal to rho times --

this is equal to rho times the sine of this angle.

See, the point is that this has the right angle.

This is a triangle -- this is a triangle which has a right

angle, and this is phi, so you can find this distance, which

is r of the cylindrical coordinate system, by taking

rho and multiplying by the sine of this angle.

So this is the row times sine of phi.

Now I substitute this in here and I get rho sine

phi equals sine theta.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Which axis is phi measured from?

It's measured from z.

So, I did not put labels, but I should, it's

xyz, the standard way.

Another question.

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: The rho does not belong to any plane of priori,

it's in space somewhere.

It's just that -- I have a point--

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: kind of, it flows around like

this, wiggles around.

It's not doesn't belong to any particular plane.

It's free to move, right, because our point is free to

movein free space, and so the segment connecting our point

to the origin it doesn't not necessarily belong

to any plane.

Which one?

This triangle?

It has a right angle here, because this is a projection.

I don't know if it looks like a projection -- it looks a

little bit, it's a little bit crooked I guess.

It should be a little bit more parallel to the z plane than

what I made it look like.

Is that better?

Slightly.

So this is a perpendicular, which we drop from this

point or our point, this is our point.

We drop it onto the xy plane.

So, I guess the rest should be clear.

Any other questions?

Yes?

Is theta always taken from the x-axis?

That's right and theta always taken from the x-axis

for the projection.

Everything is a little bit not straight on this picture.

This is r -- this is old r, which is not -- this r is

not part of the spherical coordinate system, it's

part of a cylindrical coordinate system.

The reason I have drawn it, I have written it here, put this

label is to connect, to make a connection between the

spherical and the cylindrical coordinates, and also to

simplify the derivation of the formula for xy and z.

Because I do it in two steps.

I first recall the formula for the cylindrical coordinates,

which is really the same as for polar coordinates, and

then I just substitute r equals row sine phi.

And then I do the same for the y variable, and

then z is, what is z?

To find z, we have to complete this to a rectangle--.

I'm sorry?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: It should be sine theta, that's right.

Thank you.

And to find z I have to use another right triangle, which

is this one -- I guess again I didn't draw it very well

-- maybe more like this.

So this is right triangle, this is phi, so z is

rho times cosine phi.

So, if you visualize -- if you visualize this picture, you see

that we don't have to memorize or write on your cheat sheet

the formulas for the cylindrical coordinates,

it's very easy.

You have to remember the polar coordinates, of course, but I

mean, if you don't remember polar coordinates, then it's

like remembering your phone number in some sense I think.

So, if you remember polar coordinates then, then the

spherical coordinates are

very easy by memorizing this picture or visualizing

this picture.

OK, what are the ranges?

So, a row is like r, rho is a non-negative number, although

for r if you remember, we had certain convention in

polar coordinates.

We sometimes allowed r to be negative, but in spherical

coordinate system, that would make things too complicated

so we don't do that.

So rho is non-negative.

Theta is just like in a polar coordinate system or in

the cylindrical, it's betweenm zero and 2 pi.

And what about phi?

What are the ranges for phi?

Zero and pi, exactly, because you can go, so if this is a

vertical the z-axis, you can go all the way down to pi, but if

you go more than pi, then you can sort of approach it from a

different direction and it will be less, from that direction

will be less than pi.

So that's what we do.

So it's going to be from zero to pi.

Do you see what I mean?

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: That's right, from the vertical.

So, in other words, what I'm saying is let's say, so this is

z-axis and this is the origin, so you have a -- let's suppose

the our point actualy lives on the yz plane.

So, and let's start rotating it.

So this is phi for this point.

This is phi for this point, this is phi for this point.

Still less than pi, right.

So we get here, it's still -- what I'm trying to say is this,

this, this, this, right -- we measure angle from here.

But finally when we go like this, we should not measure

it like this, but we should measure it this way.

You see what I mean?

So this is wrong and this is right.

This is phi.

So you choose whichever is shorter, whichever's smaller,

and that's going to always be between zero and pi.

So pi would be this -- this is pi, which is points on the

negative -- phi equals pi corresponds to point on the

negative part of the z-axis.

Do you have a question?

Now, what are these coordinates good for?

Well, the first answer has already been given.

In this coordinate system, it's much easier to

describe the sphere.

So the sphere is now given by the equation rho equals r,

instead of x squared plus y squared plus z squared equals

r squared of the Cartesian coordinate system, or r squared

plus z squared equals r squared of the cylindrical coordinates.

It's just the simplest possible equation, rho

equals the constant.

So that's, in a way, justification for using

this coordinate system and for the name.

But that's not the only object which has a nice representation

in this coordinate system.

We can also represent a cone by the equation phi equals

some fixed angle, phi zero.

So that's the cone.

Let's look at this picture again.

So let's suppose I fix some phi and I look at all the points

which have a given angle like this, phi zero, 6

to the z-axis.

So then it's like this is the z-axis and my

point is like this.

I don't know, I don't have enough visual aide, so I have

to trust your imagination and your intuition.

Maybe like this is better, I don't know.

So the point is that this is the angle, so the angle is

fixed so it has to be the same angle between your points

and the same z-axis.

So, the result is a cone.

And because I don't specify what row is, I don't specify

the distance, it can go as far away as I want, so it's an

infinite cone -- kind of an upper cone going to infinity,

so it's not bounded, unbounded cone.

So don't think that I just drew one level curve for it,

but it goes to inifinity.

So the cone

is given by the formula, which is again, much simpler than z

equals square root of x squared plus y squared of the Cartesian

coordinate system, or z equals r of the cylindrical coordinate

system, which is not so bad but this is better, this

is easier to handle.

So these are the things you should remember.

The cone and the sphere have a very nice expression in terms

of this coordinate system.

Now what about the integration?

What about integration in spherical coordinate system.

So once again, we have to figure out what is the volume

of the elementary object with respect to the spherical

coordinate system.

So these are the elementary for the Cartesian and the

cylindrical coordinate systems, which give us the usual

formulas, dxdydz and rdrd theta dz.

So now we have to do exactly the same for the spherical

coordinate system.

So what is the elementary object now?

And again, by elementary object, I mean simply that you

impose the condition that row is between some row zero and

row zero plus delta rho and -- I must have lost my

second variable phi.

Phi is between phi zero and phi zero plus delta phi, and theta

is between some theta zero and theta zero plus delta theta.

So let's draw this, an object which corresponds

to this inequality.

Like this.

This whole thing should be kind of dotted line.

OK, and like this and like this.

OK, you see, you see what I mean?

Well, in fact this -- do it like this, this may be better.

So here, what happens here?

First of all, this is delta phi, right, this is rho zero.

So this thing is actually between two spheres.

Imagine two spheres which have very close radii, very similar

radii, so this is something which is between those spheres,

and first of all, we cut it by saying that the angle goes

between, you know, it's confined within delta phi, and

also when we project this down onto the xy plane, this angle

-- I'm not going to draw it all the way -- so the projection's

going to be delta theta.

So this is a projection onto xy plane.

So now I have to evaluate the volume of this, just like I

have to evaluate the volume of this and of this in my

other coordinate systems.

So what is it like?

What is this volume equal to.

Well, we don't calclate exactly, we approximate it, and

we approximate it by the volume of the cube -- sorry, not the

cube, but a box -- a box whose sides are given by the lengths

of this kind of circular sements -- they are going to

be all circular segments.

So this one is very easy to find -- well, first of all,

this one is very easy to find.

This is just delta rho this is just delta row.

That's one side.

OK.

Now this one is also easy to find, this is -- you see this

is rho zero, and the angle here is delta phi, so this

is rho times delta phi.

It's similar to the way we found that to be r zero

times delta theta.

So it remains to find this one, which I guess you can not

see this color, right.

Actually it's better than the other one, it's

kind of greenish.

Can you see this one?

Oh, good.

So this one is a tricky one because you would be tempted

to take rho zero times delta theta, but this is not correct.

This is only one, there is only one subtle point here.

The point is that this is the same as this.

In other words, this actually is not, this is

not part of a circle of radius rho zero and confined within

the angle detla theta.

On the contrary, it is actually part of a circle on the xy

plane, but the radius is not -- the radius of that circle is

this, which is not rho zero, but it is r.

It is row zero times sine phi.

So that's the only point that you have to realize

to get the right answer.

And so -- and then you've got delta theta.

I'm trying to draw this like this.

This is on the xy plane.

And so this segment is actually, its length is

rho zero times sine phi delta theta.

So the answer is for the volume it's approximated by rho zero

sine phi delta theta times delta rho times delta times rho

zero again -- this was rho zero -- and times delta phi.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: I am trying to calculate the volume of this,

and I am approximating it by the volume of a box like this,

which has the following sides.

This side is rho zero sine phi delta theta, that's this, times

delta rho, which is this side, times rho zero times delta

phi is the yellow side.

So the end result is row zero squared sine phi times delta

rho delta phi delta theta.

So, the upshot of this is when you compute a triple integral

and spherical coordinates, you have to introduce the fact that

rho squared times sine phi.

So it's a little bit more complicated than

for cylindrical and polar coordinates.

And triple integral over some solid is rho phi theta times--

let's write it here-- rho squared sine phi -- well, since

I write it in red before, let me write in red again.

A row squared sine phi -- that's the factor to remember

-- zero d phi d theta.

OK?

So we do this calculation once and for all, then we just

memorize it, row square times sine phi, and after that, we

can do this triple integrals by using iterated integrals for

these three variables, rho phi and theta.

Let's see how it works in practice.

So here is an example.

Compute the integral of x squared plus y squared plus d

squared dz where e of the region is bounded by the xz

plane and the spheres. x squared plus y squared plus z

squared equals 9, and x squared plus y squared

plus z squared equals 16.

So within two spheres and also the xz plane.

So first of all, I have to say just a general comment.

On a -- next week we'll have a mid-term.

On the mid-term I will not tell you which coordinate system to

use, you have to decide youself.

So the question is you have to look for clues --

which coordinate system, spherical, cylindrical.

You have to remember one thing, that this expression, x squared

plus y squared plus z squared simplifies in the spherical

coordinate system.

So if you see x squared plus y squared plus z

squared how many times?

One, two, three times, chances are it's best to use, in

the case, spherical coordinate system.

You see what I mean?

Because this is rho squared, right, this is row squared.

If you see x squared plus y squared, that's r squared,

r being the r of the polar or cylindrical coordinate.

That's how you know that you probably should do, use polar

coordinates or cylindrical coordinate, depending whether

you use -- are doing a two dimentional, a double integral,

or a three dimensional triple integral.

But in this case it's fairly clear that you'd be better off

using spherical coordinates.

Now we can visualize this.

Think of a sphere, first of all.

Let me draw just a quarter of this picture.

Actually, yeah, about a quarter.

So that's a part of the sphere -- this is a sphere of radius

three, because 9 is 3 squared, right, and this is sphere of

radius 4 because 16 is 4 squared.

So let's say this is a sphere of radius three, and then this

is the inner sphere, and then there is an outer sphere

which has radius 4.

So see this one is inside the other, and what you are looking

for is -- what you want to look at is the three dimensional

shell which is in between the two -- in between

the two spheres.

It's like an [? analise ?]

but three dimensional. [? Analise ?]

between two circles and this is between two spheres.

And also you have, you bind -- the other bound is xz plane,

so this is the xz plane.

So actually that's not the whole thing.

This is one quarter of the whole thing, because you will

also have something in this quadrant and in those two

quadrants, which are behind the zy plane, so this is one

quarter, so-to-speak, of the picture.

If I try to draw the whole picture it will

be more confusing.

So you just have to draw the same thing but three more

times, and that's what you need to calculate.

Now, actually here, the way it's phrased, it sounds almost

as though it is not -- it is ambigiously defined, because

it doesn't say on which side of the xz plane.

It could be on this -- I draw it on this side where y

is positive, but it doesn't say that.

So in principle I could also look at the other half.

But the point is that the answer does not depend on which

half you choose, because the function which you use does not

change if you flip y, if you replace y by negative y.

So, if that were not the case, the problem would not be

well-defined, would not have been well-defined, but actually

it is well-defined because the answer is the same, no matter

on which side of the xz plane you look.

Now,

to compute this you write -- see, this is one case, this is

one, coordinate system where you can not so easy visualize

the way I explained in a previous discussion for

cylindrical and Cartesian coordinates where I was

talking about projections.

Because these coordinates do not involve any of the

variables of the Cartesian coordinate system.

Does not involve x, it doesn't involve y, it

does not involve z.

So you have to be a little bit, use your imagination, and you

have to just try to describe this by inequalities in the

coordinate of the spherical coordinate system.

So the inequalities will be, in this case, the row is between

3 and 4, we know this, right.

Then what about phi?

Well, phi actually can be anything from zero to pi.

It takes the maximum range, and then you have to --

the condition is on theta.

If there was no condition about the xz plane, if we didn't say

that -- if the problem didn't say that one of the bounds was

the zx plane, we would write that theta is between zero and

2 pi, which is the maximum possible value, right, this

is a total range of theta.

But because we are bounded by xz plane, that means that theta

can go do from zero to only up to here, but it can not go

further because we are on this side.

You see what I mean?

So that means that theta actually goes from zero to pi.

And that's what this extra bound gives you.

So these are the ranges.

And so actually in the spherical coordinate system,

this is very complicated domain looks very simple.

It looks like a box, it looks like a box, and so the integral

that you get describing this trip integral is actually going

to be the simplest possible triple integral with respect

to this coordinate system.

And you can choose any order you like for the integration

because the bounds do not depend on other variables.

You see, they are fixed.

So it is like the basic integral.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Very good question.

Is it going to be like this in all coordinates for all

problems for spherical coordinates or just for this

one-- just for this one-- of course, because you can imagine

more complicated domains in which the bounds will depend on

other variables, right, so I'm just getting the simplest

possible example.

OK.

So, here you're going to have say row from 3 to 4, and then

phi from zero to pi, and theta from zero to phi, and then

don't forget your function which is rho squared, and don't

forget the extra factor, which is rho squared -- this one --

rho squared sine phi, so times row squared sine phi d

theta d phi zero--.

Row squared times phi.

Sorry?

[UNINTELLIGIBLE]

Why is it rho squared times rho squared?

It's called rho.

So that's actually a good question on the mid-term

would be what is this variable called?

And so if you write for t is the this, then you

get a point deducted.

So this is row, right?

So this rho squared comes from the integrand, which is x

squared plus y squared plus z squared, which is rho squared,

and this rho squared comes from the extra factor.

So, the net effect is that you got rho to the fourth sine phi.

So then you get a very simple manageable integral.

Hold on, hold on, there's one minute left.

Yes?

Row squared sine phi is always there, and then

you have the function.

OK?

So we'll continue on Thursday.

How are you guys?

Good?

Did you have a good week?

I'm good, yeah.

So, but it looks like there are fewer of you.

I hope we haven't lost anyone, or maybe it's just the subject

is too easy for some people.

Well, we also have this wonderful resource.

So, for those watching online, welcome.

But do come to visit us sometimes.

All right, so we are talking about integrals, right.

And I know that last week we discussed double integrals and

triple integrals, and for double integrals you talked

about integration using polar coordinates.

And so this is actually a very -- it's a very useful tool for

integration that sometimes thing simplify when you use

-- some interesting stuff on the floor.

When you use a different coordinate system -- we saw

that when we talked about as early as like first or

second week of the class.

Are you guys lacking discipline after not having me for a week?

I hope not.

So, remember that I'm really strict, but fair.

All right.

So, already at the very beginning we saw that many --

our solutions simplify when we use polar coordinates, for

example, in describing curves on the plane.

And you saw last week that integration also simplifies,

double integrals simplify if you use polar coordinates.

So now the next question is what about triple integrals?

Is there an analog of polar coordinate system for triple

integrals and how can we use such a coordinate system to

effectively evaluate triple integrals.

And in fact, because now we are in three dimension, there is

more than one choice to make, there are more options, and

there are two particular coordinate systems, which are

very convenient in three dimension, which are called

cylindrical and spherical coordinate system.

So we will dicuss them in turn.

First we'll talk about cylindrical.

Cylindrical coordinate system.

And cylindrical coordinate system is really just an

offshot of the polar coordinate system in two dimensions.

In a way we don't introduce anything new in triple

integrals, we use the same tool, which we already use

effectively in two dimension.

So the way it works is like this.

The usual coordinate system, xyz is replaced by another

coordinate system where on the xy plane you use polar

coordinates on the xy plane.

And we just add the z variable , the way we

just added z to x and y.

So the result is that instead of the xyz coordinate system,

which we normally use, we pass through the coordinate

system, which

will have r, theta and z.

So x and y it get replaced by r and theta in the same way as

for polar coordinates and z is just thrown in as

an extra variable.

So the formula's expressing this coordinates in terms of

this new coordinates are exactly the same as for polars.

They are r cosine theta and r sine theta, and then we have

this z, and so z we have on both sides and it's

the same variable.

OK?

So what is this coordinate system good for?

This coordinate system is good for describing object

which are cylindrical.

Well, hence the name.

What do I mean by cylindrical?

A cylinder in general is something which comes from an

object on the xy plane by kind of sweeping a surface by using

that let's say a curve on the xy plane.

So a cylinder itself, a cylinder itself is like that.

We take a circle and if we just move it up and down vertically,

we sweep a surface and that is what we normally

call a cylinder.

So a circle, the equation of the circle on the plane

simplifies in polar coordinates, it's r equal r

capital where this is a number, and this is one of the

variables, one of the coordinates that's in the polar

coordinate system, and this number is just the

radius of the circle.

So in

other words it's a circle over radius r.

So when we move it up and down like this along the z-axis, we

create the surface in which the equation is the same in 3D, the

same equation means that we also have an arbitrary value of

z, and therefore, we get the cylinder, the cylinder

of radius r.

And so just like in integrating on the plane in double

integrals, it is beneficial to use polar coordinates in

describing things related to the circle or to the disk or

sectors of the disk or annuli and things like that.

It is beneficial to use the cylindrical coordinate system

for triple integrals in describing things -- in

integrating things related to cylinders and various things

that you can obtain from the cylinder.

So what does the triple integral look like in the

cylindrical coordinate?

There is an important point which we should not forget,

which is that when we pass to a new coordinate system, there is

some advantage which is that some of the formala simplify.

But there is a sort of a price to pay for it, which is that we

have to insert an additional factor in the integral.

So for polar coordinates, something that was discussed

last week: if you have an integral of a function f to the

a, you can write it as double integral in polar coordinates,

and have if of r and theta, but then instead of simply putting

the rd theta, which you would normally put, you insert an

additional factor, which is r.

So you get rdrd theta.

When instead of the usual dfdy, which we would have in the

polar coordinate, in the normal coordinate system.

What is the reason for this, just to go over it one more

time because we will now see how it works for cylindrical

and then for spherical coordinates.

Everything boils down to the area of an elementary object

with respect to this coordinate system.

An elementary object is a rectangle, which you obtain by

saying that x is -- in this case two coordinates, x and y,

x between some fixed value for x zero, and x zero plus delta

x, and y is between some fixed value and that value plus

some increment delta y.

For the xy coordinates for that rectangular coordinate system,

this inequalities describe a rectangle in which one of the

sides is delta x and the other side is delta y.

And so the area of this rectangle is simply

delta x times delta y.

And that is what gives us the expression for dadxdy.

And so it leads to a description of a double

integral as an iterated integral in x and then y

or first in y and then x, like Fubini's theorem.

But the key point is the area of the elementary rectangle,

which is given by this formula.

Now, when we pass to polar coordinates, the analog of

this elementary rectangle is the following.

We have to look at all points which are given

by this inequality.

Say r between r zero and r zero plus delta r, and theta is

between some fixed value theta zero, and theta zero

plus delta theta.

And the picture will be different, right because this

gives us a sector of angle delta theta, and then the

condition for r to be between r zero and r zero plus delta r,

let's assume that this is our zero, right, just like I am

assuming that this is theta zero.

Of course, in this calculation, eventually you would like delta

r and delta theta to become very small, whereas r zero and

theta zero are fixed, but I'm just drawing this picture.

I can magnify everything, so it looks like delta theta looks

like sort of the same magnitude as theta zero, but in fact

it should be much smaller.

And so then if this is delta r, and this is confined within

delta theta, and the elementary object instead of this one, we

get a kind of a -- it sort of looks like a rectangle but not

exactly because first of all there's a certain angle here

and these are not straight lines, but these are segments

of a circle, right.

So what's the area of this?

The area of this one -- actually we can not -- I mean

it will be -- we can but it will be different -- it's not

so simple to calculate, to give exact answer, but because if

I'm going to take the limit anyway, eventually, we only

need a sort of a good approximation to this.

And what's a good approximation to this?

We can think of this -- we can think that when delta r and

delta theta are very small, this will look like in a

rectangle with a side delta r, and what's the other side?

Well, this actually has length, r zero times delta theta.

The segment of a circle of radius r zero confined within

the angle delta theta is given by this formula, r zero

times delta theta.

And so we approximate this by the area of the rectangle with

the sides delta r and r zero times delta theta, and the

result of this is just the product of this and we get r

zero delta theta delta r.

And so that's where -- that's the r which shows up in the

formula for the integral.

Here I denote it by r zero, because I wanted to emphasize

that for calculating this particular area, I fix r zero,

I fix this length, but when we do a general calculation, this

will just become r, and delta theta will get replaced by d

theta and delta r by dr in the limit, right.

And so that's how we end up with this formula.

So this only sort of non trivial thing to remember

about polar coordinates.

Don't forget to put this factor r, which actually

could be a good thing.

I made it sound like it's a price to pay that we are being

taxed for sort of for the convenience of using these

coordinates, but actually sometimes, and oftentimes, this

could be a good thing, and the typical example is let's say

you have an integral like this, something like 1 minus x

squared minus y squared dfdy.

So you see this is complicated, right, this is actually

something we discussed, it's kind of similar to something

we discussed earlier.

This will be very difficult to take just as an integral in x

and y, but when you pass two polar coordinates this will

become square -- I'm not writing the limit, but

I'm just sort of giving a rough sketch here.

This becomes 1 minus r squared.

And now we get this extra factor r, and then

we get drd theta.

And so you see, this actually becomes a much better integral

because I could introduce a new variable, let's say t, which is

r squared, and if I do that then dt becomes two times rdr.

And then this actually gives me 1/2 of dt.

And this becomes a square root of 1 minus t instead of 1 minus

r squared, and this is much easier to calculate, because

the anti-derivative of this is just 1 minus t to the power

3/2 times 2/3, minus 2/3.

Whereas for this one it's much more complicated.

Well, imagine if it were like exponential function or

something, so then it becomes even worse, right.

So sometimes actually having this factor actually

works to our advantage.

But in any case we have to remember to insert it to

get the right answer.

Now, so this was all about the polar coordinate, but

cylindrical are not that far away, not that far apart,

because for cylindrical, the only thing that changes is that

we now throw in an extra variable.

Namely variable z, so we add the variable z, and now instead

of a double integral we have a triple integral over some

solid, some region in the three dimensional space, and so we

have some fda where this da, usually in the normal, in the

Cartesian coordinate system, we simply write as dxdydz.

But now if we want to use polar coordinate system, we'll have

to write this function as a function of r theta and z, and

then we'll have to write drd theta dz, and the question is

what should we put in front?

There has to be some factor, and that factor is the factor

responsible for the area of the elementary domain like this.

In the usual coordinate system in the three dimensional space

in the -- when I say usual I mean Cartesian

coordinate system.

The elementary object is a box of sides delta x,

delta y, and delta z.

So it's -- actually no one corrected me.

What is it mistake here? dv, that's right.

See, I'm a little rusty after break, but do correct me

when I make these mistakes.

So dv, because we emphasize that this is elementary

volume, its volume, it's not area anymore.

The a, of course, was for area and v is for volume.

So the elementary volume for this guy is just delta x, delta

y, delta z, which gives rise to dxdydz in the integral and

allows us to calculate this integral as it iterates

integral first x then y then z or any other order

if you choose.

But now we choose this coordinate system and so we

have to be careful, we have to now look at the elementary

object in this coordinate system, and what is this

elementary object?

Well, we just define it in the same way as before by simply --

I should have done it this way.

I much be jumping ahead.

Let's just do it like this.

It really should be cylindrical, so it

should be like this.

So it is, in a way, it's a kind of a -- it's what you get by

combining the two things.

In this phase, this side of this object looks like this.

It's--.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Why is it called cylindrical?

This is a cylinder, right.

He saved you, by the way, because I was kind of

approaching to count how many seconds it will take for you to

notice me, but he saved you so you should thank him.

So, cylindrical is called because you see the, you make a

change on the xy plane and then we just throw in

the z variable.

In other words, you change the coordinates just on x plane and

then you kind of let the third variable the same as before.

So it's the same idea as the idea for say taking the circle

and kind of just moving it up and down to get this object.

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: No, no, the circle is fixed, right,

it doesn't grow.

The radius of the circle is fixed, it's this r capital.

Just think of a ring which is made of metal or something and

you just move it up and down vertically along the --

parallel to the z-axis.

right.

So what's the result?

You sweep a surface, which is the surface of the cylinder.

You can only use it -- no, you don't use it only for cylinder,

but, well, what's in a name, you know.

It's a name, so it is derived from this, right, but it's not

supposed to explain everything we can do with this

coordinate system.

Any other questions?

So, what's the elementary volume of this?

Well, we already know the approximate area of this that

rdr, r delta r delta theta, and we know that this is delta z.

And so the volume is rdr delta -- sorry, r delta

r delta theta delta z.

So it's the same r which we had -- or maybe if you want r zero,

if you want this to be r zero.

The same r which we had in the calculation of the

polar coordinates.

So the bottom line is this factor is exactly the same as

in the polar coordinate system.

It's not more complicated.

So that's the formula we're

going to use.

And now let's do an example.

Let's do an example.

So, evaluate triple integral over e of the function e to the

z where e is enclosed by the paraboloid, z equals 1 plus

x squared plus y squared.

The cylinder x squared plus y squared equals

5, and the xy plane.

So, actually let me get -- let me draw a bigger picture.

So the paraboloid, what does a paraboloid look like?

It starts at the point 1 on the z-axis and it opens up like

this, and the cylinder is OK, I just erased it.

So that's a cylinder, let's assume that this is radius.

This is going to be radius square with a 5, right, because

we have 5 is the square of the radius, as always.

So the radius itself is square of the 5, and so that's a

circle at the base of the cylinder, and we want to figure

which is a solid, which is confined by this by the

paraboloid, by the cylinder and by the xy plane.

So it is the inside of this, so it's kind of concave, the

surface kind of concave, it goes inside.

It's like a fancy glass.

So what we need to do is to calculate the integral over

this -- this is our region, e, of the function e to the z.

So, when you do triple integrals, you have to --

so then you have to write as an iterated integral.

So you have to choose in which order you're

going to integrate.

So you're going to end up with an integral, so you have e to

the z, here actually z is one of the variables, so we just

write it like this if this is good, right, because these one

of the variables in the new cylindrical system.

And then you're going to have the r -- let me emphasize this

r again, that's the same r as before -- and then we

have rdrd theta and dz.

So now you want to write it as an iterated integral, so you

have to choose in which order to integrate.

So what's the -- what's the best way to do it?

For the cylindrical coordinate system it's always -- you see

the point is that you have to choose what the base of this --

actually this is a general approach to when you have

to choose the order.

You have to see whether, what's a good projection of your

object -- is there a good projection onto say xy plane or

yz plane or zx plane -- that's the first question, and once

you have a good projection, everything projects onto

something nicely, right, so in this particular case, it

certainly does projejct nicely onto the xy plane, it projects

just onto the disk, onto z which is r less than or

equal to square root of 5.

The disk of radius square root of 5.

So that's this, that's this yellow -- let's

draw it with yellow.

And then for each point in this disk, we'll have to integrate,

so first of all, we'll have to integrate over the disk, and

this will be the -- I'm talking about the outer integration,

I'm starting from outside.

When you project onto this, you are choosing the first two

variables, which normally if we were using x and y and z

coordinate, that would be integrating first x and y.

Then you decide, you decide between x and y later, but

first we choose the first two, and then you choose which

one goes first, right.

So that's a strategy which I propose that you're going

to end up with three different integrations.

So the last one will be the integration over the remaining

variable with respect to this projection.

So here I project onto the xy plane, so this will be dz.

But let's now see what happens here.

So this is going to be in the usual coordinate system, in the

Cartesian coordinates system, this would have been just dxdy

which you would have to decide, dxdy or dydx.

And once you decide, then in the last integration you have

the freedom to choose what x and y are, so that means that

you have a point here, xy inside, and then the limits

will be -- the limits you have to sort of draw the segment

starting from the bottom of this, of the object to the top

of this object, which in this case from this disk, which is

part of the xy plane to this paraboloid, and you would have

to integrate, in the last integration you would have to

integrate from the bottom value to the top value, which in

this particular case what does it mean?

The z goes from zero, right, to this value.

But what is this value?

This value's is going to be 1 plus this is going to be the

value of z on the paraboloid.

So that's -- in the Cartesian coordinate system, this

would have been 1 plus x squared plus y squared.

But now we're replacing this with r squared.

So, because we are doing the integral in the polar

coordinate system, we will actually write it as 1 plus r

squared where it is understood that r squared comprises x

squared and y squared -- the sum of x squared and y squared.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Well, up and down because I have chosen the

projection onto the xy plane.

Right, so the way I suggest to do the integral

is the following.

First let's choose the two outer integrations, the first

and the second, right.

So geometrically it means that you are projecting your three

dimensional region, three dimensional solid, onto one of

the planes, one of the coordinate planes.

I'm talking now about really the Cartesian coordinate

system, but cylindrical coordinate is not that far away

from Cartesian, so the same analysis sort of applies here.

So the projection here is going to be the disk, right, but in

general it could be square, it could be something else.

So that will be taken care of by these two integrations,

by the limits here.

STUDENT:

PROFESSOR: That's right.

The projection goes in those two.

And the inner integral is the remaining variable,

which in this case is z.

So to put the limit here, you have to pick one of the points

in the projection, in

the image of the projection, which is this point, say xy,

which actually because we are doing cylindrical coordinates,

this would have to be recorded as r theta.

So this point will be r theta.

And we'll have to take the segment along the third

variable, which is the z variable going from the bottom

to the top, from sort of the bottom lid to the top lid of

this figure, of this region.

Which, as I explained, the bottom is on the xy plane, so

the z is equal to zero at the bottom, and at the top, we have

to look at the equation of the paraboloid, it's 1 plus x

squared plus y squared, which because we are working with

cylindrical coordinates will be written as 1 plus r squared.

Let me finish the integral and then you ask me if

you have more questions.

So then what remains to be done?

What remains to be done is that we have to put also here r and

theta, so I have to choose some order in which we

will do r and theta.

So let's say we put here dr and put here d theta, because now

we are actually, see, once we are here in these two outer

integrals, we're actually doing a double integral.

We are deciding how to split the double integral

into iterated integral.

So, and what we are working with is already the image of

this three dimensional object onto the xy plane, which

is nothing but the disk.

So for the disk we know what the limits are -- first of all,

we know that it doesn't matter in which order to take r and

theta, and second of all we know that r goes from zero to

the radius, which is in the case square root of 5, and

theta goes from zero to 2 pi.

That's because the disk, this disk described by -- well,

I don't even have to write one more time.

I just have to add here theta from zero to 2 pi.

And finally we should not forget to put the function,

which is e to the z, and we should not forget to

put this factor r.

So I'm introducing a slightly different notation from before.

Normally we would write this as e to the z times r times dr d

theta and then we put sort of brackets and we think about

integrating first this with respect to r, and of this with

respect to theta -- sorry, integrating this with respect

to z then this with respect to theta and this with

respect to r.

But I'm writing it in this way, which I

think is a little bit more suggestive that you write the

dr, the d theta, dz, the d of the variable you're integrating

right next to the integral, so you remember which one it is.

You see because in the old way you would have to

put dr the last one.

So the last one will get paired with the first integral, the

next to last will get paired with the second

integral and so on.

This way it's a little bit more intuitive, so I prefer to write

in this way, but you can choose whichever way you like.

But I hope that it's clear what I mean.

Well, this certainly is unclear because it says onto and then

it says zero, so let me write it here.

I think now it should be clear.

OK, any questions about this?

No?

Yes?

[UNINTELLIGIBLE]

So, in fact, I could have put r all the way here.

[UNINTELLIGIBLE]

Oh, yeah, yeah, yeah, you're right.

[UNINTELLIGIBLE]

Right.

So I actually did it in the wrong way, in

the wrong quarter.

So--.

Did I do it in the wrong quarter?

Does the order matter?

No, the order, of course, matters, because see the

point is that -- no, no I did it correctly.

Who thinks that I did it correctly?

Who thinks that I didn't do it correctly?

OK, you guys, you fail the course, no.

We are here to find the truth.

But see here's the way -- I thought about it this

afternoon, actually, and I convinced myself this is right.

But see here's the point.

The point is--.

[UNINTELLIGIBLE]

Right.

So the point is that we are -- let me -- this is actually a

very -- it could be very confusing.

So let's actually do it slowly.

You should not worry so much about the fact that here

there is a dependence on r in the inner integral.

What you should worry about is the fact that there is a

dependence on r in the limit.

So you certainly don't want to put this outside.

Because you want to end up with a number.

In other words, we can disagree on many things, but there is

one thing we should certainly agree on, which is that

the integral is a number.

It should not depend on r, it should not depend on theta,

it should not depend on anything, it's a number.

You can not possibly get a number if the last integration

as the limit which depends on the variable.

That's not going the work.

But more conceptually, the way I'd like to think

about this is as follows.

And I would like to think of maybe in a slightly -- you

know, it could be some more complicated domain.

So what I'm doing is I'm projecting --

so there is some e.

Let's say I want to project it onto the xy plane.

But right now let's talk about just the Cartesian coordinate

system, let's not worry for now about cylindrical or spherical.

Even in the Cartesian coordinate system, the way I

would like to think about integration is as follows.

That I have a region in a three dimensional space, right, and

what I do is I project it onto the xy variable.

So now if I have a triple integral over e where I

have f dv, first of all I split it like this.

It's going to be a double integral over z, over the z.

Of the single integral with respect to the

remaining variable, dz.

You see where for each x and y in d, you will have some lower

limits, let's say l of xy, and you have some upper

limits, u of xy.

So that's what it's going to look like.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: So then I would just write, and here,

yeah, that's right.

So then here I would put da, like this.

Yes.

[UNINTELLIGIBLE]

That's right.

In the old way.

Right, so you want me to write this in the old way.

So this in the old way would be zero to square to 5, zero to 2

pi, zero to 1 plus r squared, e to the z, dz, d theta, dr.

Oh,

I'm sorry, rdr -- yes, very good, thank you.

See because it's red so I don't, I didn't see it. rdr --

well, usually we'd write rdr -- yeah, rdr, OK, fine, whatever.

We can put r inside here, maybe it's better, kind

of keep it up over here.

So then normally we would put brackets like this.

Is that OK?

Does it make sense?

So what I'm doing is just I'm trying to avoid using the

brackets and instead jut putting the differentials right

next to the integral, so it becomes a little bit more clear

to me anyway, but you are free to use whichever way you like.

So this is sort of a new way to write, and the old way to write

would be f dz -- well, it's almost the same, za over

-- maybe like this.

So but the main point which I would like to make is that the

first two integrations in the formula -- see, it's very

confusing because when I say the first integration, do I

mean the first integral or do I mean the actual

integration you perform.

This is opposite, right?

The first integral you write is the last one, it's the last

integral you write is the first you calculate.

That's right.

That's right, so this is the outer, the outer is

written first, so that's why it goes last.

So you have to figure this out on your own.

I guess at some point you should have to think about

it and have a clear picture because like I said

it's very confusing.

What do you call first, what do you call second.

Who's on first and, right.

So the last integration, is outer one, and the last

integration corresponds to the projection of your three

dimensional region onto the plane.

OK, That's the point.

So, having established that let's evaluate finally

this integral.

So what do we get?

We get zero square to 5, 2 pi, and then here we get r times e

to the 1 plus r squared minus 1, and you get dr -- this

integration just gives you 2 pi because the integrand does not

depend on theta, so you just get 2 pi times r e to the

1 plus r squared minus 1.

And finally you integrate over r, and so here you want to use

change variables, so I'll just write the answers so you've got

2 pi as an overall factor, and then you get 1/2 e to the 1

plus r squared minus 1/2 r squared between zero and square

root of 5, and so the answer is 2 pi times e to the 6 minus e

minus 5. 1/2 -- well, because 1/2 and 2 get canceled, so

you've got e to the 6 minus e and this guys gives you 5.

So that's correct.

OK?

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: It's pi that's right, so -- thanks.

OK, cool.

So we good with this?

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Is it universally accepted that--?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Are you talking about this formula

or this formula?

[UNINTELLIGIBLE]

You mean this integral?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: But I'm not sure I understand

what the question is.

Which line, first of all, this one or--?

This one, yes.

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Are you talking about the left hand side

or the right hand side?

The question is it OK to put dr before the function of r.

OK.

The answer is yes, it is OK.

I am suggesting this notation.

In the book it's not used, right, but as a practicing

mathematician I can assure you that a lot of people use that.

It is certainly interchangeable.

They do commute, in other words, it doesn't matter in

which order you write them.

It's a more interesting point whether this guy's commute, you

know, whether it is important how you write drd theta or d

theta dr, this is a much more subtle point, which I'm going

to get into too much, but certainly dr you can put on

the left or on the right of the function of r.

So no worries here.

Let's move on, so the next one is called spherical coordinates

and it's a little bit more interesting.

Sorry?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Spherical coordinates, and, of course,

the question is right away why are they called the spherical

coordinates, because they should remind you of a

sphere, as you will see.

So, let me actually keep this picture.

So, spherical coordinates work in the following way.

Suppose you want to invent a coordinate system in which

the sphere -- it is a sphere which gets the simplest

possible equation.

So, in the cylindrical coordinate system, it is a

circle that gets the -- say in a polar coordinate system, is a

circle which gets the simplest equation, r equals a constant.

That's the circle of a given radius.

In a cylindrical coordinate system, the simplest equation

is the equation for the cylinder, OK? r equals -- r is

a circle in polar coordinates because r is a cylinder in the

cylindrical coordinate system, in 3D, this is in 2D.

And now I want to have some variable rho such that this

equation will give me a sphere, and that's called the spherical

coordinate system and that's in 3D.

So that means that this rho should really be the distance

from the origin of my coordinate system to my point,

because a sphere -- well, sphere with the center

at the origin.

The sphere with a center at the origin of radius r capital,

this one, is going to be the set of all points such that the

distance from the origin to my point is given by this number

r, let's say square root of 5 like in the previous example.

So, note that this is not the same as r in the cylindrical

coordinate system, because r in the cylindrical coordinate

system is not the distance from the origin to this, but rather

it's the distance from the origin to the projection of my

point onto the xy plane, so that's not the same thing.

So you see right away that this is not the same coordinate

system, and sure enough, this simplest equation rho equals a

constant describes not a cylinder but a sphere.

But now, so now I got the first variable of my spherical

coordinate system, but it's not enough, I need three variables,

right, because I'm in three dimensional space.

So I need to complete this rho by two additional variables,

two additional degrees of freedom, if you will, so that I

could give each point a unique address by using those

three coordinates.

What are they?

So the standard -- there are different ways actually to do

it, but the standard convention which we are using in this

class and in this book is the following, that we

measure two angles.

In addition to rho we measure two angles, and the first

angle is exactly the same as before, theta.

It's the theta of the cylindrical or polar

coordinate systems.

So we need one more, and we find this one more by

measuring this angle, and we call this phi.

So the spherical coordinates in a three coordinates rho, phi

and theta where only one of them is part of the cylindrical

coordinate system, which theta is part of the cylindrical

coordinate system.

But the others are new.

So, by the way, notationally phi sometimes is also,

is written like this, it's the same thing.

So I find it easier to write phi like this, but when you

type, for example, and in the book it's written like this.

So you can use whichever notation you like.

So a row of phi and theta.

The first step, of course, is to express the usual

coordinates, the Cartesian coordinates in terms of a row

of phi and theta and to see that indeed these coordinates

are -- determine the points in space.

And also to see what the ranges are.

So what is the formula?

Well, what we can do is we can use the old formula, which is r

cosine theta, and then note that r is equal to rho times --

this is equal to rho times the sine of this angle.

See, the point is that this has the right angle.

This is a triangle -- this is a triangle which has a right

angle, and this is phi, so you can find this distance, which

is r of the cylindrical coordinate system, by taking

rho and multiplying by the sine of this angle.

So this is the row times sine of phi.

Now I substitute this in here and I get rho sine

phi equals sine theta.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Which axis is phi measured from?

It's measured from z.

So, I did not put labels, but I should, it's

xyz, the standard way.

Another question.

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: The rho does not belong to any plane of priori,

it's in space somewhere.

It's just that -- I have a point--

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: kind of, it flows around like

this, wiggles around.

It's not doesn't belong to any particular plane.

It's free to move, right, because our point is free to

movein free space, and so the segment connecting our point

to the origin it doesn't not necessarily belong

to any plane.

Which one?

This triangle?

It has a right angle here, because this is a projection.

I don't know if it looks like a projection -- it looks a

little bit, it's a little bit crooked I guess.

It should be a little bit more parallel to the z plane than

what I made it look like.

Is that better?

Slightly.

So this is a perpendicular, which we drop from this

point or our point, this is our point.

We drop it onto the xy plane.

So, I guess the rest should be clear.

Any other questions?

Yes?

Is theta always taken from the x-axis?

That's right and theta always taken from the x-axis

for the projection.

Everything is a little bit not straight on this picture.

This is r -- this is old r, which is not -- this r is

not part of the spherical coordinate system, it's

part of a cylindrical coordinate system.

The reason I have drawn it, I have written it here, put this

label is to connect, to make a connection between the

spherical and the cylindrical coordinates, and also to

simplify the derivation of the formula for xy and z.

Because I do it in two steps.

I first recall the formula for the cylindrical coordinates,

which is really the same as for polar coordinates, and

then I just substitute r equals row sine phi.

And then I do the same for the y variable, and

then z is, what is z?

To find z, we have to complete this to a rectangle--.

I'm sorry?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: It should be sine theta, that's right.

Thank you.

And to find z I have to use another right triangle, which

is this one -- I guess again I didn't draw it very well

-- maybe more like this.

So this is right triangle, this is phi, so z is

rho times cosine phi.

So, if you visualize -- if you visualize this picture, you see

that we don't have to memorize or write on your cheat sheet

the formulas for the cylindrical coordinates,

it's very easy.

You have to remember the polar coordinates, of course, but I

mean, if you don't remember polar coordinates, then it's

like remembering your phone number in some sense I think.

So, if you remember polar coordinates then, then the

spherical coordinates are

very easy by memorizing this picture or visualizing

this picture.

OK, what are the ranges?

So, a row is like r, rho is a non-negative number, although

for r if you remember, we had certain convention in

polar coordinates.

We sometimes allowed r to be negative, but in spherical

coordinate system, that would make things too complicated

so we don't do that.

So rho is non-negative.

Theta is just like in a polar coordinate system or in

the cylindrical, it's betweenm zero and 2 pi.

And what about phi?

What are the ranges for phi?

Zero and pi, exactly, because you can go, so if this is a

vertical the z-axis, you can go all the way down to pi, but if

you go more than pi, then you can sort of approach it from a

different direction and it will be less, from that direction

will be less than pi.

So that's what we do.

So it's going to be from zero to pi.

Do you see what I mean?

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: That's right, from the vertical.

So, in other words, what I'm saying is let's say, so this is

z-axis and this is the origin, so you have a -- let's suppose

the our point actualy lives on the yz plane.

So, and let's start rotating it.

So this is phi for this point.

This is phi for this point, this is phi for this point.

Still less than pi, right.

So we get here, it's still -- what I'm trying to say is this,

this, this, this, right -- we measure angle from here.

But finally when we go like this, we should not measure

it like this, but we should measure it this way.

You see what I mean?

So this is wrong and this is right.

This is phi.

So you choose whichever is shorter, whichever's smaller,

and that's going to always be between zero and pi.

So pi would be this -- this is pi, which is points on the

negative -- phi equals pi corresponds to point on the

negative part of the z-axis.

Do you have a question?

Now, what are these coordinates good for?

Well, the first answer has already been given.

In this coordinate system, it's much easier to

describe the sphere.

So the sphere is now given by the equation rho equals r,

instead of x squared plus y squared plus z squared equals

r squared of the Cartesian coordinate system, or r squared

plus z squared equals r squared of the cylindrical coordinates.

It's just the simplest possible equation, rho

equals the constant.

So that's, in a way, justification for using

this coordinate system and for the name.

But that's not the only object which has a nice representation

in this coordinate system.

We can also represent a cone by the equation phi equals

some fixed angle, phi zero.

So that's the cone.

Let's look at this picture again.

So let's suppose I fix some phi and I look at all the points

which have a given angle like this, phi zero, 6

to the z-axis.

So then it's like this is the z-axis and my

point is like this.

I don't know, I don't have enough visual aide, so I have

to trust your imagination and your intuition.

Maybe like this is better, I don't know.

So the point is that this is the angle, so the angle is

fixed so it has to be the same angle between your points

and the same z-axis.

So, the result is a cone.

And because I don't specify what row is, I don't specify

the distance, it can go as far away as I want, so it's an

infinite cone -- kind of an upper cone going to infinity,

so it's not bounded, unbounded cone.

So don't think that I just drew one level curve for it,

but it goes to inifinity.

So the cone

is given by the formula, which is again, much simpler than z

equals square root of x squared plus y squared of the Cartesian

coordinate system, or z equals r of the cylindrical coordinate

system, which is not so bad but this is better, this

is easier to handle.

So these are the things you should remember.

The cone and the sphere have a very nice expression in terms

of this coordinate system.

Now what about the integration?

What about integration in spherical coordinate system.

So once again, we have to figure out what is the volume

of the elementary object with respect to the spherical

coordinate system.

So these are the elementary for the Cartesian and the

cylindrical coordinate systems, which give us the usual

formulas, dxdydz and rdrd theta dz.

So now we have to do exactly the same for the spherical

coordinate system.

So what is the elementary object now?

And again, by elementary object, I mean simply that you

impose the condition that row is between some row zero and

row zero plus delta rho and -- I must have lost my

second variable phi.

Phi is between phi zero and phi zero plus delta phi, and theta

is between some theta zero and theta zero plus delta theta.

So let's draw this, an object which corresponds

to this inequality.

Like this.

This whole thing should be kind of dotted line.

OK, and like this and like this.

OK, you see, you see what I mean?

Well, in fact this -- do it like this, this may be better.

So here, what happens here?

First of all, this is delta phi, right, this is rho zero.

So this thing is actually between two spheres.

Imagine two spheres which have very close radii, very similar

radii, so this is something which is between those spheres,

and first of all, we cut it by saying that the angle goes

between, you know, it's confined within delta phi, and

also when we project this down onto the xy plane, this angle

-- I'm not going to draw it all the way -- so the projection's

going to be delta theta.

So this is a projection onto xy plane.

So now I have to evaluate the volume of this, just like I

have to evaluate the volume of this and of this in my

other coordinate systems.

So what is it like?

What is this volume equal to.

Well, we don't calclate exactly, we approximate it, and

we approximate it by the volume of the cube -- sorry, not the

cube, but a box -- a box whose sides are given by the lengths

of this kind of circular sements -- they are going to

be all circular segments.

So this one is very easy to find -- well, first of all,

this one is very easy to find.

This is just delta rho this is just delta row.

That's one side.

OK.

Now this one is also easy to find, this is -- you see this

is rho zero, and the angle here is delta phi, so this

is rho times delta phi.

It's similar to the way we found that to be r zero

times delta theta.

So it remains to find this one, which I guess you can not

see this color, right.

Actually it's better than the other one, it's

kind of greenish.

Can you see this one?

Oh, good.

So this one is a tricky one because you would be tempted

to take rho zero times delta theta, but this is not correct.

This is only one, there is only one subtle point here.

The point is that this is the same as this.

In other words, this actually is not, this is

not part of a circle of radius rho zero and confined within

the angle detla theta.

On the contrary, it is actually part of a circle on the xy

plane, but the radius is not -- the radius of that circle is

this, which is not rho zero, but it is r.

It is row zero times sine phi.

So that's the only point that you have to realize

to get the right answer.

And so -- and then you've got delta theta.

I'm trying to draw this like this.

This is on the xy plane.

And so this segment is actually, its length is

rho zero times sine phi delta theta.

So the answer is for the volume it's approximated by rho zero

sine phi delta theta times delta rho times delta times rho

zero again -- this was rho zero -- and times delta phi.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: I am trying to calculate the volume of this,

and I am approximating it by the volume of a box like this,

which has the following sides.

This side is rho zero sine phi delta theta, that's this, times

delta rho, which is this side, times rho zero times delta

phi is the yellow side.

So the end result is row zero squared sine phi times delta

rho delta phi delta theta.

So, the upshot of this is when you compute a triple integral

and spherical coordinates, you have to introduce the fact that

rho squared times sine phi.

So it's a little bit more complicated than

for cylindrical and polar coordinates.

And triple integral over some solid is rho phi theta times--

let's write it here-- rho squared sine phi -- well, since

I write it in red before, let me write in red again.

A row squared sine phi -- that's the factor to remember

-- zero d phi d theta.

OK?

So we do this calculation once and for all, then we just

memorize it, row square times sine phi, and after that, we

can do this triple integrals by using iterated integrals for

these three variables, rho phi and theta.

Let's see how it works in practice.

So here is an example.

Compute the integral of x squared plus y squared plus d

squared dz where e of the region is bounded by the xz

plane and the spheres. x squared plus y squared plus z

squared equals 9, and x squared plus y squared

plus z squared equals 16.

So within two spheres and also the xz plane.

So first of all, I have to say just a general comment.

On a -- next week we'll have a mid-term.

On the mid-term I will not tell you which coordinate system to

use, you have to decide youself.

So the question is you have to look for clues --

which coordinate system, spherical, cylindrical.

You have to remember one thing, that this expression, x squared

plus y squared plus z squared simplifies in the spherical

coordinate system.

So if you see x squared plus y squared plus z

squared how many times?

One, two, three times, chances are it's best to use, in

the case, spherical coordinate system.

You see what I mean?

Because this is rho squared, right, this is row squared.

If you see x squared plus y squared, that's r squared,

r being the r of the polar or cylindrical coordinate.

That's how you know that you probably should do, use polar

coordinates or cylindrical coordinate, depending whether

you use -- are doing a two dimentional, a double integral,

or a three dimensional triple integral.

But in this case it's fairly clear that you'd be better off

using spherical coordinates.

Now we can visualize this.

Think of a sphere, first of all.

Let me draw just a quarter of this picture.

Actually, yeah, about a quarter.

So that's a part of the sphere -- this is a sphere of radius

three, because 9 is 3 squared, right, and this is sphere of

radius 4 because 16 is 4 squared.

So let's say this is a sphere of radius three, and then this

is the inner sphere, and then there is an outer sphere

which has radius 4.

So see this one is inside the other, and what you are looking

for is -- what you want to look at is the three dimensional

shell which is in between the two -- in between

the two spheres.

It's like an [? analise ?]

but three dimensional. [? Analise ?]

between two circles and this is between two spheres.

And also you have, you bind -- the other bound is xz plane,

so this is the xz plane.

So actually that's not the whole thing.

This is one quarter of the whole thing, because you will

also have something in this quadrant and in those two

quadrants, which are behind the zy plane, so this is one

quarter, so-to-speak, of the picture.

If I try to draw the whole picture it will

be more confusing.

So you just have to draw the same thing but three more

times, and that's what you need to calculate.

Now, actually here, the way it's phrased, it sounds almost

as though it is not -- it is ambigiously defined, because

it doesn't say on which side of the xz plane.

It could be on this -- I draw it on this side where y

is positive, but it doesn't say that.

So in principle I could also look at the other half.

But the point is that the answer does not depend on which

half you choose, because the function which you use does not

change if you flip y, if you replace y by negative y.

So, if that were not the case, the problem would not be

well-defined, would not have been well-defined, but actually

it is well-defined because the answer is the same, no matter

on which side of the xz plane you look.

Now,

to compute this you write -- see, this is one case, this is

one, coordinate system where you can not so easy visualize

the way I explained in a previous discussion for

cylindrical and Cartesian coordinates where I was

talking about projections.

Because these coordinates do not involve any of the

variables of the Cartesian coordinate system.

Does not involve x, it doesn't involve y, it

does not involve z.

So you have to be a little bit, use your imagination, and you

have to just try to describe this by inequalities in the

coordinate of the spherical coordinate system.

So the inequalities will be, in this case, the row is between

3 and 4, we know this, right.

Then what about phi?

Well, phi actually can be anything from zero to pi.

It takes the maximum range, and then you have to --

the condition is on theta.

If there was no condition about the xz plane, if we didn't say

that -- if the problem didn't say that one of the bounds was

the zx plane, we would write that theta is between zero and

2 pi, which is the maximum possible value, right, this

is a total range of theta.

But because we are bounded by xz plane, that means that theta

can go do from zero to only up to here, but it can not go

further because we are on this side.

You see what I mean?

So that means that theta actually goes from zero to pi.

And that's what this extra bound gives you.

So these are the ranges.

And so actually in the spherical coordinate system,

this is very complicated domain looks very simple.

It looks like a box, it looks like a box, and so the integral

that you get describing this trip integral is actually going

to be the simplest possible triple integral with respect

to this coordinate system.

And you can choose any order you like for the integration

because the bounds do not depend on other variables.

You see, they are fixed.

So it is like the basic integral.

Yes?

STUDENT: [UNINTELLIGIBLE]

PROFESSOR: Very good question.

Is it going to be like this in all coordinates for all

problems for spherical coordinates or just for this

one-- just for this one-- of course, because you can imagine

more complicated domains in which the bounds will depend on

other variables, right, so I'm just getting the simplest

possible example.

OK.

So, here you're going to have say row from 3 to 4, and then

phi from zero to pi, and theta from zero to phi, and then

don't forget your function which is rho squared, and don't

forget the extra factor, which is rho squared -- this one --

rho squared sine phi, so times row squared sine phi d

theta d phi zero--.

Row squared times phi.

Sorry?

[UNINTELLIGIBLE]

Why is it rho squared times rho squared?

It's called rho.

So that's actually a good question on the mid-term

would be what is this variable called?

And so if you write for t is the this, then you

get a point deducted.

So this is row, right?

So this rho squared comes from the integrand, which is x

squared plus y squared plus z squared, which is rho squared,

and this rho squared comes from the extra factor.

So, the net effect is that you got rho to the fourth sine phi.

So then you get a very simple manageable integral.

Hold on, hold on, there's one minute left.

Yes?

Row squared sine phi is always there, and then

you have the function.

OK?

So we'll continue on Thursday.