Mathematics - Multivariable Calculus - Lecture 15

Uploaded by UCBerkeley on 18.11.2009

How are you guys?
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.
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.

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.
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.
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.

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.
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?
So, in fact, I could have put r all the way here.
Oh, yeah, yeah, yeah, you're right.
So I actually did it in the wrong way, in
the wrong quarter.

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--.
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.
PROFESSOR: So then I would just write, and here,
yeah, that's right.
So then here I would put da, like this.
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.

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.
PROFESSOR: It's pi that's right, so -- thanks.
OK, cool.
So we good with this?
PROFESSOR: Is it universally accepted that--?
PROFESSOR: Are you talking about this formula
or this formula?
You mean this integral?
PROFESSOR: But I'm not sure I understand
what the question is.
Which line, first of all, this one or--?
This one, yes.
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.
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.
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.
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.
PROFESSOR: The rho does not belong to any plane of priori,
it's in space somewhere.
It's just that -- I have a point--
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?
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?
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?
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?
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.
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.
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.
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.
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.
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.
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.
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.

Row squared sine phi is always there, and then
you have the function.
So we'll continue on Thursday.