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Yesterday we saw how to define
double integrals and how to

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00:00:28,000 --> 00:00:33,000
start computing them in terms of
x and y coordinates.

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00:00:33,000 --> 00:00:41,000
We have defined the double
integral over a region R and

10
00:00:41,000 --> 00:00:45,000
plane of a function f of x,
y dA.

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00:00:45,000 --> 00:00:51,000
You cannot hear me?
Is the sound working?

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00:00:51,000 --> 00:00:57,000
Can you hear me in the back now?
Can we make the sound louder?

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00:00:57,000 --> 00:01:01,000
Does this work?
People are not hearing me in

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00:01:01,000 --> 00:01:05,000
the back.
Is it better?

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00:01:05,000 --> 00:01:09,000
People are still saying make it
louder.

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00:01:09,000 --> 00:01:11,000
Is it better?
OK.

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00:01:11,000 --> 00:01:18,000
Great.
Thanks.

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00:01:18,000 --> 00:01:22,000
That's not a reason to start
chatting with your friends.

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00:01:22,000 --> 00:01:27,000
Thanks.
When we have a region in the x,

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y plane and we have a function
of x and y,

21
00:01:31,000 --> 00:01:36,000
we are defining the double
integral of f over this region

22
00:01:36,000 --> 00:01:40,000
by taking basically the sum of
the values of a function

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00:01:40,000 --> 00:01:44,000
everywhere in here times the
area element.

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00:01:44,000 --> 00:01:48,000
And the definition, actually,
is we split the region into

25
00:01:48,000 --> 00:01:52,000
lots of tiny little pieces,
we multiply the value of a

26
00:01:52,000 --> 00:01:55,000
function at the point times the
area of a little piece and we

27
00:01:55,000 --> 00:01:59,000
sum that everywhere.
And we have seen,

28
00:01:59,000 --> 00:02:07,000
actually, how to compute these
things as iterated integrals.

29
00:02:07,000 --> 00:02:14,000
First, integrating over dy and
then over dx,

30
00:02:14,000 --> 00:02:21,000
or the other way around.
One example that we did,

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00:02:21,000 --> 00:02:25,000
in particular,
was to compute the double

32
00:02:25,000 --> 00:02:29,000
integral of a quarter of a unit
disk.

33
00:02:29,000 --> 00:02:35,000
That was the region where x
squared plus y squared is less

34
00:02:35,000 --> 00:02:40,000
than one and x and y are
positive, of one minus x squared

35
00:02:40,000 --> 00:02:43,000
minus y squared dA.
Well, hopefully,

36
00:02:43,000 --> 00:02:48,000
I kind of convinced you that we
can do it using enough trig and

37
00:02:48,000 --> 00:02:52,000
substitutions and so on,
but it is not very pleasant.

38
00:02:52,000 --> 00:02:56,000
And the reason for that is that
using x and y coordinates here

39
00:02:56,000 --> 00:03:04,000
does not seem very appropriate.
In fact, we can use polar

40
00:03:04,000 --> 00:03:16,000
coordinates instead to compute
this double integral.

41
00:03:16,000 --> 00:03:22,000
Remember that polar coordinates
are about replacing x and y as

42
00:03:22,000 --> 00:03:28,000
coordinates for a point on a
plane by instead r,

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00:03:28,000 --> 00:03:31,000
which is the distance from the
origin to a point,

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00:03:31,000 --> 00:03:35,000
and theta,
which is the angle measured

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00:03:35,000 --> 00:03:40,000
counterclockwise from the
positive x-axis.

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00:03:40,000 --> 00:03:48,000
In terms of r and theta,
you have x equals r cosine

47
00:03:48,000 --> 00:03:54,000
theta, y equals r sine theta.
The claim is we are able,

48
00:03:54,000 --> 00:03:59,000
actually, to do double
integrals in polar coordinates.

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00:03:59,000 --> 00:04:06,000
We just have to learn how to.
Just to draw a quick picture --

50
00:04:06,000 --> 00:04:12,000
When we were integrating in x,
y coordinates,

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00:04:12,000 --> 00:04:16,000
in rectangular coordinates,
we were slicing our region by

52
00:04:16,000 --> 00:04:20,000
gridlines that were either
horizontal or vertical.

53
00:04:20,000 --> 00:04:23,000
And we used that to set up the
iterated integral.

54
00:04:23,000 --> 00:04:28,000
And we said dA became dx dy or
dy dx.

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00:04:28,000 --> 00:04:35,000
Now we are going to actually
integrate, in terms of the polar

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00:04:35,000 --> 00:04:41,000
coordinates, r and theta.
Let's say we will integrate in

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00:04:41,000 --> 00:04:45,000
the order with r first and then
theta.

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00:04:45,000 --> 00:04:50,000
That is the order that makes
the most sense usually when you

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00:04:50,000 --> 00:04:53,000
do polar coordinates.
What does that mean?

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00:04:53,000 --> 00:04:58,000
It means that we will first
focus on a slice where we fix

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00:04:58,000 --> 00:05:02,000
the value of theta and we will
let r vary.

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00:05:02,000 --> 00:05:06,000
That means we fix a direction,
we fix a ray out from the

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00:05:06,000 --> 00:05:11,000
origin in a certain direction.
And we will travel along this

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00:05:11,000 --> 00:05:15,000
ray and see which part of it,
which values of r are in our

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00:05:15,000 --> 00:05:18,000
region.
Here it will be actually pretty

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00:05:18,000 --> 00:05:24,000
easy because r will just start
at zero, and you will have to

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00:05:24,000 --> 00:05:28,000
stop when you exit this quarter
disk.

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00:05:28,000 --> 00:05:31,000
Well, what is the equation of
this circle in polar

69
00:05:31,000 --> 00:05:34,000
coordinates?
It is just r equals one.

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00:05:34,000 --> 00:05:40,000
So, we will stop when r reaches
one.

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00:05:40,000 --> 00:05:43,000
But what about theta?
Well, the first ray that we

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00:05:43,000 --> 00:05:48,000
might want to consider is the
one that goes along the x-axis.

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00:05:48,000 --> 00:05:51,000
That is when theta equals zero.
And we will stop when theta

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00:05:51,000 --> 00:05:55,000
reaches pi over two because we
don't care about the rest of the

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00:05:55,000 --> 00:06:00,000
disk.
We only care about the first

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00:06:00,000 --> 00:06:05,000
quadrant.
We will stop at pi over two.

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00:06:05,000 --> 00:06:11,000
Now, there is a catch,
though, which is that dA is not

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00:06:11,000 --> 00:06:15,000
dr d theta.
Let me explain to you why.

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00:06:15,000 --> 00:06:19,000
Let's say that we are slicing.
What it means is we are cutting

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00:06:19,000 --> 00:06:24,000
our region into little pieces
that are the elementary,

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00:06:24,000 --> 00:06:26,000
you know,
what corresponds to a small

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00:06:26,000 --> 00:06:28,000
rectangle in the x,
y coordinate system,

83
00:06:28,000 --> 00:06:36,000
here would be actually a little
piece of circle between a given

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00:06:36,000 --> 00:06:42,000
radius r and r plus delta r.
And given between an angle

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00:06:42,000 --> 00:06:44,000
theta and theta plus delta
theta.

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00:06:44,000 --> 00:06:48,000
I need to draw,
actually, a bigger picture of

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00:06:48,000 --> 00:06:53,000
that because it makes it really
hard to read.

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00:06:53,000 --> 00:06:58,000
Let's say that I fix an angle
theta and a slightly different

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00:06:58,000 --> 00:07:02,000
one where I have added delta
theta to it.

90
00:07:02,000 --> 00:07:11,000
And let's say that I have a
radius r and I add delta r to

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00:07:11,000 --> 00:07:14,000
it.
Then I will have a little piece

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00:07:14,000 --> 00:07:20,000
of x, y plane that is in here.
And I have to figure out what

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00:07:20,000 --> 00:07:26,000
is its area?
What is delta A for this guy?

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00:07:26,000 --> 00:07:29,000
Well, let's see.
This guy actually,

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00:07:29,000 --> 00:07:33,000
you know, if my delta r and
delta theta are small enough,

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00:07:33,000 --> 00:07:35,000
it will almost look like a
rectangle.

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00:07:35,000 --> 00:07:37,000
It is rotated,
but it is basically a

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00:07:37,000 --> 00:07:39,000
rectangle.
I mean these sides,

99
00:07:39,000 --> 00:07:44,000
of course, are curvy,
but they are short enough and

100
00:07:44,000 --> 00:07:50,000
it is almost straight.
The area here should be this

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00:07:50,000 --> 00:07:55,000
length times that length.
Well, what is this length?

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00:07:55,000 --> 00:08:00,000
That one is easy.
It is delta r.

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00:08:00,000 --> 00:08:03,000
What about that length?
Well, it is not delta theta.

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00:08:03,000 --> 00:08:05,000
It is something slightly
different.

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00:08:05,000 --> 00:08:13,000
It is a piece of a circle of
radius r corresponding to angle

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00:08:13,000 --> 00:08:18,000
delta theta, so it is r delta
theta.

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00:08:18,000 --> 00:08:26,000
So, times r delta theta.
That means now,

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00:08:26,000 --> 00:08:32,000
even if we shrink things and
take smaller and smaller

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00:08:32,000 --> 00:08:37,000
regions, dA is going to be r dr
d theta.

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00:08:37,000 --> 00:08:38,000
That is an important thing to
remember.

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00:08:38,000 --> 00:08:44,000
When you integrate in polar
coordinates, you just set up

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00:08:44,000 --> 00:08:50,000
your bounds in terms of r and
theta, but you replace dA by r

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00:08:50,000 --> 00:08:55,000
dr d theta, not just dr d theta.
And then, of course,

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00:08:55,000 --> 00:08:57,000
we have some function that we
are integrating.

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00:08:57,000 --> 00:09:06,000
Let's say that I call that
thing f then it is the same f

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00:09:06,000 --> 00:09:12,000
that I put up here.
Concretely, how do I do it here?

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00:09:12,000 --> 00:09:18,000
Well, my function f was given
as one minus x squared minus y

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00:09:18,000 --> 00:09:20,000
squared.
And I would like to switch that

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00:09:20,000 --> 00:09:24,000
to polar coordinates.
I want to put r and theta in

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00:09:24,000 --> 00:09:26,000
there.
Well, I have formulas for x and

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00:09:26,000 --> 00:09:30,000
y in polar coordinates so I
could just replace x squared by

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00:09:30,000 --> 00:09:34,000
r squared cosine squared theta,
y squared by r squared sine

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00:09:34,000 --> 00:09:37,000
squared theta.
And that works just fine.

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00:09:37,000 --> 00:09:44,000
But maybe you can observe that
this is x squared plus y

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00:09:44,000 --> 00:09:46,000
squared.
It is just the square of a

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00:09:46,000 --> 00:09:49,000
distance from the origin,
so that is just r squared.

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00:09:49,000 --> 00:09:52,000
That is a useful thing.
You don't strictly need it,

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00:09:52,000 --> 00:09:56,000
but it is much faster if you
see this right away.

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00:09:56,000 --> 00:10:07,000
It saves you writing down a
sine and a cosine.

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00:10:07,000 --> 00:10:13,000
Now we just end up with the
integral from zero to pi over

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00:10:13,000 --> 00:10:20,000
two, integral from zero to one
of one minus r squared r dr d

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00:10:20,000 --> 00:10:24,000
theta.
Now, if I want to compute this

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00:10:24,000 --> 00:10:30,000
integral, so let's first do the
inner integral.

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00:10:30,000 --> 00:10:37,000
If I integrate r minus r cubed,
I will get r squared over two

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00:10:37,000 --> 00:10:43,000
minus r squared over four
between zero and one.

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00:10:43,000 --> 00:10:47,000
And then I will integrate d
theta.

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00:10:47,000 --> 00:10:51,000
What is this equal to?
Well, for r equals one you get

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00:10:51,000 --> 00:10:54,000
one-half minus one-quarter,
which is going to be just

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00:10:54,000 --> 00:10:56,000
one-quarter.
And when you plug in zero you

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00:10:56,000 --> 00:10:59,000
get zero.
So, it is the integral from

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00:10:59,000 --> 00:11:02,000
zero to pi over two of
one-quarter d theta.

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00:11:02,000 --> 00:11:11,000
And that just integrates to
one-quarter times pi over two,

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00:11:11,000 --> 00:11:18,000
which is pi over eight.
That is a lot easier than the

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00:11:18,000 --> 00:11:23,000
way we did it yesterday.
Well, here we were lucky.

145
00:11:23,000 --> 00:11:26,000
I mean usually you will switch
to polar coordinates either

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00:11:26,000 --> 00:11:28,000
because the region is easier to
set up.

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00:11:28,000 --> 00:11:31,000
Here it is indeed easier to set
up because the bounds became

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00:11:31,000 --> 00:11:34,000
very simple.
We don't have that square root

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00:11:34,000 --> 00:11:38,000
of one minus x squared anymore.
Or because the integrant

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00:11:38,000 --> 00:11:40,000
becomes much simpler.
Here our function,

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00:11:40,000 --> 00:11:43,000
well, it is not very
complicated in x,

152
00:11:43,000 --> 00:11:46,000
y coordinates,
but it is even simpler in r

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00:11:46,000 --> 00:11:50,000
theta coordinates.
Here we were very lucky.

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00:11:50,000 --> 00:11:52,000
In general, there is maybe a
trade off.

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00:11:52,000 --> 00:11:55,000
Maybe it will be easier to set
up bounds but maybe the function

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00:11:55,000 --> 00:11:58,000
will become harder because it
will have all these sines and

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00:11:58,000 --> 00:12:01,000
cosines in it.
If our function had been just

158
00:12:01,000 --> 00:12:04,000
x, x is very easy in x,
y coordinates.

159
00:12:04,000 --> 00:12:08,000
Here it becomes r cosine theta.
That means you will have a

160
00:12:08,000 --> 00:12:10,000
little bit of trig to do in the
integral.

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00:12:10,000 --> 00:12:14,000
Not a very big one,
not a very complicated

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00:12:14,000 --> 00:12:21,000
integral, but imagine it could
get potentially much harder.

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00:12:21,000 --> 00:12:25,000
Anyway, that is double
integrals in polar coordinates.

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00:12:25,000 --> 00:12:30,000
And the way you set up the
bounds in general,

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00:12:30,000 --> 00:12:37,000
well, in 99% of the cases you
will integrate over r first.

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00:12:37,000 --> 00:12:40,000
What you will do is you will
look for a given theta what are

167
00:12:40,000 --> 00:12:42,000
the bounds of r to be in the
region.

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00:12:42,000 --> 00:12:46,000
What is the portion of my ray
that is in the given region?

169
00:12:46,000 --> 00:12:49,000
And then you will put bounds
for theta.

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00:12:49,000 --> 00:12:51,000
But conceptually it is the same
as before.

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00:12:51,000 --> 00:12:55,000
Instead of slicing horizontally
or vertically,

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00:12:55,000 --> 00:12:59,000
we slice radially.
We will do more examples in a

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00:12:59,000 --> 00:13:03,000
bit.
Any questions about this or the

174
00:13:03,000 --> 00:13:21,000
general method?
Yes?

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00:13:21,000 --> 00:13:27,000
That is a very good question.
Why do I measure the length

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00:13:27,000 --> 00:13:30,000
inside instead of outside?
Which one do I want?

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00:13:30,000 --> 00:13:34,000
This one.
Here I said this side is r

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00:13:34,000 --> 00:13:36,000
delta theta.
I could have said,

179
00:13:36,000 --> 00:13:38,000
actually, r delta theta is the
length here.

180
00:13:38,000 --> 00:13:41,000
Here it is slightly more,
r plus delta r times delta

181
00:13:41,000 --> 00:13:43,000
theta.
But, if delta r is very small

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00:13:43,000 --> 00:13:46,000
compared to r,
then that is almost the same

183
00:13:46,000 --> 00:13:48,000
thing.
And this is an approximation

184
00:13:48,000 --> 00:13:51,000
anyway.
I took this one because it

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00:13:51,000 --> 00:13:56,000
gives me the simpler formula.
If you take the limit as delta

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00:13:56,000 --> 00:14:01,000
r turns to zero then the two
things become the same anyway.

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00:14:01,000 --> 00:14:04,000
The length, whether you put r
or r plus delta r in here,

188
00:14:04,000 --> 00:14:09,000
doesn't matter anymore.
If you imagine that this guy is

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00:14:09,000 --> 00:14:15,000
infinitely small then,
really, the lengths become the

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00:14:15,000 --> 00:14:17,000
same.
We will also see another proof

191
00:14:17,000 --> 00:14:20,000
of this formula,
using changes of variables,

192
00:14:20,000 --> 00:14:23,000
next week.
But, I mean,

193
00:14:23,000 --> 00:14:28,000
hopefully this is at least
slightly convincing.

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00:14:28,000 --> 00:14:34,000
More questions?
No.

195
00:14:34,000 --> 00:14:40,000
OK.
Let's see.

196
00:14:40,000 --> 00:14:42,000
We have seen how to compute
double integrals.

197
00:14:42,000 --> 00:14:49,000
I have to tell you what they
are good for as well.

198
00:14:49,000 --> 00:14:53,000
The definition we saw yesterday
and the motivation was in terms

199
00:14:53,000 --> 00:14:57,000
of finding volumes,
but that is not going to be our

200
00:14:57,000 --> 00:15:00,000
main preoccupation.
Because finding volumes is fun

201
00:15:00,000 --> 00:15:02,000
but that is not all there is to
life.

202
00:15:02,000 --> 00:15:05,000
I mean, you are doing single
integrals.

203
00:15:05,000 --> 00:15:08,000
When you do single integrals it
is usually not to find the area

204
00:15:08,000 --> 00:15:13,000
of some region of a plane.
It is for something else

205
00:15:13,000 --> 00:15:16,000
usually.
The way we actually think of

206
00:15:16,000 --> 00:15:19,000
the double integral is really as
summing the values of a function

207
00:15:19,000 --> 00:15:22,000
all around this region.
We can use that to get

208
00:15:22,000 --> 00:15:26,000
information about maybe the
region or about the average

209
00:15:26,000 --> 00:15:29,000
value of a function in that
region and so on.

210
00:15:29,000 --> 00:15:39,000
Let's think about various uses
of double integrals.

211
00:15:39,000 --> 00:15:43,000
The first one that I will
mention is actually something

212
00:15:43,000 --> 00:15:47,000
you thought maybe you could do
with a single integral,

213
00:15:47,000 --> 00:15:51,000
but it is useful very often to
do it as a double integral.

214
00:15:51,000 --> 00:15:59,000
It is to find the area of a
given region r.

215
00:15:59,000 --> 00:16:06,000
I give you some region in the
plane and you want to know just

216
00:16:06,000 --> 00:16:08,000
its area.
In various cases,

217
00:16:08,000 --> 00:16:12,000
you could set this up as a
single integral,

218
00:16:12,000 --> 00:16:16,000
but often it could be useful to
set it up as a double integral.

219
00:16:16,000 --> 00:16:20,000
How do you express the area as
a double integral?

220
00:16:20,000 --> 00:16:22,000
Well, the area of this region
is the sum of the areas of all

221
00:16:22,000 --> 00:16:28,000
the little pieces.
It means you want to sum one dA

222
00:16:28,000 --> 00:16:37,000
of the entire region.
The area R is the double

223
00:16:37,000 --> 00:16:46,000
integral over R of a function
one.

224
00:16:46,000 --> 00:16:48,000
One way to think about it,
if you are really still

225
00:16:48,000 --> 00:16:51,000
attached to the idea of double
integral as a volume,

226
00:16:51,000 --> 00:16:54,000
what this measures is the
volume below the graph of a

227
00:16:54,000 --> 00:16:56,000
function one.
The graph of a function one is

228
00:16:56,000 --> 00:16:59,000
just a horizontal plane at
height one.

229
00:16:59,000 --> 00:17:07,000
What you would be measuring is
the volume of a prism with base

230
00:17:07,000 --> 00:17:11,000
r and height one.
And the volume of that would

231
00:17:11,000 --> 00:17:12,000
be, of course,
base times height.

232
00:17:12,000 --> 00:17:16,000
It would just be the area of r
again.

233
00:17:16,000 --> 00:17:18,000
But we don't actually need to
think about it that way.

234
00:17:18,000 --> 00:17:24,000
Really, what we are doing is
summing dA over the entire

235
00:17:24,000 --> 00:17:28,000
region.
A related thing we can do,

236
00:17:28,000 --> 00:17:33,000
imagine that,
actually, this is some physical

237
00:17:33,000 --> 00:17:35,000
object.
I mean, it has to be a flat

238
00:17:35,000 --> 00:17:38,000
object because we are just
dealing with things in the plane

239
00:17:38,000 --> 00:17:41,000
so far.
But you have a flat metal plate

240
00:17:41,000 --> 00:17:45,000
or something and you would like
to know its mass.

241
00:17:45,000 --> 00:17:50,000
Well, its mass is the sum of
the masses of every single

242
00:17:50,000 --> 00:17:52,000
little piece.
You would get that by

243
00:17:52,000 --> 00:17:57,000
integrating the density.
The density for a flat object

244
00:17:57,000 --> 00:18:09,000
would be the mass per unit area.
So, you can get the mass of a

245
00:18:09,000 --> 00:18:23,000
flat object with density.
Let's use delta for density,

246
00:18:23,000 --> 00:18:29,000
which is the mass per unit
area.

247
00:18:29,000 --> 00:18:34,000
Each little piece of your
object will have a mass,

248
00:18:34,000 --> 00:18:40,000
which will be just the density,
times its area for each small

249
00:18:40,000 --> 00:18:45,000
piece.
And you will get the total mass

250
00:18:45,000 --> 00:18:51,000
by summing these things.
The mass will be the double

251
00:18:51,000 --> 00:18:56,000
integral of the density times
the area element.

252
00:18:56,000 --> 00:18:58,000
Now, if it has constant
density,

253
00:18:58,000 --> 00:19:00,000
if it is always the same
material then,

254
00:19:00,000 --> 00:19:03,000
of course,
you could just take the density

255
00:19:03,000 --> 00:19:07,000
out and you will get density
times the total area if you know

256
00:19:07,000 --> 00:19:10,000
that it is always the same
material.

257
00:19:10,000 --> 00:19:13,000
But if, actually,
it has varying density maybe

258
00:19:13,000 --> 00:19:17,000
because it is some metallic
thing with various metals or

259
00:19:17,000 --> 00:19:21,000
with varying thickness or
something then you can still get

260
00:19:21,000 --> 00:19:24,000
the mass by integrating the
density.

261
00:19:24,000 --> 00:19:26,000
Of course, looking at flat
objects might be a little bit

262
00:19:26,000 --> 00:19:28,000
strange.
That is because we are only

263
00:19:28,000 --> 00:19:30,000
doing double integrals so far.
In a few weeks,

264
00:19:30,000 --> 00:19:33,000
we will be triple integrals.
And then we will be able to do

265
00:19:33,000 --> 00:19:36,000
solids in space,
but one thing at a time.

266
00:19:55,000 --> 00:20:08,000
Another useful application is
to find the average value of

267
00:20:08,000 --> 00:20:16,000
some quantity in a region.
What does it mean to take the

268
00:20:16,000 --> 00:20:19,000
average value of some function f
in this region r?

269
00:20:19,000 --> 00:20:22,000
Well, you know what the average
of a finite set of data is.

270
00:20:22,000 --> 00:20:24,000
For example,
if I asked you to compute your

271
00:20:24,000 --> 00:20:26,000
average score on 18.02 problem
sets,

272
00:20:26,000 --> 00:20:30,000
you would just take the scores,
add them and divide by the

273
00:20:30,000 --> 00:20:33,000
number of problem sets.
What if there are infinitely

274
00:20:33,000 --> 00:20:35,000
many things?
Say I ask you to find the

275
00:20:35,000 --> 00:20:37,000
average temperature in this
room.

276
00:20:37,000 --> 00:20:39,000
Well, you would have to measure
the temperature everywhere.

277
00:20:39,000 --> 00:20:42,000
And then add all of these
together and divide by the

278
00:20:42,000 --> 00:20:45,000
number of data points.
But, depending on how careful

279
00:20:45,000 --> 00:20:47,000
you are, actually,
there are potentially

280
00:20:47,000 --> 00:20:49,000
infinitely many points to look
at.

281
00:20:49,000 --> 00:20:54,000
The mathematical way to define
the average of a continuous set

282
00:20:54,000 --> 00:20:58,000
of data is that you actually
integrate the function over the

283
00:20:58,000 --> 00:21:02,000
entire set of data,
and then you divide by the size

284
00:21:02,000 --> 00:21:06,000
of the sample,
which is just the area of the

285
00:21:06,000 --> 00:21:10,000
region.
In fact, the average of f,

286
00:21:10,000 --> 00:21:17,000
the notation we will use
usually for that is f with a bar

287
00:21:17,000 --> 00:21:23,000
on top to tell us it is the
average f.

288
00:21:23,000 --> 00:21:31,000
We say we will take the
integral of f and we will divide

289
00:21:31,000 --> 00:21:38,000
by the area of the region.
You can really think of it as

290
00:21:38,000 --> 00:21:44,000
the sum of the values of f
everywhere divided by the number

291
00:21:44,000 --> 00:21:48,000
of points everywhere.
And so that is an average where

292
00:21:48,000 --> 00:21:51,000
everything is,
actually, equally likely.

293
00:21:51,000 --> 00:21:55,000
That is a uniform average where
all the points on the region,

294
00:21:55,000 --> 00:21:59,000
all the little points of the
region are equally likely.

295
00:21:59,000 --> 00:22:02,000
But maybe if want to do,
say, an average of some solid

296
00:22:02,000 --> 00:22:06,000
with variable density or if you
want to somehow give more

297
00:22:06,000 --> 00:22:10,000
importance to certain parts than
to others then you can actually

298
00:22:10,000 --> 00:22:14,000
do a weighted average.
What is a weighted average?

299
00:22:14,000 --> 00:22:21,000
Well,
in the case of taking the

300
00:22:21,000 --> 00:22:23,000
average your problem sets,
if I tell you problem set one

301
00:22:23,000 --> 00:22:25,000
is worth twice as much as the
others,

302
00:22:25,000 --> 00:22:29,000
then you would count twice that
score in the sum and then you

303
00:22:29,000 --> 00:22:33,000
would count it as two,
of course, when you divide.

304
00:22:33,000 --> 00:22:36,000
The weighted average is the sum
of the values,

305
00:22:36,000 --> 00:22:39,000
but each weighted by a certain
coefficient.

306
00:22:39,000 --> 00:22:43,000
And then you will divide by the
sum of the weight.

307
00:22:43,000 --> 00:22:48,000
It is a bit the same idea as
when we replace area by some

308
00:22:48,000 --> 00:22:53,000
mass that tells you how
important a given piece.

309
00:22:53,000 --> 00:23:02,000
We will actually have a density.
Let's call it delta again.

310
00:23:02,000 --> 00:23:07,000
We will see what we divide by,
but what we will take is the

311
00:23:07,000 --> 00:23:13,000
integral of a function times the
density times the area element.

312
00:23:13,000 --> 00:23:18,000
Because this would correspond
to the mass element telling us

313
00:23:18,000 --> 00:23:22,000
how to weight the various points
of our region.

314
00:23:22,000 --> 00:23:27,000
And then we would divide by the
total weight,

315
00:23:27,000 --> 00:23:34,000
which is the mass of a region,
as defined up there.

316
00:23:34,000 --> 00:23:39,000
If a density is uniform then,
of course, the density gets out

317
00:23:39,000 --> 00:23:44,000
and you can simplify and reduce
to that if all the points are

318
00:23:44,000 --> 00:23:47,000
equally likely.
Why is that important?

319
00:23:47,000 --> 00:23:49,000
Well, that is important for
various applications.

320
00:23:49,000 --> 00:23:53,000
But one that you might have
seen in physics,

321
00:23:53,000 --> 00:23:58,000
we care about maybe where is
the center of mass of a given

322
00:23:58,000 --> 00:24:01,000
object?
The center of mass is basically

323
00:24:01,000 --> 00:24:05,000
a point that you would say is
right in the middle of the

324
00:24:05,000 --> 00:24:06,000
object.
But, of course,

325
00:24:06,000 --> 00:24:10,000
if the object has a very
strange shape or if somehow part

326
00:24:10,000 --> 00:24:14,000
of it is heavier than the rest
then that takes a very different

327
00:24:14,000 --> 00:24:17,000
meaning.
Strictly speaking,

328
00:24:17,000 --> 00:24:20,000
the center of mass of a solid
is the point where you would

329
00:24:20,000 --> 00:24:24,000
have to concentrate all the mass
if you wanted it to behave

330
00:24:24,000 --> 00:24:28,000
equivalently from a point of
view of mechanics,

331
00:24:28,000 --> 00:24:31,000
if you are trying to do
translations of that object.

332
00:24:31,000 --> 00:24:37,000
If you are going to push that
object that would be really

333
00:24:37,000 --> 00:24:42,000
where the equivalent point mass
would lie.

334
00:24:42,000 --> 00:24:44,000
The other way to think about
it,

335
00:24:44,000 --> 00:24:47,000
if I had a flat object then the
center of mass would basically

336
00:24:47,000 --> 00:24:50,000
be the point where I would need
to hold it so it is perfectly

337
00:24:50,000 --> 00:24:52,000
balanced.
And, of course,

338
00:24:52,000 --> 00:24:56,000
I cannot do this.
Well, you get the idea.

339
00:24:56,000 --> 00:24:59,000
And the center of mass of this
eraser is somewhere in the

340
00:24:59,000 --> 00:25:00,000
middle.
And so, in principle,

341
00:25:00,000 --> 00:25:03,000
that is where I would have to
put my finger for it to stay.

342
00:25:03,000 --> 00:25:11,000
Well, it doesn't work.
But that is where the center of

343
00:25:11,000 --> 00:25:19,000
mass should be.
I think it should be in the

344
00:25:19,000 --> 00:25:22,000
middle.
Maybe I shouldn't call this

345
00:25:22,000 --> 00:25:25,000
three.
I should call this 2a,

346
00:25:25,000 --> 00:25:31,000
because it is really a special
case of the average value.

347
00:25:31,000 --> 00:25:47,000
How do we find the center of
mass of a flat object with

348
00:25:47,000 --> 00:25:57,000
density delta.
If you have your object in the

349
00:25:57,000 --> 00:26:00,000
x,
y plane then its center of mass

350
00:26:00,000 --> 00:26:03,000
will be at positions that are
actually just the coordinates of

351
00:26:03,000 --> 00:26:09,000
a center of mass,
will just be weighted averages

352
00:26:09,000 --> 00:26:14,000
of x and y on the solid.
So, the center of mass will be

353
00:26:14,000 --> 00:26:17,000
a position that I will call x
bar, y bar.

354
00:26:17,000 --> 00:26:21,000
And these are really just the
averages, the average values of

355
00:26:21,000 --> 00:26:26,000
x and of y in the solid.
Just to give you the formulas

356
00:26:26,000 --> 00:26:33,000
again, x bar would be one over
the mass times the double

357
00:26:33,000 --> 00:26:42,000
integral of x times density dA.
And the same thing with y.

358
00:26:42,000 --> 00:26:53,000
y bar is the weighted average
of a y coordinate in your

359
00:26:53,000 --> 00:26:56,000
region.
You see, if you take a region

360
00:26:56,000 --> 00:26:59,000
that is symmetric and has
uniform density that will just

361
00:26:59,000 --> 00:27:01,000
give you the center of the
region.

362
00:27:01,000 --> 00:27:05,000
But if the region has a strange
shape or if a density is not

363
00:27:05,000 --> 00:27:08,000
homogeneous,
if parts of it are heavier then

364
00:27:08,000 --> 00:27:12,000
you will get whatever the
weighted average will be.

365
00:27:12,000 --> 00:27:15,000
And that will be the point
where this thing would be

366
00:27:15,000 --> 00:27:19,000
balanced if you were trying to
balance it on a pole or on your

367
00:27:19,000 --> 00:27:20,000
finger.

368
00:27:56,000 --> 00:28:12,000
Any questions so far?
Yes.

369
00:28:12,000 --> 00:28:18,000
No.
Here I didn't set this up as a

370
00:28:18,000 --> 00:28:23,000
iterated integral yet.
The function that I am

371
00:28:23,000 --> 00:28:28,000
integrating is x times delta
where density will be given to

372
00:28:28,000 --> 00:28:30,000
me maybe as a function of x and
y.

373
00:28:30,000 --> 00:28:33,000
And then I will integrate this
dA.

374
00:28:33,000 --> 00:28:36,000
And dA could mean dx over dy,
it could mean dy over dx,

375
00:28:36,000 --> 00:28:40,000
it could be mean r dr d theta.
I will choose how to set it up

376
00:28:40,000 --> 00:28:43,000
depending maybe on the shape of
the region.

377
00:28:43,000 --> 00:28:46,000
If my solid is actually just
going to be round then I might

378
00:28:46,000 --> 00:28:49,000
want to use polar coordinates.
If it is a square,

379
00:28:49,000 --> 00:28:51,000
I might want to use x,
y coordinates.

380
00:28:51,000 --> 00:28:56,000
If it is more complicated,
well, I will choose depending

381
00:28:56,000 --> 00:29:01,000
on how I feel about it.
Yes?

382
00:29:01,000 --> 00:29:05,000
Delta is the density.
In general, it is a function of

383
00:29:05,000 --> 00:29:08,000
x and y.
If you imagine that your solid

384
00:29:08,000 --> 00:29:11,000
is not homogenous then its
density will depend on which

385
00:29:11,000 --> 00:29:15,000
piece of it you are looking at.
Of course, to compute this,

386
00:29:15,000 --> 00:29:18,000
you need to know the density.
If you have a problem asking

387
00:29:18,000 --> 00:29:20,000
you to find the center of mass
of something and you have no

388
00:29:20,000 --> 00:29:23,000
information about the density,
assume it is uniform.

389
00:29:23,000 --> 00:29:26,000
Take the density to be a
constant.

390
00:29:26,000 --> 00:29:29,000
Even take it to be a one.
That is even easier.

391
00:29:29,000 --> 00:29:30,000
I mean it is a general fact of
math.

392
00:29:30,000 --> 00:29:34,000
We don't care about units.
If density is constant,

393
00:29:34,000 --> 00:29:36,000
we might as well take it to be
one.

394
00:29:36,000 --> 00:29:41,000
That just means our mass unit
becomes the area unit.

395
00:29:41,000 --> 00:29:53,000
Yes?
That is a good question.

396
00:29:53,000 --> 00:29:57,000
No, I don't think we could
actually find the center of mass

397
00:29:57,000 --> 00:30:00,000
in polar coordinates by finding
the average of R or the average

398
00:30:00,000 --> 00:30:02,000
of theta.
For example,

399
00:30:02,000 --> 00:30:05,000
take a disk center at the
origin, well,

400
00:30:05,000 --> 00:30:09,000
the center of mass should be at
the origin.

401
00:30:09,000 --> 00:30:12,000
But the average of R is
certainly not zero because R is

402
00:30:12,000 --> 00:30:14,000
positive everywhere.
So, that doesn't work.

403
00:30:14,000 --> 00:30:18,000
You cannot get the polar
coordinates of a center of mass

404
00:30:18,000 --> 00:30:21,000
just by taking the average of R
and the average of theta.

405
00:30:21,000 --> 00:30:23,000
By the way, what is the average
of theta?

406
00:30:23,000 --> 00:30:26,000
If you take theta to from zero
to 2pi, the average theta will

407
00:30:26,000 --> 00:30:28,000
be pi.
If you take it to go from minus

408
00:30:28,000 --> 00:30:30,000
pi to pi, the average theta will
be zero.

409
00:30:30,000 --> 00:30:34,000
So, there is a problem there.
That actually just doesn't

410
00:30:34,000 --> 00:30:38,000
work, so we really have to
compute x bar and y bar.

411
00:30:38,000 --> 00:30:41,000
But still we could set this up
and then switch to polar

412
00:30:41,000 --> 00:30:44,000
coordinates to evaluate this
integral.

413
00:30:44,000 --> 00:30:58,000
But we still would be computing
the average values of x and y.

414
00:30:58,000 --> 00:31:04,000
We are basically re-exploring
mechanics and motion of solids

415
00:31:04,000 --> 00:31:10,000
here.
The next thing is moment of

416
00:31:10,000 --> 00:31:15,000
inertia.
Just to remind you or in case

417
00:31:15,000 --> 00:31:18,000
you somehow haven't seen it in
physics yet,

418
00:31:18,000 --> 00:31:23,000
the moment of inertia is
basically to rotation of a solid

419
00:31:23,000 --> 00:31:26,000
where the mass is to
translation.

420
00:31:26,000 --> 00:31:30,000
In the following sense,
the mass of a solid is what

421
00:31:30,000 --> 00:31:34,000
makes it hard to push it.
How hard it is to throw

422
00:31:34,000 --> 00:31:36,000
something is related to its
mass.

423
00:31:36,000 --> 00:31:41,000
How hard it is to spin
something, on the other hand,

424
00:31:41,000 --> 00:31:44,000
is given by its moment of
inertia.

425
00:31:44,000 --> 00:31:51,000
Maybe I should write this down.
Mass is how hard it is to

426
00:31:51,000 --> 00:31:59,000
impart a translation motion to a
solid.

427
00:31:59,000 --> 00:32:06,000
I am using fancy words today.
And the moment of inertia --

428
00:32:06,000 --> 00:32:13,000
The difference with a mass is
that the moment of inertia is

429
00:32:13,000 --> 00:32:17,000
defined about some axis.
You choose an axis.

430
00:32:17,000 --> 00:32:19,000
Then you would try to measure
how hard it is to spin your

431
00:32:19,000 --> 00:32:21,000
object around that axis.
For example,

432
00:32:21,000 --> 00:32:24,000
you can try to measure how hard
it is to spin this sheet of

433
00:32:24,000 --> 00:32:27,000
paper about an axis that is in
the center of it.

434
00:32:27,000 --> 00:32:30,000
We would try to spin it light
that and see how much effort I

435
00:32:30,000 --> 00:32:35,000
would have to make.
Well, for a sheet of paper not

436
00:32:35,000 --> 00:32:43,000
very much.
That would measure the same

437
00:32:43,000 --> 00:32:58,000
thing but it would be rotation
motion about that axis.

438
00:32:58,000 --> 00:33:02,000
Maybe some of you know the
definition but I am going to try

439
00:33:02,000 --> 00:33:05,000
to derive it again.
I am sorry but it won't be as

440
00:33:05,000 --> 00:33:07,000
quite as detailed as the way you
have probably seen it in

441
00:33:07,000 --> 00:33:09,000
physics, but I am not trying to
replace your physics teachers.

442
00:33:09,000 --> 00:33:16,000
I am sure they are doing a
great job.

443
00:33:16,000 --> 00:33:19,000
What is the idea for the
definition to find a formula for

444
00:33:19,000 --> 00:33:21,000
moment of inertia?
The idea is to think about

445
00:33:21,000 --> 00:33:24,000
kinetic energy.
Kinetic energy is really when

446
00:33:24,000 --> 00:33:28,000
you push something or when you
try to make it move and you have

447
00:33:28,000 --> 00:33:32,000
to put some inertia to it.
Then it has kinetic energy.

448
00:33:32,000 --> 00:33:38,000
And then, if you have the right
device, you can convert back

449
00:33:38,000 --> 00:33:41,000
that kinetic energy into
something else.

450
00:33:41,000 --> 00:33:46,000
If you try to look at the
kinetic energy of a point mass,

451
00:33:46,000 --> 00:33:53,000
so you have something with mass
m going at the velocity v,

452
00:33:53,000 --> 00:33:57,000
well, that will be one-half of
a mass times the square of the

453
00:33:57,000 --> 00:34:00,000
speed.
I hope you have all seen that

454
00:34:00,000 --> 00:34:04,000
formula some time before.
Now, let's say instead of just

455
00:34:04,000 --> 00:34:07,000
trying to push this mass,
I am going to make it spin

456
00:34:07,000 --> 00:34:12,000
around something.
Instead of just somewhere,

457
00:34:12,000 --> 00:34:20,000
maybe I will have the origin,
and I am trying to make it go

458
00:34:20,000 --> 00:34:29,000
around the origin in a circle at
a certain angular velocity.

459
00:34:29,000 --> 00:34:40,000
For a mass m at distance r,
let's call r this distance.

460
00:34:40,000 --> 00:34:47,000
And angular velocity,
let's call the angular velocity

461
00:34:47,000 --> 00:34:50,000
omega.
I think that is what physicists

462
00:34:50,000 --> 00:34:53,000
call it.
Remember angular velocity is

463
00:34:53,000 --> 00:34:57,000
just the rate of the change of
the angle over time.

464
00:34:57,000 --> 00:35:02,000
It is d theta dt, if you want.
Well, what is the kinetic

465
00:35:02,000 --> 00:35:05,000
energy now?
Well, first we have to find out

466
00:35:05,000 --> 00:35:07,000
what the speed is.
What is the speed?

467
00:35:07,000 --> 00:35:10,000
Well,
if we are going on a circle of

468
00:35:10,000 --> 00:35:16,000
radius r at angular velocity
omega that means that in unit

469
00:35:16,000 --> 00:35:22,000
time we rotate by omega and we
go by a distance of r times

470
00:35:22,000 --> 00:35:26,000
omega.
The actual speed is the radius

471
00:35:26,000 --> 00:35:32,000
times angular velocity.
And so the kinetic energy is

472
00:35:32,000 --> 00:35:38,000
one-half mv squared,
which is one-half m r squared

473
00:35:38,000 --> 00:35:41,000
omega squared.
And so,

474
00:35:41,000 --> 00:35:47,000
by similarity with that
formula,

475
00:35:47,000 --> 00:35:51,000
the coefficient of v squared is
the mass,

476
00:35:51,000 --> 00:35:53,000
and here we will say the
coefficient of omega squared,

477
00:35:53,000 --> 00:35:57,000
so this thing is the moment of
inertia.

478
00:35:57,000 --> 00:36:16,000
That is how we define moment of
inertia.

479
00:36:16,000 --> 00:36:20,000
Now, that is only for a point
mass.

480
00:36:20,000 --> 00:36:23,000
And it is kind of fun to spin
just a small bowl,

481
00:36:23,000 --> 00:36:26,000
but maybe you would like to
spin actually a larger solid and

482
00:36:26,000 --> 00:36:29,000
try to define this moment of
inertia.

483
00:36:29,000 --> 00:36:33,000
Well, the moment inertia of a
solid will be just the sum of

484
00:36:33,000 --> 00:36:36,000
the moments of inertia of all
the little pieces.

485
00:36:36,000 --> 00:36:45,000
What we will do is just cut our
solid into little chunks and

486
00:36:45,000 --> 00:36:51,000
will sum this thing for each
little piece.

487
00:36:51,000 --> 00:37:00,000
For a solid with density delta,
each little piece has mass

488
00:37:00,000 --> 00:37:07,000
which is the density times the
amount of area.

489
00:37:07,000 --> 00:37:12,000
This is equal actually.
And the moment of inertia of

490
00:37:12,000 --> 00:37:16,000
that small portion of a solid
will be delta m,

491
00:37:16,000 --> 00:37:18,000
the small mass,
times r squared,

492
00:37:18,000 --> 00:37:25,000
the square of a distance to the
center of the axis along which I

493
00:37:25,000 --> 00:37:29,000
am spinning.
That means if I sum these

494
00:37:29,000 --> 00:37:35,000
things together,
well, it has moment of inertia

495
00:37:35,000 --> 00:37:42,000
delta m times r squared,
which is r squared times the

496
00:37:42,000 --> 00:37:48,000
density times delta A.
And so I will be summing these

497
00:37:48,000 --> 00:37:52,000
things together.
And so, the moment of inertia

498
00:37:52,000 --> 00:37:56,000
about the origin will be the
double integral of r squared

499
00:37:56,000 --> 00:37:59,000
times density times dA.

500
00:38:28,000 --> 00:38:36,000
The final formula for the
moment of inertia about the

501
00:38:36,000 --> 00:38:46,000
origin is the double integral of
a region of r squared density

502
00:38:46,000 --> 00:38:48,000
dA.
If you are going to do it in x,

503
00:38:48,000 --> 00:38:51,000
y coordinates,
of course, r squared becomes x

504
00:38:51,000 --> 00:38:56,000
squared plus y squared,
it is the square of the

505
00:38:56,000 --> 00:39:02,000
distance from the origin.
When you integrate this,

506
00:39:02,000 --> 00:39:05,000
that tells you how hard it is
to spin that solid about the

507
00:39:05,000 --> 00:39:09,000
origin.
The motion that we try to do --

508
00:39:09,000 --> 00:39:15,000
We keep this fixed and then we
just rotate around the origin.

509
00:39:15,000 --> 00:39:20,000
Sorry.
That is a pretty bad picture,

510
00:39:20,000 --> 00:39:26,000
but hopefully you know what I
mean.

511
00:39:26,000 --> 00:39:29,000
And the name we use for that is
I0.

512
00:39:29,000 --> 00:39:37,000
And then the rotational kinetic
energy is one-half times this

513
00:39:37,000 --> 00:39:46,000
moment of inertia times the
square of the angular velocity.

514
00:39:46,000 --> 00:39:54,000
So that shows as that this
replaces the mass for rotation

515
00:39:54,000 --> 00:39:57,000
motions.
OK.

516
00:39:57,000 --> 00:40:03,000
What about other kinds of
rotations?

517
00:40:03,000 --> 00:40:06,000
In particular,
we have been rotating things

518
00:40:06,000 --> 00:40:13,000
about just a point in the plane.
What you could imagine also is

519
00:40:13,000 --> 00:40:19,000
instead you have your solid.
What I have done so far is I

520
00:40:19,000 --> 00:40:22,000
have skewered it this way,
and I am rotating around the

521
00:40:22,000 --> 00:40:25,000
axis.
Instead, I could skewer it

522
00:40:25,000 --> 00:40:27,000
through, say,
the horizontal axis.

523
00:40:27,000 --> 00:40:36,000
And then I could try to spin
about the horizontal axis so

524
00:40:36,000 --> 00:40:46,000
then it would rotate in space in
that direction like that.

525
00:40:46,000 --> 00:40:51,000
Let's say we do rotation about
the x-axis.

526
00:40:51,000 --> 00:40:53,000
Well, the idea would still be
the same.

527
00:40:53,000 --> 00:40:58,000
The moment of inertia for any
small piece of a solid would be

528
00:40:58,000 --> 00:41:02,000
its mass element times the
square of a distance to the x

529
00:41:02,000 --> 00:41:06,000
axes because that will be the
radius of a trajectory.

530
00:41:06,000 --> 00:41:12,000
If you take this point here,
it is going to go in a circle

531
00:41:12,000 --> 00:41:16,000
like that centered on the
x-axis.

532
00:41:16,000 --> 00:41:21,000
So the radius will just be this
distance here.

533
00:41:21,000 --> 00:41:24,000
Well, what is this distance?
It is just y,

534
00:41:24,000 --> 00:41:34,000
or maybe absolute value of y.
Distance to x-axis is absolute

535
00:41:34,000 --> 00:41:39,000
value of y.
What we actually care about is

536
00:41:39,000 --> 00:41:44,000
the square of a distance,
so it will just be y squared.

537
00:41:44,000 --> 00:41:51,000
The moment of inertia about the
x-axis is going to be obtained

538
00:41:51,000 --> 00:41:57,000
by integrating y squared times
the mass element.

539
00:41:57,000 --> 00:42:00,000
It is slightly strange but I
have y in inertia about the

540
00:42:00,000 --> 00:42:03,000
x-axis.
But, if you think about it,

541
00:42:03,000 --> 00:42:07,000
y tells me how far I am from
the x-axis, so how hard it will

542
00:42:07,000 --> 00:42:11,000
be to spin around the x-axis.
And I could do the same about

543
00:42:11,000 --> 00:42:17,000
any axis that I want.
Just I would have to sum the

544
00:42:17,000 --> 00:42:23,000
square of a distance to the axis
of rotation.

545
00:42:23,000 --> 00:42:31,000
Maybe I should do an example.
Yes?

546
00:42:31,000 --> 00:42:36,000
Same thing as above,
distance to the x-axis,

547
00:42:36,000 --> 00:42:39,000
because that is what we care
about.

548
00:42:39,000 --> 00:42:47,000
For the moment of inertia,
we want the square of a

549
00:42:47,000 --> 00:42:52,000
distance to the axis of
rotation.

550
00:42:52,000 --> 00:42:57,000
Let's do an example.
Let's try to figure out if we

551
00:42:57,000 --> 00:43:03,000
have just a uniform disk how
hard it is to spin it around its

552
00:43:03,000 --> 00:43:08,000
center.
That shouldn't be very hard to

553
00:43:08,000 --> 00:43:16,000
figure out.
Say that we have a disk of

554
00:43:16,000 --> 00:43:29,000
radius a and we want to rotate
it about its center.

555
00:43:29,000 --> 00:43:32,000
And let's say that it is of
uniform density.

556
00:43:32,000 --> 00:43:36,000
And let's take just the density
to be a one so that we don't

557
00:43:36,000 --> 00:43:40,000
really care about the density.
What is the moment of inertia

558
00:43:40,000 --> 00:43:45,000
of that?
Well, we have to integrate of

559
00:43:45,000 --> 00:43:51,000
our disk r squared times the
density, which is one,

560
00:43:51,000 --> 00:43:55,000
times dA.
What is r squared?

561
00:43:55,000 --> 00:43:58,000
You have here to resist the
urge to say the radius is just

562
00:43:58,000 --> 00:44:00,000
a.
We know the radius is a.

563
00:44:00,000 --> 00:44:05,000
No, it is not a because we are
looking at rotation of any point

564
00:44:05,000 --> 00:44:07,000
inside this disk.
And, when you are inside the

565
00:44:07,000 --> 00:44:09,000
disk, the distance to the origin
is not a.

566
00:44:09,000 --> 00:44:13,000
It is less than a.
It is actually anything between

567
00:44:13,000 --> 00:44:16,000
zero and a.
Just to point out a pitfall,

568
00:44:16,000 --> 00:44:18,000
r here is really a function on
this disk.

569
00:44:18,000 --> 00:44:20,000
And we are going to integrate
this function.

570
00:44:20,000 --> 00:44:28,000
Don't plug r equals a just yet.
What coordinates do we use to

571
00:44:28,000 --> 00:44:31,000
compute this integral?
They are probably polar

572
00:44:31,000 --> 00:44:35,000
coordinates, unless you want a
repeat of what happened already

573
00:44:35,000 --> 00:44:39,000
with x and y.
That will tell us we want to

574
00:44:39,000 --> 00:44:42,000
integrate r squared time r dr d
theta.

575
00:44:42,000 --> 00:44:47,000
And the bounds for r,
well, r will go from zero to a.

576
00:44:47,000 --> 00:44:51,000
No matter which direction I go
from the origin,

577
00:44:51,000 --> 00:44:56,000
if I fixed it,
r goes from zero to r equals a.

578
00:44:56,000 --> 00:45:02,000
The part of this ray that lives
inside the disk is always from

579
00:45:02,000 --> 00:45:05,000
zero to a.
And theta goes from,

580
00:45:05,000 --> 00:45:11,000
well, zero to 2 pi for example.
And now you can compute this

581
00:45:11,000 --> 00:45:14,000
integral.
Well, I will let you figure it

582
00:45:14,000 --> 00:45:18,000
out.
But the inner integral becomes

583
00:45:18,000 --> 00:45:25,000
a to the four over four and the
outer multiplies things by 2pi,

584
00:45:25,000 --> 00:45:30,000
so you get pi a to the four
over two.

585
00:45:30,000 --> 00:45:33,000
OK.
That is how hard it is to spin

586
00:45:33,000 --> 00:45:37,000
this disk.
Now, what about instead of

587
00:45:37,000 --> 00:45:43,000
spinning it about the center we
decided to spin it about a point

588
00:45:43,000 --> 00:45:46,000
on a second point.
For example, think of a Frisbee.

589
00:45:46,000 --> 00:45:50,000
A Frisbee has this rim so you
can actually try to make it

590
00:45:50,000 --> 00:45:55,000
rotate around the point on the
circumference by holding it near

591
00:45:55,000 --> 00:45:59,000
the rim and spinning it there.
How much harder is that than

592
00:45:59,000 --> 00:46:02,000
around the center?
Well, we will try to compute

593
00:46:02,000 --> 00:46:05,000
now the moment of inertia about
this point.

594
00:46:05,000 --> 00:46:08,000
We have two options.
One is we keep the system of

595
00:46:08,000 --> 00:46:12,000
coordinates centers here.
But then the formula for

596
00:46:12,000 --> 00:46:15,000
distance to this point becomes
harder.

597
00:46:15,000 --> 00:46:18,000
The other option,
which is the one I will choose,

598
00:46:18,000 --> 00:46:21,000
is to change the coordinate so
that this point become the

599
00:46:21,000 --> 00:46:23,000
origin.
Let's do that.

600
00:46:50,000 --> 00:46:58,000
About a point on the
circumference,

601
00:46:58,000 --> 00:47:13,000
what I would have to do maybe
is set up my region like that.

602
00:47:13,000 --> 00:47:17,000
I have moved the origin so that
it is on the circumference of a

603
00:47:17,000 --> 00:47:21,000
disk,
and I will again try to find

604
00:47:21,000 --> 00:47:27,000
the moment of inertia of this
disk about the origin.

605
00:47:27,000 --> 00:47:31,000
It is still,
for the the double integral of

606
00:47:31,000 --> 00:47:36,000
r squared dA.
But now I want to find out how

607
00:47:36,000 --> 00:47:40,000
to set up the integral.
I could try to use x,

608
00:47:40,000 --> 00:47:43,000
y coordinates and it would
work.

609
00:47:43,000 --> 00:47:47,000
Or I can use polar coordinates,
and it works a little bit

610
00:47:47,000 --> 00:47:52,000
better that way.
But both are doable.

611
00:47:52,000 --> 00:47:56,000
Let's say I do it this way.
I have to figure out how to set

612
00:47:56,000 --> 00:48:00,000
up my bounds.
What are the bounds for r?

613
00:48:00,000 --> 00:48:06,000
Well, if I fix a value for
theta, which means I chose an

614
00:48:06,000 --> 00:48:12,000
angle here, now I am shooting a
ray from the origin in that

615
00:48:12,000 --> 00:48:16,000
direction.
I enter my region at r equals

616
00:48:16,000 --> 00:48:19,000
zero.
That hasn't changed.

617
00:48:19,000 --> 00:48:23,000
The question is where do I exit
the region?

618
00:48:23,000 --> 00:48:33,000
What is that distance?
Maybe you have seen it in

619
00:48:33,000 --> 00:48:37,000
recitation, maybe not.
Let's see.

620
00:48:37,000 --> 00:48:40,000
Actually, I should have written
down the radius of a circle is

621
00:48:40,000 --> 00:48:47,000
a.
So this distance here is 2a.

622
00:48:47,000 --> 00:48:52,000
If you draw this segment in
here, you know that here you

623
00:48:52,000 --> 00:48:56,000
have a right angle.
You have a right triangle.

624
00:48:56,000 --> 00:48:59,000
The hypotenuse here has length
2a.

625
00:48:59,000 --> 00:49:08,000
This angle is theta.
Well, this length is 2a cosine

626
00:49:08,000 --> 00:49:14,000
theta.
The polar coordinates equation

627
00:49:14,000 --> 00:49:22,000
of this circle passing through
the origin is r equals 2a cosine

628
00:49:22,000 --> 00:49:27,000
theta.
So, r will go from zero to 2a

629
00:49:27,000 --> 00:49:33,000
cosine theta.
That is the distance here.

630
00:49:33,000 --> 00:49:37,000
Now, what are the bounds for
theta?

631
00:49:37,000 --> 00:49:39,000
It is not quite zero to 2pi
because, actually,

632
00:49:39,000 --> 00:49:42,000
you see in this direction,
if I shoot a ray in this

633
00:49:42,000 --> 00:49:44,000
direction I will never meet my
region.

634
00:49:44,000 --> 00:49:47,000
We have to actually think a bit
more.

635
00:49:47,000 --> 00:49:52,000
Well, the directions in which I
will actually hit my circle are

636
00:49:52,000 --> 00:49:56,000
all the directions in the right
half of a plane.

637
00:49:56,000 --> 00:49:58,000
I mean, of course,
if I shoot very close to the

638
00:49:58,000 --> 00:50:00,000
axis, you might think,
oh, I won't be in there.

639
00:50:00,000 --> 00:50:03,000
But, actually,
that is not true because here

640
00:50:03,000 --> 00:50:05,000
the circle is tangent to the
axis.

641
00:50:05,000 --> 00:50:09,000
No matter which direction I
take, I will still have a little

642
00:50:09,000 --> 00:50:13,000
tiny piece.
The angle actually goes from

643
00:50:13,000 --> 00:50:15,000
minus pi over two to pi over
two.

644
00:50:15,000 --> 00:50:20,000
If you compute that you will
get,

645
00:50:20,000 --> 00:50:25,000
well, the inner integral will
be r to the four over four

646
00:50:25,000 --> 00:50:28,000
between zero and 2a cosine
theta,

647
00:50:28,000 --> 00:50:34,000
which will turn out to be 4a to
the four cosine to the four

648
00:50:34,000 --> 00:50:39,000
theta.
And now you will integrate that

649
00:50:39,000 --> 00:50:43,000
for minus pi over two to pi over
two.

650
00:50:43,000 --> 00:50:47,000
And that is,
again, the evil integral that

651
00:50:47,000 --> 00:50:50,000
we had yesterday.
Either we remember the method

652
00:50:50,000 --> 00:50:53,000
from yesterday or we remember
from yesterday that actually

653
00:50:53,000 --> 00:50:56,000
there are formulas in the notes
to help you.

654
00:50:56,000 --> 00:50:58,000
On homework,
you can use these formulas.

655
00:50:58,000 --> 00:51:04,000
In the notes at the beginning
of section 3b there are formulas

656
00:51:04,000 --> 00:51:08,000
for these particular kinds of
integrals.

657
00:51:08,000 --> 00:51:13,000
And that will end up being
three-halves of pi a to the

658
00:51:13,000 --> 00:51:15,000
four.
In case you wanted to know,

659
00:51:15,000 --> 00:51:18,000
it is three times harder to
spin a Frisbee about a point on

660
00:51:18,000 --> 00:51:20,000
a circumference than around the
center.

661
00:51:20,000 --> 00:51:26,000
We got three times the moment
of inertia about the center.

662
00:51:26,000 --> 00:51:27,000
OK.
That is it.

663
00:51:27,000 --> 00:51:29,000
Have a nice weekend.