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Yesterday we saw how to define
double integrals and how to
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start computing them in terms of
x and y coordinates.
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We have defined the double
integral over a region R and
00:00:41.000 --> 00:00:45.000
plane of a function f of x,
y dA.
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You cannot hear me?
Is the sound working?
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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Does this work?
People are not hearing me in
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the back.
Is it better?
00:01:05.000 --> 00:01:09.000
People are still saying make it
louder.
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Is it better?
OK.
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Great.
Thanks.
00:01:18.000 --> 00:01:22.000
That's not a reason to start
chatting with your friends.
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Thanks.
When we have a region in the x,
00:01:27.000 --> 00:01:31.000
y plane and we have a function
of x and y,
00:01:31.000 --> 00:01:36.000
we are defining the double
integral of f over this region
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by taking basically the sum of
the values of a function
00:01:40.000 --> 00:01:44.000
everywhere in here times the
area element.
00:01:44.000 --> 00:01:48.000
And the definition, actually,
is we split the region into
00:01:48.000 --> 00:01:52.000
lots of tiny little pieces,
we multiply the value of a
00:01:52.000 --> 00:01:55.000
function at the point times the
area of a little piece and we
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sum that everywhere.
And we have seen,
00:01:59.000 --> 00:02:07.000
actually, how to compute these
things as iterated integrals.
00:02:07.000 --> 00:02:14.000
First, integrating over dy and
then over dx,
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or the other way around.
One example that we did,
00:02:21.000 --> 00:02:25.000
in particular,
was to compute the double
00:02:25.000 --> 00:02:29.000
integral of a quarter of a unit
disk.
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That was the region where x
squared plus y squared is less
00:02:35.000 --> 00:02:40.000
than one and x and y are
positive, of one minus x squared
00:02:40.000 --> 00:02:43.000
minus y squared dA.
Well, hopefully,
00:02:43.000 --> 00:02:48.000
I kind of convinced you that we
can do it using enough trig and
00:02:48.000 --> 00:02:52.000
substitutions and so on,
but it is not very pleasant.
00:02:52.000 --> 00:02:56.000
And the reason for that is that
using x and y coordinates here
00:02:56.000 --> 00:03:04.000
does not seem very appropriate.
In fact, we can use polar
00:03:04.000 --> 00:03:16.000
coordinates instead to compute
this double integral.
00:03:16.000 --> 00:03:22.000
Remember that polar coordinates
are about replacing x and y as
00:03:22.000 --> 00:03:28.000
coordinates for a point on a
plane by instead r,
00:03:28.000 --> 00:03:31.000
which is the distance from the
origin to a point,
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and theta,
which is the angle measured
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counterclockwise from the
positive x-axis.
00:03:40.000 --> 00:03:48.000
In terms of r and theta,
you have x equals r cosine
00:03:48.000 --> 00:03:54.000
theta, y equals r sine theta.
The claim is we are able,
00:03:54.000 --> 00:03:59.000
actually, to do double
integrals in polar coordinates.
00:03:59.000 --> 00:04:06.000
We just have to learn how to.
Just to draw a quick picture --
00:04:06.000 --> 00:04:12.000
When we were integrating in x,
y coordinates,
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in rectangular coordinates,
we were slicing our region by
00:04:16.000 --> 00:04:20.000
gridlines that were either
horizontal or vertical.
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And we used that to set up the
iterated integral.
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And we said dA became dx dy or
dy dx.
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Now we are going to actually
integrate, in terms of the polar
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coordinates, r and theta.
Let's say we will integrate in
00:04:41.000 --> 00:04:45.000
the order with r first and then
theta.
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That is the order that makes
the most sense usually when you
00:04:50.000 --> 00:04:53.000
do polar coordinates.
What does that mean?
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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the value of theta and we will
let r vary.
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That means we fix a direction,
we fix a ray out from the
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origin in a certain direction.
And we will travel along this
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ray and see which part of it,
which values of r are in our
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region.
Here it will be actually pretty
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easy because r will just start
at zero, and you will have to
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stop when you exit this quarter
disk.
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Well, what is the equation of
this circle in polar
00:05:31.000 --> 00:05:34.000
coordinates?
It is just r equals one.
00:05:34.000 --> 00:05:40.000
So, we will stop when r reaches
one.
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But what about theta?
Well, the first ray that we
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might want to consider is the
one that goes along the x-axis.
00:05:48.000 --> 00:05:51.000
That is when theta equals zero.
And we will stop when theta
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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disk.
We only care about the first
00:06:00.000 --> 00:06:05.000
quadrant.
We will stop at pi over two.
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Now, there is a catch,
though, which is that dA is not
00:06:11.000 --> 00:06:15.000
dr d theta.
Let me explain to you why.
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Let's say that we are slicing.
What it means is we are cutting
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our region into little pieces
that are the elementary,
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you know,
what corresponds to a small
00:06:26.000 --> 00:06:28.000
rectangle in the x,
y coordinate system,
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here would be actually a little
piece of circle between a given
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radius r and r plus delta r.
And given between an angle
00:06:42.000 --> 00:06:44.000
theta and theta plus delta
theta.
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I need to draw,
actually, a bigger picture of
00:06:48.000 --> 00:06:53.000
that because it makes it really
hard to read.
00:06:53.000 --> 00:06:58.000
Let's say that I fix an angle
theta and a slightly different
00:06:58.000 --> 00:07:02.000
one where I have added delta
theta to it.
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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it.
Then I will have a little piece
00:07:14.000 --> 00:07:20.000
of x, y plane that is in here.
And I have to figure out what
00:07:20.000 --> 00:07:26.000
is its area?
What is delta A for this guy?
00:07:26.000 --> 00:07:29.000
Well, let's see.
This guy actually,
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you know, if my delta r and
delta theta are small enough,
00:07:33.000 --> 00:07:35.000
it will almost look like a
rectangle.
00:07:35.000 --> 00:07:37.000
It is rotated,
but it is basically a
00:07:37.000 --> 00:07:39.000
rectangle.
I mean these sides,
00:07:39.000 --> 00:07:44.000
of course, are curvy,
but they are short enough and
00:07:44.000 --> 00:07:50.000
it is almost straight.
The area here should be this
00:07:50.000 --> 00:07:55.000
length times that length.
Well, what is this length?
00:07:55.000 --> 00:08:00.000
That one is easy.
It is delta r.
00:08:00.000 --> 00:08:03.000
What about that length?
Well, it is not delta theta.
00:08:03.000 --> 00:08:05.000
It is something slightly
different.
00:08:05.000 --> 00:08:13.000
It is a piece of a circle of
radius r corresponding to angle
00:08:13.000 --> 00:08:18.000
delta theta, so it is r delta
theta.
00:08:18.000 --> 00:08:26.000
So, times r delta theta.
That means now,
00:08:26.000 --> 00:08:32.000
even if we shrink things and
take smaller and smaller
00:08:32.000 --> 00:08:37.000
regions, dA is going to be r dr
d theta.
00:08:37.000 --> 00:08:38.000
That is an important thing to
remember.
00:08:38.000 --> 00:08:44.000
When you integrate in polar
coordinates, you just set up
00:08:44.000 --> 00:08:50.000
your bounds in terms of r and
theta, but you replace dA by r
00:08:50.000 --> 00:08:55.000
dr d theta, not just dr d theta.
And then, of course,
00:08:55.000 --> 00:08:57.000
we have some function that we
are integrating.
00:08:57.000 --> 00:09:06.000
Let's say that I call that
thing f then it is the same f
00:09:06.000 --> 00:09:12.000
that I put up here.
Concretely, how do I do it here?
00:09:12.000 --> 00:09:18.000
Well, my function f was given
as one minus x squared minus y
00:09:18.000 --> 00:09:20.000
squared.
And I would like to switch that
00:09:20.000 --> 00:09:24.000
to polar coordinates.
I want to put r and theta in
00:09:24.000 --> 00:09:26.000
there.
Well, I have formulas for x and
00:09:26.000 --> 00:09:30.000
y in polar coordinates so I
could just replace x squared by
00:09:30.000 --> 00:09:34.000
r squared cosine squared theta,
y squared by r squared sine
00:09:34.000 --> 00:09:37.000
squared theta.
And that works just fine.
00:09:37.000 --> 00:09:44.000
But maybe you can observe that
this is x squared plus y
00:09:44.000 --> 00:09:46.000
squared.
It is just the square of a
00:09:46.000 --> 00:09:49.000
distance from the origin,
so that is just r squared.
00:09:49.000 --> 00:09:52.000
That is a useful thing.
You don't strictly need it,
00:09:52.000 --> 00:09:56.000
but it is much faster if you
see this right away.
00:09:56.000 --> 00:10:07.000
It saves you writing down a
sine and a cosine.
00:10:07.000 --> 00:10:13.000
Now we just end up with the
integral from zero to pi over
00:10:13.000 --> 00:10:20.000
two, integral from zero to one
of one minus r squared r dr d
00:10:20.000 --> 00:10:24.000
theta.
Now, if I want to compute this
00:10:24.000 --> 00:10:30.000
integral, so let's first do the
inner integral.
00:10:30.000 --> 00:10:37.000
If I integrate r minus r cubed,
I will get r squared over two
00:10:37.000 --> 00:10:43.000
minus r squared over four
between zero and one.
00:10:43.000 --> 00:10:47.000
And then I will integrate d
theta.
00:10:47.000 --> 00:10:51.000
What is this equal to?
Well, for r equals one you get
00:10:51.000 --> 00:10:54.000
one-half minus one-quarter,
which is going to be just
00:10:54.000 --> 00:10:56.000
one-quarter.
And when you plug in zero you
00:10:56.000 --> 00:10:59.000
get zero.
So, it is the integral from
00:10:59.000 --> 00:11:02.000
zero to pi over two of
one-quarter d theta.
00:11:02.000 --> 00:11:11.000
And that just integrates to
one-quarter times pi over two,
00:11:11.000 --> 00:11:18.000
which is pi over eight.
That is a lot easier than the
00:11:18.000 --> 00:11:23.000
way we did it yesterday.
Well, here we were lucky.
00:11:23.000 --> 00:11:26.000
I mean usually you will switch
to polar coordinates either
00:11:26.000 --> 00:11:28.000
because the region is easier to
set up.
00:11:28.000 --> 00:11:31.000
Here it is indeed easier to set
up because the bounds became
00:11:31.000 --> 00:11:34.000
very simple.
We don't have that square root
00:11:34.000 --> 00:11:38.000
of one minus x squared anymore.
Or because the integrant
00:11:38.000 --> 00:11:40.000
becomes much simpler.
Here our function,
00:11:40.000 --> 00:11:43.000
well, it is not very
complicated in x,
00:11:43.000 --> 00:11:46.000
y coordinates,
but it is even simpler in r
00:11:46.000 --> 00:11:50.000
theta coordinates.
Here we were very lucky.
00:11:50.000 --> 00:11:52.000
In general, there is maybe a
trade off.
00:11:52.000 --> 00:11:55.000
Maybe it will be easier to set
up bounds but maybe the function
00:11:55.000 --> 00:11:58.000
will become harder because it
will have all these sines and
00:11:58.000 --> 00:12:01.000
cosines in it.
If our function had been just
00:12:01.000 --> 00:12:04.000
x, x is very easy in x,
y coordinates.
00:12:04.000 --> 00:12:08.000
Here it becomes r cosine theta.
That means you will have a
00:12:08.000 --> 00:12:10.000
little bit of trig to do in the
integral.
00:12:10.000 --> 00:12:14.000
Not a very big one,
not a very complicated
00:12:14.000 --> 00:12:21.000
integral, but imagine it could
get potentially much harder.
00:12:21.000 --> 00:12:25.000
Anyway, that is double
integrals in polar coordinates.
00:12:25.000 --> 00:12:30.000
And the way you set up the
bounds in general,
00:12:30.000 --> 00:12:37.000
well, in 99% of the cases you
will integrate over r first.
00:12:37.000 --> 00:12:40.000
What you will do is you will
look for a given theta what are
00:12:40.000 --> 00:12:42.000
the bounds of r to be in the
region.
00:12:42.000 --> 00:12:46.000
What is the portion of my ray
that is in the given region?
00:12:46.000 --> 00:12:49.000
And then you will put bounds
for theta.
00:12:49.000 --> 00:12:51.000
But conceptually it is the same
as before.
00:12:51.000 --> 00:12:55.000
Instead of slicing horizontally
or vertically,
00:12:55.000 --> 00:12:59.000
we slice radially.
We will do more examples in a
00:12:59.000 --> 00:13:03.000
bit.
Any questions about this or the
00:13:03.000 --> 00:13:21.000
general method?
Yes?
00:13:21.000 --> 00:13:27.000
That is a very good question.
Why do I measure the length
00:13:27.000 --> 00:13:30.000
inside instead of outside?
Which one do I want?
00:13:30.000 --> 00:13:34.000
This one.
Here I said this side is r
00:13:34.000 --> 00:13:36.000
delta theta.
I could have said,
00:13:36.000 --> 00:13:38.000
actually, r delta theta is the
length here.
00:13:38.000 --> 00:13:41.000
Here it is slightly more,
r plus delta r times delta
00:13:41.000 --> 00:13:43.000
theta.
But, if delta r is very small
00:13:43.000 --> 00:13:46.000
compared to r,
then that is almost the same
00:13:46.000 --> 00:13:48.000
thing.
And this is an approximation
00:13:48.000 --> 00:13:51.000
anyway.
I took this one because it
00:13:51.000 --> 00:13:56.000
gives me the simpler formula.
If you take the limit as delta
00:13:56.000 --> 00:14:01.000
r turns to zero then the two
things become the same anyway.
00:14:01.000 --> 00:14:04.000
The length, whether you put r
or r plus delta r in here,
00:14:04.000 --> 00:14:09.000
doesn't matter anymore.
If you imagine that this guy is
00:14:09.000 --> 00:14:15.000
infinitely small then,
really, the lengths become the
00:14:15.000 --> 00:14:17.000
same.
We will also see another proof
00:14:17.000 --> 00:14:20.000
of this formula,
using changes of variables,
00:14:20.000 --> 00:14:23.000
next week.
But, I mean,
00:14:23.000 --> 00:14:28.000
hopefully this is at least
slightly convincing.
00:14:28.000 --> 00:14:34.000
More questions?
No.
00:14:34.000 --> 00:14:40.000
OK.
Let's see.
00:14:40.000 --> 00:14:42.000
We have seen how to compute
double integrals.
00:14:42.000 --> 00:14:49.000
I have to tell you what they
are good for as well.
00:14:49.000 --> 00:14:53.000
The definition we saw yesterday
and the motivation was in terms
00:14:53.000 --> 00:14:57.000
of finding volumes,
but that is not going to be our
00:14:57.000 --> 00:15:00.000
main preoccupation.
Because finding volumes is fun
00:15:00.000 --> 00:15:02.000
but that is not all there is to
life.
00:15:02.000 --> 00:15:05.000
I mean, you are doing single
integrals.
00:15:05.000 --> 00:15:08.000
When you do single integrals it
is usually not to find the area
00:15:08.000 --> 00:15:13.000
of some region of a plane.
It is for something else
00:15:13.000 --> 00:15:16.000
usually.
The way we actually think of
00:15:16.000 --> 00:15:19.000
the double integral is really as
summing the values of a function
00:15:19.000 --> 00:15:22.000
all around this region.
We can use that to get
00:15:22.000 --> 00:15:26.000
information about maybe the
region or about the average
00:15:26.000 --> 00:15:29.000
value of a function in that
region and so on.
00:15:29.000 --> 00:15:39.000
Let's think about various uses
of double integrals.
00:15:39.000 --> 00:15:43.000
The first one that I will
mention is actually something
00:15:43.000 --> 00:15:47.000
you thought maybe you could do
with a single integral,
00:15:47.000 --> 00:15:51.000
but it is useful very often to
do it as a double integral.
00:15:51.000 --> 00:15:59.000
It is to find the area of a
given region r.
00:15:59.000 --> 00:16:06.000
I give you some region in the
plane and you want to know just
00:16:06.000 --> 00:16:08.000
its area.
In various cases,
00:16:08.000 --> 00:16:12.000
you could set this up as a
single integral,
00:16:12.000 --> 00:16:16.000
but often it could be useful to
set it up as a double integral.
00:16:16.000 --> 00:16:20.000
How do you express the area as
a double integral?
00:16:20.000 --> 00:16:22.000
Well, the area of this region
is the sum of the areas of all
00:16:22.000 --> 00:16:28.000
the little pieces.
It means you want to sum one dA
00:16:28.000 --> 00:16:37.000
of the entire region.
The area R is the double
00:16:37.000 --> 00:16:46.000
integral over R of a function
one.
00:16:46.000 --> 00:16:48.000
One way to think about it,
if you are really still
00:16:48.000 --> 00:16:51.000
attached to the idea of double
integral as a volume,
00:16:51.000 --> 00:16:54.000
what this measures is the
volume below the graph of a
00:16:54.000 --> 00:16:56.000
function one.
The graph of a function one is
00:16:56.000 --> 00:16:59.000
just a horizontal plane at
height one.
00:16:59.000 --> 00:17:07.000
What you would be measuring is
the volume of a prism with base
00:17:07.000 --> 00:17:11.000
r and height one.
And the volume of that would
00:17:11.000 --> 00:17:12.000
be, of course,
base times height.
00:17:12.000 --> 00:17:16.000
It would just be the area of r
again.
00:17:16.000 --> 00:17:18.000
But we don't actually need to
think about it that way.
00:17:18.000 --> 00:17:24.000
Really, what we are doing is
summing dA over the entire
00:17:24.000 --> 00:17:28.000
region.
A related thing we can do,
00:17:28.000 --> 00:17:33.000
imagine that,
actually, this is some physical
00:17:33.000 --> 00:17:35.000
object.
I mean, it has to be a flat
00:17:35.000 --> 00:17:38.000
object because we are just
dealing with things in the plane
00:17:38.000 --> 00:17:41.000
so far.
But you have a flat metal plate
00:17:41.000 --> 00:17:45.000
or something and you would like
to know its mass.
00:17:45.000 --> 00:17:50.000
Well, its mass is the sum of
the masses of every single
00:17:50.000 --> 00:17:52.000
little piece.
You would get that by
00:17:52.000 --> 00:17:57.000
integrating the density.
The density for a flat object
00:17:57.000 --> 00:18:09.000
would be the mass per unit area.
So, you can get the mass of a
00:18:09.000 --> 00:18:23.000
flat object with density.
Let's use delta for density,
00:18:23.000 --> 00:18:29.000
which is the mass per unit
area.
00:18:29.000 --> 00:18:34.000
Each little piece of your
object will have a mass,
00:18:34.000 --> 00:18:40.000
which will be just the density,
times its area for each small
00:18:40.000 --> 00:18:45.000
piece.
And you will get the total mass
00:18:45.000 --> 00:18:51.000
by summing these things.
The mass will be the double
00:18:51.000 --> 00:18:56.000
integral of the density times
the area element.
00:18:56.000 --> 00:18:58.000
Now, if it has constant
density,
00:18:58.000 --> 00:19:00.000
if it is always the same
material then,
00:19:00.000 --> 00:19:03.000
of course,
you could just take the density
00:19:03.000 --> 00:19:07.000
out and you will get density
times the total area if you know
00:19:07.000 --> 00:19:10.000
that it is always the same
material.
00:19:10.000 --> 00:19:13.000
But if, actually,
it has varying density maybe
00:19:13.000 --> 00:19:17.000
because it is some metallic
thing with various metals or
00:19:17.000 --> 00:19:21.000
with varying thickness or
something then you can still get
00:19:21.000 --> 00:19:24.000
the mass by integrating the
density.
00:19:24.000 --> 00:19:26.000
Of course, looking at flat
objects might be a little bit
00:19:26.000 --> 00:19:28.000
strange.
That is because we are only
00:19:28.000 --> 00:19:30.000
doing double integrals so far.
In a few weeks,
00:19:30.000 --> 00:19:33.000
we will be triple integrals.
And then we will be able to do
00:19:33.000 --> 00:19:36.000
solids in space,
but one thing at a time.
00:19:55.000 --> 00:20:08.000
Another useful application is
to find the average value of
00:20:08.000 --> 00:20:16.000
some quantity in a region.
What does it mean to take the
00:20:16.000 --> 00:20:19.000
average value of some function f
in this region r?
00:20:19.000 --> 00:20:22.000
Well, you know what the average
of a finite set of data is.
00:20:22.000 --> 00:20:24.000
For example,
if I asked you to compute your
00:20:24.000 --> 00:20:26.000
average score on 18.02 problem
sets,
00:20:26.000 --> 00:20:30.000
you would just take the scores,
add them and divide by the
00:20:30.000 --> 00:20:33.000
number of problem sets.
What if there are infinitely
00:20:33.000 --> 00:20:35.000
many things?
Say I ask you to find the
00:20:35.000 --> 00:20:37.000
average temperature in this
room.
00:20:37.000 --> 00:20:39.000
Well, you would have to measure
the temperature everywhere.
00:20:39.000 --> 00:20:42.000
And then add all of these
together and divide by the
00:20:42.000 --> 00:20:45.000
number of data points.
But, depending on how careful
00:20:45.000 --> 00:20:47.000
you are, actually,
there are potentially
00:20:47.000 --> 00:20:49.000
infinitely many points to look
at.
00:20:49.000 --> 00:20:54.000
The mathematical way to define
the average of a continuous set
00:20:54.000 --> 00:20:58.000
of data is that you actually
integrate the function over the
00:20:58.000 --> 00:21:02.000
entire set of data,
and then you divide by the size
00:21:02.000 --> 00:21:06.000
of the sample,
which is just the area of the
00:21:06.000 --> 00:21:10.000
region.
In fact, the average of f,
00:21:10.000 --> 00:21:17.000
the notation we will use
usually for that is f with a bar
00:21:17.000 --> 00:21:23.000
on top to tell us it is the
average f.
00:21:23.000 --> 00:21:31.000
We say we will take the
integral of f and we will divide
00:21:31.000 --> 00:21:38.000
by the area of the region.
You can really think of it as
00:21:38.000 --> 00:21:44.000
the sum of the values of f
everywhere divided by the number
00:21:44.000 --> 00:21:48.000
of points everywhere.
And so that is an average where
00:21:48.000 --> 00:21:51.000
everything is,
actually, equally likely.
00:21:51.000 --> 00:21:55.000
That is a uniform average where
all the points on the region,
00:21:55.000 --> 00:21:59.000
all the little points of the
region are equally likely.
00:21:59.000 --> 00:22:02.000
But maybe if want to do,
say, an average of some solid
00:22:02.000 --> 00:22:06.000
with variable density or if you
want to somehow give more
00:22:06.000 --> 00:22:10.000
importance to certain parts than
to others then you can actually
00:22:10.000 --> 00:22:14.000
do a weighted average.
What is a weighted average?
00:22:14.000 --> 00:22:21.000
Well,
in the case of taking the
00:22:21.000 --> 00:22:23.000
average your problem sets,
if I tell you problem set one
00:22:23.000 --> 00:22:25.000
is worth twice as much as the
others,
00:22:25.000 --> 00:22:29.000
then you would count twice that
score in the sum and then you
00:22:29.000 --> 00:22:33.000
would count it as two,
of course, when you divide.
00:22:33.000 --> 00:22:36.000
The weighted average is the sum
of the values,
00:22:36.000 --> 00:22:39.000
but each weighted by a certain
coefficient.
00:22:39.000 --> 00:22:43.000
And then you will divide by the
sum of the weight.
00:22:43.000 --> 00:22:48.000
It is a bit the same idea as
when we replace area by some
00:22:48.000 --> 00:22:53.000
mass that tells you how
important a given piece.
00:22:53.000 --> 00:23:02.000
We will actually have a density.
Let's call it delta again.
00:23:02.000 --> 00:23:07.000
We will see what we divide by,
but what we will take is the
00:23:07.000 --> 00:23:13.000
integral of a function times the
density times the area element.
00:23:13.000 --> 00:23:18.000
Because this would correspond
to the mass element telling us
00:23:18.000 --> 00:23:22.000
how to weight the various points
of our region.
00:23:22.000 --> 00:23:27.000
And then we would divide by the
total weight,
00:23:27.000 --> 00:23:34.000
which is the mass of a region,
as defined up there.
00:23:34.000 --> 00:23:39.000
If a density is uniform then,
of course, the density gets out
00:23:39.000 --> 00:23:44.000
and you can simplify and reduce
to that if all the points are
00:23:44.000 --> 00:23:47.000
equally likely.
Why is that important?
00:23:47.000 --> 00:23:49.000
Well, that is important for
various applications.
00:23:49.000 --> 00:23:53.000
But one that you might have
seen in physics,
00:23:53.000 --> 00:23:58.000
we care about maybe where is
the center of mass of a given
00:23:58.000 --> 00:24:01.000
object?
The center of mass is basically
00:24:01.000 --> 00:24:05.000
a point that you would say is
right in the middle of the
00:24:05.000 --> 00:24:06.000
object.
But, of course,
00:24:06.000 --> 00:24:10.000
if the object has a very
strange shape or if somehow part
00:24:10.000 --> 00:24:14.000
of it is heavier than the rest
then that takes a very different
00:24:14.000 --> 00:24:17.000
meaning.
Strictly speaking,
00:24:17.000 --> 00:24:20.000
the center of mass of a solid
is the point where you would
00:24:20.000 --> 00:24:24.000
have to concentrate all the mass
if you wanted it to behave
00:24:24.000 --> 00:24:28.000
equivalently from a point of
view of mechanics,
00:24:28.000 --> 00:24:31.000
if you are trying to do
translations of that object.
00:24:31.000 --> 00:24:37.000
If you are going to push that
object that would be really
00:24:37.000 --> 00:24:42.000
where the equivalent point mass
would lie.
00:24:42.000 --> 00:24:44.000
The other way to think about
it,
00:24:44.000 --> 00:24:47.000
if I had a flat object then the
center of mass would basically
00:24:47.000 --> 00:24:50.000
be the point where I would need
to hold it so it is perfectly
00:24:50.000 --> 00:24:52.000
balanced.
And, of course,
00:24:52.000 --> 00:24:56.000
I cannot do this.
Well, you get the idea.
00:24:56.000 --> 00:24:59.000
And the center of mass of this
eraser is somewhere in the
00:24:59.000 --> 00:25:00.000
middle.
And so, in principle,
00:25:00.000 --> 00:25:03.000
that is where I would have to
put my finger for it to stay.
00:25:03.000 --> 00:25:11.000
Well, it doesn't work.
But that is where the center of
00:25:11.000 --> 00:25:19.000
mass should be.
I think it should be in the
00:25:19.000 --> 00:25:22.000
middle.
Maybe I shouldn't call this
00:25:22.000 --> 00:25:25.000
three.
I should call this 2a,
00:25:25.000 --> 00:25:31.000
because it is really a special
case of the average value.
00:25:31.000 --> 00:25:47.000
How do we find the center of
mass of a flat object with
00:25:47.000 --> 00:25:57.000
density delta.
If you have your object in the
00:25:57.000 --> 00:26:00.000
x,
y plane then its center of mass
00:26:00.000 --> 00:26:03.000
will be at positions that are
actually just the coordinates of
00:26:03.000 --> 00:26:09.000
a center of mass,
will just be weighted averages
00:26:09.000 --> 00:26:14.000
of x and y on the solid.
So, the center of mass will be
00:26:14.000 --> 00:26:17.000
a position that I will call x
bar, y bar.
00:26:17.000 --> 00:26:21.000
And these are really just the
averages, the average values of
00:26:21.000 --> 00:26:26.000
x and of y in the solid.
Just to give you the formulas
00:26:26.000 --> 00:26:33.000
again, x bar would be one over
the mass times the double
00:26:33.000 --> 00:26:42.000
integral of x times density dA.
And the same thing with y.
00:26:42.000 --> 00:26:53.000
y bar is the weighted average
of a y coordinate in your
00:26:53.000 --> 00:26:56.000
region.
You see, if you take a region
00:26:56.000 --> 00:26:59.000
that is symmetric and has
uniform density that will just
00:26:59.000 --> 00:27:01.000
give you the center of the
region.
00:27:01.000 --> 00:27:05.000
But if the region has a strange
shape or if a density is not
00:27:05.000 --> 00:27:08.000
homogeneous,
if parts of it are heavier then
00:27:08.000 --> 00:27:12.000
you will get whatever the
weighted average will be.
00:27:12.000 --> 00:27:15.000
And that will be the point
where this thing would be
00:27:15.000 --> 00:27:19.000
balanced if you were trying to
balance it on a pole or on your
00:27:19.000 --> 00:27:20.000
finger.
00:27:56.000 --> 00:28:12.000
Any questions so far?
Yes.
00:28:12.000 --> 00:28:18.000
No.
Here I didn't set this up as a
00:28:18.000 --> 00:28:23.000
iterated integral yet.
The function that I am
00:28:23.000 --> 00:28:28.000
integrating is x times delta
where density will be given to
00:28:28.000 --> 00:28:30.000
me maybe as a function of x and
y.
00:28:30.000 --> 00:28:33.000
And then I will integrate this
dA.
00:28:33.000 --> 00:28:36.000
And dA could mean dx over dy,
it could mean dy over dx,
00:28:36.000 --> 00:28:40.000
it could be mean r dr d theta.
I will choose how to set it up
00:28:40.000 --> 00:28:43.000
depending maybe on the shape of
the region.
00:28:43.000 --> 00:28:46.000
If my solid is actually just
going to be round then I might
00:28:46.000 --> 00:28:49.000
want to use polar coordinates.
If it is a square,
00:28:49.000 --> 00:28:51.000
I might want to use x,
y coordinates.
00:28:51.000 --> 00:28:56.000
If it is more complicated,
well, I will choose depending
00:28:56.000 --> 00:29:01.000
on how I feel about it.
Yes?
00:29:01.000 --> 00:29:05.000
Delta is the density.
In general, it is a function of
00:29:05.000 --> 00:29:08.000
x and y.
If you imagine that your solid
00:29:08.000 --> 00:29:11.000
is not homogenous then its
density will depend on which
00:29:11.000 --> 00:29:15.000
piece of it you are looking at.
Of course, to compute this,
00:29:15.000 --> 00:29:18.000
you need to know the density.
If you have a problem asking
00:29:18.000 --> 00:29:20.000
you to find the center of mass
of something and you have no
00:29:20.000 --> 00:29:23.000
information about the density,
assume it is uniform.
00:29:23.000 --> 00:29:26.000
Take the density to be a
constant.
00:29:26.000 --> 00:29:29.000
Even take it to be a one.
That is even easier.
00:29:29.000 --> 00:29:30.000
I mean it is a general fact of
math.
00:29:30.000 --> 00:29:34.000
We don't care about units.
If density is constant,
00:29:34.000 --> 00:29:36.000
we might as well take it to be
one.
00:29:36.000 --> 00:29:41.000
That just means our mass unit
becomes the area unit.
00:29:41.000 --> 00:29:53.000
Yes?
That is a good question.
00:29:53.000 --> 00:29:57.000
No, I don't think we could
actually find the center of mass
00:29:57.000 --> 00:30:00.000
in polar coordinates by finding
the average of R or the average
00:30:00.000 --> 00:30:02.000
of theta.
For example,
00:30:02.000 --> 00:30:05.000
take a disk center at the
origin, well,
00:30:05.000 --> 00:30:09.000
the center of mass should be at
the origin.
00:30:09.000 --> 00:30:12.000
But the average of R is
certainly not zero because R is
00:30:12.000 --> 00:30:14.000
positive everywhere.
So, that doesn't work.
00:30:14.000 --> 00:30:18.000
You cannot get the polar
coordinates of a center of mass
00:30:18.000 --> 00:30:21.000
just by taking the average of R
and the average of theta.
00:30:21.000 --> 00:30:23.000
By the way, what is the average
of theta?
00:30:23.000 --> 00:30:26.000
If you take theta to from zero
to 2pi, the average theta will
00:30:26.000 --> 00:30:28.000
be pi.
If you take it to go from minus
00:30:28.000 --> 00:30:30.000
pi to pi, the average theta will
be zero.
00:30:30.000 --> 00:30:34.000
So, there is a problem there.
That actually just doesn't
00:30:34.000 --> 00:30:38.000
work, so we really have to
compute x bar and y bar.
00:30:38.000 --> 00:30:41.000
But still we could set this up
and then switch to polar
00:30:41.000 --> 00:30:44.000
coordinates to evaluate this
integral.
00:30:44.000 --> 00:30:58.000
But we still would be computing
the average values of x and y.
00:30:58.000 --> 00:31:04.000
We are basically re-exploring
mechanics and motion of solids
00:31:04.000 --> 00:31:10.000
here.
The next thing is moment of
00:31:10.000 --> 00:31:15.000
inertia.
Just to remind you or in case
00:31:15.000 --> 00:31:18.000
you somehow haven't seen it in
physics yet,
00:31:18.000 --> 00:31:23.000
the moment of inertia is
basically to rotation of a solid
00:31:23.000 --> 00:31:26.000
where the mass is to
translation.
00:31:26.000 --> 00:31:30.000
In the following sense,
the mass of a solid is what
00:31:30.000 --> 00:31:34.000
makes it hard to push it.
How hard it is to throw
00:31:34.000 --> 00:31:36.000
something is related to its
mass.
00:31:36.000 --> 00:31:41.000
How hard it is to spin
something, on the other hand,
00:31:41.000 --> 00:31:44.000
is given by its moment of
inertia.
00:31:44.000 --> 00:31:51.000
Maybe I should write this down.
Mass is how hard it is to
00:31:51.000 --> 00:31:59.000
impart a translation motion to a
solid.
00:31:59.000 --> 00:32:06.000
I am using fancy words today.
And the moment of inertia --
00:32:06.000 --> 00:32:13.000
The difference with a mass is
that the moment of inertia is
00:32:13.000 --> 00:32:17.000
defined about some axis.
You choose an axis.
00:32:17.000 --> 00:32:19.000
Then you would try to measure
how hard it is to spin your
00:32:19.000 --> 00:32:21.000
object around that axis.
For example,
00:32:21.000 --> 00:32:24.000
you can try to measure how hard
it is to spin this sheet of
00:32:24.000 --> 00:32:27.000
paper about an axis that is in
the center of it.
00:32:27.000 --> 00:32:30.000
We would try to spin it light
that and see how much effort I
00:32:30.000 --> 00:32:35.000
would have to make.
Well, for a sheet of paper not
00:32:35.000 --> 00:32:43.000
very much.
That would measure the same
00:32:43.000 --> 00:32:58.000
thing but it would be rotation
motion about that axis.
00:32:58.000 --> 00:33:02.000
Maybe some of you know the
definition but I am going to try
00:33:02.000 --> 00:33:05.000
to derive it again.
I am sorry but it won't be as
00:33:05.000 --> 00:33:07.000
quite as detailed as the way you
have probably seen it in
00:33:07.000 --> 00:33:09.000
physics, but I am not trying to
replace your physics teachers.
00:33:09.000 --> 00:33:16.000
I am sure they are doing a
great job.
00:33:16.000 --> 00:33:19.000
What is the idea for the
definition to find a formula for
00:33:19.000 --> 00:33:21.000
moment of inertia?
The idea is to think about
00:33:21.000 --> 00:33:24.000
kinetic energy.
Kinetic energy is really when
00:33:24.000 --> 00:33:28.000
you push something or when you
try to make it move and you have
00:33:28.000 --> 00:33:32.000
to put some inertia to it.
Then it has kinetic energy.
00:33:32.000 --> 00:33:38.000
And then, if you have the right
device, you can convert back
00:33:38.000 --> 00:33:41.000
that kinetic energy into
something else.
00:33:41.000 --> 00:33:46.000
If you try to look at the
kinetic energy of a point mass,
00:33:46.000 --> 00:33:53.000
so you have something with mass
m going at the velocity v,
00:33:53.000 --> 00:33:57.000
well, that will be one-half of
a mass times the square of the
00:33:57.000 --> 00:34:00.000
speed.
I hope you have all seen that
00:34:00.000 --> 00:34:04.000
formula some time before.
Now, let's say instead of just
00:34:04.000 --> 00:34:07.000
trying to push this mass,
I am going to make it spin
00:34:07.000 --> 00:34:12.000
around something.
Instead of just somewhere,
00:34:12.000 --> 00:34:20.000
maybe I will have the origin,
and I am trying to make it go
00:34:20.000 --> 00:34:29.000
around the origin in a circle at
a certain angular velocity.
00:34:29.000 --> 00:34:40.000
For a mass m at distance r,
let's call r this distance.
00:34:40.000 --> 00:34:47.000
And angular velocity,
let's call the angular velocity
00:34:47.000 --> 00:34:50.000
omega.
I think that is what physicists
00:34:50.000 --> 00:34:53.000
call it.
Remember angular velocity is
00:34:53.000 --> 00:34:57.000
just the rate of the change of
the angle over time.
00:34:57.000 --> 00:35:02.000
It is d theta dt, if you want.
Well, what is the kinetic
00:35:02.000 --> 00:35:05.000
energy now?
Well, first we have to find out
00:35:05.000 --> 00:35:07.000
what the speed is.
What is the speed?
00:35:07.000 --> 00:35:10.000
Well,
if we are going on a circle of
00:35:10.000 --> 00:35:16.000
radius r at angular velocity
omega that means that in unit
00:35:16.000 --> 00:35:22.000
time we rotate by omega and we
go by a distance of r times
00:35:22.000 --> 00:35:26.000
omega.
The actual speed is the radius
00:35:26.000 --> 00:35:32.000
times angular velocity.
And so the kinetic energy is
00:35:32.000 --> 00:35:38.000
one-half mv squared,
which is one-half m r squared
00:35:38.000 --> 00:35:41.000
omega squared.
And so,
00:35:41.000 --> 00:35:47.000
by similarity with that
formula,
00:35:47.000 --> 00:35:51.000
the coefficient of v squared is
the mass,
00:35:51.000 --> 00:35:53.000
and here we will say the
coefficient of omega squared,
00:35:53.000 --> 00:35:57.000
so this thing is the moment of
inertia.
00:35:57.000 --> 00:36:16.000
That is how we define moment of
inertia.
00:36:16.000 --> 00:36:20.000
Now, that is only for a point
mass.
00:36:20.000 --> 00:36:23.000
And it is kind of fun to spin
just a small bowl,
00:36:23.000 --> 00:36:26.000
but maybe you would like to
spin actually a larger solid and
00:36:26.000 --> 00:36:29.000
try to define this moment of
inertia.
00:36:29.000 --> 00:36:33.000
Well, the moment inertia of a
solid will be just the sum of
00:36:33.000 --> 00:36:36.000
the moments of inertia of all
the little pieces.
00:36:36.000 --> 00:36:45.000
What we will do is just cut our
solid into little chunks and
00:36:45.000 --> 00:36:51.000
will sum this thing for each
little piece.
00:36:51.000 --> 00:37:00.000
For a solid with density delta,
each little piece has mass
00:37:00.000 --> 00:37:07.000
which is the density times the
amount of area.
00:37:07.000 --> 00:37:12.000
This is equal actually.
And the moment of inertia of
00:37:12.000 --> 00:37:16.000
that small portion of a solid
will be delta m,
00:37:16.000 --> 00:37:18.000
the small mass,
times r squared,
00:37:18.000 --> 00:37:25.000
the square of a distance to the
center of the axis along which I
00:37:25.000 --> 00:37:29.000
am spinning.
That means if I sum these
00:37:29.000 --> 00:37:35.000
things together,
well, it has moment of inertia
00:37:35.000 --> 00:37:42.000
delta m times r squared,
which is r squared times the
00:37:42.000 --> 00:37:48.000
density times delta A.
And so I will be summing these
00:37:48.000 --> 00:37:52.000
things together.
And so, the moment of inertia
00:37:52.000 --> 00:37:56.000
about the origin will be the
double integral of r squared
00:37:56.000 --> 00:37:59.000
times density times dA.
00:38:28.000 --> 00:38:36.000
The final formula for the
moment of inertia about the
00:38:36.000 --> 00:38:46.000
origin is the double integral of
a region of r squared density
00:38:46.000 --> 00:38:48.000
dA.
If you are going to do it in x,
00:38:48.000 --> 00:38:51.000
y coordinates,
of course, r squared becomes x
00:38:51.000 --> 00:38:56.000
squared plus y squared,
it is the square of the
00:38:56.000 --> 00:39:02.000
distance from the origin.
When you integrate this,
00:39:02.000 --> 00:39:05.000
that tells you how hard it is
to spin that solid about the
00:39:05.000 --> 00:39:09.000
origin.
The motion that we try to do --
00:39:09.000 --> 00:39:15.000
We keep this fixed and then we
just rotate around the origin.
00:39:15.000 --> 00:39:20.000
Sorry.
That is a pretty bad picture,
00:39:20.000 --> 00:39:26.000
but hopefully you know what I
mean.
00:39:26.000 --> 00:39:29.000
And the name we use for that is
I0.
00:39:29.000 --> 00:39:37.000
And then the rotational kinetic
energy is one-half times this
00:39:37.000 --> 00:39:46.000
moment of inertia times the
square of the angular velocity.
00:39:46.000 --> 00:39:54.000
So that shows as that this
replaces the mass for rotation
00:39:54.000 --> 00:39:57.000
motions.
OK.
00:39:57.000 --> 00:40:03.000
What about other kinds of
rotations?
00:40:03.000 --> 00:40:06.000
In particular,
we have been rotating things
00:40:06.000 --> 00:40:13.000
about just a point in the plane.
What you could imagine also is
00:40:13.000 --> 00:40:19.000
instead you have your solid.
What I have done so far is I
00:40:19.000 --> 00:40:22.000
have skewered it this way,
and I am rotating around the
00:40:22.000 --> 00:40:25.000
axis.
Instead, I could skewer it
00:40:25.000 --> 00:40:27.000
through, say,
the horizontal axis.
00:40:27.000 --> 00:40:36.000
And then I could try to spin
about the horizontal axis so
00:40:36.000 --> 00:40:46.000
then it would rotate in space in
that direction like that.
00:40:46.000 --> 00:40:51.000
Let's say we do rotation about
the x-axis.
00:40:51.000 --> 00:40:53.000
Well, the idea would still be
the same.
00:40:53.000 --> 00:40:58.000
The moment of inertia for any
small piece of a solid would be
00:40:58.000 --> 00:41:02.000
its mass element times the
square of a distance to the x
00:41:02.000 --> 00:41:06.000
axes because that will be the
radius of a trajectory.
00:41:06.000 --> 00:41:12.000
If you take this point here,
it is going to go in a circle
00:41:12.000 --> 00:41:16.000
like that centered on the
x-axis.
00:41:16.000 --> 00:41:21.000
So the radius will just be this
distance here.
00:41:21.000 --> 00:41:24.000
Well, what is this distance?
It is just y,
00:41:24.000 --> 00:41:34.000
or maybe absolute value of y.
Distance to x-axis is absolute
00:41:34.000 --> 00:41:39.000
value of y.
What we actually care about is
00:41:39.000 --> 00:41:44.000
the square of a distance,
so it will just be y squared.
00:41:44.000 --> 00:41:51.000
The moment of inertia about the
x-axis is going to be obtained
00:41:51.000 --> 00:41:57.000
by integrating y squared times
the mass element.
00:41:57.000 --> 00:42:00.000
It is slightly strange but I
have y in inertia about the
00:42:00.000 --> 00:42:03.000
x-axis.
But, if you think about it,
00:42:03.000 --> 00:42:07.000
y tells me how far I am from
the x-axis, so how hard it will
00:42:07.000 --> 00:42:11.000
be to spin around the x-axis.
And I could do the same about
00:42:11.000 --> 00:42:17.000
any axis that I want.
Just I would have to sum the
00:42:17.000 --> 00:42:23.000
square of a distance to the axis
of rotation.
00:42:23.000 --> 00:42:31.000
Maybe I should do an example.
Yes?
00:42:31.000 --> 00:42:36.000
Same thing as above,
distance to the x-axis,
00:42:36.000 --> 00:42:39.000
because that is what we care
about.
00:42:39.000 --> 00:42:47.000
For the moment of inertia,
we want the square of a
00:42:47.000 --> 00:42:52.000
distance to the axis of
rotation.
00:42:52.000 --> 00:42:57.000
Let's do an example.
Let's try to figure out if we
00:42:57.000 --> 00:43:03.000
have just a uniform disk how
hard it is to spin it around its
00:43:03.000 --> 00:43:08.000
center.
That shouldn't be very hard to
00:43:08.000 --> 00:43:16.000
figure out.
Say that we have a disk of
00:43:16.000 --> 00:43:29.000
radius a and we want to rotate
it about its center.
00:43:29.000 --> 00:43:32.000
And let's say that it is of
uniform density.
00:43:32.000 --> 00:43:36.000
And let's take just the density
to be a one so that we don't
00:43:36.000 --> 00:43:40.000
really care about the density.
What is the moment of inertia
00:43:40.000 --> 00:43:45.000
of that?
Well, we have to integrate of
00:43:45.000 --> 00:43:51.000
our disk r squared times the
density, which is one,
00:43:51.000 --> 00:43:55.000
times dA.
What is r squared?
00:43:55.000 --> 00:43:58.000
You have here to resist the
urge to say the radius is just
00:43:58.000 --> 00:44:00.000
a.
We know the radius is a.
00:44:00.000 --> 00:44:05.000
No, it is not a because we are
looking at rotation of any point
00:44:05.000 --> 00:44:07.000
inside this disk.
And, when you are inside the
00:44:07.000 --> 00:44:09.000
disk, the distance to the origin
is not a.
00:44:09.000 --> 00:44:13.000
It is less than a.
It is actually anything between
00:44:13.000 --> 00:44:16.000
zero and a.
Just to point out a pitfall,
00:44:16.000 --> 00:44:18.000
r here is really a function on
this disk.
00:44:18.000 --> 00:44:20.000
And we are going to integrate
this function.
00:44:20.000 --> 00:44:28.000
Don't plug r equals a just yet.
What coordinates do we use to
00:44:28.000 --> 00:44:31.000
compute this integral?
They are probably polar
00:44:31.000 --> 00:44:35.000
coordinates, unless you want a
repeat of what happened already
00:44:35.000 --> 00:44:39.000
with x and y.
That will tell us we want to
00:44:39.000 --> 00:44:42.000
integrate r squared time r dr d
theta.
00:44:42.000 --> 00:44:47.000
And the bounds for r,
well, r will go from zero to a.
00:44:47.000 --> 00:44:51.000
No matter which direction I go
from the origin,
00:44:51.000 --> 00:44:56.000
if I fixed it,
r goes from zero to r equals a.
00:44:56.000 --> 00:45:02.000
The part of this ray that lives
inside the disk is always from
00:45:02.000 --> 00:45:05.000
zero to a.
And theta goes from,
00:45:05.000 --> 00:45:11.000
well, zero to 2 pi for example.
And now you can compute this
00:45:11.000 --> 00:45:14.000
integral.
Well, I will let you figure it
00:45:14.000 --> 00:45:18.000
out.
But the inner integral becomes
00:45:18.000 --> 00:45:25.000
a to the four over four and the
outer multiplies things by 2pi,
00:45:25.000 --> 00:45:30.000
so you get pi a to the four
over two.
00:45:30.000 --> 00:45:33.000
OK.
That is how hard it is to spin
00:45:33.000 --> 00:45:37.000
this disk.
Now, what about instead of
00:45:37.000 --> 00:45:43.000
spinning it about the center we
decided to spin it about a point
00:45:43.000 --> 00:45:46.000
on a second point.
For example, think of a Frisbee.
00:45:46.000 --> 00:45:50.000
A Frisbee has this rim so you
can actually try to make it
00:45:50.000 --> 00:45:55.000
rotate around the point on the
circumference by holding it near
00:45:55.000 --> 00:45:59.000
the rim and spinning it there.
How much harder is that than
00:45:59.000 --> 00:46:02.000
around the center?
Well, we will try to compute
00:46:02.000 --> 00:46:05.000
now the moment of inertia about
this point.
00:46:05.000 --> 00:46:08.000
We have two options.
One is we keep the system of
00:46:08.000 --> 00:46:12.000
coordinates centers here.
But then the formula for
00:46:12.000 --> 00:46:15.000
distance to this point becomes
harder.
00:46:15.000 --> 00:46:18.000
The other option,
which is the one I will choose,
00:46:18.000 --> 00:46:21.000
is to change the coordinate so
that this point become the
00:46:21.000 --> 00:46:23.000
origin.
Let's do that.
00:46:50.000 --> 00:46:58.000
About a point on the
circumference,
00:46:58.000 --> 00:47:13.000
what I would have to do maybe
is set up my region like that.
00:47:13.000 --> 00:47:17.000
I have moved the origin so that
it is on the circumference of a
00:47:17.000 --> 00:47:21.000
disk,
and I will again try to find
00:47:21.000 --> 00:47:27.000
the moment of inertia of this
disk about the origin.
00:47:27.000 --> 00:47:31.000
It is still,
for the the double integral of
00:47:31.000 --> 00:47:36.000
r squared dA.
But now I want to find out how
00:47:36.000 --> 00:47:40.000
to set up the integral.
I could try to use x,
00:47:40.000 --> 00:47:43.000
y coordinates and it would
work.
00:47:43.000 --> 00:47:47.000
Or I can use polar coordinates,
and it works a little bit
00:47:47.000 --> 00:47:52.000
better that way.
But both are doable.
00:47:52.000 --> 00:47:56.000
Let's say I do it this way.
I have to figure out how to set
00:47:56.000 --> 00:48:00.000
up my bounds.
What are the bounds for r?
00:48:00.000 --> 00:48:06.000
Well, if I fix a value for
theta, which means I chose an
00:48:06.000 --> 00:48:12.000
angle here, now I am shooting a
ray from the origin in that
00:48:12.000 --> 00:48:16.000
direction.
I enter my region at r equals
00:48:16.000 --> 00:48:19.000
zero.
That hasn't changed.
00:48:19.000 --> 00:48:23.000
The question is where do I exit
the region?
00:48:23.000 --> 00:48:33.000
What is that distance?
Maybe you have seen it in
00:48:33.000 --> 00:48:37.000
recitation, maybe not.
Let's see.
00:48:37.000 --> 00:48:40.000
Actually, I should have written
down the radius of a circle is
00:48:40.000 --> 00:48:47.000
a.
So this distance here is 2a.
00:48:47.000 --> 00:48:52.000
If you draw this segment in
here, you know that here you
00:48:52.000 --> 00:48:56.000
have a right angle.
You have a right triangle.
00:48:56.000 --> 00:48:59.000
The hypotenuse here has length
2a.
00:48:59.000 --> 00:49:08.000
This angle is theta.
Well, this length is 2a cosine
00:49:08.000 --> 00:49:14.000
theta.
The polar coordinates equation
00:49:14.000 --> 00:49:22.000
of this circle passing through
the origin is r equals 2a cosine
00:49:22.000 --> 00:49:27.000
theta.
So, r will go from zero to 2a
00:49:27.000 --> 00:49:33.000
cosine theta.
That is the distance here.
00:49:33.000 --> 00:49:37.000
Now, what are the bounds for
theta?
00:49:37.000 --> 00:49:39.000
It is not quite zero to 2pi
because, actually,
00:49:39.000 --> 00:49:42.000
you see in this direction,
if I shoot a ray in this
00:49:42.000 --> 00:49:44.000
direction I will never meet my
region.
00:49:44.000 --> 00:49:47.000
We have to actually think a bit
more.
00:49:47.000 --> 00:49:52.000
Well, the directions in which I
will actually hit my circle are
00:49:52.000 --> 00:49:56.000
all the directions in the right
half of a plane.
00:49:56.000 --> 00:49:58.000
I mean, of course,
if I shoot very close to the
00:49:58.000 --> 00:50:00.000
axis, you might think,
oh, I won't be in there.
00:50:00.000 --> 00:50:03.000
But, actually,
that is not true because here
00:50:03.000 --> 00:50:05.000
the circle is tangent to the
axis.
00:50:05.000 --> 00:50:09.000
No matter which direction I
take, I will still have a little
00:50:09.000 --> 00:50:13.000
tiny piece.
The angle actually goes from
00:50:13.000 --> 00:50:15.000
minus pi over two to pi over
two.
00:50:15.000 --> 00:50:20.000
If you compute that you will
get,
00:50:20.000 --> 00:50:25.000
well, the inner integral will
be r to the four over four
00:50:25.000 --> 00:50:28.000
between zero and 2a cosine
theta,
00:50:28.000 --> 00:50:34.000
which will turn out to be 4a to
the four cosine to the four
00:50:34.000 --> 00:50:39.000
theta.
And now you will integrate that
00:50:39.000 --> 00:50:43.000
for minus pi over two to pi over
two.
00:50:43.000 --> 00:50:47.000
And that is,
again, the evil integral that
00:50:47.000 --> 00:50:50.000
we had yesterday.
Either we remember the method
00:50:50.000 --> 00:50:53.000
from yesterday or we remember
from yesterday that actually
00:50:53.000 --> 00:50:56.000
there are formulas in the notes
to help you.
00:50:56.000 --> 00:50:58.000
On homework,
you can use these formulas.
00:50:58.000 --> 00:51:04.000
In the notes at the beginning
of section 3b there are formulas
00:51:04.000 --> 00:51:08.000
for these particular kinds of
integrals.
00:51:08.000 --> 00:51:13.000
And that will end up being
three-halves of pi a to the
00:51:13.000 --> 00:51:15.000
four.
In case you wanted to know,
00:51:15.000 --> 00:51:18.000
it is three times harder to
spin a Frisbee about a point on
00:51:18.000 --> 00:51:20.000
a circumference than around the
center.
00:51:20.000 --> 00:51:26.000
We got three times the moment
of inertia about the center.
00:51:26.000 --> 00:51:27.000
OK.
That is it.
00:51:27.000 --> 00:51:29.000
Have a nice weekend.