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Yesterday we saw how to define
double integrals and how to
8
00:00:28 --> 00:00:33
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
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00:00:41 --> 00:00:45
plane of a function f of x,
y dA.
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00:00:45 --> 00:00:51
You cannot hear me?
Is the sound working?
12
00:00:51 --> 00:00:57
Can you hear me in the back now?
Can we make the sound louder?
13
00:00:57 --> 00:01:01
Does this work?
People are not hearing me in
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00:01:01 --> 00:01:05
the back.
Is it better?
15
00:01:05 --> 00:01:09
People are still saying make it
louder.
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00:01:09 --> 00:01:11
Is it better?
OK.
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00:01:11 --> 00:01:18
Great.
Thanks.
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00:01:18 --> 00:01:22
That's not a reason to start
chatting with your friends.
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00:01:22 --> 00:01:27
Thanks.
When we have a region in the x,
20
00:01:27 --> 00:01:31
y plane and we have a function
of x and y,
21
00:01:31 --> 00:01:36
we are defining the double
integral of f over this region
22
00:01:36 --> 00:01:40
by taking basically the sum of
the values of a function
23
00:01:40 --> 00:01:44
everywhere in here times the
area element.
24
00:01:44 --> 00:01:48
And the definition, actually,
is we split the region into
25
00:01:48 --> 00:01:52
lots of tiny little pieces,
we multiply the value of a
26
00:01:52 --> 00:01:55
function at the point times the
area of a little piece and we
27
00:01:55 --> 00:01:59
sum that everywhere.
And we have seen,
28
00:01:59 --> 00:02:07
actually, how to compute these
things as iterated integrals.
29
00:02:07 --> 00:02:14
First, integrating over dy and
then over dx,
30
00:02:14 --> 00:02:21
or the other way around.
One example that we did,
31
00:02:21 --> 00:02:25
in particular,
was to compute the double
32
00:02:25 --> 00:02:29
integral of a quarter of a unit
disk.
33
00:02:29 --> 00:02:35
That was the region where x
squared plus y squared is less
34
00:02:35 --> 00:02:40
than one and x and y are
positive, of one minus x squared
35
00:02:40 --> 00:02:43
minus y squared dA.
Well, hopefully,
36
00:02:43 --> 00:02:48
I kind of convinced you that we
can do it using enough trig and
37
00:02:48 --> 00:02:52
substitutions and so on,
but it is not very pleasant.
38
00:02:52 --> 00:02:56
And the reason for that is that
using x and y coordinates here
39
00:02:56 --> 00:03:04
does not seem very appropriate.
In fact, we can use polar
40
00:03:04 --> 00:03:16
coordinates instead to compute
this double integral.
41
00:03:16 --> 00:03:22
Remember that polar coordinates
are about replacing x and y as
42
00:03:22 --> 00:03:28
coordinates for a point on a
plane by instead r,
43
00:03:28 --> 00:03:31
which is the distance from the
origin to a point,
44
00:03:31 --> 00:03:35
and theta,
which is the angle measured
45
00:03:35 --> 00:03:40
counterclockwise from the
positive x-axis.
46
00:03:40 --> 00:03:48
In terms of r and theta,
you have x equals r cosine
47
00:03:48 --> 00:03:54
theta, y equals r sine theta.
The claim is we are able,
48
00:03:54 --> 00:03:59
actually, to do double
integrals in polar coordinates.
49
00:03:59 --> 00:04:06
We just have to learn how to.
Just to draw a quick picture --
50
00:04:06 --> 00:04:12
When we were integrating in x,
y coordinates,
51
00:04:12 --> 00:04:16
in rectangular coordinates,
we were slicing our region by
52
00:04:16 --> 00:04:20
gridlines that were either
horizontal or vertical.
53
00:04:20 --> 00:04:23
And we used that to set up the
iterated integral.
54
00:04:23 --> 00:04:28
And we said dA became dx dy or
dy dx.
55
00:04:28 --> 00:04:35
Now we are going to actually
integrate, in terms of the polar
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00:04:35 --> 00:04:41
coordinates, r and theta.
Let's say we will integrate in
57
00:04:41 --> 00:04:45
the order with r first and then
theta.
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00:04:45 --> 00:04:50
That is the order that makes
the most sense usually when you
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00:04:50 --> 00:04:53
do polar coordinates.
What does that mean?
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00:04:53 --> 00:04:58
It means that we will first
focus on a slice where we fix
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00:04:58 --> 00:05:02
the value of theta and we will
let r vary.
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00:05:02 --> 00:05:06
That means we fix a direction,
we fix a ray out from the
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00:05:06 --> 00:05:11
origin in a certain direction.
And we will travel along this
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00:05:11 --> 00:05:15
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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00:05:18 --> 00:05:24
easy because r will just start
at zero, and you will have to
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00:05:24 --> 00:05:28
stop when you exit this quarter
disk.
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00:05:28 --> 00:05:31
Well, what is the equation of
this circle in polar
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00:05:31 --> 00:05:34
coordinates?
It is just r equals one.
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00:05:34 --> 00:05:40
So, we will stop when r reaches
one.
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00:05:40 --> 00:05:43
But what about theta?
Well, the first ray that we
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00:05:43 --> 00:05:48
might want to consider is the
one that goes along the x-axis.
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00:05:48 --> 00:05:51
That is when theta equals zero.
And we will stop when theta
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00:05:51 --> 00:05:55
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
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00:06:00 --> 00:06:05
quadrant.
We will stop at pi over two.
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00:06:05 --> 00:06:11
Now, there is a catch,
though, which is that dA is not
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00:06:11 --> 00:06:15
dr d theta.
Let me explain to you why.
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00:06:15 --> 00:06:19
Let's say that we are slicing.
What it means is we are cutting
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00:06:19 --> 00:06:24
our region into little pieces
that are the elementary,
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00:06:24 --> 00:06:26
you know,
what corresponds to a small
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00:06:26 --> 00:06:28
rectangle in the x,
y coordinate system,
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00:06:28 --> 00:06:36
here would be actually a little
piece of circle between a given
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00:06:36 --> 00:06:42
radius r and r plus delta r.
And given between an angle
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00:06:42 --> 00:06:44
theta and theta plus delta
theta.
86
00:06:44 --> 00:06:48
I need to draw,
actually, a bigger picture of
87
00:06:48 --> 00:06:53
that because it makes it really
hard to read.
88
00:06:53 --> 00:06:58
Let's say that I fix an angle
theta and a slightly different
89
00:06:58 --> 00:07:02
one where I have added delta
theta to it.
90
00:07:02 --> 00:07:11
And let's say that I have a
radius r and I add delta r to
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00:07:11 --> 00:07:14
it.
Then I will have a little piece
92
00:07:14 --> 00:07:20
of x, y plane that is in here.
And I have to figure out what
93
00:07:20 --> 00:07:26
is its area?
What is delta A for this guy?
94
00:07:26 --> 00:07:29
Well, let's see.
This guy actually,
95
00:07:29 --> 00:07:33
you know, if my delta r and
delta theta are small enough,
96
00:07:33 --> 00:07:35
it will almost look like a
rectangle.
97
00:07:35 --> 00:07:37
It is rotated,
but it is basically a
98
00:07:37 --> 00:07:39
rectangle.
I mean these sides,
99
00:07:39 --> 00:07:44
of course, are curvy,
but they are short enough and
100
00:07:44 --> 00:07:50
it is almost straight.
The area here should be this
101
00:07:50 --> 00:07:55
length times that length.
Well, what is this length?
102
00:07:55 --> 00:08:00
That one is easy.
It is delta r.
103
00:08:00 --> 00:08:03
What about that length?
Well, it is not delta theta.
104
00:08:03 --> 00:08:05
It is something slightly
different.
105
00:08:05 --> 00:08:13
It is a piece of a circle of
radius r corresponding to angle
106
00:08:13 --> 00:08:18
delta theta, so it is r delta
theta.
107
00:08:18 --> 00:08:26
So, times r delta theta.
That means now,
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00:08:26 --> 00:08:32
even if we shrink things and
take smaller and smaller
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00:08:32 --> 00:08:37
regions, dA is going to be r dr
d theta.
110
00:08:37 --> 00:08:38
That is an important thing to
remember.
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00:08:38 --> 00:08:44
When you integrate in polar
coordinates, you just set up
112
00:08:44 --> 00:08:50
your bounds in terms of r and
theta, but you replace dA by r
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00:08:50 --> 00:08:55
dr d theta, not just dr d theta.
And then, of course,
114
00:08:55 --> 00:08:57
we have some function that we
are integrating.
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00:08:57 --> 00:09:06
Let's say that I call that
thing f then it is the same f
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00:09:06 --> 00:09:12
that I put up here.
Concretely, how do I do it here?
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00:09:12 --> 00:09:18
Well, my function f was given
as one minus x squared minus y
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00:09:18 --> 00:09:20
squared.
And I would like to switch that
119
00:09:20 --> 00:09:24
to polar coordinates.
I want to put r and theta in
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00:09:24 --> 00:09:26
there.
Well, I have formulas for x and
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00:09:26 --> 00:09:30
y in polar coordinates so I
could just replace x squared by
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00:09:30 --> 00:09:34
r squared cosine squared theta,
y squared by r squared sine
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00:09:34 --> 00:09:37
squared theta.
And that works just fine.
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00:09:37 --> 00:09:44
But maybe you can observe that
this is x squared plus y
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00:09:44 --> 00:09:46
squared.
It is just the square of a
126
00:09:46 --> 00:09:49
distance from the origin,
so that is just r squared.
127
00:09:49 --> 00:09:52
That is a useful thing.
You don't strictly need it,
128
00:09:52 --> 00:09:56
but it is much faster if you
see this right away.
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00:09:56 --> 00:10:07
It saves you writing down a
sine and a cosine.
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00:10:07 --> 00:10:13
Now we just end up with the
integral from zero to pi over
131
00:10:13 --> 00:10:20
two, integral from zero to one
of one minus r squared r dr d
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00:10:20 --> 00:10:24
theta.
Now, if I want to compute this
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00:10:24 --> 00:10:30
integral, so let's first do the
inner integral.
134
00:10:30 --> 00:10:37
If I integrate r minus r cubed,
I will get r squared over two
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00:10:37 --> 00:10:43
minus r squared over four
between zero and one.
136
00:10:43 --> 00:10:47
And then I will integrate d
theta.
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00:10:47 --> 00:10:51
What is this equal to?
Well, for r equals one you get
138
00:10:51 --> 00:10:54
one-half minus one-quarter,
which is going to be just
139
00:10:54 --> 00:10:56
one-quarter.
And when you plug in zero you
140
00:10:56 --> 00:10:59
get zero.
So, it is the integral from
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00:10:59 --> 00:11:02
zero to pi over two of
one-quarter d theta.
142
00:11:02 --> 00:11:11
And that just integrates to
one-quarter times pi over two,
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00:11:11 --> 00:11:18
which is pi over eight.
That is a lot easier than the
144
00:11:18 --> 00:11:23
way we did it yesterday.
Well, here we were lucky.
145
00:11:23 --> 00:11:26
I mean usually you will switch
to polar coordinates either
146
00:11:26 --> 00:11:28
because the region is easier to
set up.
147
00:11:28 --> 00:11:31
Here it is indeed easier to set
up because the bounds became
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00:11:31 --> 00:11:34
very simple.
We don't have that square root
149
00:11:34 --> 00:11:38
of one minus x squared anymore.
Or because the integrant
150
00:11:38 --> 00:11:40
becomes much simpler.
Here our function,
151
00:11:40 --> 00:11:43
well, it is not very
complicated in x,
152
00:11:43 --> 00:11:46
y coordinates,
but it is even simpler in r
153
00:11:46 --> 00:11:50
theta coordinates.
Here we were very lucky.
154
00:11:50 --> 00:11:52
In general, there is maybe a
trade off.
155
00:11:52 --> 00:11:55
Maybe it will be easier to set
up bounds but maybe the function
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00:11:55 --> 00:11:58
will become harder because it
will have all these sines and
157
00:11:58 --> 00:12:01
cosines in it.
If our function had been just
158
00:12:01 --> 00:12:04
x, x is very easy in x,
y coordinates.
159
00:12:04 --> 00:12:08
Here it becomes r cosine theta.
That means you will have a
160
00:12:08 --> 00:12:10
little bit of trig to do in the
integral.
161
00:12:10 --> 00:12:14
Not a very big one,
not a very complicated
162
00:12:14 --> 00:12:21
integral, but imagine it could
get potentially much harder.
163
00:12:21 --> 00:12:25
Anyway, that is double
integrals in polar coordinates.
164
00:12:25 --> 00:12:30
And the way you set up the
bounds in general,
165
00:12:30 --> 00:12:37
well, in 99% of the cases you
will integrate over r first.
166
00:12:37 --> 00:12:40
What you will do is you will
look for a given theta what are
167
00:12:40 --> 00:12:42
the bounds of r to be in the
region.
168
00:12:42 --> 00:12:46
What is the portion of my ray
that is in the given region?
169
00:12:46 --> 00:12:49
And then you will put bounds
for theta.
170
00:12:49 --> 00:12:51
But conceptually it is the same
as before.
171
00:12:51 --> 00:12:55
Instead of slicing horizontally
or vertically,
172
00:12:55 --> 00:12:59
we slice radially.
We will do more examples in a
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00:12:59 --> 00:13:03
bit.
Any questions about this or the
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00:13:03 --> 00:13:21
general method?
Yes?
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00:13:21 --> 00:13:27
That is a very good question.
Why do I measure the length
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00:13:27 --> 00:13:30
inside instead of outside?
Which one do I want?
177
00:13:30 --> 00:13:34
This one.
Here I said this side is r
178
00:13:34 --> 00:13:36
delta theta.
I could have said,
179
00:13:36 --> 00:13:38
actually, r delta theta is the
length here.
180
00:13:38 --> 00:13:41
Here it is slightly more,
r plus delta r times delta
181
00:13:41 --> 00:13:43
theta.
But, if delta r is very small
182
00:13:43 --> 00:13:46
compared to r,
then that is almost the same
183
00:13:46 --> 00:13:48
thing.
And this is an approximation
184
00:13:48 --> 00:13:51
anyway.
I took this one because it
185
00:13:51 --> 00:13:56
gives me the simpler formula.
If you take the limit as delta
186
00:13:56 --> 00:14:01
r turns to zero then the two
things become the same anyway.
187
00:14:01 --> 00:14:04
The length, whether you put r
or r plus delta r in here,
188
00:14:04 --> 00:14:09
doesn't matter anymore.
If you imagine that this guy is
189
00:14:09 --> 00:14:15
infinitely small then,
really, the lengths become the
190
00:14:15 --> 00:14:17
same.
We will also see another proof
191
00:14:17 --> 00:14:20
of this formula,
using changes of variables,
192
00:14:20 --> 00:14:23
next week.
But, I mean,
193
00:14:23 --> 00:14:28
hopefully this is at least
slightly convincing.
194
00:14:28 --> 00:14:34
More questions?
No.
195
00:14:34 --> 00:14:40
OK.
Let's see.
196
00:14:40 --> 00:14:42
We have seen how to compute
double integrals.
197
00:14:42 --> 00:14:49
I have to tell you what they
are good for as well.
198
00:14:49 --> 00:14:53
The definition we saw yesterday
and the motivation was in terms
199
00:14:53 --> 00:14:57
of finding volumes,
but that is not going to be our
200
00:14:57 --> 00:15:00
main preoccupation.
Because finding volumes is fun
201
00:15:00 --> 00:15:02
but that is not all there is to
life.
202
00:15:02 --> 00:15:05
I mean, you are doing single
integrals.
203
00:15:05 --> 00:15:08
When you do single integrals it
is usually not to find the area
204
00:15:08 --> 00:15:13
of some region of a plane.
It is for something else
205
00:15:13 --> 00:15:16
usually.
The way we actually think of
206
00:15:16 --> 00:15:19
the double integral is really as
summing the values of a function
207
00:15:19 --> 00:15:22
all around this region.
We can use that to get
208
00:15:22 --> 00:15:26
information about maybe the
region or about the average
209
00:15:26 --> 00:15:29
value of a function in that
region and so on.
210
00:15:29 --> 00:15:39
Let's think about various uses
of double integrals.
211
00:15:39 --> 00:15:43
The first one that I will
mention is actually something
212
00:15:43 --> 00:15:47
you thought maybe you could do
with a single integral,
213
00:15:47 --> 00:15:51
but it is useful very often to
do it as a double integral.
214
00:15:51 --> 00:15:59
It is to find the area of a
given region r.
215
00:15:59 --> 00:16:06
I give you some region in the
plane and you want to know just
216
00:16:06 --> 00:16:08
its area.
In various cases,
217
00:16:08 --> 00:16:12
you could set this up as a
single integral,
218
00:16:12 --> 00:16:16
but often it could be useful to
set it up as a double integral.
219
00:16:16 --> 00:16:20
How do you express the area as
a double integral?
220
00:16:20 --> 00:16:22
Well, the area of this region
is the sum of the areas of all
221
00:16:22 --> 00:16:28
the little pieces.
It means you want to sum one dA
222
00:16:28 --> 00:16:37
of the entire region.
The area R is the double
223
00:16:37 --> 00:16:46
integral over R of a function
one.
224
00:16:46 --> 00:16:48
One way to think about it,
if you are really still
225
00:16:48 --> 00:16:51
attached to the idea of double
integral as a volume,
226
00:16:51 --> 00:16:54
what this measures is the
volume below the graph of a
227
00:16:54 --> 00:16:56
function one.
The graph of a function one is
228
00:16:56 --> 00:16:59
just a horizontal plane at
height one.
229
00:16:59 --> 00:17:07
What you would be measuring is
the volume of a prism with base
230
00:17:07 --> 00:17:11
r and height one.
And the volume of that would
231
00:17:11 --> 00:17:12
be, of course,
base times height.
232
00:17:12 --> 00:17:16
It would just be the area of r
again.
233
00:17:16 --> 00:17:18
But we don't actually need to
think about it that way.
234
00:17:18 --> 00:17:24
Really, what we are doing is
summing dA over the entire
235
00:17:24 --> 00:17:28
region.
A related thing we can do,
236
00:17:28 --> 00:17:33
imagine that,
actually, this is some physical
237
00:17:33 --> 00:17:35
object.
I mean, it has to be a flat
238
00:17:35 --> 00:17:38
object because we are just
dealing with things in the plane
239
00:17:38 --> 00:17:41
so far.
But you have a flat metal plate
240
00:17:41 --> 00:17:45
or something and you would like
to know its mass.
241
00:17:45 --> 00:17:50
Well, its mass is the sum of
the masses of every single
242
00:17:50 --> 00:17:52
little piece.
You would get that by
243
00:17:52 --> 00:17:57
integrating the density.
The density for a flat object
244
00:17:57 --> 00:18:09
would be the mass per unit area.
So, you can get the mass of a
245
00:18:09 --> 00:18:23
flat object with density.
Let's use delta for density,
246
00:18:23 --> 00:18:29
which is the mass per unit
area.
247
00:18:29 --> 00:18:34
Each little piece of your
object will have a mass,
248
00:18:34 --> 00:18:40
which will be just the density,
times its area for each small
249
00:18:40 --> 00:18:45
piece.
And you will get the total mass
250
00:18:45 --> 00:18:51
by summing these things.
The mass will be the double
251
00:18:51 --> 00:18:56
integral of the density times
the area element.
252
00:18:56 --> 00:18:58
Now, if it has constant
density,
253
00:18:58 --> 00:19:00
if it is always the same
material then,
254
00:19:00 --> 00:19:03
of course,
you could just take the density
255
00:19:03 --> 00:19:07
out and you will get density
times the total area if you know
256
00:19:07 --> 00:19:10
that it is always the same
material.
257
00:19:10 --> 00:19:13
But if, actually,
it has varying density maybe
258
00:19:13 --> 00:19:17
because it is some metallic
thing with various metals or
259
00:19:17 --> 00:19:21
with varying thickness or
something then you can still get
260
00:19:21 --> 00:19:24
the mass by integrating the
density.
261
00:19:24 --> 00:19:26
Of course, looking at flat
objects might be a little bit
262
00:19:26 --> 00:19:28
strange.
That is because we are only
263
00:19:28 --> 00:19:30
doing double integrals so far.
In a few weeks,
264
00:19:30 --> 00:19:33
we will be triple integrals.
And then we will be able to do
265
00:19:33 --> 00:19:36
solids in space,
but one thing at a time.
266
00:19:36 --> 00:19:55
267
00:19:55 --> 00:20:08
Another useful application is
to find the average value of
268
00:20:08 --> 00:20:16
some quantity in a region.
What does it mean to take the
269
00:20:16 --> 00:20:19
average value of some function f
in this region r?
270
00:20:19 --> 00:20:22
Well, you know what the average
of a finite set of data is.
271
00:20:22 --> 00:20:24
For example,
if I asked you to compute your
272
00:20:24 --> 00:20:26
average score on 18.02 problem
sets,
273
00:20:26 --> 00:20:30
you would just take the scores,
add them and divide by the
274
00:20:30 --> 00:20:33
number of problem sets.
What if there are infinitely
275
00:20:33 --> 00:20:35
many things?
Say I ask you to find the
276
00:20:35 --> 00:20:37
average temperature in this
room.
277
00:20:37 --> 00:20:39
Well, you would have to measure
the temperature everywhere.
278
00:20:39 --> 00:20:42
And then add all of these
together and divide by the
279
00:20:42 --> 00:20:45
number of data points.
But, depending on how careful
280
00:20:45 --> 00:20:47
you are, actually,
there are potentially
281
00:20:47 --> 00:20:49
infinitely many points to look
at.
282
00:20:49 --> 00:20:54
The mathematical way to define
the average of a continuous set
283
00:20:54 --> 00:20:58
of data is that you actually
integrate the function over the
284
00:20:58 --> 00:21:02
entire set of data,
and then you divide by the size
285
00:21:02 --> 00:21:06
of the sample,
which is just the area of the
286
00:21:06 --> 00:21:10
region.
In fact, the average of f,
287
00:21:10 --> 00:21:17
the notation we will use
usually for that is f with a bar
288
00:21:17 --> 00:21:23
on top to tell us it is the
average f.
289
00:21:23 --> 00:21:31
We say we will take the
integral of f and we will divide
290
00:21:31 --> 00:21:38
by the area of the region.
You can really think of it as
291
00:21:38 --> 00:21:44
the sum of the values of f
everywhere divided by the number
292
00:21:44 --> 00:21:48
of points everywhere.
And so that is an average where
293
00:21:48 --> 00:21:51
everything is,
actually, equally likely.
294
00:21:51 --> 00:21:55
That is a uniform average where
all the points on the region,
295
00:21:55 --> 00:21:59
all the little points of the
region are equally likely.
296
00:21:59 --> 00:22:02
But maybe if want to do,
say, an average of some solid
297
00:22:02 --> 00:22:06
with variable density or if you
want to somehow give more
298
00:22:06 --> 00:22:10
importance to certain parts than
to others then you can actually
299
00:22:10 --> 00:22:14
do a weighted average.
What is a weighted average?
300
00:22:14 --> 00:22:21
Well,
in the case of taking the
301
00:22:21 --> 00:22:23
average your problem sets,
if I tell you problem set one
302
00:22:23 --> 00:22:25
is worth twice as much as the
others,
303
00:22:25 --> 00:22:29
then you would count twice that
score in the sum and then you
304
00:22:29 --> 00:22:33
would count it as two,
of course, when you divide.
305
00:22:33 --> 00:22:36
The weighted average is the sum
of the values,
306
00:22:36 --> 00:22:39
but each weighted by a certain
coefficient.
307
00:22:39 --> 00:22:43
And then you will divide by the
sum of the weight.
308
00:22:43 --> 00:22:48
It is a bit the same idea as
when we replace area by some
309
00:22:48 --> 00:22:53
mass that tells you how
important a given piece.
310
00:22:53 --> 00:23:02
We will actually have a density.
Let's call it delta again.
311
00:23:02 --> 00:23:07
We will see what we divide by,
but what we will take is the
312
00:23:07 --> 00:23:13
integral of a function times the
density times the area element.
313
00:23:13 --> 00:23:18
Because this would correspond
to the mass element telling us
314
00:23:18 --> 00:23:22
how to weight the various points
of our region.
315
00:23:22 --> 00:23:27
And then we would divide by the
total weight,
316
00:23:27 --> 00:23:34
which is the mass of a region,
as defined up there.
317
00:23:34 --> 00:23:39
If a density is uniform then,
of course, the density gets out
318
00:23:39 --> 00:23:44
and you can simplify and reduce
to that if all the points are
319
00:23:44 --> 00:23:47
equally likely.
Why is that important?
320
00:23:47 --> 00:23:49
Well, that is important for
various applications.
321
00:23:49 --> 00:23:53
But one that you might have
seen in physics,
322
00:23:53 --> 00:23:58
we care about maybe where is
the center of mass of a given
323
00:23:58 --> 00:24:01
object?
The center of mass is basically
324
00:24:01 --> 00:24:05
a point that you would say is
right in the middle of the
325
00:24:05 --> 00:24:06
object.
But, of course,
326
00:24:06 --> 00:24:10
if the object has a very
strange shape or if somehow part
327
00:24:10 --> 00:24:14
of it is heavier than the rest
then that takes a very different
328
00:24:14 --> 00:24:17
meaning.
Strictly speaking,
329
00:24:17 --> 00:24:20
the center of mass of a solid
is the point where you would
330
00:24:20 --> 00:24:24
have to concentrate all the mass
if you wanted it to behave
331
00:24:24 --> 00:24:28
equivalently from a point of
view of mechanics,
332
00:24:28 --> 00:24:31
if you are trying to do
translations of that object.
333
00:24:31 --> 00:24:37
If you are going to push that
object that would be really
334
00:24:37 --> 00:24:42
where the equivalent point mass
would lie.
335
00:24:42 --> 00:24:44
The other way to think about
it,
336
00:24:44 --> 00:24:47
if I had a flat object then the
center of mass would basically
337
00:24:47 --> 00:24:50
be the point where I would need
to hold it so it is perfectly
338
00:24:50 --> 00:24:52
balanced.
And, of course,
339
00:24:52 --> 00:24:56
I cannot do this.
Well, you get the idea.
340
00:24:56 --> 00:24:59
And the center of mass of this
eraser is somewhere in the
341
00:24:59 --> 00:25:00
middle.
And so, in principle,
342
00:25:00 --> 00:25:03
that is where I would have to
put my finger for it to stay.
343
00:25:03 --> 00:25:11
Well, it doesn't work.
But that is where the center of
344
00:25:11 --> 00:25:19
mass should be.
I think it should be in the
345
00:25:19 --> 00:25:22
middle.
Maybe I shouldn't call this
346
00:25:22 --> 00:25:25
three.
I should call this 2a,
347
00:25:25 --> 00:25:31
because it is really a special
case of the average value.
348
00:25:31 --> 00:25:47
How do we find the center of
mass of a flat object with
349
00:25:47 --> 00:25:57
density delta.
If you have your object in the
350
00:25:57 --> 00:26:00
x,
y plane then its center of mass
351
00:26:00 --> 00:26:03
will be at positions that are
actually just the coordinates of
352
00:26:03 --> 00:26:09
a center of mass,
will just be weighted averages
353
00:26:09 --> 00:26:14
of x and y on the solid.
So, the center of mass will be
354
00:26:14 --> 00:26:17
a position that I will call x
bar, y bar.
355
00:26:17 --> 00:26:21
And these are really just the
averages, the average values of
356
00:26:21 --> 00:26:26
x and of y in the solid.
Just to give you the formulas
357
00:26:26 --> 00:26:33
again, x bar would be one over
the mass times the double
358
00:26:33 --> 00:26:42
integral of x times density dA.
And the same thing with y.
359
00:26:42 --> 00:26:53
y bar is the weighted average
of a y coordinate in your
360
00:26:53 --> 00:26:56
region.
You see, if you take a region
361
00:26:56 --> 00:26:59
that is symmetric and has
uniform density that will just
362
00:26:59 --> 00:27:01
give you the center of the
region.
363
00:27:01 --> 00:27:05
But if the region has a strange
shape or if a density is not
364
00:27:05 --> 00:27:08
homogeneous,
if parts of it are heavier then
365
00:27:08 --> 00:27:12
you will get whatever the
weighted average will be.
366
00:27:12 --> 00:27:15
And that will be the point
where this thing would be
367
00:27:15 --> 00:27:19
balanced if you were trying to
balance it on a pole or on your
368
00:27:19 --> 00:27:20
finger.
369
00:27:20 --> 00:27:56
370
00:27:56 --> 00:28:12
Any questions so far?
Yes.
371
00:28:12 --> 00:28:18
No.
Here I didn't set this up as a
372
00:28:18 --> 00:28:23
iterated integral yet.
The function that I am
373
00:28:23 --> 00:28:28
integrating is x times delta
where density will be given to
374
00:28:28 --> 00:28:30
me maybe as a function of x and
y.
375
00:28:30 --> 00:28:33
And then I will integrate this
dA.
376
00:28:33 --> 00:28:36
And dA could mean dx over dy,
it could mean dy over dx,
377
00:28:36 --> 00:28:40
it could be mean r dr d theta.
I will choose how to set it up
378
00:28:40 --> 00:28:43
depending maybe on the shape of
the region.
379
00:28:43 --> 00:28:46
If my solid is actually just
going to be round then I might
380
00:28:46 --> 00:28:49
want to use polar coordinates.
If it is a square,
381
00:28:49 --> 00:28:51
I might want to use x,
y coordinates.
382
00:28:51 --> 00:28:56
If it is more complicated,
well, I will choose depending
383
00:28:56 --> 00:29:01
on how I feel about it.
Yes?
384
00:29:01 --> 00:29:05
Delta is the density.
In general, it is a function of
385
00:29:05 --> 00:29:08
x and y.
If you imagine that your solid
386
00:29:08 --> 00:29:11
is not homogenous then its
density will depend on which
387
00:29:11 --> 00:29:15
piece of it you are looking at.
Of course, to compute this,
388
00:29:15 --> 00:29:18
you need to know the density.
If you have a problem asking
389
00:29:18 --> 00:29:20
you to find the center of mass
of something and you have no
390
00:29:20 --> 00:29:23
information about the density,
assume it is uniform.
391
00:29:23 --> 00:29:26
Take the density to be a
constant.
392
00:29:26 --> 00:29:29
Even take it to be a one.
That is even easier.
393
00:29:29 --> 00:29:30
I mean it is a general fact of
math.
394
00:29:30 --> 00:29:34
We don't care about units.
If density is constant,
395
00:29:34 --> 00:29:36
we might as well take it to be
one.
396
00:29:36 --> 00:29:41
That just means our mass unit
becomes the area unit.
397
00:29:41 --> 00:29:53
Yes?
That is a good question.
398
00:29:53 --> 00:29:57
No, I don't think we could
actually find the center of mass
399
00:29:57 --> 00:30:00
in polar coordinates by finding
the average of R or the average
400
00:30:00 --> 00:30:02
of theta.
For example,
401
00:30:02 --> 00:30:05
take a disk center at the
origin, well,
402
00:30:05 --> 00:30:09
the center of mass should be at
the origin.
403
00:30:09 --> 00:30:12
But the average of R is
certainly not zero because R is
404
00:30:12 --> 00:30:14
positive everywhere.
So, that doesn't work.
405
00:30:14 --> 00:30:18
You cannot get the polar
coordinates of a center of mass
406
00:30:18 --> 00:30:21
just by taking the average of R
and the average of theta.
407
00:30:21 --> 00:30:23
By the way, what is the average
of theta?
408
00:30:23 --> 00:30:26
If you take theta to from zero
to 2pi, the average theta will
409
00:30:26 --> 00:30:28
be pi.
If you take it to go from minus
410
00:30:28 --> 00:30:30
pi to pi, the average theta will
be zero.
411
00:30:30 --> 00:30:34
So, there is a problem there.
That actually just doesn't
412
00:30:34 --> 00:30:38
work, so we really have to
compute x bar and y bar.
413
00:30:38 --> 00:30:41
But still we could set this up
and then switch to polar
414
00:30:41 --> 00:30:44
coordinates to evaluate this
integral.
415
00:30:44 --> 00:30:58
But we still would be computing
the average values of x and y.
416
00:30:58 --> 00:31:04
We are basically re-exploring
mechanics and motion of solids
417
00:31:04 --> 00:31:10
here.
The next thing is moment of
418
00:31:10 --> 00:31:15
inertia.
Just to remind you or in case
419
00:31:15 --> 00:31:18
you somehow haven't seen it in
physics yet,
420
00:31:18 --> 00:31:23
the moment of inertia is
basically to rotation of a solid
421
00:31:23 --> 00:31:26
where the mass is to
translation.
422
00:31:26 --> 00:31:30
In the following sense,
the mass of a solid is what
423
00:31:30 --> 00:31:34
makes it hard to push it.
How hard it is to throw
424
00:31:34 --> 00:31:36
something is related to its
mass.
425
00:31:36 --> 00:31:41
How hard it is to spin
something, on the other hand,
426
00:31:41 --> 00:31:44
is given by its moment of
inertia.
427
00:31:44 --> 00:31:51
Maybe I should write this down.
Mass is how hard it is to
428
00:31:51 --> 00:31:59
impart a translation motion to a
solid.
429
00:31:59 --> 00:32:06
I am using fancy words today.
And the moment of inertia --
430
00:32:06 --> 00:32:13
The difference with a mass is
that the moment of inertia is
431
00:32:13 --> 00:32:17
defined about some axis.
You choose an axis.
432
00:32:17 --> 00:32:19
Then you would try to measure
how hard it is to spin your
433
00:32:19 --> 00:32:21
object around that axis.
For example,
434
00:32:21 --> 00:32:24
you can try to measure how hard
it is to spin this sheet of
435
00:32:24 --> 00:32:27
paper about an axis that is in
the center of it.
436
00:32:27 --> 00:32:30
We would try to spin it light
that and see how much effort I
437
00:32:30 --> 00:32:35
would have to make.
Well, for a sheet of paper not
438
00:32:35 --> 00:32:43
very much.
That would measure the same
439
00:32:43 --> 00:32:58
thing but it would be rotation
motion about that axis.
440
00:32:58 --> 00:33:02
Maybe some of you know the
definition but I am going to try
441
00:33:02 --> 00:33:05
to derive it again.
I am sorry but it won't be as
442
00:33:05 --> 00:33:07
quite as detailed as the way you
have probably seen it in
443
00:33:07 --> 00:33:09
physics, but I am not trying to
replace your physics teachers.
444
00:33:09 --> 00:33:16
I am sure they are doing a
great job.
445
00:33:16 --> 00:33:19
What is the idea for the
definition to find a formula for
446
00:33:19 --> 00:33:21
moment of inertia?
The idea is to think about
447
00:33:21 --> 00:33:24
kinetic energy.
Kinetic energy is really when
448
00:33:24 --> 00:33:28
you push something or when you
try to make it move and you have
449
00:33:28 --> 00:33:32
to put some inertia to it.
Then it has kinetic energy.
450
00:33:32 --> 00:33:38
And then, if you have the right
device, you can convert back
451
00:33:38 --> 00:33:41
that kinetic energy into
something else.
452
00:33:41 --> 00:33:46
If you try to look at the
kinetic energy of a point mass,
453
00:33:46 --> 00:33:53
so you have something with mass
m going at the velocity v,
454
00:33:53 --> 00:33:57
well, that will be one-half of
a mass times the square of the
455
00:33:57 --> 00:34:00
speed.
I hope you have all seen that
456
00:34:00 --> 00:34:04
formula some time before.
Now, let's say instead of just
457
00:34:04 --> 00:34:07
trying to push this mass,
I am going to make it spin
458
00:34:07 --> 00:34:12
around something.
Instead of just somewhere,
459
00:34:12 --> 00:34:20
maybe I will have the origin,
and I am trying to make it go
460
00:34:20 --> 00:34:29
around the origin in a circle at
a certain angular velocity.
461
00:34:29 --> 00:34:40
For a mass m at distance r,
let's call r this distance.
462
00:34:40 --> 00:34:47
And angular velocity,
let's call the angular velocity
463
00:34:47 --> 00:34:50
omega.
I think that is what physicists
464
00:34:50 --> 00:34:53
call it.
Remember angular velocity is
465
00:34:53 --> 00:34:57
just the rate of the change of
the angle over time.
466
00:34:57 --> 00:35:02
It is d theta dt, if you want.
Well, what is the kinetic
467
00:35:02 --> 00:35:05
energy now?
Well, first we have to find out
468
00:35:05 --> 00:35:07
what the speed is.
What is the speed?
469
00:35:07 --> 00:35:10
Well,
if we are going on a circle of
470
00:35:10 --> 00:35:16
radius r at angular velocity
omega that means that in unit
471
00:35:16 --> 00:35:22
time we rotate by omega and we
go by a distance of r times
472
00:35:22 --> 00:35:26
omega.
The actual speed is the radius
473
00:35:26 --> 00:35:32
times angular velocity.
And so the kinetic energy is
474
00:35:32 --> 00:35:38
one-half mv squared,
which is one-half m r squared
475
00:35:38 --> 00:35:41
omega squared.
And so,
476
00:35:41 --> 00:35:47
by similarity with that
formula,
477
00:35:47 --> 00:35:51
the coefficient of v squared is
the mass,
478
00:35:51 --> 00:35:53
and here we will say the
coefficient of omega squared,
479
00:35:53 --> 00:35:57
so this thing is the moment of
inertia.
480
00:35:57 --> 00:36:16
That is how we define moment of
inertia.
481
00:36:16 --> 00:36:20
Now, that is only for a point
mass.
482
00:36:20 --> 00:36:23
And it is kind of fun to spin
just a small bowl,
483
00:36:23 --> 00:36:26
but maybe you would like to
spin actually a larger solid and
484
00:36:26 --> 00:36:29
try to define this moment of
inertia.
485
00:36:29 --> 00:36:33
Well, the moment inertia of a
solid will be just the sum of
486
00:36:33 --> 00:36:36
the moments of inertia of all
the little pieces.
487
00:36:36 --> 00:36:45
What we will do is just cut our
solid into little chunks and
488
00:36:45 --> 00:36:51
will sum this thing for each
little piece.
489
00:36:51 --> 00:37:00
For a solid with density delta,
each little piece has mass
490
00:37:00 --> 00:37:07
which is the density times the
amount of area.
491
00:37:07 --> 00:37:12
This is equal actually.
And the moment of inertia of
492
00:37:12 --> 00:37:16
that small portion of a solid
will be delta m,
493
00:37:16 --> 00:37:18
the small mass,
times r squared,
494
00:37:18 --> 00:37:25
the square of a distance to the
center of the axis along which I
495
00:37:25 --> 00:37:29
am spinning.
That means if I sum these
496
00:37:29 --> 00:37:35
things together,
well, it has moment of inertia
497
00:37:35 --> 00:37:42
delta m times r squared,
which is r squared times the
498
00:37:42 --> 00:37:48
density times delta A.
And so I will be summing these
499
00:37:48 --> 00:37:52
things together.
And so, the moment of inertia
500
00:37:52 --> 00:37:56
about the origin will be the
double integral of r squared
501
00:37:56 --> 00:37:59
times density times dA.
502
00:37:59 --> 00:38:28
503
00:38:28 --> 00:38:36
The final formula for the
moment of inertia about the
504
00:38:36 --> 00:38:46
origin is the double integral of
a region of r squared density
505
00:38:46 --> 00:38:48
dA.
If you are going to do it in x,
506
00:38:48 --> 00:38:51
y coordinates,
of course, r squared becomes x
507
00:38:51 --> 00:38:56
squared plus y squared,
it is the square of the
508
00:38:56 --> 00:39:02
distance from the origin.
When you integrate this,
509
00:39:02 --> 00:39:05
that tells you how hard it is
to spin that solid about the
510
00:39:05 --> 00:39:09
origin.
The motion that we try to do --
511
00:39:09 --> 00:39:15
We keep this fixed and then we
just rotate around the origin.
512
00:39:15 --> 00:39:20
Sorry.
That is a pretty bad picture,
513
00:39:20 --> 00:39:26
but hopefully you know what I
mean.
514
00:39:26 --> 00:39:29
And the name we use for that is
I0.
515
00:39:29 --> 00:39:37
And then the rotational kinetic
energy is one-half times this
516
00:39:37 --> 00:39:46
moment of inertia times the
square of the angular velocity.
517
00:39:46 --> 00:39:54
So that shows as that this
replaces the mass for rotation
518
00:39:54 --> 00:39:57
motions.
OK.
519
00:39:57 --> 00:40:03
What about other kinds of
rotations?
520
00:40:03 --> 00:40:06
In particular,
we have been rotating things
521
00:40:06 --> 00:40:13
about just a point in the plane.
What you could imagine also is
522
00:40:13 --> 00:40:19
instead you have your solid.
What I have done so far is I
523
00:40:19 --> 00:40:22
have skewered it this way,
and I am rotating around the
524
00:40:22 --> 00:40:25
axis.
Instead, I could skewer it
525
00:40:25 --> 00:40:27
through, say,
the horizontal axis.
526
00:40:27 --> 00:40:36
And then I could try to spin
about the horizontal axis so
527
00:40:36 --> 00:40:46
then it would rotate in space in
that direction like that.
528
00:40:46 --> 00:40:51
Let's say we do rotation about
the x-axis.
529
00:40:51 --> 00:40:53
Well, the idea would still be
the same.
530
00:40:53 --> 00:40:58
The moment of inertia for any
small piece of a solid would be
531
00:40:58 --> 00:41:02
its mass element times the
square of a distance to the x
532
00:41:02 --> 00:41:06
axes because that will be the
radius of a trajectory.
533
00:41:06 --> 00:41:12
If you take this point here,
it is going to go in a circle
534
00:41:12 --> 00:41:16
like that centered on the
x-axis.
535
00:41:16 --> 00:41:21
So the radius will just be this
distance here.
536
00:41:21 --> 00:41:24
Well, what is this distance?
It is just y,
537
00:41:24 --> 00:41:34
or maybe absolute value of y.
Distance to x-axis is absolute
538
00:41:34 --> 00:41:39
value of y.
What we actually care about is
539
00:41:39 --> 00:41:44
the square of a distance,
so it will just be y squared.
540
00:41:44 --> 00:41:51
The moment of inertia about the
x-axis is going to be obtained
541
00:41:51 --> 00:41:57
by integrating y squared times
the mass element.
542
00:41:57 --> 00:42:00
It is slightly strange but I
have y in inertia about the
543
00:42:00 --> 00:42:03
x-axis.
But, if you think about it,
544
00:42:03 --> 00:42:07
y tells me how far I am from
the x-axis, so how hard it will
545
00:42:07 --> 00:42:11
be to spin around the x-axis.
And I could do the same about
546
00:42:11 --> 00:42:17
any axis that I want.
Just I would have to sum the
547
00:42:17 --> 00:42:23
square of a distance to the axis
of rotation.
548
00:42:23 --> 00:42:31
Maybe I should do an example.
Yes?
549
00:42:31 --> 00:42:36
Same thing as above,
distance to the x-axis,
550
00:42:36 --> 00:42:39
because that is what we care
about.
551
00:42:39 --> 00:42:47
For the moment of inertia,
we want the square of a
552
00:42:47 --> 00:42:52
distance to the axis of
rotation.
553
00:42:52 --> 00:42:57
Let's do an example.
Let's try to figure out if we
554
00:42:57 --> 00:43:03
have just a uniform disk how
hard it is to spin it around its
555
00:43:03 --> 00:43:08
center.
That shouldn't be very hard to
556
00:43:08 --> 00:43:16
figure out.
Say that we have a disk of
557
00:43:16 --> 00:43:29
radius a and we want to rotate
it about its center.
558
00:43:29 --> 00:43:32
And let's say that it is of
uniform density.
559
00:43:32 --> 00:43:36
And let's take just the density
to be a one so that we don't
560
00:43:36 --> 00:43:40
really care about the density.
What is the moment of inertia
561
00:43:40 --> 00:43:45
of that?
Well, we have to integrate of
562
00:43:45 --> 00:43:51
our disk r squared times the
density, which is one,
563
00:43:51 --> 00:43:55
times dA.
What is r squared?
564
00:43:55 --> 00:43:58
You have here to resist the
urge to say the radius is just
565
00:43:58 --> 00:44:00
a.
We know the radius is a.
566
00:44:00 --> 00:44:05
No, it is not a because we are
looking at rotation of any point
567
00:44:05 --> 00:44:07
inside this disk.
And, when you are inside the
568
00:44:07 --> 00:44:09
disk, the distance to the origin
is not a.
569
00:44:09 --> 00:44:13
It is less than a.
It is actually anything between
570
00:44:13 --> 00:44:16
zero and a.
Just to point out a pitfall,
571
00:44:16 --> 00:44:18
r here is really a function on
this disk.
572
00:44:18 --> 00:44:20
And we are going to integrate
this function.
573
00:44:20 --> 00:44:28
Don't plug r equals a just yet.
What coordinates do we use to
574
00:44:28 --> 00:44:31
compute this integral?
They are probably polar
575
00:44:31 --> 00:44:35
coordinates, unless you want a
repeat of what happened already
576
00:44:35 --> 00:44:39
with x and y.
That will tell us we want to
577
00:44:39 --> 00:44:42
integrate r squared time r dr d
theta.
578
00:44:42 --> 00:44:47
And the bounds for r,
well, r will go from zero to a.
579
00:44:47 --> 00:44:51
No matter which direction I go
from the origin,
580
00:44:51 --> 00:44:56
if I fixed it,
r goes from zero to r equals a.
581
00:44:56 --> 00:45:02
The part of this ray that lives
inside the disk is always from
582
00:45:02 --> 00:45:05
zero to a.
And theta goes from,
583
00:45:05 --> 00:45:11
well, zero to 2 pi for example.
And now you can compute this
584
00:45:11 --> 00:45:14
integral.
Well, I will let you figure it
585
00:45:14 --> 00:45:18
out.
But the inner integral becomes
586
00:45:18 --> 00:45:25
a to the four over four and the
outer multiplies things by 2pi,
587
00:45:25 --> 00:45:30
so you get pi a to the four
over two.
588
00:45:30 --> 00:45:33
OK.
That is how hard it is to spin
589
00:45:33 --> 00:45:37
this disk.
Now, what about instead of
590
00:45:37 --> 00:45:43
spinning it about the center we
decided to spin it about a point
591
00:45:43 --> 00:45:46
on a second point.
For example, think of a Frisbee.
592
00:45:46 --> 00:45:50
A Frisbee has this rim so you
can actually try to make it
593
00:45:50 --> 00:45:55
rotate around the point on the
circumference by holding it near
594
00:45:55 --> 00:45:59
the rim and spinning it there.
How much harder is that than
595
00:45:59 --> 00:46:02
around the center?
Well, we will try to compute
596
00:46:02 --> 00:46:05
now the moment of inertia about
this point.
597
00:46:05 --> 00:46:08
We have two options.
One is we keep the system of
598
00:46:08 --> 00:46:12
coordinates centers here.
But then the formula for
599
00:46:12 --> 00:46:15
distance to this point becomes
harder.
600
00:46:15 --> 00:46:18
The other option,
which is the one I will choose,
601
00:46:18 --> 00:46:21
is to change the coordinate so
that this point become the
602
00:46:21 --> 00:46:23
origin.
Let's do that.
603
00:46:23 --> 00:46:50
604
00:46:50 --> 00:46:58
About a point on the
circumference,
605
00:46:58 --> 00:47:13
what I would have to do maybe
is set up my region like that.
606
00:47:13 --> 00:47:17
I have moved the origin so that
it is on the circumference of a
607
00:47:17 --> 00:47:21
disk,
and I will again try to find
608
00:47:21 --> 00:47:27
the moment of inertia of this
disk about the origin.
609
00:47:27 --> 00:47:31
It is still,
for the the double integral of
610
00:47:31 --> 00:47:36
r squared dA.
But now I want to find out how
611
00:47:36 --> 00:47:40
to set up the integral.
I could try to use x,
612
00:47:40 --> 00:47:43
y coordinates and it would
work.
613
00:47:43 --> 00:47:47
Or I can use polar coordinates,
and it works a little bit
614
00:47:47 --> 00:47:52
better that way.
But both are doable.
615
00:47:52 --> 00:47:56
Let's say I do it this way.
I have to figure out how to set
616
00:47:56 --> 00:48:00
up my bounds.
What are the bounds for r?
617
00:48:00 --> 00:48:06
Well, if I fix a value for
theta, which means I chose an
618
00:48:06 --> 00:48:12
angle here, now I am shooting a
ray from the origin in that
619
00:48:12 --> 00:48:16
direction.
I enter my region at r equals
620
00:48:16 --> 00:48:19
zero.
That hasn't changed.
621
00:48:19 --> 00:48:23
The question is where do I exit
the region?
622
00:48:23 --> 00:48:33
What is that distance?
Maybe you have seen it in
623
00:48:33 --> 00:48:37
recitation, maybe not.
Let's see.
624
00:48:37 --> 00:48:40
Actually, I should have written
down the radius of a circle is
625
00:48:40 --> 00:48:47
a.
So this distance here is 2a.
626
00:48:47 --> 00:48:52
If you draw this segment in
here, you know that here you
627
00:48:52 --> 00:48:56
have a right angle.
You have a right triangle.
628
00:48:56 --> 00:48:59
The hypotenuse here has length
2a.
629
00:48:59 --> 00:49:08
This angle is theta.
Well, this length is 2a cosine
630
00:49:08 --> 00:49:14
theta.
The polar coordinates equation
631
00:49:14 --> 00:49:22
of this circle passing through
the origin is r equals 2a cosine
632
00:49:22 --> 00:49:27
theta.
So, r will go from zero to 2a
633
00:49:27 --> 00:49:33
cosine theta.
That is the distance here.
634
00:49:33 --> 00:49:37
Now, what are the bounds for
theta?
635
00:49:37 --> 00:49:39
It is not quite zero to 2pi
because, actually,
636
00:49:39 --> 00:49:42
you see in this direction,
if I shoot a ray in this
637
00:49:42 --> 00:49:44
direction I will never meet my
region.
638
00:49:44 --> 00:49:47
We have to actually think a bit
more.
639
00:49:47 --> 00:49:52
Well, the directions in which I
will actually hit my circle are
640
00:49:52 --> 00:49:56
all the directions in the right
half of a plane.
641
00:49:56 --> 00:49:58
I mean, of course,
if I shoot very close to the
642
00:49:58 --> 00:50:00
axis, you might think,
oh, I won't be in there.
643
00:50:00 --> 00:50:03
But, actually,
that is not true because here
644
00:50:03 --> 00:50:05
the circle is tangent to the
axis.
645
00:50:05 --> 00:50:09
No matter which direction I
take, I will still have a little
646
00:50:09 --> 00:50:13
tiny piece.
The angle actually goes from
647
00:50:13 --> 00:50:15
minus pi over two to pi over
two.
648
00:50:15 --> 00:50:20
If you compute that you will
get,
649
00:50:20 --> 00:50:25
well, the inner integral will
be r to the four over four
650
00:50:25 --> 00:50:28
between zero and 2a cosine
theta,
651
00:50:28 --> 00:50:34
which will turn out to be 4a to
the four cosine to the four
652
00:50:34 --> 00:50:39
theta.
And now you will integrate that
653
00:50:39 --> 00:50:43
for minus pi over two to pi over
two.
654
00:50:43 --> 00:50:47
And that is,
again, the evil integral that
655
00:50:47 --> 00:50:50
we had yesterday.
Either we remember the method
656
00:50:50 --> 00:50:53
from yesterday or we remember
from yesterday that actually
657
00:50:53 --> 00:50:56
there are formulas in the notes
to help you.
658
00:50:56 --> 00:50:58
On homework,
you can use these formulas.
659
00:50:58 --> 00:51:04
In the notes at the beginning
of section 3b there are formulas
660
00:51:04 --> 00:51:08
for these particular kinds of
integrals.
661
00:51:08 --> 00:51:13
And that will end up being
three-halves of pi a to the
662
00:51:13 --> 00:51:15
four.
In case you wanted to know,
663
00:51:15 --> 00:51:18
it is three times harder to
spin a Frisbee about a point on
664
00:51:18 --> 00:51:20
a circumference than around the
center.
665
00:51:20 --> 00:51:26
We got three times the moment
of inertia about the center.
666
00:51:26 --> 00:51:27
OK.
That is it.
667
00:51:27 --> 00:51:29
Have a nice weekend.
668
00:51:29 --> 00:51:34