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