Welcome to the sixth week of the course. So
if you look at the course plan, if I go by
the course plan, then there is this whole
change of variable formula was supposed to
be the first lecture of the sixth week, but
we have already spoken about change of variable
formula in the last class. So I think I will
not bore you on that again. Rather this week,
because I have already exceeded as per the
norms, I have exceeded the number of hours
by quite a heavy margin.
I do not know whether it is good or bad, but
I have done that. So I think the sixth week
will have four lectures, which will also I
believe, would exceed the given fixed time
of two and a half hours, 30 minutes, five
lectures 30 minutes each, but we will start
with the second one and we will have four
lectures in this session, in the sixth week
session.
That is multiple integral and mechanics, which
is which is which I wrote as application of
integration to mechanics. Line integrals 1,
which we will do and then line integrals 2.
These are two two important things which we
need to know to do further things about Gauss
and Stokes theorem, line integrals would come;
Gauss theorem, Green’s theorem.
Then we will talk about parametrized surfaces,
means how can we look at surfaces which cannot
be viewed as functions z = f (x, y). How can
we view them mathematically. Can they can
functions be used to represent them. So in
that case we need to parametrize and that
is what we will do here in this sixth week.
So we will start today with application of
multiple integrals.
See one application of multiple integral I
have already told you is that of finding average
temperature. In the last class, class before
that when we spoke about triple integrals,
we spoke about why triple integrals are needed.
For example, we spoke about finding the average
temperature in a, say a rectangular room.
Now if you go back to the notion of average,
the notion of average plays a very, very fundamental
role in mechanics.
The notion of average, for example, uh takes
on important roles for determining things
like center of mass, moments of inertia. So
all these things has a notion of average.
So average means summing up things and then
dividing by it by the number of things.
So in general if you have a continuous function,
or even a bounded function and you want to
integrate, then the average of a function,
continuous function over a given interval,
the integration represents a sum of the values
divided by the total x’s that you have,
that is captured by the expression b – a.
And this is often called the average of f
and this symbol of average of f is something
which I have borrowed from the physicist.
Physicists use this sort of thing.
So what happens if I have a scenario where
I have now an area to handle? That is two
variable for when I am talking about functions
of two variables, and I am talking about double
integrals, then what I need, which way I can
define the average and how can I generalize
this idea that I already know. So again, I
am just taking the same thing. Here my f is
no longer single valued function, but a multiple
valued function.
In that case, sorry is a function of two variables.
And in that case if I am talking about a domain
W and I am doing a double integral. So that
is the sum of the function over the whole
area divided by the whole area. So that is
the average that you have when you have a
function of more than one variable. So you
the general this idea has now got generalized.
You can of course find the average, I am not
going to keep on discussing this. I am not
going to do computations here. Because why
to unnecessarily take time doing simple computations.
We will now discuss the notion of center of
mass 
of a body or a collection of masses was already
known to Archimedes; Archimedes that great
Greek geometer and scientist. So what did
Archimedes do? Archimedes had a seminal contribution
to science. Not only he who found the area
of circles, area of segments of parabola,
but his key idea rested on something called
the law of levers.
The law of levers is that okay, here is a
lever. So we are talking about the law of
levers. So here is a mass m 1 and here is
a mass m 2. If I just put a, if I try to put
my hands anywhere or put a support anywhere
there is a high chance, if I put the support
here more towards m 2 this m 1 mass will create
a torque around that supporting point and
pull the whole thing down, pull m 2 up and
pull, put the m 1 down.
So this is a kind of seesaw type of thing.
So Archimedes’ question is, which way where
should I place a pivot, this pivot so that
this lever remains in equilibrium, it does
not move. Where shall I place the pivot? So
let me now look at it in a much more simpler
way. So let me now put along the lever I draw
the x axis. And suppose this is the origin.
From the origin this is at the point x 1,
m 2 is at the point x 2 and here the pivoting
is at the point x bar. What do I expect? What
does the law of lever says? The law of lever
says, as the distance of the pivot from m
1 multiplied by m 1 should be equal to the
distance of the pivot from m 2 and multiplied
by the mass m 2. If these two quantities are
equal, so what I do? What is the distance
of the pivot from m 1?
It is x bar minus x 1. This multiplied with
m 1 must equal, so this is the law of levers.
So this is this is finally the law of levers.
This is what Archimedes did. Proving it is
slightly difficult. There is a whole book
by David Spivak, who also has a beautiful
book on calculus. He has actually tried to
explain why it is not so easy to prove the
law of lever of Archimedes, right?
And that very simple mechanics problem can
be very hard and there are logical issues.
So, okay let us assume that we accept the
law of levers. If I take the law of levers
and because in practice that is what happens.
Please understand, even the area of the circle
that which which Archimedes developed or the
area of the segments of the parabola which
was known to Archimedes was based on the idea
of the law of levers.
So it says that if the lever need not move
it has to remain at rest then it must have
m 2 into the distance from the pivot that
is x 2. So what do I have? I have m 1 x bar
minus m 1 x 1 is equal to m 2 x 2 minus m
2 x bar. So this gives me m 1 x bar plus m
2 x bar because I transferred this here and
take this part m 1 x 1 to the other side;
m 1 x 1 plus m 2 x 2.
So x bar, the point that I want to really
locate because x 1 and x 2 are known points.
The point that I want to locate is m 1 x 1
+ m 2 x 2 divided by m 1 + m 2. So it is as
if this x bar is often called the center of
mass; x bar is often called the center of
mass as if the both the masses, the sum of
both the masses is located at the point x
bar. So this portion of this x bar can be
obtained by the law of lever of Archimedes.
x bar is called the center of mass. So here
I have made up a system of two discrete masses.
What if I just look at a continuous body,
a rigid body and then if I try to talk about
a center of mass, what will happen, okay?
If I then try to talk about a center of mass,
a body which is lying along the x axis.
Suppose here is a body which lies along the
x axis and I know that the density per unit
length is rho x, okay. Then if I want to,
so far a distance of x, the mass would be
that distance into the density, right? So
here volume is given by the length basically.
So in that case, the upper part that summation
m i x i , the m i is now x into that rho x,
rho x is the density per unit length.
So if I come the distance x from the origin,
so if I am here x from the origin okay and
I am as you let me let me assume that the
origin itself lies at one end of the rod.
Then if I come a distance x on the origin,
the mass is nothing but x into rho x, where
rho x is the density per unit length. So in
that case, your x bar is just trying to write
down a continuous version of the thing that
we obtained from the Archimedes’ law of
lever.
I then want to write a continuous version
of x bar equal to m 1 x 1 + m 2 x 2 by m 1
+ m 2. So if I had more masses it will be
summation m i x i by summation m i. So that
that idea can be now progressed with the use
of integral in in this continuous mass. So
here I have this x, this m i instead of m
i the role is now played by m, m dx, mx dx,
at every x what is the mass.
So mx rho x dx integration of that whole length
basically the length l if I say an integral
l of rho x dx is the total mass, summation
m i that is the total mass. So now if I am
in two dimension, see if I am in two dimension,
what are the coordinates of the center of
mass? In this case the density cannot be per
unit length but has to be per unit area. So
the density has to be like this.
So if you take a small unit area around x,
y rho x, y is the density per unit area. Here
similarly, the coordinates would be given
in the following way; x bar has to deal with
x part only. So could be a region W x rho
(x, y) dx dy integral over the total area.
So density per unit area. So the area of the,
so the volume or the mass of dx dy the unit
elemental areas the mass of the elemental
area is actually rho (x, y) into dx dy.
And y bar is replacing the same thing, but
x is replaced by y. That is what we are expecting.
Of course, you can definitely talk about I
am not going to do integrations here. I am
just giving you the conceptual issues here
because it is unnecessary to do same kind
of integration. But I will show you some type
of integration which will be useful. I will
do one temperature thing. So you see here,
what happens.
Can I take it to the notion of, can I take
this whole thing to three dimensional system.
So in a three dimensional body now, I have
a point x bar y bar z bar which has to be
the center of mass. So in a three dimensional
body, so what is now three dimensional case,
three dimensional body.
So let us take a body which is a closed bounded
nice three dimensional body, body actually
means a closed and bounded set like this.
So we are trying to find a kind of point where
all the mass would seem to be concentrated.
So x bar y bar z bar. So what is the volume
of this body? So we will call this body as
V to differentiate everything. So volume integral,
integral, integral V dx dy dz.
Now if for every unit volume rho x, y, z at
a point x, y, z if the density is x, y, z.
Then if you take a x, y, z point lies in a
elemental volume dx dy dz and if the density
does not change much much. If you take the
density remains constant over unit volume
that is what is rho x, y, z. Then in a very
small elemental volume the density would also
remain as rho x, y, z. That will remain constant,
do not vary.
So the mass in an elemental volume is rho
x, y, z into dx dy dz so density into volume.
So the total mass is to integrate over the
volume, total mass of this body V of the density
rho x, y, z. So the center of mass, which
is called the center of mass, the center of
mass x bar coordinates. The same story which
goes on, do not worry is V x rho x, y, z dx
dy dz divided by the mass.
If I have to write for y bar, so y bar will
be the same story but x now replaced by y.
So it is a repetitive writing which I do not
want to do or for completeness sake, people
might just ask me that why did you not write
it in detail, I would complete the writing;
rho x, y, z into dx dy dz. Of course, you
can talk about z bar, z bar is the same story
where we have z instead of x or y.
Obviously, got V and then we write down over
V the total mass. Now if you want to find
the average, what is f average in this case?
f average is integrating over V f (x, y, z)
dx dy dz and then dividing by the total volume.
That is it. That is neatly summed up for the
three dimensional body, for example. So let
us go to the next page. And then here we start
discussing a small problem.
So let the average, so given any, finding
average temperature in a body. In this case
my body V is the cube, so unit cube with center
at 000. So you can immediately visualize what
kind of a body we are looking at. It is a
kind of room and we are trying to get the
temperature or the average temperature, okay?
So let me draw, draw the, so it is a cube
or other better things to draw cube.
It will be a cube of course, you can visualize
it much better than me. Maybe I should draw
a cube. This is also another way of drawing
a cube. You know that 00 is at the center.
So this could be the x axis, this is the y
axis, this is the z axis. Now I am telling
that okay at any point x, y, z in this cube
the temperature T is proportional to x square
plus y square plus z square.
So basically T is some c time some proportionality
constant into x square plus y square plus
z square, okay? Now I have to find the average
temperature in the room. So how will I find
the average temperature? So what I will do
is exactly in the same way. T average is T
is integral, integral, integral, where x,
y, z are all varying between -1, +1, c into
x square plus y square plus x square. Basically
this is T.
So T I can write as a, now I can write it
like a function T (x, y, z). Here also you
can put T as T (x, y, z) it does not matter,
into dx dy dz or dz dy dz it does not matter,
whatever you want into the volume of the cube.
And volume the cube is this length is 2 2
2, so it is 8. So 1 by 8, I think c should
be also be taken out, of -1 -1 to +1, -1 to
+1 x square plus y square plus z square dx
dy dz.
You see here I have actually three separate
integrals, x square dx dy dz plus y square
dx dy dz plus z square dx dy dz, and all of
them will have the same answer. Because we
can iterate accordingly. All of them would
have the same answer, all of the three. So
instead, we can write this as so say at the
same integrals 3c by 8 -1 +1 -1 +1, I just
do the z, z thing. So and multiply, they are
all the same.
So I can do one of them and multiply them
with 3. So that is exactly what I am doing.
-1 +1 -1 no not +1. I should write this 1.
1 -1 1 -1 1 z square dx dy dz okay. You immediately
see what is happening, you immediately see.
So basically what I can write here is 3c by
8 -1 z square. I can actually take this whole
integral, you see how interesting it is. -1
here also -1 +1 dx dy.
Basically I am looking at the area of a square
of side 2 which will be 4 into dz. So basically
it is 3c 8 -1 to +1 4z square dz. So that
will be 3c by 2 z square dz is z cube by 3.
And I am now writing the answer. The answer
to this everything is finally z cube means
here one third minus two third. So basically
it will be c, the answer would be c.
So the average the proportional so what see,
by calculating the average we learn a very
very important physical fact that the average
temperature is nothing but the constant of
proportionality, average temperature. So what
is the physical fact that we have learned?
Average temperature is the constraint of proportionality.
See, so you have some information which the
mathematics is giving you which by by just
looking at the proportional of, constant of
proportionality you cannot say anything about,
you may think that average temperature and
constant of proportionality is different.
But you see in this particular scenario, it
is the same. It is amazing that here in this
particular for this particular box, the average
temperature is same as the constant the proportionality.
Of course if we change the box things will
be different. But average temperature is same
as constant of proportionality. So that is
a physical fact that we derived from here.
Well done. So here we end our discussion.
We have been trying to maintain time, because
the next class will take a slightly much more
time. We will be talking about line integrals,
and which we will start next. So maybe I should
take a break.
And then I will talk to you about line integrals.
Okay, so this is what you get that these are
the some ways you can of course, you can talk
on moment of inertia. The book also talks
about how to use all these to find the gravitational
potential and all those things. But we are
not getting into so much of details about
physics at this moment because of the variety
of students that are there.
I am sure economists would be unhappy if I
just keep on talking about physics or biologists
might be unhappy. So we will go over to line
integrals, which we will start in the next
class. Thank you.
