DAVID JORDAN: Hello, and welcome back to recitation.
So the problem that I want to work with you now is to
compute some integrals, but we want to compute them in the
presence of a density function.
So the region that we're considering is very simple.
It's just the unit square.
So we have the origin here, we have the line x equals 1, we
have the line y equals 1, and we just want to compute in
this region.
And so we want to use this density function to find
various sort of physical
characteristics of this region.
So first, we want to find its mass, and so we are going to
need to recall how you get mass from density.
We want to find the center of mass.
That is, where is the point on which we could balance this if
we cut it out?
If we tried to balance it on our fingers, where is the
average mass concentrated?
We want to find the moment of inertia about the origin, and
we want to find the moment of inertia about the x-axis.
So we're going to have to remember our formulas for
moments of inertia.
So why don't you pause the video and work on this for a
little bit.
Check back with me and I'll show you how I solved it.
Hi.
Welcome back.
Why don't we start by finding the mass.
So the mass is the most straightforward of these, and
I find it helpful to use the language of differentials.
So what I want us to do is I want us to take a little
square here, and this little square has area dA.
OK?
And what we want to do is we want to sum up the masses of
all the little squares dA here.
So what we want to know is what is the little bit of mass
dM which corresponds to this little bit of area dA.
And more or less by definition, this is delta dA.
So delta is the ratio of area to mass.
And so this little contribution of mass is just
delta times the little contribution of area.
OK?
Now, once we write it this way, then our total mass for
the entire square is just the integral over the region of
all the little contributions of dM.
And so in particular, this is just the integral from x
equals 0 to 1, y equals 0 to 1.
We have xy--
that's our density--
and then we have dy dx.
OK.
And this is an integral which we can just compute.
So why don't we compute this one all the way through and
see what we get.
OK.
So we have integral x equals 0 to 1.
So we have xy, and we need to integrate that in y.
So we have xy squared over 2.
And then y ranges from 1 to 0, dx.
So this is the integral from x equals 0 to 1 of x over 2, dx.
And this is just x squared over 4 from 1 to 0.
This is just 1/4.
So that tells us that the total mass of this
unit square is 1/4.
OK.
So now, we need to do similar, we have a similar challenge
for the other physical quantities.
We just need to figure out what is the appropriate
differential quantity, and then we just need
to integrate that.
For b, we need to compute the center of mass.
So remember that the center of mass involves finding the
average x-coordinate and the average y-coordinate.
And I wanted to remind you what the formula is for the
center of mass, and remind you how I remind myself of it.
So in the formula for the center of mass, we need to
take the average of x times dM divided by the integral of dM.
So this is our formula for the center of mass.
And I just wanted to say that the way that I remember this
is by thinking about seesaws.
So if you think about it-- and if we were not doing multiple
variables but a single variable--
if I had a seesaw, and I had some weights.
So I had m1 and m2 and m3 and m4--
I had some weights--
and these were at positions x1 and x2 and x3 and x4.
Well, the fact that the scale would be balanced would be to
say that this point x here, where the fulcrum is located,
is exactly the weighted average of these points.
That's what's going to guarantee that there's the
same amount of torque pushing this way and this way.
So if we were in one variable and we just had some discrete
weights, then we would want to take the average of all of
these positions, and we would want to
weight it with the masses.
So we would want to take the sum of xi mi and divide
by the sum of mi.
This would be the average coordinates in this kind of
toy example.
And now if you look at the formula for the center of
mass, it's really the same thing, isn't it?
Because integrals are just a continuous version of the sum.
We have x as a function instead of xi--
as a list. And the mi's are just the little
infinitesimal dM's here.
And then the bottom here is just the total mass of the
system, and so is this.
OK, so that's how I think about this
center of mass formula.
And it's actually pretty easy to compute.
So we have the integral from x equals 0 to 1,
y equals 0 to 1.
So now we have x times delta times dx dy.
So altogether we get x squared y dy dx.
So one of those x's is because we're averaging x and the
other one is from the density function.
So we have this whole integral.
And then we divide by this integral of the mass, but we
already computed this in part a, and we found it to be 1/4.
OK.
So this numerator here is fairly
straightforward to compute.
And if you do this you'll get--
let me double check--
I believe we got 1/6.
So you should get 1/6 when you compute this integral.
So we have 1/6 over 1/4, and so cancelling
off, this is 2/3.
OK.
So that was just the x center of mass.
But now I want to make an important point, which is that
this density function is symmetric in x and y.
It was just x times y.
It wasn't something more complicated.
And so the center of mass in the x-direction is just equal
to the center of mass in the y-direction, so these are both
equal to 2/3.
OK.
So that depended on the fact that our density was
symmetric, and also on the fact that our region was
symmetric about switching x and y.
So we could save ourselves some trouble here.
OK, very good.
So now to do c, again we need to recall what is the
infinitesimal moment of inertia.
So let me draw this picture again.
So here's our little dA here.
And we want to know the infinitesimal moment of
inertia around the origin.
So we tie a string to this little piece of mass, and we
start spinning it, and we want to know what is our moment of
inertia corresponding to that little mass.
And I'll just remind you from lecture that the formula is r
squared dM.
So this is r squared times xy dx dy.
And so the r squared here is saying that as you get farther
and farther out, your moment of inertia is getting larger
and larger.
And this makes sense in terms of the physical idea that
you're moving a longer distance if
you're farther out.
So anyway.
So this is our formula r squared dM.
And so that tells us that I is just the integral of dI.
And so this is the integral from x goes from 0 to 1, y
goes from 0 to 1.
And then we have x squared plus y squared--
that's just r squared--
times xy dx dy.
And so we can rewrite this as x cubed y plus xy cubed dx dy.
And this is a computation that we can do.
Let me just check my notes real quick.
So this is 1/4.
I'll skip the computation, but this is just integrating some
polynomials, so we can do that.
All right.
And now finally, we want to compute the moment of inertia.
So remember, d asked us to compute the moment of inertia
around the x-axis.
So instead of around the origin,
it's around the x-axis.
So the idea here is the same.
So again, dI is a factor times dM.
And again, it's the radius, but now it's the radius about
which we're spinning.
So we're not anymore spinning around the origin as we were
doing before.
Now we're spinning around--
sort of out of the board-- around the x-axis here.
But we still have the same formula, and now our radius is
the height y.
Because we're not spinning around the origin anymore,
we're spinning around this rod here.
And so if you think about it, that's the radius about which
we're spinning is just the height y.
So this is just y squared delta.
OK.
And so that tells us that I --the total
inertia about the x-axis--
is just the integral of dI.
And so we get the integral from x equals 0 to 1, integral
y equals 0 to 1.
And then we just have y squared xy dy dx.
And this, again, we could compute-- and let me just
check my notes--
and find that it's 1/8.
So in each of these problems, the most important thing to
have been able to do is to argue, what is this sort of
infinitesimal contribution to the physical quantity that you
want to compute?
And eventually, you want to express it in terms of the
quantity dA, because dA is what we
actually can integrate.
And so all the other physical quantities that we need to
study are going to be an integral of some infinitesimal
element, and that infinitesimal element is going
to be some coefficient times dA.
So here, we had that this was--
oh, dear.
This is a mistake.
So this should have said y squared dM, and that's y
squared delta dA.
So I wrote the delta implicitly.
I wrote it twice.
So what we meant to say is dI is y squared delta dA.
And so in all these examples, the infinitesimal quantity
that we're after is some straightforward coefficient
times the infinitesimal area.
And so once we know that, then we can just do a
straightforward integral.
OK, and I'll leave it at that.