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DAVID JORDAN: Hello, and
welcome back to recitation.

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So the problem that I
want to work with you

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now is to compute
some integrals,

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but we want to compute them
in the presence of a density

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function.

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So the region that we're
considering is very simple.

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It's just the unit square.

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So we have the origin here,
we have the line x equals 1,

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we have the line y
equals 1, and we just

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want to compute in this region.

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And so we want to use
this density function

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to find various sort of physical
characteristics of this region.

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So first, we want
to find its mass,

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and so we are going
to need to recall

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how you get mass from density.

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We want to find
the center of mass.

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That is, where is the point
on which we could balance this

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if we cut it out?

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If we tried to balance
it on our fingers,

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where is the average
mass concentrated?

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We want to find the moment
of inertia about the origin,

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and we want to find the moment
of inertia about the x-axis.

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So we're going to have
to remember our formulas

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for moments of inertia.

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So why don't you pause
the video and work

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on this for a little bit.

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Check back with me and I'll
show you how I solved it.

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Hi.

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Welcome back.

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Why don't we start
by finding the mass.

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So the mass is the most
straightforward of these,

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and I find it helpful to use
the language of differentials.

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So what I want us
to do is I want

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us to take a little square
here, and this little square

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has area dA.

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OK?

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And what we want
to do is we want

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to sum up the masses of all
the little squares dA here.

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So what we want to know is what
is the little bit of mass dM

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which corresponds to this
little bit of area dA.

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And more or less by
definition, this is delta dA.

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So delta is the ratio
of area to mass.

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And so this little
contribution of mass

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is just delta times the
little contribution of area.

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OK?

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Now, once we write
it this way, then

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our total mass for
the entire square

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is just the integral
over the region of all

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the little contributions of dM.

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And so in particular, this
is just the integral from x

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equals 0 to 1, y equals 0 to 1.

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We have x*y--
that's our density--

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and then we have dy dx.

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OK.

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And this is an integral
which we can just compute.

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So why don't we compute
this one all the way through

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and see what we get.

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OK.

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So we have integral
x equals 0 to 1.

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So we have x*y, and we need
to integrate that in y.

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So we have x y squared over 2.

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And then y ranges
from 1 to 0, dx.

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So this is the integral from x
equals 0 to 1 of x over 2, dx.

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And this is just x squared
over 4 from 1 to 0.

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This is just 1/4.

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So that tells us that the
total mass of this unit square

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is 1/4.

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OK.

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So now, we need
to do similar-- we

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have a similar challenge for
the other physical quantities.

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We just need to figure out what
is the appropriate differential

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quantity, and then we just
need to integrate that.

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For b, we need to compute
the center of mass.

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So remember that
the center of mass

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involves finding the
average x-coordinate

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and the average y-coordinate.

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And I wanted to remind
you what the formula is

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for the center of
mass, and remind you

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how I remind myself of it.

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So in the formula for
the center of mass,

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we need to take the
average of x times dM

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divided by the integral of dM.

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So this is our formula
for the center of mass.

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And I just wanted
to say that the way

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that I remember this is
by thinking about seesaws.

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So if you think about
it-- and if we were not

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doing multiple variables
but a single variable--

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if I had a seesaw, and
I had some weights.

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So I had m_1 and m_2 and m_3
and m_4-- I had some weights--

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and these were at positions
x_1 and x_2 and x_3 and x_4.

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Well, the fact that the
scale would be balanced

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would be to say that this point
x here, where the fulcrum is

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located, is exactly the weighted
average of these points.

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That's what's going to
guarantee that there's

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the same amount of torque
pushing this way and this way.

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So if we were in one
variable and we just

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had some discrete
weights, then we

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would want to take the average
of all of these positions,

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and we would want to
weight it with the masses.

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So we would want to take the
sum of x_i*m_i and divide

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by the sum of m_i.

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This would be the
average coordinates

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in this kind of toy example.

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And now if you look at the
formula for the center of mass,

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it's really the same
thing, isn't it?

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Because integrals are just a
continuous version of the sum.

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We have x as a function
instead of x_i as a list.

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And the m_i's are just the
little infinitesimal dM's here.

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And then the bottom here is just
the total mass of the system,

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and so is this.

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OK, so that's how I think about
this center of mass formula.

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And it's actually
pretty easy to compute.

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So we have the integral from x
equals 0 to 1, y equals 0 to 1.

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So now we have x times
delta times dx dy.

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So altogether we get
x squared y dy dx.

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So one of those x's
is because we're

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averaging x and the other one
is from the density function.

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So we have this whole integral.

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And then we divide by
this integral of the mass,

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but we already computed
this in part a,

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and we found it to be 1/4.

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OK.

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So this numerator here is fairly
straightforward to compute.

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And if you do this
you'll get-- let

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me double check-- I
believe we got 1/6.

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So you should get 1/6 when
you compute this integral.

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So we have 1/6 over 1/4, and
so cancelling off, this is 2/3.

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OK.

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So that was just the
x center of mass.

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But now I want to make
an important point, which

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is that this density function
is symmetric in x and y.

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It was just x times y.

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It wasn't something
more complicated.

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And so the center of
mass in the x-direction

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is just equal to the center
of mass in the y-direction,

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so these are both equal to 2/3.

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OK.

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So that depended on the fact
that our density was symmetric,

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and also on the fact
that our region was

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symmetric about
switching x and y.

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So we could save ourselves
some trouble here.

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OK, very good.

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So now to do c, again
we need to recall

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what is the infinitesimal
moment of inertia.

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So let me draw
this picture again.

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So here's our little dA here.

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And we want to know the
infinitesimal moment of inertia

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around the origin.

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So we tie a string to
this little piece of mass,

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and we start spinning
it, and we want

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to know what is our moment
of inertia corresponding

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to that little mass.

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And I'll just remind
you from lecture

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that the formula
is r squared dM.

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So this is r squared
times x*y dx dy.

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And so the r squared
here is saying

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that as you get farther
and farther out,

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your moment of inertia is
getting larger and larger.

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And this makes sense in
terms of the physical idea

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that you're moving a longer
distance if you're farther out.

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So anyway.

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So this is our
formula, r squared dM.

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And so that tells us that I
is just the integral of dI.

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And so this is the integral
from x goes from 0 to 1,

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y goes from 0 to 1.

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And then we have x
squared plus y squared--

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that's just r squared--
times x*y dx dy.

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And so we can rewrite this as
x cubed y plus x y cubed dx dy.

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And this is a computation
that we can do.

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Let me just check
my notes real quick.

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So this is 1/4.

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I'll skip the computation,
but this is just

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integrating some polynomials,
so we can do that.

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All right.

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And now finally, we want to
compute the moment of inertia--

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so remember, d asked us to
compute the moment of inertia

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around the x-axis.

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So instead of around the
origin, it's around the x-axis.

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So the idea here is the same.

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So again, dI is a
factor times dM.

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And again, it's the
radius, but now it's

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the radius about
which we're spinning.

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So we're not anymore
spinning around the origin

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as we were doing before.

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Now we're spinning
around-- sort of out

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of the board-- around
the x-axis here.

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But we still have
the same formula,

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and now our radius
is the height y.

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Because we're not spinning
around the origin anymore,

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we're spinning
around this rod here.

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And so if you think
about it, that's

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the radius about which we're
spinning is just the height y.

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So this is just y squared delta.

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OK.

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And so that tells us that
I-- the total inertia

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about the x-axis-- is
just the integral of dI.

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And so we get the integral
from x equals 0 to 1,

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integral y equals 0 to 1.

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And then we just have
y squared x*y dy dx.

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And this, again, we
could compute-- and let

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me just check my notes--
and find that it's 1/8.

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So in each of these problems,
the most important thing

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to have been able
to do is to argue,

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what is this sort of
infinitesimal contribution

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to the physical quantity
that you want to compute?

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And eventually, you
want to express it

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in terms of the quantity dA,
because dA is what we actually

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can integrate.

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And so all the other
physical quantities

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that we need to
study are going to be

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an integral of some
infinitesimal element,

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and that infinitesimal
element is

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going to be some
coefficient times dA.

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So here, we had that
this was-- oh, dear.

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This is a mistake.

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So this should have
said y squared dM,

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and that's y squared delta dA.

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So I wrote the delta implicitly.

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I wrote it twice.

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So what we meant to say is
dI is y squared delta dA.

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And so in all these examples,
the infinitesimal quantity

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that we're after is some
straightforward coefficient

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times the infinitesimal area.

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And so once we know
that, then we can just

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do a straightforward integral.

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OK, and I'll leave it at that.