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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 now is to

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compute some integrals, but we
want to compute them in the

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presence of a density
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, we

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have the line y equals 1, and
we just want to compute in

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this region.

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And so we want to use this
density function to find

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various sort of physical

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characteristics of this region.

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So first, we want to find its
mass, and so we are going to

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need to recall 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 if

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we cut it out?

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If we tried to balance it on
our fingers, where is the

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average mass concentrated?

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

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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 for

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moments of inertia.

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So why don't you pause the video
and work on this for a

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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, and

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I find it helpful to use the
language of differentials.

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So what I want us to do is I
want us to take a little

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square here, and this little
square has area dA.

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OK?

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And what we want to do is we
want to sum up the masses of

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all the little squares
dA here.

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

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dM 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 is just

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delta times the little
contribution of area.

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OK?

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Now, once we write it this way,
then our total mass for

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the entire square is just the
integral over the region of

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all 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 xy--

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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 and

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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 xy, and we need
to integrate that in y.

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So we have xy 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

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unit square is 1/4.

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OK.

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So now, we need to do similar,
we have a similar challenge

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for the other physical
quantities.

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We just need to figure out
what is the appropriate

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differential quantity,
and then we just need

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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 involves finding the

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average x-coordinate and the
average y-coordinate.

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And I wanted to remind you what
the formula is for the

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center of mass, and remind you
how I remind myself of it.

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So in the formula for the center
of mass, we need to

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take the average of x times dM
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 that I remember this

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is by thinking about seesaws.

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

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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 m1 and m2
and m3 and m4--

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I had some weights--

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and these were at positions
x1 and x2 and x3 and x4.

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Well, the fact that the scale
would be balanced would be to

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say that this point x here,
where the fulcrum is located,

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is exactly the weighted average
of these points.

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

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same amount of torque pushing
this way and this way.

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So if we were in one variable
and we just had some discrete

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weights, then we would want to
take the average of all of

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these positions, and
we would want to

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weight it with the masses.

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So we would want to take the
sum of xi mi and divide

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by the sum of mi.

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This would be the average
coordinates in this kind of

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toy example.

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

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mass, 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 xi--

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as a list. And the mi's
are just the little

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infinitesimal dM's here.

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

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system, and so is this.

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OK, so that's how I
think about this

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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,

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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 averaging x and the

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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, but we

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already computed this in part a,
and we found it to be 1/4.

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OK.

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So this numerator
here is fairly

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straightforward to compute.

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And if you do this
you'll get--

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let me double check--

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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

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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 is that

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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 is just equal

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to the center of mass in the
y-direction, so these are both

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equal to 2/3.

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OK.

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So that depended on the fact
that our density was

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symmetric, 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 what is the

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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

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inertia around the origin.

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

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start spinning it, and we want
to know what is our moment of

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inertia corresponding
to that little mass.

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And I'll just remind you from
lecture that the formula is r

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squared dM.

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So this is r squared
times xy dx dy.

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And so the r squared here is
saying that as you get farther

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and farther out, your moment of
inertia is getting larger

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and larger.

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And this makes sense in terms
of the physical idea that

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you're moving a longer
distance if

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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, y

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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--

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times xy dx dy.

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And so we can rewrite this as x
cubed y plus xy 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 integrating some

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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,

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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 the radius about

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which we're spinning.

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So we're not anymore spinning
around the origin as we were

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doing before.

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Now we're spinning around--

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

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But we still have the same
formula, and now our radius is

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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 the radius about which

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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

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inertia about the x-axis--

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is just the integral of dI.

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

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y equals 0 to 1.

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And then we just have
y squared xy dy dx.

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

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check my notes--

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and find that it's 1/8.

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00:14:07,860 --> 00:14:11,140

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So in each of these problems,
the most important thing to

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have been able to do is to
argue, what is this sort of

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infinitesimal contribution to
the physical quantity that you

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want to compute?

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And eventually, you want to
express it in terms of the

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quantity dA, because
dA is what we

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actually can integrate.

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And so all the other physical
quantities that we need to

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study are going to be an
integral of some infinitesimal

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element, and that infinitesimal
element is going

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to be some coefficient
times dA.

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So here, we had that
this was--

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oh, dear.

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This is a mistake.

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So this should have said y
squared dM, and that's y

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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 do a

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straightforward integral.

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OK, and I'll leave it at that.

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