PROFESSOR: Hi, welcome
back to recitation.
I have a nice
exercise here for you
that tests your knowledge
of triple integration.
So in particular, I've
got for you a cylinder.
And my cylinder has height
h and it has radius b.
And this is the kind
of cylinder I like.
It's a constant
density cylinder.
So its density is
just 1 everywhere.
So what I'd like you to
do is, for the cylinder,
I'd like you to compute
its moment of inertia
around its central axis.
So why don't you pause the
video, have a go at that,
come back, and you can check
your work against mine.
Hopefully you had some luck
working on this problem.
Let's talk about it.
So the first thing to
notice is that there
aren't any coordinates
in this problem.
I've given you a cylinder
but it's up to you
to choose coordinates, a way to
arrange your cylinder in space
or a way to arrange your
coordinates with respect
to your cylinder.
So a convenient thing
to do in this case
is going to be-- you
know, we're working
with respect to the central
axis of the cylinder.
So let's make that
one of our axes.
So in particular, why don't
we make it our z-axis.
That seems like a natural
sort of thing to do.
So let me try and draw
a little picture here.
So we've got our cylinder.
And there it is with our
three coordinate axes.
I guess it's got radius
b and it's got height h.
Now we've arranged it,
now we have coordinates,
so now we want to see what it
is we're trying to do with it.
So we're trying to compute
a moment of inertia.
So we have to remember what
a moment of inertia means.
So let me think.
So a moment of inertia,
when you have a solid--
so your moment of inertia I with
respect to an axis is what you
get when you take
the triple integral--
so let's say your solid is D.
Your solid D. So you take D.
So you take a triple
integral over D
and you're integrating
r squared with respect
to the element of mass.
OK.
So r squared here,
this is the distance
from the axis around
which you're computing
the moment of inertia.
And in our case, so in any
case, this little moment of mass
is-- sorry, little
element of mass--
is density times a
little element of volume.
So we can also write this
as the triple integral
over our region of r squared
times delta times dV.
OK, so this is what
this is in general.
So now let's think
about it in our case.
Well, in our case,
we've cleverly
chosen our central
axis to be the z-axis.
So this r is just the
distance from the z-axis.
This I was kind
enough to give you
a constant-density cylinder,
so this delta is just 1.
That's going to be easy.
And then we have
this triple integral
over the whole cylinder.
So we want a triple integral
over the whole cylinder
of some integrand
that involves r.
So a natural thing to
do in this situation--
the natural sort
of thing to do when
you have a cylinder or anything
that's rotationally symmetric,
and you have an integrand that
behaves nicely with respect
to rotation, that
can be written easily
in terms of r, or
r and theta-- is
to do cylindrical coordinates.
Is to think of
cylindrical coordinates.
So in our case
that means we just
need to figure out-- at this
point-- we need to figure out
how do we integrate
over the cylinder
in cylindrical coordinates?
So let's do it.
So in our case-- so it doesn't
matter too much what order
we do things in.
So we need dV.
We need to write that in terms
of the cylindrical coordinates.
So that's dz, dr, and d theta.
And so we know that
dV is r dz dr d theta.
You might want some
other order there,
but that's a good, nice order.
It usually is the simplest
order to consider.
So this moment of
inertia, in our case,
is going to be this
triple integral.
OK, so we said r squared
delta, r squared times density,
density is 1.
So that's just r squared.
And r, the distance
to the axis is
r, the distance to the z-axis.
So that's just r squared
times r dz dr d theta.
So this is the integral
we're trying to compute,
but we need bounds, right?
It's a triple integral,
it's a definite integral,
we need to figure out what
the bounds are to evaluate it
as an iterated integral.
So let's go look at this
little picture we drew.
So I guess I didn't
discuss this,
but I made a choice
just to put the bottom
of the cylinder in the xy-plane
and the top at height h.
It's not going to matter.
If you had made some other
choice, it would work out fine.
So that means the z
is going from 0 to h,
regardless of r and theta.
So z is going from 0 to h.
That's nice, let's
put that over here.
z is the inside one.
It's going from 0 to h.
Then r is next.
Well, this is also--
you know, cylinders
are great for
cylindrical coordinates.
Shocker, right, given the name.
I know.
So r is going from 0
to what's the radius?
Our radius was b.
So r goes from 0
to b, and that's
true regardless of theta.
So back over here, so we have r
going from 0 to b and theta is
just going from 0 to 2*pi;
we're doing a full rotation all
the way around the cylinder.
So this is what our
moment of inertia is,
and now we just
have to compute it.
So we've got our inner integral
here is with respect to z.
So the inner integral
is the integral from 0
to h of r cubed dz.
And r cubed doesn't
have any z's in it.
Fabulous.
So that's just going
to be r cubed z,
where z goes from 0 to h.
So that's h r cubed minus 0.
So h r cubed.
All right.
So that's the inner.
Now let's look at
the middle integral.
So this is going
to be the integral
as-- that's our r integral.
So that's going from 0 to b.
And what are we integrating?
We're integrating
the inner integral.
So the inner integral
was h r cubed.
So we're integrating h
r cubed dr from 0 to b.
All right, this is
not quite as easy.
But h as a constant,
we're integrating r cubed.
I've done worse.
You've done worse.
So that's going to be h r to the
fourth over 4 between 0 and b.
So OK.
So that's h b to the fourth
over 4 minus h times 0
to the fourth over 4.
So the second term's 0.
So this is just equal to h
times b to the fourth over 4.
And finally, we have
our outermost integral.
So what was that integral?
Well, that was the integral from
0 to 2*pi d theta of the second
integral.
So this is of the
middle integral.
So it's the integral from 0
to 2*pi d theta of the middle
integral which was h b
to the fourth over 4.
And this is just
a constant again.
Great.
So this is h b to the
fourth over 4 times 2*pi.
So what does that work out to?
That's h b to the
fourth pi over 2.
All right.
So there you go.
Now if you wanted to, you could
also rewrite this a little bit,
because you could note that
this is pi h b squared,
that's your volume
of your cylinder.
And in fact, it's
not just your volume,
it's your mass of your
cylinder, because it
had constant density 1.
So you also could've
written this as mass times
a squared over 2.
Sorry. b squared over 2.
I don't know where a came from.
Mass times b squared.
So you have some
other options for how
you could write this
answer by involving
the volume and mass and so on.
So let's just recap very
quickly why we did what we did.
We had a cylinder.
And so really,
given a cylinder, it
was a natural choice to look
at cylindrical coordinates.
And once we had
cylindrical coordinates,
everything was easy.
So we just took our general
form of the moment of inertia,
took the region in
question, in cylindrical
coordinates it was very,
very easy to describe
this entire region.
And then our
integrals were pretty
easy to compute after
we made that choice.
After we made that choice they
were nice and easy to compute.
So I'll stop there.
