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PROFESSOR: Hi, welcome
back to recitation.

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I have a nice
exercise here for you

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that tests your knowledge
of triple integration.

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So in particular, I've
got for you a cylinder.

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And my cylinder has height
h and it has radius b.

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And this is the kind
of cylinder I like.

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It's a constant
density cylinder.

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So its density is
just 1 everywhere.

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So what I'd like you to
do is, for the cylinder,

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I'd like you to compute
its moment of inertia

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around its central axis.

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So why don't you pause the
video, have a go at that,

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come back, and you can check
your work against mine.

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Hopefully you had some luck
working on this problem.

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Let's talk about it.

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So the first thing to
notice is that there

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aren't any coordinates
in this problem.

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I've given you a cylinder
but it's up to you

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to choose coordinates, a way to
arrange your cylinder in space

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or a way to arrange your
coordinates with respect

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to your cylinder.

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So a convenient thing
to do in this case

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is going to be-- you
know, we're working

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with respect to the central
axis of the cylinder.

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So let's make that
one of our axes.

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So in particular, why don't
we make it our z-axis.

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That seems like a natural
sort of thing to do.

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So let me try and draw
a little picture here.

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So we've got our cylinder.

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And there it is with our
three coordinate axes.

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I guess it's got radius
b and it's got height h.

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Now we've arranged it,
now we have coordinates,

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so now we want to see what it
is we're trying to do with it.

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So we're trying to compute
a moment of inertia.

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So we have to remember what
a moment of inertia means.

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So let me think.

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So a moment of inertia,
when you have a solid--

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so your moment of inertia I with
respect to an axis is what you

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get when you take
the triple integral--

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so let's say your solid is D.
Your solid D. So you take D.

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So you take a triple
integral over D

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and you're integrating
r squared with respect

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to the element of mass.

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OK.

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So r squared here,
this is the distance

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from the axis around
which you're computing

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the moment of inertia.

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And in our case, so in any
case, this little moment of mass

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is-- sorry, little
element of mass--

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is density times a
little element of volume.

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So we can also write this
as the triple integral

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over our region of r squared
times delta times dV.

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OK, so this is what
this is in general.

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So now let's think
about it in our case.

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Well, in our case,
we've cleverly

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chosen our central
axis to be the z-axis.

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So this r is just the
distance from the z-axis.

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This I was kind
enough to give you

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a constant-density cylinder,
so this delta is just 1.

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That's going to be easy.

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And then we have
this triple integral

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over the whole cylinder.

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So we want a triple integral
over the whole cylinder

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of some integrand
that involves r.

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So a natural thing to
do in this situation--

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the natural sort
of thing to do when

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you have a cylinder or anything
that's rotationally symmetric,

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and you have an integrand that
behaves nicely with respect

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to rotation, that
can be written easily

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in terms of r, or
r and theta-- is

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to do cylindrical coordinates.

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Is to think of
cylindrical coordinates.

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So in our case
that means we just

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need to figure out-- at this
point-- we need to figure out

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how do we integrate
over the cylinder

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in cylindrical coordinates?

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So let's do it.

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So in our case-- so it doesn't
matter too much what order

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we do things in.

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So we need dV.

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We need to write that in terms
of the cylindrical coordinates.

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So that's dz, dr, and d theta.

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And so we know that
dV is r dz dr d theta.

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You might want some
other order there,

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but that's a good, nice order.

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It usually is the simplest
order to consider.

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So this moment of
inertia, in our case,

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is going to be this
triple integral.

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OK, so we said r squared
delta, r squared times density,

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density is 1.

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So that's just r squared.

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And r, the distance
to the axis is

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r, the distance to the z-axis.

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So that's just r squared
times r dz dr d theta.

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So this is the integral
we're trying to compute,

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but we need bounds, right?

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It's a triple integral,
it's a definite integral,

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we need to figure out what
the bounds are to evaluate it

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as an iterated integral.

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So let's go look at this
little picture we drew.

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So I guess I didn't
discuss this,

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but I made a choice
just to put the bottom

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of the cylinder in the xy-plane
and the top at height h.

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It's not going to matter.

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If you had made some other
choice, it would work out fine.

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So that means the z
is going from 0 to h,

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regardless of r and theta.

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So z is going from 0 to h.

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That's nice, let's
put that over here.

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z is the inside one.

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It's going from 0 to h.

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Then r is next.

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Well, this is also--
you know, cylinders

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are great for
cylindrical coordinates.

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Shocker, right, given the name.

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I know.

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So r is going from 0
to what's the radius?

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Our radius was b.

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So r goes from 0
to b, and that's

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true regardless of theta.

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So back over here, so we have r
going from 0 to b and theta is

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just going from 0 to 2*pi;
we're doing a full rotation all

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the way around the cylinder.

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So this is what our
moment of inertia is,

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and now we just
have to compute it.

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So we've got our inner integral
here is with respect to z.

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So the inner integral
is the integral from 0

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to h of r cubed dz.

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And r cubed doesn't
have any z's in it.

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Fabulous.

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So that's just going
to be r cubed z,

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where z goes from 0 to h.

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So that's h r cubed minus 0.

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So h r cubed.

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All right.

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So that's the inner.

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Now let's look at
the middle integral.

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So this is going
to be the integral

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as-- that's our r integral.

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So that's going from 0 to b.

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And what are we integrating?

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We're integrating
the inner integral.

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So the inner integral
was h r cubed.

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So we're integrating h
r cubed dr from 0 to b.

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All right, this is
not quite as easy.

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But h as a constant,
we're integrating r cubed.

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I've done worse.

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You've done worse.

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So that's going to be h r to the
fourth over 4 between 0 and b.

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So OK.

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So that's h b to the fourth
over 4 minus h times 0

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to the fourth over 4.

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So the second term's 0.

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So this is just equal to h
times b to the fourth over 4.

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And finally, we have
our outermost integral.

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So what was that integral?

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Well, that was the integral from
0 to 2*pi d theta of the second

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integral.

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So this is of the
middle integral.

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So it's the integral from 0
to 2*pi d theta of the middle

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integral which was h b
to the fourth over 4.

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And this is just
a constant again.

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Great.

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So this is h b to the
fourth over 4 times 2*pi.

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So what does that work out to?

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That's h b to the
fourth pi over 2.

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All right.

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So there you go.

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Now if you wanted to, you could
also rewrite this a little bit,

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because you could note that
this is pi h b squared,

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that's your volume
of your cylinder.

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And in fact, it's
not just your volume,

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it's your mass of your
cylinder, because it

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had constant density 1.

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So you also could've
written this as mass times

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a squared over 2.

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Sorry. b squared over 2.

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I don't know where a came from.

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Mass times b squared.

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So you have some
other options for how

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you could write this
answer by involving

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the volume and mass and so on.

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So let's just recap very
quickly why we did what we did.

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We had a cylinder.

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And so really,
given a cylinder, it

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was a natural choice to look
at cylindrical coordinates.

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And once we had
cylindrical coordinates,

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everything was easy.

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So we just took our general
form of the moment of inertia,

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took the region in
question, in cylindrical

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coordinates it was very,
very easy to describe

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this entire region.

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And then our
integrals were pretty

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easy to compute after
we made that choice.

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After we made that choice they
were nice and easy to compute.

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So I'll stop there.