WEBVTT
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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.