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PROFESSOR: Welcome
back to recitation.

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In this video I want us to
work on the following problem.

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A very shallow circular
reflecting pool

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has uniform depth D and radius
R. And this is in meters.

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And a disinfecting chemical
is released at its center.

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After a few hours of
symmetrical diffusion outward,

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the concentration at a point
little r meters from the center

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is k over 1 plus r squared
grams per cubic meter.

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The k here is a constant.

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So we drop some chemical
into the middle of the pool.

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And then it diffuses outward.

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That's the idea there.

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Now the question
is the following.

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What amount of chemical
was released into the pool?

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And obviously the
amount will be in grams.

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Now to give you a
hint, probably you

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want to draw a picture of this.

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And then maybe
get some estimates

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or some approximations
using shells.

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Or you might just say,
you know, some strips,

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some shells probably is
a better word for it.

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And then you should
think about the fact

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that, as you get better
and better estimates,

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your approximation should
tend toward an integral.

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So with that hint, I'll give you
a little time to work on this.

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And when we come back,
I'll show you how I do it.

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OK, welcome back.

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Hopefully you were able
to make some headway.

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And so let me start off by
doing what I asked you to do,

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which is draw a picture.

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Now the first picture
is fairly simple.

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The first picture is my pool.

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Right?

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But that's not quite enough
to tell me some estimates.

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But what do we have?

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So the pool has radius
capital R and depth D.

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And when I said we want to
do some approximation, what

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I really meant is we
want to-- let me actually

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get another color here-- we want
to say, take some fixed radius

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out and assume that--
let me actually

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draw this behind--
some fixed radius out,

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and assume that the diffusion
is constant for some small bit.

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Some small strip like this.

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Which, actually, you notice
it's going to be rotated around.

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Because all of this is relative
to distance from the center.

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So, I'm hoping that
you can see kind

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of what this drawing is doing.

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Essentially what we
have here is if we

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open that up, this blue
cylindrical shell is

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approximately a piece-- oh it
doesn't quite look flat-- but,

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a piece that looks
like-- oh well

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we'll just stick with
that-- a little prism here.

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So that's
approximately, if I were

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to cut this blue cylindrical
shell here open and lay it

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down flat, it would be
approximately a piece

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kind of like this.

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Right?

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And so, what we're going
to do is estimate first

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what amount of chemical
is released based

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on pieces that look like this.

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And then we're going to let
those pieces get very, very

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narrow and get more
and more of them.

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And this should remind
you of Riemann sums.

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And how Riemann sums, as
you let the number of things

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you're summing over
tend to 0-- sorry--

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tend to infinity, so that
the little pieces are getting

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narrower and narrower,
you're actually

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going to end up
with an integral.

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So this is where we're headed.

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So let's just make sure
we understand kind of all

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the pieces that are happening.

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What we're going to
do is we're going

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to take a bunch of
these cylinders,

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and let's just determine
that we'll take n of them.

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I shouldn't say cylinder.

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Sorry, I should say shells.

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We're going to take n of
these shell-type things.

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So I'm going to say
the radii-- I'm going

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to start with r_0 equals 0.

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And I'm going to take
n different radii.

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So r_0, r_1, up to--
sorry this is 0--

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so r_n is equal to capital R.
So I guess I'm taking n plus 1,

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but 0 is not really a radius.

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But I have n
different partitions.

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For each partition
of this big cylinder,

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I get a piece like this.

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Right?

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I get a piece that,
when I open it up,

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looks approximately like this.

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Now, what I want
is a total amount.

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I want grams.

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Right?

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And what I'm given-- if we come
back over here-- what I'm given

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is the concentration
at a certain radius.

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It's k over 1 plus r squared
grams per cubic meter.

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Now, if nothing else, you should
have looked at this problem

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and seen, well if I want
grams and I have something

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in grams per cubic
meter, somewhere I'm

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going to need something
with cubic meters

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to cancel this unit,
so I end up with grams.

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So if nothing else, then
maybe you can you can think,

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oh I need to understand
volume of something

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in order to solve this problem.

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Right?

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Now if we have n different
partitions-- so n shells--

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that all started off as
this sort of blue-type thing

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and I open up and
look like this.

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Then I want to figure out what
is the volume of these shells.

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Once I know the
volume of that shell,

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I can figure out
the amount, roughly,

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of chemical in that
piece by multiplying

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by the concentration.

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So let's figure out
what this volume is

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in terms of these little
radii I'm looking at.

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Well, when I open it
up, what do I get?

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We're assuming this here is
this little segment here,

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and so that's our delta r,
that's our change in radius.

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That's how much I'm changing.

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So this would be some r
subscript i and this would

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be some r subscript i plus 1.

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And let's assume that
we're taking everything

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from the smaller radius.

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We're going to do everything
from the smaller radius.

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So then, when I open
this up, this circle

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is going to be my length.

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So my length is 2 pi r sub i.

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And then the height is easy.

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The height is constant.

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The height is just
capital D. Right?

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So the volume of each shell--
let me come over here--

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the volume of a shell is
something like 2 pi r sub i

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times D times delta r.

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And so then, the amount
of chemical in the shell

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is going to be the volume
times the concentration.

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Right?

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The volume times
the concentration.

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So the volume is, again, this.

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I'm going to put the D
in front of the r sub i.

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2 pi D r sub i delta r times
the concentration, which

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if we come back over
here, the concentration is

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k divided by 1 plus r squared
grams per cubic meter.

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The r in this case
is the r sub i.

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I'm assuming, because
I'm approximating

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this, that everywhere
in the shell

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has the same concentration,
has the concentration

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of the interior radius.

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So if we come back over here,
we're going to write k over 1

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plus r sub i squared.

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And now what do I do to estimate
the amount in the entire pool?

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Well I add all of these up.

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So let me come over
here and write down

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what the sum will look like.

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So I'm going to
be summing from i

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equals-- I said I was
taking the interior radius,

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I think-- i equals 0 to n
minus 1 of this quantity.

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2*pi*D-- let me put the
k in there as well-- k.

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And then r sub i over 1 plus
r sub i squared delta r.

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So this is our
approximation of the amount

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of chemical in the pool.

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And again, we always want
to check and make sure.

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I didn't write in any units,
but do the units make sense?

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Well we know the
units make sense

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because when I did the
amounts in the shell,

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I did volume times
concentration.

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And volume times concentration
is going to be in grams.

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This is in cubic meters.

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This is in grams
per cubic meter.

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So I know I have the right unit.

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So that's a good way to check.

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It doesn't guarantee
you've done it correctly.

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But at least you can check and
make sure you didn't do it,

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you know that-- how
would I say this?

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I would say that if the
units are not in grams,

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you know you did
something wrong.

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So at least now we know, OK,
it passes the first smell test.

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Now what do I do to
find the exact value?

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Well what I want
to do is, I want

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to come back over to
the picture I have here

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and I want to let these shells
get smaller and smaller.

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And how do I let those shells
get smaller and smaller?

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Narrower and narrower,
I should say.

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I let them get
narrower by increasing

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the number of radii on which
I do this kind of operation.

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So I'm coming over
here and now I'm

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letting the n get
bigger and bigger.

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And as n gets bigger
and bigger, these values

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are still determined
the same way.

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But over here this n is
getting larger and larger.

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So I can take the limit,
as n goes to infinity,

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of this quantity to
get the exact amount.

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What is the limit as n
goes to infinity of this?

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This is actually an integral.

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It's the integral
from 0 to capital R--

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because my radius
is ranging from 0

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to that big R-- of this
exact function of R. Right?

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So I'm going to put
the 2*pi*D*k out here.

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And then I get little
r over 1 plus r squared

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and the delta r becomes our
dr. So this is, in fact,

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going to be the amount, in
grams, of the chemical that

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was released into the pool.

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So we've set up our integral.

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I think I'll stop here.

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If you want to go further
and determine it, you can.

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And you may want to think
about what strategy,

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00:10:39,909 --> 00:10:41,950
obviously, what strategy
you want to use in order

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00:10:41,950 --> 00:10:42,890
to solve this problem.

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00:10:42,890 --> 00:10:44,150
I'll give you a hint.

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00:10:44,150 --> 00:10:46,360
Maybe the best way
to solve this problem

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00:10:46,360 --> 00:10:48,510
is the fact that when you
take the derivative of 1

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00:10:48,510 --> 00:10:52,180
plus r squared, you
actually get 2r.

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00:10:52,180 --> 00:10:54,330
And so that derivative
is almost up here.

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00:10:54,330 --> 00:10:57,220
So maybe this is a good
hint to give you how you

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00:10:57,220 --> 00:10:58,640
would continue this problem.

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00:10:58,640 --> 00:10:59,490
But I'll stop there.

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00:10:59,490 --> 00:11:01,355
So let me just go
back one more time

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00:11:01,355 --> 00:11:03,890
and remind you
what we were doing.

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00:11:03,890 --> 00:11:07,500
What we were doing, if
we come back over here,

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00:11:07,500 --> 00:11:09,590
is we were given
a situation where

219
00:11:09,590 --> 00:11:13,860
we knew a certain function
of the radius, the distance

220
00:11:13,860 --> 00:11:15,230
from the center.

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00:11:15,230 --> 00:11:19,360
And we wanted to determine the
total amount of chemical that

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00:11:19,360 --> 00:11:20,920
was released into the pool.

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00:11:20,920 --> 00:11:22,130
And so we estimated.

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00:11:22,130 --> 00:11:24,110
We figured out a
way to estimate it

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00:11:24,110 --> 00:11:27,190
in terms of splitting
up the radii.

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00:11:27,190 --> 00:11:29,884
We had the radius
from 0 to big R.

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00:11:29,884 --> 00:11:32,550
And we just divided up the radii
and assumed certain things were

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constant in these regions.

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And we determined
the right function

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00:11:37,110 --> 00:11:41,370
to find-- if we move over here,
over here-- we would determine

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00:11:41,370 --> 00:11:43,730
the right function to find
the amount of chemical

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00:11:43,730 --> 00:11:46,810
in a certain shell, assuming
that the concentration was

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00:11:46,810 --> 00:11:49,250
constant throughout that shell.

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00:11:49,250 --> 00:11:52,940
And then, what we do is we know
that if we let those shells get

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00:11:52,940 --> 00:11:55,437
arbitrarily narrow,
that means that we're

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00:11:55,437 --> 00:11:57,270
letting the number of
radii over which we're

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00:11:57,270 --> 00:11:59,000
doing this go to infinity.

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00:11:59,000 --> 00:12:02,050
And we know that this
summation that we have here,

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00:12:02,050 --> 00:12:05,540
as n goes to infinity,
becomes an integral.

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00:12:05,540 --> 00:12:08,690
So that's really, we exploited
what we know about this sum

241
00:12:08,690 --> 00:12:11,830
and letting this partition,
letting the delta

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00:12:11,830 --> 00:12:14,270
r get arbitrarily small.

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00:12:14,270 --> 00:12:16,160
That that, in the limit,
goes to an integral.

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00:12:16,160 --> 00:12:19,090
And then that's something we
can definitively calculate.

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00:12:19,090 --> 00:12:21,308
So I think I will stop there.