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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 has uniform

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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

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center 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. Now to give you a

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hint, probably you want to
draw a picture of this.

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And then maybe get some
estimates or some

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approximations using shells.

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

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probably is a better
word for it.

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And then you should think about
the fact that, as you

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get better and better estimates,
your approximation

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should tend toward
an integral.

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So with that hint, I'll
give you a little

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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. And

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

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meant is we want to-- let me
actually get another color

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here-- we want to say, take
some fixed radius out and

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assume that--

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let me actually draw this
behind-- some fixed radius

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out, and assume that the
diffusion is constant

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for some small bit.

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Some small strip like this.

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Which, actually, you
notice it's going

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to be rotated around.

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Because all of this
is relative to

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distance from the center.

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So, I'm hoping that you
can see kind of what

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this drawing is doing.

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Essentially what we have here
is if we open that up, this

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blue cylindrical shell is
approximately a piece--

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oh it doesn't quite
look flat--

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but, a piece that looks like--

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oh well we'll just
stick with that--

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a little prism here.

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So that's approximately, if
I were to cut this blue

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cylindrical shell here open and
lay it down flat, it would

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be approximately a piece
kind of like this.

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Right?

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

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of chemical is released based 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. And how Riemann

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sums, as you let the number of
things you're summing over

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tend to 0-- sorry-- tend to
infinity, and so that the

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little pieces are getting
narrower and narrower, you're

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actually 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 the

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pieces that are happening.

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What we're going to do is we're
going to take a bunch of

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these cylinders, and let's
just determine that

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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--

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I'm going to start
with r0 equals 0.

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And I'm going to take
n different radii.

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So r0, r1, up to--

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sorry this is 0--

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so rn is equal to capital R. So
I guess I'm taking n plus

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1, 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, I get

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a piece like this.

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Right?

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

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approximately like this.

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Now, what I want is
a total amount.

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I want grams. Right?

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And what I'm given--

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

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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

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

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something in grams per cubic
meter, somewhere I'm going to

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need something with cubic meters
to cancel this unit, so

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I end up with grams. So if
nothing else, then maybe you

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can you can think, oh I need
to understand volume of

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something in order to
solve this problem.

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Right?

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Now if we have n different
partitions--

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so n shells--

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

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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, I can figure out the

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amount, roughly, 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 in terms of these

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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, and

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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 be

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some r subscript i plus 1.

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And let's assume that we're
taking everything from the

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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 is

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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--

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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 is going

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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 if we

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come back over here, the
concentration is k divided by

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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 this, that

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everywhere in the shell has the
same concentration, has

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the concentration 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 get, do to
estimate the amount in the

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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 what the

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sum will look like.

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So I'm going to be summing
from i equals--

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I said I was taking the interior
radius, I think-- i

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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--

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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 of

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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 because when I did the

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amounts in the shell, I did
volume times concentration.

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And volume times concentration
is going to be in grams. This

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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--

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how would I say this?

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

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did something wrong.

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

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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 to come back over to the

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picture I have here and I want
to let these shells get

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smaller and smaller.

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And how do I let these 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 the number of radii

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on which I do this kind
of operation.

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So I'm coming over here and
now I'm letting the n get

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bigger and bigger.

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And as n gets bigger and bigger,
these values are still

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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, of this

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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. Because my radius

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is ranging from 0 to that
big R. Of this exact

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function of R. Right?

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So I'm going to put the
2 pi Dk out here.

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And then I get little r over 1
plus r squared and the delta r

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becomes our dr. So this is, in
fact, going to be the amount,

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in grams, of the chemical that
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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obviously, what strategy you
want to use in order to solve

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this problem.

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I'll give you a hint.

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Maybe the best way to solve this
problem is the fact that

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when you take the derivative
of 1 plus r squared, you

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actually get 2r.

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And so that derivative
is almost up here.

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So maybe this is a good hint
to give you how you would

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continue this problem.

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But I'll stop there.

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So let me just go back one more
time and remind you what

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we were doing.

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What we were doing, if we come
back over here, is we were

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given a situation where we knew
a certain function of the

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radius, the distance
from the center.

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And we wanted to determine the
total amount of chemical that

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was released into the pool.

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And so we estimated.

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We figured out a way to estimate
it in terms of

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splitting up the radii.

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We had the radius from 0 to big
R. And we just divided up

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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 to find--

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if we move over here,
over here--

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we would determine the right
function to find the amount of

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chemical in a certain shell,
assuming that the

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concentration was constant
throughout that shell.

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And then, what we do, is we
know that if we let those

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shells get arbitrarily narrow,
that means that we're letting

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the number of radii
over which we're

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doing this, go to infinity.

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And we know that this summation
that we have here,

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as n goes to infinity,
becomes an integral.

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So that's really, we exploited
what we know about this sum

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and letting this partition,
letting the delta r get

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arbitrarily small.

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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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So I think I will stop there.

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