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PROFESSOR: Welcome
back to recitation.

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Today in this video,
what I'd like us to do

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is work on some solid
of revolution problems.

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Well, one solid of
revolution problem.

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So what I want us to
do in this one is,

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find the volume of the
solid generated by rotating

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the region bounded by these
three curves, x equals 0,

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y equals 5, and y
equals x squared

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plus 1, about the y-axis.

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And I'm forcing you--
or I'm requesting--

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that you use the disk method.

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So I want us to work on the
disk method in this case.

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So why don't you take
a crack at the problem?

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And then I'll be back.

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OK, welcome back.

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So what we're going
to do is we're

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going to work on finding the
volume of a solid generated

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by rotating this region
that's bounded by the three

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curves mentioned
about the y-axis.

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And as I said before, we're
going to use the disk method.

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So when you're solving
these problems,

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the first thing you want
to do is give yourself

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at least a rough sketch of
what this region looks like,

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of with the region bounded by
these three curves actually

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looks like.

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So the first thing I'm going to
do before I do anything else,

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is at least get a rough sketch
of what that region looks like.

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And I'll do that over
here on the right.

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So, let's see.

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We know we've got the region--
let me actually even use

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some different colored chalk.

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We'll say the x equals 0
is actually the y-axis.

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That's x equals 0.

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And then y equals 5-- actually
we'll put that in last.

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But y equals x squared plus
1 is the parabola y equals

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x squared shifted up by 1.

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So if this is y
equals 1, then it's

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just roughly
something like this.

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That might not look
like the best parabola

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I've ever seen, but well.

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Something roughly like that.

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Maybe, as I said, rough sketch.

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Maybe more like that.

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And then, we want to also put in
the y equals 5, which I'll say

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is roughly about here.

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So I could-- notice I've
got these three curves.

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This curve, this
curve, this curve.

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So I'm interested in
this region right here.

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Right?

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Notice that if I also
include this region, when

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I rotated that around, I'd
actually get the same solid,

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whether I included
this region or not.

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But in terms of doing
the integration,

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I want to make sure I don't
count anything doubly.

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So I'm just looking at what
happens when I take this piece

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and I rotate it
around the y-axis.

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So what do we want to do?

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We said we want to
use the disk method.

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Which means ultimately,
I'm looking for finding out

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what the radius is.

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Right?

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I'm doing pi r squared
when I integrate.

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So I'm interested in
finding this length-- that's

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going to be our
radius-- and then

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we rotate-- you
probably can't see

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that, so let me just say that's
going to represent our radius.

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And then when we rotate
that around the y-axis

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we get our disk.

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Right?

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We've seen some
pictures of this before.

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So I'm interested
in this length here.

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And because I want to
use the disk method,

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I'm forcing us to
figure out what

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is this value in terms of y?

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So we know we have,
well we know that y

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equals x squared plus 1 there.

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So in terms of y, this is
y minus 1 equals x squared.

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So the square root of y
minus 1 is equal to x.

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So this is the expression I'm
interested in as our radius,

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as my radius.

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Right?

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And now, what I need
to do is figure out,

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I know if this is the
radius I need to figure out,

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since I'm doing
this in terms of y,

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what this lower bound is for
y and what the upper bound is

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for y.

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And then I'm going
to be integrating

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pi times the radius squared
over y, over the changes in y.

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Now let's see.

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This one's easy. y
equals 5 up here.

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And what is this value here?

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Well, the function
that I had here was y

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equals x squared plus 1.

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So when x is 0, y is 1.

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So this is the value
where y equals 1.

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So I know what I'm
interested in doing.

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Again, let me write it down.

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Is I'm integrating from 1
to 5 of pi r squared dy.

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And now, I need to write
r as a function of y.

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So r is square
root of y minus 1.

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When I square that,
I just get y minus 1.

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So what I'm really
interested in integrating is,

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the integral from
1 to 5 of-- let

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me put the pi outside, because
who cares-- pi times that

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integral, y minus 1 dy.

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And if I actually
evaluate this, figure out,

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if I actually take the integral,
use my rules for polynomials,

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I can get a numerical value.

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I'm going to stop evaluating
here, because from here on I

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know you guys can do it.

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But, the point of
this problem, what

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I want to mention, the
point of this problem

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is the main thing we had
to do is figure out what

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is the radius in terms of y.

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That was the main thing
we had to figure out

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in order to do this problem.

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I said you had to use disks.

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So we had to figure out what
the radius was in terms of y.

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And then we had
to figure out what

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the bounds were in terms of y.

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And once we do that,
it's a simple matter

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of evaluating this integral,
or finding the actual value

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for what this integral is.

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So I think that is
where I'll stop.