WEBVTT

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PROFESSOR: Welcome back
to recitation.

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Today in this video, what I'd
like us to do is work on some

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solid of revolution problems.
Well, one solid

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of revolution problem.

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

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of the solid generated by
rotating the region bounded by

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these three curves, x equals 0,
y equals 5, and y equals x

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squared 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 going to work on finding

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the volume of a solid generated
by rotating this

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region that's bounded by the
three curves mentioned about

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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, the first thing you

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want to do is give yourself at
least a rough sketch of what

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this region looks like, of with
the region bounded by

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these three curves actually
looks like.

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

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else, is at least get a rough
sketch of what that region

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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--

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let me actually even use 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--

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actually we'll put that in last.
But y equals x squared

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plus 1 is the parabola
y equals x squared

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shifted up by 1.

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

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something like this.

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That might not look like
the best parabola

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I've ever seen, but.

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Well, 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

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say is roughly about here.

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So I could--

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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 I rotated

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

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included this region or not.

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But in terms of doing the
integration, I want to make

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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 what

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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--

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that's going to be
our radius--

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and then we rotate--

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you probably can't see that;
so let me just say that's

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going to represent our radius.

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And then when we rotate
that around the

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y-axis 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, I'm forcing us to

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figure out what is this
value in terms of y?

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So we know we have, well we know
that y equals x squared

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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, I know if this is

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the radius I need to figure out,
since I'm doing this in

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terms of y, what this lower
bound is for y and what the

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upper bound is for y.

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And then I'm going to be
integrating pi times the

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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 equals x

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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, the

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integral from 1 to 5 of--

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let me put the pi outside,
because who cares--

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pi times that integral,
y minus 1 dy.

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And if I actually evaluate
this, figure out, if I

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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 I want to

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mention, the point of this
problem is the main thing we

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had to do is figure out what is
the radius in terms of y.

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That was the main thing we had
to figure out in order to do

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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 the bounds

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were in terms of y.

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

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this integral, or finding the
actual value for what this

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integral is.

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So I think that is
where I'll stop.