WEBVTT

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PROFESSOR: Welcome back
to recitation.

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In this video we're going
to do another solid

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of revolution problem.

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So what I'd like to do in this
problem is to find the volume

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of the solid generated by
rotating the region bounded by

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the following curves: y equals
0, x equal 4, and y equals

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square root of x, around
the line x equals 6.

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And you can choose your favorite
method to do this.

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What I would like first, when
you're doing this kind of

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problem, is get a rough
sketch of the region.

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So you have some picture
of what's

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actually going to happen.

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You don't necessarily need a
three-dimensional picture.

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But at least have the
two-dimensional region and

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understand the where the
rotation line is, with respect

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to that region.

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So I will give you a little
bit of time to

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work on that problem.

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

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OK, welcome back.

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So again, what we're doing in
this video is we're going to

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be looking, finding the volume
of a solid of revolution.

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And as I mentioned, the first
thing you want to do is get a

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rough picture of what the
region looks like

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that's being rotated.

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So I'm going to draw a
rough sketch of that.

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Right here.

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So my axes, here's my x-axis
and my y-axis, and I will

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draw, first the curve
y equals 0.

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That's y equals 0.

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And I'm next going to draw the
curve y equals square root of

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x to make this a little easier
for myself to do that one

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first. So y equals square root
of x looks something like

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this, roughly.

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And then, x equal 4, because
it's a very rough sketch, I

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can just draw this line right
here and say that

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line's x equal 4.

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And I want the region bounded
by those three curves.

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So, this region right here.

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And my rotating line, in this
case, is x equals 6.

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So to give myself some
perspective, I remember this

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is x equals 4, here's
x equals 0.

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So x equals 6, which may be
right about here, I'll draw a

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dotted line.

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And I'm going to draw a little
arrow around there so I know

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that's the line I'm
rotating about.

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Now we notice there's a little
bit of a gap here.

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But you can actually work on
this problem quite simply.

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If we use the shell method, we
won't have to worry about the

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gap at all.

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We will have to be careful in
how we determine our radius.

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That's actually where
you'll see this sort

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of gap coming up.

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So when I rotate this around,
I'm going to get some object

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that's going to have
a nice curve here.

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And there's going to be a little
cut-out, a cylinder

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missing in the middle
when I rotate it.

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So I'm going to use
the shell method.

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I like the shell method
for this problem.

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You could use the Washer method
for this problem, but I

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like the shells and that's
what I'm going to use.

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So to do the shell method,
remember we are going to be

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interested in 2 pi rh.

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That's really what we need.

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And now what we want to figure
out is, do I want to do this

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in terms of x or in terms of y?

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All right?

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When I talk about these shells
is it better to think about

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them as functions of x or y?

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And it will, you'll see we sort
of have to think about

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them in a certain way.

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So if I draw, let's think about
if I draw a straight

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line down and I rotate it around
the line x equals 6,

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which I should have labeled
here, then I get my shell.

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So I need these lines
to go from 0 to 4.

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I need to have those lines
go from 0 to 4.

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And so I see that it's the
x value I'm varying over.

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So I know this is all going to
be an integral in terms of x.

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So let's keep that in mind.

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So r, I should write as a
function of x and h I should

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write as a function of x.

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So if I examined this segment
again here, the height is

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fairly simple.

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Because what is this curve?

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This is the curve y equals
square root of x.

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So the height is just,
square root of x.

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So h is equal to square
root of x.

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And so now I just need
r in terms of x.

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Well let's look at this
segment again.

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It's just a generic position.

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How would I find that radius?

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What is the radius?

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Well I'm going to draw
the three-dimensional

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picture over here.

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That that's this curve, this
segment, rotated about the

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line x equals 6.

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Right?

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And the radius is not x.

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It's the distance from x
equals 6 to my segment.

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That's an important distinction
we should

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understand.

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If we come back and look
at this picture,

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this is not the radius.

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Right?

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The radius is the distance
from this value to

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this value in x.

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And how do I measure that?

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Well that's fairly
straightforward.

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That's the difference in x
values between these two.

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The x value here is 6.

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And the x value here,
this is a generic x.

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So the difference, the distance,
is just 6 minus x.

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Right?

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This is just a simple case
of the distance formula.

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Because I'm not interested in
the changes in y, at all.

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I'm just interested
in the difference

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between these two x values.

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So it's simply 6 minus x.

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That's our radius.

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So now I have the radius
in terms of x.

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And I have the height
in terms of x.

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So I can completely set
up my integral.

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So let me come a little
bit further over

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and set up the integral.

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So I know I'm integrating--

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we even said what we were

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integrating from and to already.

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I'm starting x at 0.

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And I'm stopping it
at x equal 4.

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And then I want to integrate
2 pi rh.

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So I'll put the 2
pi out in front.

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The radius is 6 minus x.

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And the height is square
root of x.

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And then I'm integrating
it all in dx.

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And at this point, solving
the problem is fairly

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straightforward.

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I can write square root of x as
x to the 1/2 to make myself

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feel a little better about it.

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I can then distribute.

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And then I just use the power
rule to do the integration and

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evaluate to get an actual
number, to get that volume.

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So I'm going to go back and
just make sure we remember

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what all the pieces are and that
we feel comfortable with

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doing this type of problem.

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And I'll let you finish and
evaluate that if you would

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like to know an actual number.

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So let's come back over.

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We were finding the volume
of a solid a generated by

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rotating a certain region around
the line x equals 6.

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The way this might be different
from some of the

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problems you've seen before, is
it's not the x or y axis.

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And the region has a gap.

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The region is not connected
to the line of rotation.

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So if we come to see
our picture, we

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can see that clearly.

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The rotating region is
the blue region.

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The line of rotation
is the dotted line.

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So there's a little bit
of a gap there.

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I let you pick your
favorite method.

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I chose shells, because I didn't
have to worry about

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subtracting off pieces, this
little piece here.

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It was inherent in the radius,
the way I determined the

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radius and the x values
that I chose.

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So I used shell method.

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And so how did I, I had to
figure out a couple things.

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I had to figure out first, what
variable I was doing this

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in, x or y.

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And then I had to figure out the
radius and height in terms

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of that variable.

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You figure out what variable you
need by saying, OK if I'm

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doing shells, I'm looking at
these kinds of segments.

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And I'm varying the x values,
then, as I move and look at

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those different segments.

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So I know that this is
a dx type of problem.

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I know I'm integrating
something in x.

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So then I have to find radius
and height in terms of x.

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The height's the freebie.

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It's just from 0 up
to the function.

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The radius is maybe
a little harder.

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But that's still just the
distance between the line of

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rotation and the point
where you are.

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And so that's 6 minus x.

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And then I just set up the
integral from there.

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So that's really what the idea
is in doing these types of

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problems. And because we get to
pick our simplest way, for

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me it was shells.

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Maybe you picked Washer's.

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But for me, this problem, the
simplest way is shells.

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So I guess that's
where I'll stop.