PROFESSOR: Welcome
back to recitation.
In this video we're going to
do another solid of revolution
problem.
So what I'd like to
do in this problem is
to find the volume of the
solid generated by rotating
the region bounded by the
following curves: y equals 0,
x equal 4, and y equals square
root of x, around the line x
equals 6.
And you can choose your
favorite method to do this.
What I would like
first, when you're
doing this kind
of problem, is get
a rough sketch of the region.
So you have some
picture of what's
actually going to happen.
You don't necessarily need
a three-dimensional picture.
But at least have the
two-dimensional region
and understand the where the
rotation line is, with respect
to that region.
So I will give you
a little bit of time
to work on that problem.
And when I come back I'll
show you how I do it.
OK, welcome back.
So again, what we're
doing in this video is
we're going to be looking,
finding the volume
of a solid of revolution.
And as I mentioned, the
first thing you want to do
is get a rough picture
of what the region looks
like that's being rotated.
So I'm going to draw a
rough sketch of that.
Right here.
So my axes, here's my x-axis
and my y-axis, and I will draw,
first the curve y equals 0.
That's y equals 0.
And I'm next going to draw the
curve y equals square root of x
to make this a little easier for
myself, to do that one first.
So y equals square root of
x looks something like this,
roughly.
And then, x equal 4, because
it's a very rough sketch,
I can just draw
this line right here
and say that line's x equal 4.
And I want the region bounded
by those three curves.
So, this region right here.
And my rotating line, in
this case, is x equals 6.
So to give myself
some perspective,
I remember this is x equals
4, here's x equals 0.
So x equals 6, which
may be right about here,
I'll draw a dotted line.
And I'm going to draw a
little arrow around there
so I know that's the
line I'm rotating about.
Now we notice there's a
little bit of a gap here.
But you can actually work on
this problem quite simply.
If we use the shell
method, we won't have
to worry about the gap at all.
We will have to be careful in
how we determine our radius.
That's actually where you'll
see this sort of gap coming up.
So when I rotate
this around, I'm
going to get some
object that's going
to have a nice curve here.
And there's going to be a little
cut-out, a cylinder missing
in the middle when I rotate it.
So I'm going to use
the shell method.
I like the shell method
for this problem.
You could use the washer
method for this problem,
but I like the shells and
that's what I'm going to use.
So to do the shell method,
remember we are going to be
interested in 2*pi*r*h.
That's really what we need.
And now what we
want to figure out
is, do I want to do this in
terms of x or in terms of y?
All right?
When I talk about
these shells is
it better to think about
them as functions of x or y?
And it will, you'll
see we sort of
have to think about
them in a certain way.
So if I draw, let's think about
if I draw a straight line down
and I rotate it
around the line x
equals 6, which I should
have labeled here,
then I get my shell.
So I need these lines
to go from 0 to 4.
I need to have those
lines go from 0 to 4.
And so I see that it's the
x-value I'm varying over.
So I know this is all going to
be an integral in terms of x.
So let's keep that in mind.
So r I should write as a
function of x, and h I should
write as a function of x.
So if I examined this
segment again here,
the height is fairly simple.
Because what is this curve?
This is the curve y
equals square root of x.
So the height is just
square root of x.
So h is equal to
square root of x.
And so now I just
need r in terms of x.
Well let's look at
this segment again.
It's just a generic position.
How would I find that radius?
What is the radius?
Well I'm going to draw the
three-dimensional picture over
here.
That that's this
curve, this segment,
rotated about the
line x equals 6.
Right?
And the radius is not x.
It's the distance from x
equals 6 to my segment.
That's an important distinction
we should understand.
If we come back and
look at this picture,
this is not the radius.
Right?
The radius is the distance from
this value to this value in x.
And how do I measure that?
Well that's fairly
straightforward.
That's the difference in x
values between these two.
The x value here is 6.
And the x value here,
this is a generic x.
So the difference, the
distance, is just 6 minus x.
Right?
This is just a simple case
of the distance formula.
Because I'm not interested
in the changes in y, at all.
I'm just interested in the
difference between these two
x values.
So it's simply 6 minus x.
That's our radius.
So now I have the
radius in terms of x.
And I have the
height in terms of x.
So I can completely
set up my integral.
So let me come a
little bit further over
and set up the integral.
So I know I'm
integrating-- we even
said what we were integrating
from and to already.
I'm starting x at 0.
And I'm stopping
it at x equal 4.
And then I want to
integrate 2*pi*r*h.
So I'll put the
2*pi out in front.
The radius is 6 minus x.
And the height is
square root of x.
And then I'm integrating
it all in dx.
And at this point,
solving the problem
is fairly straightforward.
I can write square root
of x as x to the 1/2
to make myself feel a
little better about it.
I can then distribute.
And then I just use the power
rule to do the integration
and evaluate to get an actual
number, to get that volume.
So I'm going to go
back and just make
sure we remember what
all the pieces are
and that we feel
comfortable with doing
this type of problem.
And I'll let you
finish and evaluate
that if you would like
to know an actual number.
So let's come back over.
We were finding the volume
of a solid generated
by rotating a certain region
around the line x equals 6.
The way this might be
different from some
of the problems
you've seen before,
is it's not the x or y axis.
And the region has a gap.
The region is not connected
to the line of rotation.
So if we come to
see our picture,
we can see that clearly.
The rotating region
is the blue region.
The line of rotation
is the dotted line.
So there's a little
bit of a gap there.
I let you pick your
favorite method.
I chose shells,
because I didn't have
to worry about subtracting off
pieces, this little piece here.
It was inherent in
the radius, the way
I determined the radius and
the x values that I chose.
So I used shell method.
And so how did I, I had to
figure out a couple things.
I had to figure out
first, what variable
I was doing this in, x or y.
And then I had to figure
out the radius and height
in terms of that variable.
You figure out what variable
you need by saying, OK
if I'm doing shells, I'm looking
at these kinds of segments.
And I'm varying
the x-values, then,
as I move and look at
those different segments.
So I know that this is
a dx type of problem.
I know I'm integrating
something in x.
So then I have to find radius
and height in terms of x.
The height's the freebie.
It's just from 0
up to the function.
The radius is maybe
a little harder.
But that's still
just the distance
between the line of rotation
and the point where you are.
And so that's 6 minus x.
And then I just set up
the integral from there.
So that's really
what the idea is
in doing these
types of problems.
And because we get to pick
our simplest way, for me
it was shells.
Maybe you picked washers.
But for me, this problem,
the simplest way is shells.
So I guess that's
where I'll stop.
