This right here is a solid of revolution
whose volume we were able to figure out in previous videos,
actually in a different tutorial, using the disk method
and integrating in terms of y.
What we're going to do right now is
we're going to find the same volume for the same solid
of revolution, but we're going to do it using the shell method
and integrating with respect to x.
So what we do is we have the region between these two
curves, y is equal to square root of x
and y is equal to x squared.
And we're going to rotate it around the vertical line x
equals 2.
And we're going to do that at the interval
that we're going to rotate this space between these two curves
is the interval when square root of x is greater than x squared.
And so it's between 0 and 1.
And so let's try to do it with the shell method.
And so to do that, what we do is we want to construct a shell.
Let me do this in a different color.
Let's imagine a rectangle right over here.
It has width dx.
and its height is the difference of these two functions.
And so if I were to draw it right over here,
it would look something like this.
It'd be there, and then it is a shell,
it's kind of a hollowed-out cylinder.
So it would look something like this.
Just like that.
And it has some depth, that's what the dx gives us.
So we have the depth that looks something like that.
And then let me shade it in a little bit,
just so we can see a little bit of its depth.
So when you rotate this rectangle
around the line x equals 2, you get a shell like this.
So let's think about how we can figure out
the volume of this shell.
Well, we've already done this several times.
The first thing we might want to think about
is the circumference of the top of the shell.
We know circumference is 2 pi times radius.
We just need to know what the radius of the shell is.
What is that distance going to be?
Well, it's the horizontal distance
between x equals 2 and whatever the x value is right over here.
So it's going to be 2 minus our x value.
So this radius, this distance right over here,
is going to be 2 minus x.
And so the circumference is going to be that times 2 pi.
2 pi r gives us the circumference of that circle.
So 2 pi times 2 minus x.
And then if we want the surface area
of the outside of our shell, so the area
is going to be the circumference 2 pi times 2 minus x
times the height of each shell.
Now, what is the height of each shell?
It's going to be the vertical distance expressed
as functions of y.
So it's going to be the top boundary
is y is equal to square root of x, the bottom boundary is
y is equal to x squared.
So it's going to be square root of x minus x squared.
Let me do this in the yellow.
So it's going to be square root of x minus x squared.
And so if you want the volume of a given shell--
I'll write all this in white-- it's
going to be 2 pi times 2 minus x times
square root of x minus x squared.
So this whole expression, I just rewrote it,
is the area, the outside surface area, of one of these shells.
If we want the volume, we have to get a little bit of depth,
multiply by how deep the shell is, so times dx.
And if we want the volume of this whole thing,
we just have to solve all the shells
for all of the x's in this interval
and take the limit as the dx's get smaller and smaller
and we have more and more shells.
And so, what's our interval?
Well our x's are going to go between 0 and 1.
So that right over there is the volume of this figure.