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# Shell method with two functions of x

Video transcript

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.