Professor Dave again, let’s talk about cylindrical shells.
We just learned how to find the volume of a solid of revolution by integration. Just
the way that we calculated the area under a curve by running a line segment of varying
length across the region and adding them all up, to get volume, we just ran some area across
a shape and added all of those up. Sometimes this was the area of a disk, sometimes the
area of a washer. But whatever the situation, we were able to calculate the area of a cross
section rather easily given the functions involved. Sometimes it won’t be so easy
to use this method, so let’s look at an example where another method will be a better
approach. Take for example y equals two x squared minus
x cubed. The region between this function and the x axis looks like this. Let’s say
we rotate this area around the y axis to get this bundt cake looking thing, a solid of
revolution. A cross section of this solid will be a washer, given the empty middle region,
which we can then push upwards from the base to the top of the solid. But the problem is
this. When we did this before, the functions were simple, like y equals x, or y equals
x squared. And given that we are doing things in terms of y, with the previous method we
would have to rewrite the functions in terms of y so that we can integrate. With the other
examples, that ended up being no problem, and we were able to get expressions for the
inner and outer radii of the washer that we could integrate. Here it’s not so easy.
Rather than one function per radius, here both the radii depend on this same function,
and furthermore, solving this function for x in terms of y is not a very appealing task.
But there is another technique we can try, and it is called the method of cylindrical
shells. Before we apply this to the problem at hand, let’s just look at this hollow
cylinder. This has an inner radius, r one, which ends at the inner face of the solid,
and an outer radius, r two, which ends at the outer face. This cylinder also has a height,
h. To get the volume of this figure, we can calculate the volume of the cylinder using
r two, pretending that it is solid all the way through, and then subtract from that the
volume of the inner cylinder, using r one, since that’s all empty space. The formula
for the volume of a cylinder is pi r squared h, so let’s do [pi (r two) squared h] minus
[pi (r one) squared h]. Factoring out pi and h, we can turn this into pi times the quantity
(r two) squared minus (r one) squared, times h. If we remember how to expand a difference
of squared terms from algebra, this term is equal to the quantity (r two) plus (r one),
times the quantity of (r two) minus (r one). Here’s the clever bit. Let’s take this
sum and turn it into an expression for the average value for the radius. To do that,
we just divide by two, but in order to keep the same value for the equation, let’s put
a two at the front. Since this is now the average radius, let’s just call it r. Then
over here, the difference between the radii gives us the thickness of the cylinder, so
we can call that delta r if we like. And like that, the volume of this cylindrical shell
will be equal to two pi r h delta r. One easy way to remember this is to see that two pi
r is the same as the circumference of the cylinder, so this is really just circumference
times height times thickness. The reason this is useful is that we no longer have to solve
for x in terms of y. If we picture one possible cylindrical shell, it will have a height that
is equal to the function, f of x, and a circumference equal to two pi x. As this shell spans the
volume, we then simply have to integrate (two pi x) times f of x dx. In this case, f of
x is (two x squared minus x cubed), as we said, so let’s plug that in for f of x,
and then to simplify, we can take two pi out of the integral, but the x must stay, and
it can be distributed across the other term. That leaves us with two pi times the integral
of (two x cubed minus x to the fourth), dx, evaluated from zero to two, since two is the
maximum value of the function. We integrate to get two x to the fourth over four, or one
half x to the fourth, minus one fifth x to the fifth. We evaluate at two and we get eight
minus thirty-two fifths, or forty minus thirty two over five, so we subtract and multiply
by two pi to get sixteen fifths pi. So while this may have seemed complex, part of that
is because we derived the formula for the volume of the cylinder. We won’t have to
do that every time, we just need to know that the volume of the solid with a cylindrical
shell oriented vertically will be equal to the integral of two pi x times f of x dx,
over the interval that spans the solid. So we can see that for this particular example,
this method ended up being easier than the washer method we used in the previous tutorial
as our formula for the volume of the cylindrical shell was ready to go, which meant we didn’t
have to do much to set things up. For any solid of revolution you are looking at, whether
the cross section is a disk or washer, you will just have to think critically to figure
out which approach is best, depending on whether it is easy to find the radius or radii of
the cross section with the function or functions as given, or whether it is easier to express
things another way. For a little practice with this new method, let’s check comprehension.