Some volumes of revolution are more easily computed with cylindrical shells.
More than one method
In this section, we will show you a new method for computing volumes of solids of
revolution. Consider the following region bounded by and :
If this region is rotated around the -axis, it is possible, but inconvenient, to
compute the volume of the resulting solid by the methods we have used so
far. The issue is that there are two ‘‘kinds’’ of cylindrical cross-sections:
As we see above, some of the cylindrical cross sections are defined by the line that
goes from to , and others are defined by the line that touches at both ends. To
compute the volume using accumulated cross-sections, we need to break the problem
into two integrals:
- an integral that computes the volume of the region bounded by and the
line , rotated about the -axis, and
- an integral that computes the volume of the region bounded by , and the
line , rotated about the -axis.
Since we are rotating around the -axis, we should look at and .
Write with me: On
the other hand is
not one-to-one, so we cannot invert it on the entire domain.
Nevertheless, if we restrict the domain of we may write two separate functions and
With this in mind, we can compute our volume with:
Substituting in, we find
and so
While we have successfully solved this problem, it wasn’t easy. Let’s see another,
perhaps easier method to solve this problem. If instead we consider a vertical
rectangle of height (just like we did when we computed areas of regions between
curves!) and width , and we additionally rotate this rectangle around the -axis, we
get a thin ‘‘shell’’ or hollow-tube:
Here the infinitesimal change in volume is:
Integrating will give us our desired volume.
Shells around the axes
Let’s start by actually doing our motivating example above.
Consider the region below bounded by and :
Find the volume of the solid of revolution formed by rotating this region around the
-axis.
We’ll solve this problem using accumulated shells. If we draw our shell, we see
Since the volume of each shell is we may write
Comparing our work above to our earlier work, we see that using shells not only
solves the problem with just one integral, we see that the integral itself is somewhat
easier than those in the previous calculation! Things are not always so neat, but it is
often the case that one of the two methods will be simpler than the other, so it is
worth considering both methods.
Suppose the area bounded by , the line , and the -axis is rotated around the -axis,
find the volume of this solid.
While we could use accumulated cross-sections to
compute this volume, it would require
two integrals, (the industrious young
mathematician is encouraged to attempt to solve this problem using accumulated
cross-sections). However, here we will use the method of shells. Let’s see a picture:
In this case, the volume of each shell is
Since we are integrating with respect to , we have and Hence the volume of our
shell is Thus the the total volume of our solid of revolution is given by
Shells around other lines
What if we had wanted to rotate the region from the last example about the line
instead of the -axis? In this case, we draw a picture and work much the same way as
before. Let’s see an example.
Suppose the area bounded by , the line , and the -axis is rotated around the line ,
find the volume of this solid.
Let’s see a picture:
Since we are integrating with respect to , we have and
The radius is now , so the
circumference is
Our cylindrical shells have the same width and height as in the previous problem,
but the circumference would changes, as the radius is . Hence the volume of our
shell is Thus the the total volume of our solid of revolution is given by
