In this section we will start looking at the volume of a
solid of revolution. We should first
define just what a solid of revolution is. To get a solid of revolution we start out with
a function, ,
on an interval [a,b].
We then rotate this curve about a given axis to get the
surface of the solid of revolution. For
purposes of this discussion let’s rotate the curve about the xaxis, although it could be any
vertical or horizontal axis. Doing this
for the curve above gives the following three dimensional region.
What we want to do over the course of the next two sections
is to determine the volume of this object.
In the final the Area and
Volume Formulas section of the Extras chapter we derived the following
formulas for the volume of this solid.
where, and is the crosssectional area of the solid. There are many ways to get the
crosssectional area and we’ll see two (or three depending on how you look at
it) over the next two sections. Whether
we will use or will depend upon the method and the axis of
rotation used for each problem.
One of the easier methods for getting the crosssectional
area is to cut the object perpendicular to the axis of rotation. Doing this the cross section will be either a
solid disk if the object is solid (as our above example is) or a ring if we’ve
hollowed out a portion of the solid (we will see this eventually).
In the case that we get a solid disk the area is,
where the radius will depend upon the function and the axis
of rotation.
In the case that we get a ring the area is,
where again both of the radii will depend on the functions
given and the axis of rotation. Note as
well that in the case of a solid disk we can think of the inner radius as zero
and we’ll arrive at the correct formula for a solid disk and so this is a much
more general formula to use.
Also, in both cases, whether the area is a function of x or a function of y will depend upon the axis of rotation as we will see.
This method is often called the method of disks or the method
of rings.
Let’s do an example.
Example 1 Determine
the volume of the solid obtained by rotating the region bounded by ,
,
,
and the xaxis about the xaxis.
Solution
The first thing to do is get a sketch of the bounding
region and the solid obtained by rotating the region about the xaxis. Here are both of these sketches.
Okay, to get a cross section we cut the solid at any x.
Below are a couple of sketches showing a typical cross section. The sketch on the right shows a cut away of
the object with a typical cross section without the caps. The sketch on the left shows just the curve
we’re rotating as well as its mirror image along the bottom of the solid.
In this case the radius is simply the distance from the xaxis to the curve and this is
nothing more than the function value at that particular x as shown above. The
crosssectional area is then,
Next we need to determine the limits of integration. Working from left to right the first cross
section will occur at and the last cross section will occur at . These are the limits of integration.
The volume of this solid is then,

In the above example the object was a solid object, but the
more interesting objects are those that are not solid so let’s take a look at
one of those.
Example 2 Determine
the volume of the solid obtained by rotating the portion of the region
bounded by and that lies in the first quadrant about the yaxis.
Solution
First, let’s get a graph of the bounding region and a
graph of the object. Remember that we
only want the portion of the bounding region that lies in the first
quadrant. There is a portion of the
bounding region that is in the third quadrant as well, but we don't want that
for this problem.
There are a couple of things to note with this
problem. First, we are only looking
for the volume of the “walls” of this solid, not the complete interior as we
did in the last example.
Next, we will get our cross section by cutting the object
perpendicular to the axis of rotation.
The cross section will be a ring (remember we are only looking at the
walls) for this example and it will be horizontal at some y.
This means that the inner and outer radius for the ring will be x values and so we will need to
rewrite our functions into the form . Here are the functions written in the
correct form for this example.
Here are a couple of sketches of the boundaries of the
walls of this object as well as a typical ring. The sketch on the left includes the back
portion of the object to give a little context to the figure on the right.
The inner radius in this case is the distance from the yaxis to the inner curve while the
outer radius is the distance from the yaxis
to the outer curve. Both of these are
then x distances and so are given
by the equations of the curves as shown above.
The crosssectional area is then,
Working from the bottom of the solid to the top we can see
that the first crosssection will occur at and the last crosssection will occur at . These will be the limits of
integration. The volume is then,

With these two examples out of the way we can now make a
generalization about this method. If we
rotate about a horizontal axis (the xaxis
for example) then the cross sectional area will be a function of x.
Likewise, if we rotate about a vertical axis (the yaxis for example) then the cross sectional area will be a
function of y.
The remaining two examples in this section will make sure
that we don’t get too used to the idea of always rotating about the x or yaxis.
Example 3 Determine
the volume of the solid obtained by rotating the region bounded by and about the line .
Solution
First let’s get the bounding region and the solid graphed.
Again, we are going to be looking for the volume of the
walls of this object. Also since we
are rotating about a horizontal axis we know that the crosssectional area
will be a function of x.
Here are a couple of sketches of the boundaries of the
walls of this object as well as a typical ring. The sketch on the left includes the back
portion of the object to give a little context to the figure on the right.
Now, we’re going to have to be careful here in determining
the inner and outer radius as they aren’t going to be quite as simple they
were in the previous two examples.
Let’s start with the inner radius as this one is a little
clearer. First, the inner radius is
NOT x. The distance from the xaxis to the inner edge of the ring is x, but we want the radius and that is the distance from the axis
of rotation to the inner edge of the ring.
So, we know that the distance from the axis of rotation to the xaxis is 4 and the distance from the xaxis to the inner ring is x.
The inner radius must then be the difference between these two. Or,
The outer radius works the same way. The outer radius is,
Note that given the location of the typical ring in the sketch above the formula for the outer radius may not look quite right but it is in fact correct. As sketched the
outer edge of the ring is below the xaxis and
at this point the value of the function will be negative and so when we do
the subtraction in the formula for the outer radius we’ll actually be
subtracting off a negative number which has the net effect of adding this
distance onto 4 and that gives the correct outer radius. Likewise, if the outer edge is above the xaxis, the function value will be
positive and so we’ll be doing an honest subtraction here and again we’ll get
the correct radius in this case.
The crosssectional area for this case is,
The first ring will occur at and the last ring will occur at and so these are our limits of
integration. The volume is then,

Example 4 Determine
the volume of the solid obtained by rotating the region bounded by and about the line .
Solution
As with the previous examples, let’s first graph the
bounded region and the solid.
Now, let’s notice that since we are rotating about a
vertical axis and so the crosssectional area will be a function of y.
This also means that we are going to have to rewrite the functions to
also get them in terms of y.
Here are a couple of sketches of the boundaries of the
walls of this object as well as a typical ring. The sketch on the left includes the back
portion of the object to give a little context to the figure on the right.
The inner and outer radius for this case is both similar
and different from the previous example.
This example is similar in the sense that the radii are not just the
functions. In this example the functions
are the distances from the yaxis
to the edges of the rings. The center
of the ring however is a distance of 1 from the yaxis. This means that
the distance from the center to the edges is a distance from the axis of
rotation to the yaxis (a distance
of 1) and then from the yaxis to
the edge of the rings.
So, the radii are then the functions plus 1 and that is
what makes this example different from the previous example. Here we had to add the distance to the
function value whereas in the previous example we needed to subtract the function
from this distance. Note that without
sketches the radii on these problems can be difficult to get.
So, in summary, we’ve got the following for the inner and
outer radius for this example.
The crosssectional area is then,
The first ring will occur at and the final ring will occur at and so these will be our limits of
integration.
The volume is,
