Volumes of Solids of Revolution / Method of Rings
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 x-axis, 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 cross-sectional area of the solid. There are many ways to get the
cross-sectional 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 cross-sectional
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.
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