Volumes of Solids of Revolution / Method of Cylinders
In the previous section we started looking at finding
volumes of solids of revolution. In that section we took cross sections that
were rings or disks, found the cross-sectional area and then used the following
formulas to find the volume of the solid.
In the previous section we only used cross sections that
where in the shape of a disk or a ring. This however does not always need to be
the case. We can use any shape for the cross sections as long as it can be
expanded or contracted to completely cover the solid we�re looking at. This is
a good thing because as our first example will show us we can�t always use
rings/disks.
Example
Determine the volume of the solid obtained by rotating the region
bounded by
 and
the x-axis about the y-axis.
Solution
As we did in the previous section, let�s first graph the
bounded region and solid. Note that the bounded region here is the shaded
portion shown. The curve is extended out a little past this for the purposes of
illustrating what the curve looks like.
 
So, we�ve basically got something that�s roughly doughnut
shaped. If we were to use rings on this solid here is what a typical ring would
look like.
 
This leads to several problems. First, both the inner and
outer radius are defined by the same function. This, in itself, can be dealt
with on occasion as we saw in a example in the
Area Between Curves section. However, this usually means more work than
other methods so it�s often not the best approach.
This leads to the second problem we got here. In order to
use rings we would need to put this function into the form
 .
That is NOT easy in general for a cubic polynomial and in other cases may not
even be possible to do. Even when it is possible to do this the resulting
equation is often significantly messier than the original which can also cause
problems.
The last problem with rings in this case is not so much a
problem as its just added work. If we were to use rings the limit would be y
limits and this means that we will need to know how high the graph goes. To
this point the limits of integration have always been intersection points that
were fairly easy to find. However, in this case the highest point is not an
intersection point, but instead a relative maximum. We spent several sections
in the Applications of Derivatives chapter talking about how to find maximum
values of functions. However, finding them can, on occasion, take some work.
So, we�ve seen three problems with rings in this case that
will either increase our work load or outright prevent us from using rings.
What we need to do is to find a different way to cut the
solid that will give us a cross-sectional area that we can work with. One way
to do this is to think of our solid as a lump of cookie dough and instead of
cutting it perpendicular to the axis of rotation we could instead center a
cylindrical cookie cutter on the axis of rotation and push this down into the
solid. Doing this would give the following picture,
Doing this gives us a cylinder or shell in the object and
we can easily find its surface area. The surface area of this cylinder is,
Notice as well that as we increase the radius of the
cylinder we will completely cover the solid and so we can use this in our
formula to find the volume of this solid. All we need are limits of
integration. The first cylinder will cut into the solid at
 and
as we increase x to
 we
will completely cover both sides of the solid since expanding the cylinder in
one direction will automatically expand it in the other direction as well.
The volume of this solid is then,
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