More Volume ProblemsIn this section we�re going to take a look at some more
volume problems. However, the problems we�ll be looking at here will not be
solids of revolution as we looked at in the previous two sections. There are
many solids out there that cannot be generated as solids of revolution, or at
least not easily and so we need to take a look at how to do some of these
problems.
Now, having said that these will not be solids of
revolutions they will still be worked in pretty much the same manner. For each
solid we�ll need to determine the cross-sectional area, either
 or
 ,
and they use the formulas we used in the previous two sections,
The �hard� part of these problems will be determining what
the cross-sectional area for each solid. Each problem will be different and so
each cross-sectional area will be determined by different means.
Also, before we proceed with any examples we need to
acknowledge that the integrals in this section might look a little tricky at
first. There are going to very few numbers in these problems. All of the
examples in this section are going to be more general derivation of volume
formulas for certain solids. As such we�ll be working with things like circles
of radius r and we�ll not be giving a specific value of r and
we�ll have heights of h instead of specific heights, etc.
All the letters in the integrals are going to make the
integrals look a little tricky, but all you have to remember is that the r�s
and the h�s are just letters being used to represent a fixed quantity for
the problem, i.e. it is a constant. So when we integrate we only need to
worry about the letter in the differential as that is the variable we�re
actually integrating with respect to. All other letters in the integral should
be thought of as constants. If you have trouble doing that, just think about
what you�d do if the r was a 2 or the h was a 3 for example.
Let�s start with a simple example that we don�t actually
need to do an integral that will illustrate how these problems work in general
and will get us used to seeing multiple letters in integrals.
Example
Find the volume of a cylinder of radius r and height h.
Solution
Now, as we mentioned before starting this example we really
don�t need to use an integral to find this volume, but it is a good example to
illustrate the method we�ll need to use for these types of problems.
We�ll start off with the sketch of the cylinder below.
We�ll center the cylinder on the x-axis and the
cylinder will start at
 and
end at
 as
shown. Note that we�re only choosing this particular set up to get an integral
in terms of x and to make the limits nice to deal with. There are many
other orientations that we could use.
What we need here is to get a formula for the
cross-sectional area at any x. In this case the cross-sectional area is
constant and will be a disk of radius r. Therefore, for any x
we�ll have the following cross-sectional area,
Next the limits for the integral will be
 since
that is the range of x in which the cylinder lives. Here is the integral
for the volume,
So, we get the expected formula.
Also, recall we are using r to represent the radius
of the cylinder. While r can clearly take different values it will never
change once we start the problem. Cylinders do not change their radius in the
middle of a problem and so as we move along the center of the cylinder (i.e.
the x-axis) r is a fixed number and won�t change. In other words
it is a constant that will not change as we change the x. Therefore,
because we integrated with respect to x the r will be a constant
as far as the integral is concerned. The r can then be pulled out of the
integral as shown (although that�s not required, we just did it to make the
point). At this point we�re just integrating dx and we know how to do
that.
When we evaluate the integral remember that the limits are
x values and so we plug into the x and NOT the r. Again,
remember that the r is just a letter that is being used represent the
radius of the cylinder and once we start the integral is assumed to be a fixed
constant.
As noted before we started this example if you�re having
trouble with the r just think of what you�d do if there was a 2 there
instead of an r. In this problem, because we�re integrating with respect
to x, both the 2 and the r will behave in the same manner. Note
however that you should NEVER actually replace the r with a 2 as that
WILL lead to a wrong answer. You should just think of what you would do IF the
r was a 2.
|
