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In this section we are going to look once again at solids of
revolution. We first looked at them back
in Calculus I when we found the volume
of the solid of revolution. In this
section we want to find the surface area of this region.
So, for the purposes of the derivation of the formula, let’s
look at rotating the continuous function 
in the interval about the x-axis. Below is a sketch of a function and the solid
of revolution we get by rotating the function about the x-axis.
We can derive a formula for the surface area much as we
derived the formula for arc length. We’ll start by dividing the integral into n equal subintervals of width . On each subinterval we will approximate the
function with a straight line that agrees with the function at the endpoints of
the each interval. Here is a sketch of
that for our representative function using .
Now, rotate the approximations about the x-axis and we get the following solid.
The approximation on each interval gives a distinct portion
of the solid and to make this clear each portion is colored differently. Each of these portions are called frustums and we know how
to find the surface area of frustums.
The surface area of a frustum is given by,
where,
and l is the
length of the slant of the frustum.
For the frustum on the interval we have,
and we know from the previous section that,
Before writing down the formula for the surface area we are
going to assume that is “small” and since is continuous we can then assume that,
So, the surface area of the frustum on the interval is approximately,
The surface area of the whole solid is then approximately,
and we can get the exact surface area by taking the limit as
n goes to infinity.
If we wanted to we could also derive a similar formula for
rotating on about the y-axis. This would give the following formula.
These are not the “standard” formulas however. Notice that the roots in both of these
formulas are nothing more than the two ds’s
we used in the previous section. Also,
we will replace with y
and with x. Doing this gives the following two formulas
for the surface area.
Surface Area Formulas
There are a couple of things to note about these
formulas. First, notice that the
variable in the integral itself is always the opposite variable from the one
we’re rotating about. Second, we are
allowed to use either ds in either
formula. This means that there are, in
some way, four formulas here. We will
choose the ds based upon which is the
most convenient for a given function and problem.
Now let’s work a couple of examples.
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Example 1 Determine
the surface area of the solid obtained by rotating ,
about the x-axis.
Solution
The formula that we’ll be using here is,
since we are rotating about the x-axis and we’ll use the first ds in this case because our function is in the correct form for
that ds and we won’t gain anything
by solving it for x.
Let’s first get the derivative and the root taken care of.
Here’s the integral for the surface area,
There is a problem however. The dx
means that we shouldn’t have any y’s
in the integral. So, before evaluating
the integral we’ll need to substitute in for y as well.
The surface area is then,
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Previously we made the comment that we could use either ds in the surface area formulas. Let’s work an example in which using either ds won’t create integrals that are too
difficult to evaluate and so we can check both ds’s.
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Example 2 Determine
the surface area of the solid obtained by rotating ,
about the y-axis. Use both ds’s to compute the surface area.
Solution
Note that we’ve been given the function set up for the
first ds and limits that work for
the second ds.
Solution 1
This solution will use the first ds listed above. We’ll
start with the derivative and root.
We’ll also need to get new limits. That isn’t too bad however. All we need to do is plug in the given y’s into our equation and solve to get
that the range of x’s is . The integral for the surface area is then,
Note that this time we didn’t need to substitute in for
the x as we did in the previous
example. In this case we picked up a dx from the ds and so we don’t need to do a substitution for the x.
In fact if we had substituted for x
we would have put y’s into integral
which would have caused problems.
Using the substitution
the integral becomes,
Solution 2
This time we’ll use the second ds. So, we’ll first need
to solve the equation for x. We’ll also go ahead and get the derivative
and root while we’re at it.
The surface area is then,
We used the original y
limits this time because we picked up a dy
from the ds. Also note that the presence of the dy means that this time, unlike the
first solution, we’ll need to substitute in for the x. Doing that gives,
Note that after the substitution the integral was
identical to the first solution and so the work was skipped.
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As this example has shown we can used either ds to get the surface area. It is important to point out as well that
with one ds we had to do a
substitution for the x and with the
other we didn’t. This will always work
out that way.
Note as well that in the case of the last example it was
just as easy to use either ds. That often won’t be the case. In many examples only one of the ds will be convenient to work with so
we’ll always need to determine which ds
is liable to be the easiest to work with before starting the problem.