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In this final section of looking with calculus application
with parametric equations we will take a look at determining the surface area
of a region obtained by rotating a parametric curve about the x or y-axis.
We will rotate the parametric curve given by,
about the x or y-axis.
We are going to assume that the curve is traced out exactly once as t increases from α to β.
At this point there actually isn’t all that much to do. We know that the surface area can be found by
using one of the following two formulas depending on the axis of rotation
(recall the Surface Area section of the
Applications of Integrals chapter).
All that we need is a formula for ds to use and from the previous section we have,
which is exactly what we need.
We will need to be careful with the x or y that is in the
original surface area formula. Back when
we first looked at surface area we saw that sometimes we had to substitute for
the variable in the integral and at other times we didn’t. This was dependent upon the ds that we used. In this case however, we will always have to
substitute for the variable. The ds that we use for parametric equations
introduces a dt into the integral and
that means that everything needs to be in terms of t. Therefore, we will need
to substitute the appropriate parametric equation for x or y depending on the
axis of rotation.
Let’s take a quick look at an example.
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Example 1 Determine
the surface area of the solid obtained by rotating the following parametric
curve about the x-axis.
Solution
We’ll first need the derivatives of the parametric
equations.
Before plugging into the surface area formula let’s get
the ds out of the way.
Notice that we could drop the absolute value bars since
both sine and cosine are positive in this range of θ given.
Now let’s get the surface area and don’t forget to also
plug in for the y.
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