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Section 9.5 : Surface Area with Parametric Equations

3. Determine the surface area of the object obtained by rotating the parametric curve about the given axis. You may assume that the curve traces out exactly once for the given range of \(t\)’s.

Rotate \(\displaystyle x = 3\cos \left( {\pi t} \right)\hspace{0.25in}y = 5t + 2\hspace{0.25in}0 \le t \le \frac{1}{2}\) about the \(y\)-axis.

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Start Solution

The first thing we’ll need here are the following two derivatives.

\[\frac{{dx}}{{dt}} = - 3\pi \sin \left( {\pi t} \right)\hspace{0.25in}\hspace{0.25in}\frac{{dy}}{{dt}} = 5\] Show Step 2

We’ll need the \(ds\) for this problem.

\[ds = \sqrt {{{\left[ { - 3\pi \sin \left( {\pi t} \right)} \right]}^2} + {{\left[ 5 \right]}^2}} \,dt = \sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt\] Show Step 3

The integral for the surface area is then,

\[\begin{align*}SA & = \int_{{}}^{{}}{{2\pi x\,ds}} = \int_{0}^{{\frac{1}{2}}}{{2\pi \left( {3\cos \left( {\pi t} \right)} \right)\sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt}}\\ & \hspace{0.25in}\hspace{0.25in}\,\,\,\,\, = 6\pi \int_{0}^{{\frac{1}{2}}}{{\cos \left( {\pi t} \right)\sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt}}\end{align*}\]

Remember to be careful with the formula for the surface area! The formula used is dependent upon the axis we are rotating about.

Show Step 4

Okay, this is a particularly unpleasant looking integral but we need to be able to deal with these kinds of integrals on occasion. We’ll be able to do quite a bit of simplification if we first use the following substitution.

\[u = \sin \left( {\pi t} \right)\,\,\,\,\, \to \,\,\,\,\,\,{\sin ^2}\left( {\pi t} \right) = {u^2}\hspace{0.25in}\,\,\,\,\,du = \pi \cos \left( {\pi t} \right)\] \[t = 0:\,\,\,\,\,u = \sin \left( 0 \right) = 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}t = \frac{1}{2}:\,\,\,\,\,\,u = \sin \left( {\frac{1}{2}\pi } \right) = 1\]

With this substitution the integral becomes,

\[SA = 6\int_{0}^{1}{{\sqrt {9{\pi ^2}{u^2} + 25} \,du}}\] Show Step 5

This integral can be evaluated with the following (somewhat messy…) trig substitution.

\[t = \frac{5}{{3\pi }}\tan \theta \hspace{0.25in}dt = \frac{5}{{3\pi }}{\sec ^2}\theta \,d\theta \] \[\sqrt {9{\pi ^2}{u^2} + 25} = \sqrt {25{{\tan }^2}\theta + 25} = 5\sqrt {{{\tan }^2}\theta + 1} = 5\sqrt {{{\sec }^2}\theta } = 5\left| {\sec \theta } \right|\]

To get rid of the absolute value on the secant will need to convert the limits into \(\theta \) limits.

\[\begin{align*}u & = 0:\hspace{0.25in}\hspace{0.25in}0 = \frac{5}{{3\pi }}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = 0\hspace{0.25in}\,\, \to \hspace{0.25in}\theta = 0\\ u & = 1:\hspace{0.25in}\hspace{0.25in}\,1 = \frac{5}{{3\pi }}\tan \theta \hspace{0.25in} \to \hspace{0.25in}\tan \theta = \frac{{3\pi }}{5}\hspace{0.25in}\, \to \hspace{0.25in}\theta = {\tan ^{ - 1}}\left( {\frac{{3\pi }}{5}} \right) = 1.0830\end{align*}\]

Okay, the corresponding range of\(\theta \) for this problem is \(0 \le \theta \le 1.0830\) (first quadrant) and in this range we know that secant is positive. Therefore, the root becomes,

\[\sqrt {9{\pi ^2}{u^2} + 25} = 5\sec \theta \]

The surface area is then,

\[\begin{align*}SA & = \int_{0}^{{\frac{1}{2}}}{{2\pi \left( {3\cos \left( {\pi t} \right)} \right)\sqrt {9{\pi ^2}{{\sin }^2}\left( {\pi t} \right) + 25} \,dt}}\\ & = 6\int_{0}^{1}{{\sqrt {9{\pi ^2}{u^2} + 25} \,du}}\\ & = 6\int_{0}^{{1.0830}}{{\left( {5\sec \theta } \right)\left( {\frac{5}{{3\pi }}{{\sec }^2}\theta } \right)\,d \theta}}\\ & = 6\int_{0}^{{1.0830}}{{\frac{{25}}{{3\pi }}{{\sec }^3}\theta \,d\theta}}\\ & = \left. {\frac{{25}}{\pi }\left( {\sec \theta tan\theta + ln\left| {\sec \theta + \tan \theta } \right|} \right)} \right|_0^{1.0830} = \require{bbox} \bbox[2pt,border:1px solid black]{{43.0705}}\end{align*}\]

This problem was a little messy but don’t let that make you decide that you can’t do these types of problems! They can be done and often can be simplified with some relatively simple substitutions.