Calculus II - Surface Area

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Section 8.2 : Surface Area

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5. Find the surface area of the object obtained by rotating $y = \sin \left( {2x} \right)$ , $\displaystyle 0 \le x \le \frac{\pi }{8}$ about the $x$ -axis.

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Note that we actually set this problem up in Part (a) of Problem 2. So, we’ll just summarize the steps of the set up part of the problem here. If you need to see all the details please check out the work in Problem 2.

Here is $ds$ for this problem.

$\frac{{dy}}{{dx}} = 2\cos \left( {2x} \right)\hspace{0.5in} \Rightarrow \hspace{0.5in} ds = \sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx$

The integral for the surface area is,

$SA = \int\limits_{0}^{{\frac{\pi }{8}}}{{2\pi \sin \left( {2x} \right)\,\sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx}}$ Show Step 2

In order to evaluate this integral we’ll need the following trig substitution.

$\begin{align*}\cos \left( {2x} \right) = \frac{1}{2}\tan \left( \theta \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\sin \left( {2x} \right)\,dx = \frac{1}{2}{\sec ^2}\left( \theta \right)\,d\theta \\ & \sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} = \sqrt {1 + {{\tan }^2}\left( \theta \right)} = \sqrt {{{\sec }^2}\left( \theta \right)} = \left| {\sec \left( \theta \right)} \right|\end{align*}$

In order to deal with the absolute value bars we’ll need to convert the $x$ limits to $\theta$ limits. Here’s that work.

$\begin{align*}x & = 0:\,\,\,\cos \left( 0 \right) = 1 = \frac{1}{2}\tan \left( \theta \right)\,\,\,\,\,\,\,\, \to \,\,\,\,\,\theta = {\tan ^{ - 1}}\left( 2 \right) = 1.1071\\ x & = \frac{\pi }{8}:\,\,\,\cos \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} = \frac{1}{2}\tan \left( \theta \right)\,\,\,\,\,\,\,\, \to \,\,\,\,\,\theta = {\tan ^{ - 1}}\left( {\sqrt 2 } \right) = 0.9553\end{align*}$

The corresponding range of $\theta$ is $0.9553 \le \theta \le 1.1071$ . This is in the first quadrant and secant is positive there. Therefore, we can drop the absolute value bars on the secant.

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Putting all the work from the previous step together gives,

$SA = \int\limits_{0}^{{\frac{\pi }{8}}}{{2\pi \sin \left( {2x} \right)\,\sqrt {1 + 4{{\cos }^2}\left( {2x} \right)} \,dx}} = - \frac{\pi }{2}\int_{{1.1071}}^{{0.9553}}{{{{\sec }^3}\left( \theta \right)\,d\theta }}$ Show Step 4

Step

Using the formula for the integral of ${\sec ^3}\left( \theta \right)$ we derived in the Integrals Involving Trig Functions we get,

$SA = - \frac{\pi }{2}\int_{{1.1071}}^{{0.9553}}{{{{\sec }^3}\left( \theta \right)\,d\theta }} = \left. { - \frac{\pi }{4}\left[ {\sec \left( \theta \right)\tan \left( \theta \right) + \ln \left| {\sec \left( \theta \right) + \tan \left( \theta \right)} \right|} \right]} \right|_{1.1071}^{0.9553} = \require{bbox} \bbox[2pt,border:1px solid black]{{1.8215}}$

Note that depending upon the number of decimal places you used your answer may be slightly different from that give here. The “exact” answer, obtained by computer, is 1.8222.