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I'm doing a problem involving finding the surface area of the curve for $y=\sin \left(\frac{\pi x}{6} \right)$, rotated about the $x$ axis, for $[0 < x < 6]$.

I got as far as $\frac{72}{\pi} \int^{\pi/6}_0 \sqrt{1+u^2} du$ with $u=\frac{\pi}{6}\cos(\frac{\pi x}{6})$. I'm not really sure how to precede from this point though.

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    $\begingroup$ Is that $S$ supposed to be an integral symbol? $\endgroup$ – Eleven-Eleven Feb 12 '16 at 19:47
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    $\begingroup$ See this: meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Bobson Dugnutt Feb 12 '16 at 19:50
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    $\begingroup$ Here is a link for math formatting used here. $\endgroup$ – Eleven-Eleven Feb 12 '16 at 19:50
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    $\begingroup$ To complete the integral, substitute $u=\sinh \theta$ $\endgroup$ – David Quinn Feb 12 '16 at 20:08
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    $\begingroup$ It's what's known as a hyperbolic function. If you're not familiar with them, you can use a trigonometric substitution of $u=\tan \theta$ instead. $\endgroup$ – user170231 Feb 12 '16 at 22:19
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Here's a plot of the surface of revolution:

enter image description here

The area of the surface is given by $$2\pi\int_0^6\sin\frac{\pi x}{6}\sqrt{1+\left(\frac{\mathrm{d}}{\mathrm{d}x}\left[\sin\frac{\pi x}{6}\right]\right)^2}\,\mathrm{d}x=2\pi\int_0^6\sin\frac{\pi x}{6}\sqrt{1+\frac{\pi^2}{36}\cos^2\frac{\pi x}{6}}\,\mathrm{d}x$$ Substituting $u=\frac{\pi}{6}\cos\frac{\pi x}{6}$, so that $\mathrm{d}u=-\frac{\pi^2}{36}\sin\frac{\pi x}{6}\,\mathrm{d}x$, you get the integral you're currently stuck on: $$-\frac{72}{\pi}\int_{\pi/6}^{-\pi/6}\sqrt{1+u^2}\,\mathrm{d}u=\frac{144}{\pi}\int_0^{\pi/6}\sqrt{1+u^2}\,\mathrm{d}u$$ (I believe you made a slight mistake with either the coefficient or the lower limit of the integral here.)

Now, substituting $u=\tan t$, you have that $\mathrm{d}u=\sec^2t\,\mathrm{d}t$, and so the integral is $$\frac{144}{\pi}\int_0^{\tan^{-1}(\pi/6)}\sqrt{1+\tan^2t}\sec^2t\,\mathrm{d}t=\frac{144}{\pi}\int_0^{\tan^{-1}(\pi/6)}\sec^3t\,\mathrm{d}t$$ The latter integral can be computed in a variety of ways, and the process is described in detail here.

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  • $\begingroup$ How do you get to the 144/pi and negative gone? $\endgroup$ – windy401 Feb 12 '16 at 23:23
  • $\begingroup$ @windy401 The integrand is symmetric about $u=0$ (i.e. an even function), so $$\int_{-a}^af(u)\,\mathrm{d}u=2\int_0^af(u)\,\mathrm{d}u$$ $\endgroup$ – user170231 Feb 12 '16 at 23:24
  • $\begingroup$ Ohh ok thanks.. $\endgroup$ – windy401 Feb 12 '16 at 23:25

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