Find the area of the surface obtained by rotating the curve about the x-axis.

$\displaystyle y = sin\pi x $

0 < x < 1

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- Mar 19th 2008, 09:36 PMFalconPUNCH!Area of a surface of revolution
Find the area of the surface obtained by rotating the curve about the x-axis.

$\displaystyle y = sin\pi x $

0 < x < 1 - Mar 19th 2008, 10:27 PMmr fantastic
Where exactly are you stuck? If it's finding the integral, first make the substitution $\displaystyle u = \pi \cos (\pi x)$.

Now make*either*(not both!!) of the following substitutions

1. $\displaystyle u = \sinh t$ (this is the one I prefer).

2. $\displaystyle u = \tan t$ (more work will be required). - Mar 20th 2008, 09:46 AMFalconPUNCH!
Yeah I got up to the substituting part and got confused. I haven't used the first substitution before so I don't know what to do there. I tried using tan and got lost.

- Mar 20th 2008, 11:15 AMgalactus
After making the subs, did you get it down to:

$\displaystyle \frac{2}{\pi}\int_{-\pi}^{\pi}\sqrt{u^{2}+1}du$

Now, can you do that?.

This looks similar to a Fourier series. - Mar 20th 2008, 02:45 PMFalconPUNCH!
- Mar 20th 2008, 02:55 PMmr fantastic