Area of a surface of revolution

**FalconPUNCH!** · March 19th 2008, 09:36 PM

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

$y = sin\pi x$

0 < x < 1

**mr fantastic** · March 19th 2008, 10:27 PM

Originally Posted by FalconPUNCH!

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

$y = sin\pi x$

0 < x < 1

Where exactly are you stuck? If it's finding the integral, first make the substitution $u = \pi \cos (\pi x)$ .

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

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

2. $u = \tan t$ (more work will be required).

**FalconPUNCH!** · March 20th 2008, 09:46 AM

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.

**galactus** · March 20th 2008, 11:15 AM

After making the subs, did you get it down to:

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

Now, can you do that?.

This looks similar to a Fourier series.

**FalconPUNCH!** · March 20th 2008, 02:45 PM

Originally Posted by galactus

After making the subs, did you get it down to:

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

Now, can you do that?.

This looks similar to a Fourier series.

I finished it using table of integrals.

**mr fantastic** · March 20th 2008, 02:55 PM

Originally Posted by FalconPUNCH!

I finished it using table of integrals.

Make it known if you'd like to see a solution using the substitution $u = \tan \theta$ .

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