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Math Help - Area of a surface of revolution

  1. #1
    Member FalconPUNCH!'s Avatar
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    Area of a surface of revolution

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

     y = sin\pi x

    0 < x < 1
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    Quote Originally Posted by FalconPUNCH! View Post
    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).
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  3. #3
    Member FalconPUNCH!'s Avatar
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    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.
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    Eater of Worlds
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    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.
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  5. #5
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by galactus View Post
    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.
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  6. #6
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    Quote Originally Posted by FalconPUNCH! View Post
    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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