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Math Help - area of surface of revolution

  1. #1
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    area of surface of revolution

    Hi guys. I can't think of a way to start this integration - can someone set me on the proper path? Thanks!

    What I have so far:

    find the area of the surface obtained by rotating the curve around the x-axis

    y=sin(pix), 0≤x≤1

    dy/dx = picos(pix)

    1 + (dy/dx)^2 = 1 + (pi^2)cos^2(pix)

    integral 2(pi)sin(pix)√[1 + (pi^2)cos^2(pix)] = ?
    Last edited by swicket; June 10th 2008 at 06:54 AM. Reason: Forgot to add the square for the cosine, and the rest of the integral from the area formula.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by swicket View Post
    Hi guys. I can't think of a way to start this integration - can someone set me on the proper path? Thanks!

    What I have so far:

    find the area of the surface obtained by rotating the curve around the x-axis

    y=sin(pix), 0≤x≤1

    dy/dx = picos(pix)

    1 + (dy/dx)^2 = 1 + (pi^2)cos(pix)

    integral √[1 + (pi^2)cos(pix)] = ?
    You forgot to square the cosine function.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    You forgot to square the cosine function.

    -Dan
    Whoops, that was an error in transcription. I've edited the post to include the missing information.

    Thanks!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by swicket View Post
    Whoops, that was an error in transcription. I've edited the post to include the missing information.

    Thanks!
    If you are asking how to integrate

    \int\sin(x)\sqrt{1+\cos^2(x)}dx

    The answer is with relative difficulty. This one is really really ugly, from what i have seen. But you could always try trapezoid rule

    The actual answer due to your specific definite integral is \int_0^{1}2\pi\sin(\pi{x})\sqrt{1+\cos^2(\pi{x})}d  x=\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)+2\sqrt{2}

    EDIT: sorry I did not see the coefficient on the cosē, now I reevaluated it and it is too messy to type. but it is approximately 7.778
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  5. #5
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    <br /> <br />
\int_0^{1}2\pi\sin(\pi{x})\sqrt{1+\cos^2(\pi{x})}d  x=\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)+2\sqrt{2}<br />

    Hello, I appreciate the answer, but I was hoping someone could give me a hint on the mechanics. I'd like to solve this myself, but I've gone through several different processes and I'm still having problems. What sort of substitution am I looking at here?

    Thanks,
    swicket
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  6. #6
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    Quote Originally Posted by swicket and edited by Mr F View Post
    <br /> <br />
\int_0^{1}2\pi\sin(\pi{x})\sqrt{1+ \pi^2 \cos^2(\pi{x})} \, dx= .......<br />

    Hello, I appreciate the answer, but I was hoping someone could give me a hint on the mechanics. I'd like to solve this myself, but I've gone through several different processes and I'm still having problems. What sort of substitution am I looking at here?

    Thanks,
    swicket
    First let u = \pi \cos (\pi x).

    Then S = -\frac{2}{\pi} \int_{\pi}^{-\pi} \sqrt{1 + u^2} \, du = \frac{2}{\pi} \int_{-\pi}^{\pi} \sqrt{1 + u^2} \, du ......
    Last edited by mr fantastic; June 10th 2008 at 07:51 PM. Reason: Arithmetic error correction
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  7. #7
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    Ah thank you! I was going down that road, but I must have made an error. Your guidance is much appreciated!
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