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Math Help - The area of surface of revolution

  1. #1
    Junior Member
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    The area of surface of revolution

    The loop 9y^2=x(3-x)^2 is revolved about the y-axis. Find the area of the surface generated this way.

    Okay so I solved for y, derived it and squared it. Area = integral of 2(pi)ds And ds is (square root of 1+(dy/dx)^2) dx

    Am I doing this right? Also, how would I find the limits of integration?
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  2. #2
    Super Member
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    I'd just plot it. Looks like you only want the loop which goes from 0 to 3 by setting y=0. The top is the same as the bottom, so just multiply the integral by two and use the positive term for y.
    Attached Thumbnails Attached Thumbnails The area of surface of revolution-plot13.jpg  
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  3. #3
    Junior Member
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    Ah okay. Thanks man.
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