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Math Help - Calc 2- Surface Area of Parametric Curves

  1. #1
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    Calc 2- Surface Area of Parametric Curves

    Hey guys, I'm just stuck on the manipulation part of this problem but I'll set it up for ya'll.

    Y=√(2x-x^2)
    1/2≤x≤3/2
    Rotating around the X-Axis

    I set up the problem and plugged it into the equation which I believe is,

    S=∫2πy√(1+(dy/dx))dx

    When I plug in the derivative of Y, (1-x)/√(2x-x^2) under the square root in the main equation, I get lost as what to do next. I tried a few things but I'm either missing something or just flat out messing up my manipulation.

    Thanks for your help! Audio
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  2. #2
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    Bump!
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  3. #3
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    Wink

    Hi Audio,

    I think your mistake is that you are using:
    S = \int 2 \pi y \sqrt{1+dy/dx} dx

    whereas you should be using:
    S = \int 2 \pi y \sqrt{1+(dy/dx)^2} dx

    I worked it through and it came out quite easily as 2 \pi

    Let me know if you are stuck!

    Suhada
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