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Math Help - Surface area

  1. #1
    Senior Member Twig's Avatar
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    Surface area

    Hi!

    Problem is to find the area. (See picture below, along with description of problem).

    The book says, A=2\pi\cdot \bar{x}\cdot L , where \bar{x} is the y-coordinate of the centroid of the line. And L is the length of the line.

    I donīt get correct results :/

    I did for the A circular arc: 2\pi\cdot \left(\underbrace{100-\frac{50\cdot sin(\frac{\pi}{4})}{\frac{\pi}{4}}\cdot cos(\frac{\pi}{4})}_{\text{x-coordinate}}\right)\cdot \underbrace{\frac{2\pi\cdot 50}{4}}_{\text{Length of arc}}

    Then I did the same for the B arc. And I add them.
    The result is incorrect. Even if I try to add the top area as well, I get the wrong result.

    Note: Instead of 2 inch use 50mm , and the answer should be 9.87(10^{4})

    Thx!
    Attached Thumbnails Attached Thumbnails Surface area-pic_10.jpg  
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  2. #2
    Super Member malaygoel's Avatar
    Joined
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    India
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    These articles may help you:

    Surface of Revolution -- from Wolfram MathWorld
    Pappus's Centroid Theorem -- from Wolfram MathWorld

    However, answer to your problem is 16\pi^2
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  3. #3
    Super Member
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    I'd first calculate it the regular way using the formula:

    S=2\pi \int_a^b x\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy  }{dt}\right)^2}dt

    Right?

    Me, I'd use complex coordinates. The equation for the top curve is 4+4i +2 e^{it}. You can do the bottom one. Thus, the parametric form of the top curve is x=4+2\cos(t), y=4+2\sin(t). Now, just substitute both expressions, making sure you get the starting and ending t times correctly, into the formula above.

    After I get the right answer, then I'd go on to calculate the moment of the figure and then the y-coordinate of the centroid and see if I get agreement.
    Last edited by shawsend; July 11th 2009 at 07:09 AM.
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