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Math Help - Area of a portion of a paraboloid

  1. #1
    MHF Contributor arbolis's Avatar
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    Area of a portion of a paraboloid

    Find the area of the portion of the paraboloid x=y^2+z^2 which is inside the cylinder y^2+z^2=9.

    My attempt : I notice that to be more friendly, I could change the problem as "the paraboloid x^2+y^2=z and the cylinder x^2+y^2=9", but I won't do that.
    So A=\iint _S \sqrt{ \left ( \frac{ \partial f}{\partial y} (y,z) \right ) ^2+ \left ( \frac{\partial f}{\partial z} (y,z) \right ) ^2+1} =\iint _S \sqrt {4(y^2+z^2)+1}dydz.
    The 2 surfaces cross each other when x=9. So when y^2+z^2=9.
    I converted the variables into polar coordinates, A=\int _0^{2\pi} \int _0^{9} \rho \sqrt{4\rho +1} d\rho d\theta.
    Is that good?
    Last edited by arbolis; August 4th 2009 at 08:38 PM.
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  2. #2
    Junior Member
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    Shouldn't it be

     \rho \sqrt{4\rho^2 +1} ?

    and the \rho integration limit between 0 and \sqrt{9}
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