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Thread: 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 $\displaystyle x=y^2+z^2$ which is inside the cylinder $\displaystyle y^2+z^2=9$.

    My attempt : I notice that to be more friendly, I could change the problem as "the paraboloid $\displaystyle x^2+y^2=z$ and the cylinder $\displaystyle x^2+y^2=9$", but I won't do that.
    So $\displaystyle 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}$$\displaystyle =\iint _S \sqrt {4(y^2+z^2)+1}dydz$.
    The 2 surfaces cross each other when $\displaystyle x=9$. So when $\displaystyle y^2+z^2=9$.
    I converted the variables into polar coordinates, $\displaystyle A=\int _0^{2\pi} \int _0^{9} \rho \sqrt{4\rho +1} d\rho d\theta$.
    Is that good?
    Last edited by arbolis; Aug 4th 2009 at 07:38 PM.
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  2. #2
    Junior Member
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    Shouldn't it be

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

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