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Math Help - Surface Integral

  1. #1
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    Surface Integral

    Evaluate the following surface integral of the given function on the given surface.

    g(x,y,z)=z on the first-octant part of the sphere x^2+y^2+z^2=1

    My work so far:

    z=\sqrt{1-(x^2+y^2)}

    The domain: (x^2+y^2)\leq 1

    \frac{\partial{z}}{\partial{x}}=\frac{-x}{\sqrt{1-(x^2+y^2)}}

    \frac{\partial{z}}{\partial{y}}=\frac{-y}{\sqrt{1-(x^2+y^2)}}

    \iint (z)*\frac{1}{\sqrt{1-(x^2+y^2)}}dA

    Here is the point where I get stuck. In all my previous problems my g(x,y,z) only contained x's and y's. So, if that were the case here, I would just convert everything to polar coordinates, plug in the bounds and solve the integral. In this case though, there is a z in the integrand so I'm not sure how to continue. Could someone please help me out of a jam? Thanks
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Surface Integral

    Quote Originally Posted by dbakeg00 View Post
    In this case though, there is a z in the integrand so I'm not sure how to continue. Could someone please help me out of a jam? Thanks
    Substitute z=\sqrt{1-x^2-y^2} . If S is the given surface and D its projection onto the xy plane, then \iint_Sz\;dS=\iint_D\;dxdy=\textrm{Area}\;(D)=\pi/4 .
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  3. #3
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    Re: Surface Integral

    That worked. Thanks man, I appreciate the help.
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