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Math Help - surface area ( double integral)

  1. #1
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    Question surface area ( double integral)

    Find the surface area of the portion of the cylinder y^2 + z^2 = 9 that is above the rectangle R = {(x,y) such that 0< or = x < or = 2 , -3 < or = y<or = 3}


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  2. #2
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    The portion of the cylinder which lies above the xy plane is z=\sqrt{9-y^{2}}.

    z_{x}=0, \;\ z_{y}=\frac{-y}{\sqrt{9-y^{2}}}, \;\ z_{x}^{2}+z_{y}^{2}+1=\frac{9}{9-y^{2}}

    S=\int_{0}^{2}\int_{-3}^{3}\frac{3}{\sqrt{9-y^{2}}}dydx=\int_{0}^{2}\,3{\pi}dx
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