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Thread: Surface area integral

  1. January 17th 2010, 02:29 PM #1
    qwesl
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    Surface area integral

    How do you find the surface area of the part of a circular paraboloid z=x^2+y^2 that lies inside the cylinder x^2+y^2=4
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  2. January 17th 2010, 02:39 PM #2
    galactus
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    z_{x}=2x, \;\ z_{y}=2y, \;\ z_{x}^{2}+z_{y}^{2}+1=4x^{2}+4y^{2}+1

    \int_{0}^{2\pi}\int_{0}^{2}r\sqrt{4r^{2}+1}drd{\th eta}
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