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Integration Problem

  1. Feb 13, 2007 #1
    I need to integrate
    double integral 2(sqrt(9-x^2-y^2)(-x^2+y^2-2)dxdy with the bounds determined by the fact that x^2+y^2 is greater than or equal to 2.

    This integral is impossible to calculate using cartesian coordinates. How would I do it using polar coordinates?
  2. jcsd
  3. Feb 13, 2007 #2


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    [tex] \int \!\!\! \int 2\sqrt{(9-x^2-y^2)(-x^2+y^2-2)}\,dx\,dy [/tex]

    [tex] x^2 + y^2 \geq 2 [/tex]

    Is that the integral in question?
  4. Feb 13, 2007 #3


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    That's an infinite region isn't it?
  5. Feb 13, 2007 #4
    Yes that's it, the one cepheid wrote.

    Well it's obviously part of a problem so here goes:
    Let S be the subset of the surface of the sphere x^2+y^2+z^2=9 for which x^2+y^2 is greater than or equal to 2. Let F be the vector field defined by (-y, x, xyz).

    Computer double integral(curlF)*ndS where * is the dot product and S is oriented so that the unit normal n to S points away from the enclosed volume.

    I have the vector field= -yi+xj+xjzk.
    curlF (after calculation)=xzi-yzj+2k
    G(x,y,z)=9-z^2-x^2-y^2 and gradG=-2xi-2yj-2zk
    Thus, dS=(-2xi-2yj-2zk)dxdy
    The dot product of curlF and dS=-2(x^2)z+2(y^2)z-4z
    I plug in for z=sqrt(9-x^2-y^2) and I'm in this predicament.
    Last edited by a moderator: Feb 13, 2007
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