1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [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


    User Avatar
    Staff Emeritus
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Integration Problem
  1. Integration problem (Replies: 4)

  2. Integral Problem (Replies: 31)

  3. Integration problem (Replies: 9)

  4. Integration Problem (Replies: 7)