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Double integral into the polar form

  1. Nov 10, 2007 #1
    hello i have this problem about polar form, i am aware that when you have a problem like [tex]\int\int[/tex] x^2 + y^2 dxdy you use r^2 = x^2 + y^2 but i what would you do if you had a problem like [tex]\int\int[/tex] xy dxdy?

    thanks in advance.

    edit: i know the limits if you need them plz let me know but i was more interested in the concept behind it
  2. jcsd
  3. Nov 10, 2007 #2
    If you have,
    [tex]\iint_R xy \ dA[/tex] then since [tex]x=r\cos \phi[/tex] and [tex]y = r\sin \phi[/tex] it means, [tex]xy = r^2 \sin \phi \cos \phi = \frac{1}{2} r^2 \sin (2\phi)[/tex].
  4. Nov 10, 2007 #3
    so you would just integrate 1/2r^2sin(2(teta)
  5. Nov 10, 2007 #4
    No you also have to remember the factor of [tex]r[/tex] whichs appears in the Jacobian.
  6. Nov 11, 2007 #5


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    That is, the "differential of area" in polar coordinates is [itex]r dr d\theta[/itex].
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