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Cartesian to Polar in Double Integral

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve:
    [tex]\iint_{\frac{x^2}{a^2}+\frac{y^2}{b^2} \leq 1} \sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} dx dy [/tex]

    2. Relevant equations

    Cartesian to Polar

    3. The attempt at a solution

    Well - this Integral should be solved as a polar function (the radical should be [tex]\sqrt{1-r^2}ab[/tex] when expressed in polar coordinates. I just can't get this right. Guidance please?
     
  2. jcsd
  3. Sep 12, 2009 #2

    HallsofIvy

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    Staff Emeritus
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    No, the radical is NOT of that form in polar coordinates. But you can modify the coordinate system to "elliptic coordinates". Let [itex]x= r a cos(\theta)[/itex], [itex]y= r a sin(\theta)[/itex]. Then [itex]x^2/a^2= r^2 cos^2(\theta)[/itex] and [itex]y^2/b^2= r^2 sin^2(\theta)[/itex]. You will need to use the Jacobian to get the correct differential dx dy is not just "[itex]r dr d\theta[/itex]" now.
     
  4. Sep 12, 2009 #3
    ok, so maybe the "ab" is part of the differential. I guess that it comes from the Jacobian, which I have no idea how to use.
     
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