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Conversion of cartesian coordinates to polar coordinates

  1. Aug 3, 2009 #1
    1. Was wondering if anyone could help me confirm the polar limits of integration for the below double integral problem. The question itself is straight forward in cartesian coordinates, but in polar form, i'm a bit suspect of my theta limits after having sketched the it out. any help much appreciated.



    2. Relevant equations

    [tex]\int^{6}_{0}\int^{y}_{0}xdxdy[/tex]

    in polar form:

    [tex]\int^{\frac{\pi}{2}}_{\frac{\pi}{4}}\int^{6cosec\theta}_{0} r^{2}cos\theta dr d\theta [/tex]


    3. The attempt at a solution

    Using a trig substitution over pi/2 and pi/4, i get an answer of 36. it's just when i sketch it, i get a triangle which only has half that area. am i missing something obvious? cheers
     
  2. jcsd
  3. Aug 4, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi CostasDBD! Welcome to PF! :smile:

    (have a pi: π and a theta: θ and an integral: ∫ :wink:)
    Yup! :redface:

    ∫∫ x dxdy isn't the area! :wink:
     
  4. Aug 4, 2009 #3

    Pengwuino

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    Remember, the area of integration and the actual integration are 2 separate entities. Only when the function you're integrating over is f(x,y) = 1 is the integral equal to the area of your area of a double integration.
     
  5. Aug 4, 2009 #4
    Thanks guys. funny what you pick up when you go back a few pages and have a quick read hey.
    this forum is fantastic.
     
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