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An integral paradox ?

  1. Jun 9, 2010 #1
    An integral paradox ??

    let be [tex] \int_{0}^{\infty}xdx \int_{0}^{\infty}ydy [/tex]

    changing to polar coordinates we get that the double integral above shoudl be

    [tex] 2\int_{0}^{\infty}r^{3}dr [/tex]

    althoguh they are all divergent , is this true can we ALWAYS make a change of variable to polar coordinates without any ambiguity ??
  2. jcsd
  3. Jun 9, 2010 #2


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    Science Advisor

    Re: An integral paradox ??

    One can always change from rectangular to polar coordinates. However the transformation you gave is incorrect. The coefficient is not 2 but 1/2.
  4. Jun 10, 2010 #3
    Re: An integral paradox ??

    am.. thanks a lot

    but my question is, the Area of a Circle is NOT equal to the area of an Square [tex] \frac{C}{S}= \pi [/tex]

    hence , how could we be completely sure [tex] \iint _{C} f(x,y)dxdy = \iint _{S} f(x,y)dxdy [/tex]
  5. Jun 10, 2010 #4

    Gib Z

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    Re: An integral paradox ??

    Rectangular coordinates don't necessarily trace out rectangles and Polar coordinates don't necessarily trace out Circles in the xy plane. The path they trace out is predetermined by a rule, eg To describe the path of the unit circle in rectangular coordinates we say x^2+y^2 = 1, and the same path could be described in polar coordinates with x= cos t, y= sin t, t varies from 0 to 2pi.

    It's your job to change the bounds and integrand of the integral accordingly when change coordinates so that they still sum the same overall function values over the same domain.
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