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Double Integral

  1. Dec 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Integrate [itex] f(x,y) = e^{-(x^2 + y^2)} [/itex] over the set [itex] A = \{ (x,y): x > 0, y > 0, x^2 + y^2 < a \} [/itex]



    2. Relevant equations



    3. The attempt at a solution

    The polar coordinate transformation [itex] g(r,\theta) = (r cos\theta, r sin\theta) [/itex] is a diffeomorphism from A to the set [itex] B = \{(r,\theta): 0 < r < a, 0 < \theta < \pi / 2 \} [/itex], so I can use the change of variables theorem.

    So [itex] \int_A e^{-(x^2 + y^2)} = \int_B re^{-r^2} = \int_0^a \int_0^{\pi/2} re^{-r^2} d \theta dr = (\pi / 2) \int_0^a re^{-r^2} dr [/itex].

    Let [itex] u = -r^2 [/itex]. Then [itex] -du/2 = rdr [/itex], so we have [itex] (\pi / 2) \int_0^a re^{-r^2} dr = (- \pi / 4) \int_0^{-a^2} e^u du = (- \pi / 4)(e^{-a^2} - 1) [/itex].

    I get the same answer if I reverse the order of integration. Is this answer correct?
     
  2. jcsd
  3. Dec 31, 2011 #2

    SammyS

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    It all looks fine to me.
     
  4. Jan 1, 2012 #3
    All good buddy. Why even mention diffeomorphisms? This be calculus, there be dragons there.
     
  5. Jan 1, 2012 #4
    Because as far as I know, the change of variables theorem requires g to be a C^r diffeomorphism.
     
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