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Evaluating a double integral

  1. Nov 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate over the x,y plane:
    ∫∫[itex]e^{-\sqrt{x^{2}+4y^{2}}}[/itex]dxdy
    And I know the answer SHOULD be [itex]\pi[/itex]

    2. Relevant equations
    Polar-->rectangular identities maybe?
    x--> rcos, y--> rsinθ, dxdy--> rdrdθ



    3. The attempt at a solution
    I tried using polar coordinates, but it doesn't simplify to something nice. When I did I got:
    ∫∫[itex]r*e^{-r*\sqrt{cosθ^{2}+4sinθ^{2}}}[/itex]drdθ
    But there isn't an identity that simplifies the exponent into something easier to handle
     
    Last edited: Nov 7, 2011
  2. jcsd
  3. Nov 7, 2011 #2

    SammyS

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    [itex]\displaystyle\sqrt{x^{2}+4y^{2}}=r\sqrt{\cos^{2}( \theta)+4sin^{2}(\theta)}[/itex]

    So that last integral should be [itex]\displaystyle\int\int r\, e^{r^2\sqrt{\cos^{2}( \theta)+4sin^{2}(\theta)}}dr\,d\theta[/itex]
    .
     
  4. Nov 7, 2011 #3

    HallsofIvy

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    Since the original integral is in xy coordinates, is that "r" a variable or a constant? Is r just short for [itex]\sqrt{x^2+ y^2}[/itex]?
     
  5. Nov 7, 2011 #4
    The r I put inthe original shouldn't be there, my apologies.
     
  6. Nov 7, 2011 #5
    What's with this something nice and easy to handle thing? That don't happen in the real world. Try and learn to muscle through things. Suppose I just write it as:

    [tex]\int_0^{2\pi}\left(\int_0^{\infty}re^{-ar}dr\right)d\theta,\quad a>0[/tex]

    Can you do that part in parenthesis? Alright, then switch to:

    [tex]a=\sqrt{\cos^2(t)+4\sin^2(t)}[/tex]

    and muscle through that one. We can get rid of one of them right?

    [tex]\cos^2(t)+4\sin^2(t)=1+3\sin^2(t)[/tex]

    Then what about reducing it further with other trig identities?
     
    Last edited: Nov 7, 2011
  7. Nov 7, 2011 #6
    I ultimately wound up getting [itex]\int_0^{2\pi}\frac{1}{5-3cos(2\theta}d\theta[/itex]

    Upon evaluating the integral I got [itex]\frac{1}{2}arctan(2tan(\theta))[/itex], evaluated between 0 and 2[itex]\pi[/itex], which is 0. But after splitting the boundaries at [itex]\frac{\pi}{2}[/itex] and [itex]\frac{3\pi}{2}[/itex], I got simply [itex]\pi[/itex] for an answer.

    Thank you for the help.
     
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