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Multiple integrals

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data
    By changing to polar coordinates, evaluate:

    [tex]\int[/tex][tex]\int[/tex] e ^(-[tex]\sqrt{x^2 + y^2}[/tex]) dx dy

    Both integrals go from 0 --> infinity

    2. Relevant equations

    r = [tex]\sqrt{x^2 + y^2}[/tex]

    x = r cos[tex]\theta[/tex]

    y = r sin[tex]\theta[/tex]

    Using the Jacobian to switch to polar coord we get:

    J = r d[tex]\theta[/tex] dr


    3. The attempt at a solution

    [tex]\int[/tex][tex]\int[/tex] e ^ (-r) r d[tex]\theta[/tex] dr

    I have my integral set up, but I have no clue what the integration limits are. My prof. said this is always the hardest part of multiple integrals, any help hinting in the right direction would be great!

    Thanks!
     
  2. jcsd
  3. Nov 22, 2009 #2

    tiny-tim

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    Hi Blamo_slamo! :smile:

    (have a square-root: √ and a theta: θ and an infinity: ∞ and try using the X2 tag just above the Reply box :wink:)
    x and y each go from 0 to ∞.

    So what does the region look like?

    Then where does r go from?

    And then where does θ go from? :smile:
     
  4. Nov 23, 2009 #3
    So this region has a peak at x = y = 0, and it's at 1. The region also slopes off to 0, as both x, and y go to ∞

    I'm still not entirely sure on this, but technically the original function runs off to infinity, as it gets closer and closer to 0, so would r then be from ∞ --> 1? This is an even function, so I can just say it runs over that interval and take a 2 out in front right?

    This one just completes the circle right? So it would go from 0 --> 2π
     
  5. Nov 23, 2009 #4

    tiny-tim

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    Hi Blamo_slamo! :smile:
    no no no no noforget the function, for the limits, you're only interested in the region

    the region is the whole of the first quadrant … 0 ≤ x ≤ ∞ and 0 ≤ y ≤ ∞

    so what is that in terms of r and θ ? :smile:
     
  6. Dec 1, 2009 #5
    Would it be reliable, if I just subbed in the bounds of x and y, in r?

    e.g. r2 = x2 + y2
    so as 0 ≤ x ≤ ∞ and 0 ≤ y ≤ ∞; r would then go from 0 ≤ r ≤ ∞ ?

    and as you explained, we're only looking at the first quadrant, so 0 ≤ θ ≤ π/2 ?
     
  7. Dec 2, 2009 #6

    tiny-tim

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    Hi Blamo_slamo! :smile:

    (just got up :zzz: …)
    Yup! :biggrin:

    That's exactly the way to do it! :smile:
     
  8. Dec 3, 2009 #7
    :smile: Thanks a lot tiny-tim!
     
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