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Prove the improper triple integral equals 2*Pi

  1. Mar 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex]\int^{infinity}_{-infinity}\int^{infinity}_{-infinity}\int^{infinity}_{-infinity}sqrt(x^2+y^2+z^2)e^-^(^x^2^+^y^2^+^z^2^)dxdydz = 2\pi[/tex]

    2. Relevant equations

    [tex]x^2+y^2+z^2 = \rho^2 [/tex]

    3. The attempt at a solution

    I converted to spherical coordinates to get

    [tex]\int^{2\pi}_{0}\int^{2\pi}_{0}\int^{infinity}_{0}\rho^3e^-^\rho^2sin\phi d\rho d\phi d\theta[/tex]

    But I don't think I can integrate that. Am I approaching this the right way?
     
  2. jcsd
  3. Mar 30, 2010 #2
    The bounds are independent.
    You can get rid of the (phi) and (theta) integrals (ie factor them out).

    As for the remaining integrand, you have a power of (rho) multiplied by something you can integrate (may take a second thought to get this 'something' right).
    What method does this suggest?
    Consider:
    (int) x² . x exp(-x²) dx
     
  4. Mar 30, 2010 #3
    OK, integration by parts gives

    (-1/2e^(-rho^2))(rho^2+1)

    Which is evaluated from 0 to infinity, but unless I'm missing something, the limit at infinity is undefined.

    And something else occurred to me. If I integrate with respect to phi, I'll get -cos(phi), which evaluated from 0 to 2*pi is 0. So does that mean I messed up the conversion to spherical coordinates?
     
  5. Mar 30, 2010 #4
    Kinda… In spherical coordinates, what you call phi goes from 0 to Pi, not 2*pi :)
    As for the integration over rho, it definitely converges as rho^3 Exp[-rho^2] is 0 at 0 and goes to 0 at infinity (the exponential is stronger than any power of rho) :)
     
  6. Mar 30, 2010 #5

    vela

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    You can put the exponential in the denominator and use L'Hopital's rule to evaluate the limit at infinity.
     
  7. Mar 30, 2010 #6
    Thanks a bunch guys, I got it now. And now I feel dumb for not seeing all these things I missed.
     
  8. Feb 16, 2011 #7
    I'm kind of lost. I did all the work up to the very end and have just the limit function left but can't get it to work out.

    Can someone check this and let me know what else I need to do or what I messed up on?

    I have:

    2[tex]\pi[/tex] * lim (t [tex]\rightarrow[/tex] [tex]\infty[/tex] [ -e-t^2/2 * (t2 + 1) + 1/2 ]

    I believe that the way I have this written down it is equal to some undefined number because we find that the exponential part goes to 0 while t2 becomes [tex]\infty[/tex]

    Thanks in advance, Travis
     
  9. Feb 16, 2011 #8

    vela

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    [itex]0\cdot\infty[/itex] is an indeterminate form. You can use L'Hopital's rule to evaluate the limit.
     
  10. Feb 19, 2011 #9
    Now I really feel dumb... like the person who started the thread I messed up my limits. For some reason I thought phi was [0, pi/2] so when I finished everything I came up with just pi as a result. Once I fixed this it worked out to 2*pi
     
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