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Volume of a three dimensional gaussian

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    How can I find the volume of a three dimensional gaussian [tex] exp\left [ \frac{-x^2}{\sigma_{x}} \frac{-y^2}{\sigma_{x}}\frac{-z^2}{\sigma_{z}} \right ] [/tex] ? Since it is a gaussian, the volume should actually extend to infinity. It seems like there should be a simple double or triple integral, but I can't figure out how to set it up.




    2. Relevant equations



    3. The attempt at a solution

    I multiplied the integral of each gaussian over all space (and of course each of these integrals converges)
    [tex] V = \int_{-\infty}^{\infty}exp\left [ \frac{-x^2}{\sigma_{x}} \right ]\int_{-\infty}^{\infty}exp\left [ \frac{-y^2}{\sigma_{y}} \right ]\int_{-\infty}^{\infty}exp\left [ \frac{-z^2}{\sigma_{z}} \right ] [/tex], but I'm not sure if this is right.


    Thanks
     
  2. jcsd
  3. Feb 26, 2013 #2

    mfb

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    That should be correct.
    The volume (as seen in a 4-dimensional space) extends to +- infinity in all directions, but its height (=function value) drops so quickly that the total volume remains finite.
     
  4. Feb 26, 2013 #3

    Dick

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    Hopefully you have a sum in the exp and not a product.
     
  5. Feb 26, 2013 #4
    Yes, that was meant to be a sum, not a product.

    How is the volume in 4D space? I had wanted to find the volume in 3D space, which should only requires a double integral. But I was not sure how to account for the density of the function changing in all three directions.
     
  6. Feb 26, 2013 #5

    Ray Vickson

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    No: volume in 3d requires a 3-dimensional integral (see any calculus textbook). Anyway, I think the use of the word "volume" is unnecessary and maybe misleading: you just need to compute an integral of some function over some 3-dimensional region. Period.
     
  7. Feb 26, 2013 #6

    Dick

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    You don't find the volume of a function f(x,y,z), you think of it as a mass density and find the total mass by integrating dxdydz.
     
  8. Feb 26, 2013 #7
    That makes much more sense. Thank you!
     
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