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A Mean volume of sphere (normal distributed radius)

  1. Jun 24, 2016 #1


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    I read in http://www-library.desy.de/preparch/books/vstatmp_engl.pdf page 29 (43 for pdf) that the mean volume is:
    [itex]<V> = \int_{-\infty}^\infty dr V(r) N(r| r_0,s)[/itex]
    I have two questions.
    Q1: why do they take the radius to be from -infinity to +infinity and not from 0 to infinity?
    Q2: is there an intuitive way to see/describe why the mean volume is larger than the one obtained from the mean radius?

    Last edited: Jun 24, 2016
  2. jcsd
  3. Jun 24, 2016 #2

    Charles Link

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    It would appear the normal distribution about ## r=r_o ## is an approximation they want you to use even though that allows for a finite probability that ## r<0 ##. The integral from minus infinity to plus infinity can be solved in closed form, but I don't believe you get a closed form if the limits were zero to infinity even though the answer would be nearly identical. The ## V=(4/3) \pi r^3 ## function will add more weight to the larger r's because of the 3rd power dependence, making the mean volume larger than ## V=(4/3) \pi r_o^3 ##, but I don't have a simple intuitive description for that.
  4. Jun 25, 2016 #3

    Stephen Tashi

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    You could compare the mean of ##\{1^3, 2^3, 3^3\} ## with ##2^3##.
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