Mean volume of sphere (normal distributed radius)

[ChrisVer]

I read in http://www-library.desy.de/preparch/books/vstatmp_engl.pdf page 29 (43 for pdf) that the mean volume is:
$<V> = \int_{-\infty}^\infty dr V(r) N(r| r_0,s)$
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?

Thanks

 

[Charles Link]

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.

 

[Stephen Tashi]

Q2: is there an intuitive way to see/describe why the mean volume is larger than the one obtained from the mean radius?

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