why is the content of a n-sphere given by $\int_{0}^{R} S_nr^{n-1}dr = {S_nR^n \over n} $? where $S_n$ is the hyper sphere area and $R$ its radius.
it's written on 'mathworld.wolfram' like it's a definition.
computing by hand the volume for $n= 1,2,3$ I can see that it holds, so is it just a generalization deduced from the observation of a pattern ? or is there some other kind of justification ?
 A: Not everything on Wolfram Mathworld is explained in the form of a sequence of implications. My recollection is that some formulas are not proved at all.
If you "stretch" an $n$-dimensional "solid" object (such as the $n$-ball, which is a name for the interior of a hypersphere embedded in $n$-dimensional Euclidean space) by a factor of $R$ in one dimension, you get an object that has exactly $R$ times as much content (which I'll call "volume" for analogy with the $3$-ball).
Now stretch the resulting object by a factor of $R$ in a second dimension, and its volume increases by a factor of $R$ again, to $R^2$ times the original volume.
But there are altogether $n$ separate dimensions in which we can stretch
an $n$-ball in $n$-dimensional Euclidean space.
After stretching in each of those dimensions, we have another $n$-ball that has exactly $R$ times the radius of the original ball, and $R^n$ times the volume.
In this way we can relate the volume of an $n$-ball of any radius to the
volume $B_n$ of an $n$-ball of radius $1.$ The volume of the $n$-ball of radius $R$ is
$$ B_n R^n. $$
Now take two concentric $n$-spherical shells within the $n$-ball at radii
$r$ and $r + \Delta r.$ For small $\Delta r,$ the volume of the region between the shells is approximately
$$ A(r) \Delta r,$$
where $A(r)$ is the "area" of the $n$-sphere of radius $r.$
We can confirm this by dividing the $n$-sphere into very small pieces so that the region between the shells is filled by putting an almost-prism shape on each subdivided region of the inner shell.
We can therefore integrate the volume of the $n$-ball by a method of spherical shells, which tells us that 
$$ B_n R^n = \int_0^R A(r) \, dr, \tag1 $$
and by the fundamental theorem of calculus,
it follows that
$$ A(r) = \frac{d}{dr} B_n r^n = n B_n r^{n-1}.$$
So we set $S_n = n B_n,$ interpreted as the "area" of the unit sphere in
$n$-dimensional space, and then the area of the sphere of radius $r$ is
$A(r) = S_n r^{n-1}.$
Make these substitutions for $B_n$ and $A(r)$ in Equation $(1),$
and you get
$$  \frac{S_n}{n} R^n = \int_0^R S_n r^{n-1} \, dr.$$
The author of the Wolfram Alpha page apparently considered this obvious
enough to state without proof; after that there is a derivation of 
a formula for $S_n.$
This answer gives a similar argument and generalizes it to polytopes (including hypercubes) as well as spheres.
