Surface of spheres with the transformation formula

Let $\omega_n$ the measure of the surface (as obtained with the surface measure) of the unit sphere in $n$ dimensions. E.g. $\omega_2 = 2\pi$. Now let $n\ge 3$.

I want to obtain a formula for the surface $A(r)$ of the set $S_r := \{\,x\,\,\lvert \,\,\|x\| \le r \}$ (i.e. the surface measure of the set $\partial S_r$), and I want to use the transformation formula for this. Somehow I always get confused and can't manage to find it out... The result should be $A(r) = r^{n-1}\omega_n$. Can you even use the transformation formula for this?

EDIT: By 'transformation formula', I mean this formula: http://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables

-
Just a question: what is the transformation formula, for you? –  Siminore Sep 15 '12 at 11:02
Sorry, I naively translated the german word "Transformationsformel" into English. What I mean is this: en.wikipedia.org/wiki/… –  Sh4pe Sep 15 '12 at 11:42

Note that we have the natural embedding $S^{n-1}\subset{\mathbb R}^n$; whence the unit circle $S^1\subset{\mathbb R}^2$ is the $1$-sphere. Its length serves as $\omega_1=2\pi$. Then $S^2\subset{\mathbb R}^3$ is the "ordinary" unit sphere and has $\omega_2=4\pi$.
For geometric reasons the $n$-dimensional area of the $n$-sphere of radius $r>0$ is given by $\omega_n r^n$, where $\omega_n$ is the area of the $n$-dimensional unit sphere. Therefore it suffices to compute the numbers $\omega_n$ for all $n\geq1$. When you speak about a "transformation formula" then you most probably have a parametric representation of each of these spheres in mind. Using this and the so called Gram determinant (an essential ingredient for computing higher dimensional areas) we would in fact be able to compute the numbers $\omega_n$, resp. to derive a recursion formula for the $\omega_n$.
But in the special case of spheres there is a much simpler, more "geometric" procedure available. Denote by $\kappa_n$ the volume of the $n$-dimensional unit ball $B=B_1^n$. Then an $n$-ball of radius $r>0$ will have volume $\kappa_n\, r^n$. Consider now two concentric $n$-balls of radius $1$ and $1+\epsilon$. Then $${\rm vol}\,(B_{1+\epsilon})-{\rm vol}\,(B_1)={\rm vol}\,(B_{1+\epsilon}\setminus B_1)\doteq \epsilon\,\omega_{n-1}\ ,$$ because the difference set is roughly a spherical shell of radius $1$ and thickness $\epsilon$. It follows that $$\omega_{n-1}=\lim_{\epsilon\to0+}{\kappa_n(1+\epsilon)^n - \kappa\, 1^n\over\epsilon} = n\,\kappa_n\ .$$
Therefore we need a formula for the volumes $\kappa_n$. One has $\kappa_1=2$ and $\kappa_2=\pi$; furthermore the $\kappa_n$ satisfy the the recursion $$\kappa_n={2\pi\over n}\ \kappa_{n-2}\qquad(n\geq3)\ .$$ To prove this recursion we write the points of ${\mathbb R}^n$ in the form $(x,y,{\bf z})$ with $x$, $y\in{\mathbb R}$, ${\bf z}\in{\mathbb R}^{n-2}$. Since $$B_1^n=\bigl\{(x,y,{\bf z})\ \bigm|\ (x,y)\in B_1^2, \ {\bf z}\in {B_{\sqrt{1-x^2-y^2}}^{n-2}}\bigr\}\ ,$$ we can write $$\kappa_n=\int_{B_1^n} 1\ {\rm d}{\bf x}=\int_{B_1^2}\ \int_{B_{\sqrt{1-x^2-y^2}}^{n-2}}\ {\rm d}({\bf z})\ {\rm d}(x,y)\ .$$ The inner integral has the value $\kappa_{n-2} (1-r^2)^{(n-2)/2}$, where $r:=\sqrt{x^2+y^2}$; and writing ${\rm d}(x,y)=2\pi r\ dr$ in the outer integral leads to the stated formula.
Writing $\kappa_n$, resp. $\omega_n$, in terms of the $\Gamma$-function allows of a unified formula for $\omega_n$, valid for all $n\geq0$.