Calculating Volume of surface of unit sphere I am trying to understand the proof for 
$w_n = 2\pi^{n/2}/\Gamma(n/2)$ where $w_n$ is the volume of the surface $S_n$ of the n-dimensional unit sphere $K_n$. There is stated that 
$Vol(K_n) = \int_{K_n}d^nx = \int_{(0,1)}(\oint_{\|x\|=r}dS(x))dr = \int_{(0,1)}w_nr^{n-1}dr = w_n / n$ 
follows by applying the following formula on $f(x)=\chi_{K_n}$: 
$\int_{R^n} f(x) d^nx = \int_{(0, \infty)}(\oint_{\|x\|=r}f(x)dS(x))dr = \int_{(0, \infty)}(\oint_{\|e\|=1}f(re)dS(e))r^{n-1}dr$ 
whichs holds for integrable functions $f: R^n \rightarrow R$. 
I cannot understand the second equation in the proof (and neither the third): 
If we apply as mentioned the formula, then we obtain 
$\int_{K_n}d^nx = \int_{R^n}\chi_{K_n}(x) d^nx = \int_{(0, \infty)}(\oint_{\|x\|=r}\chi_{K_n}(x)dS(x))dr$. 
How do we get from here to $\int_{(0,1)}(\oint_{\|x\|=r}dS(x))dr$? 
Thank you
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Proj}{\mathbf{P}}$I'm guessing your notation and terminology come from physics. To forestall possible confusion, I'm using mathematical notation and terminology:


*

*The unit ball $B^{n} \subset \Reals^{n}$ is the set of points lying at most one unit from the origin:
$$
B^{n} = \{x \in \Reals^{n} : \|x\|^{2} \leq 1\}.
$$

*The unit $(n-1)$-sphere $S^{n-1}$ is the boundary of $B^{n}$:
$$
S^{n-1} = \{e \in \Reals^{n} : \|e\|^{2} = 1\}.
$$
The function $f = \chi_{B^{n}}$ is, by definition, equal to $1$ on $B^{n}$ and equal to $0$ on the complement of $B^{n}$. Consequently,
$$
\int_{\Reals^{n}} f(x)\, d^{n}x
  = \int_{B^{n}} f(x)\, d^{n}x
  = \int_{B^{n}} d^{n}x
  = \operatorname{vol}(B^{n}).
$$
I'm not entirely sure which equalities are "the second [and third] equation in the proof", but here are two self-contained sketches that if $f$ is integrable, then
$$
\int_{B^{n}} f(x)\, d^{n}x = \int_{0}^{1} \int_{S^{n-1}} f(re) r^{n-1}\, dS\, dr.
$$
Analytic argument: Define a "polar coordinates" mapping $P:[0, \infty) \times S^{n-1} \to \Reals^{n}$ by $P(r, e) = re$. The unit ball is the image of the "cylinder" $[0, 1] \times S^{n-1}$, and the Jacobian is
$$
|\det DP(r, e)| = r^{n-1}.
$$
If $f:B^{n} \to \Reals$ is integrable, $dV$ denotes Lebesgue measure on $\Reals^{n}$ (i.e., the ordinary Euclidean/Cartesian volume element $d^{n}x$), $dS$ denotes the $(n - 1)$-dimensional measure on $S^{n-1}$ induced by the inclusion $S^{n-1} \subset \Reals^{n}$, and $dr$ is Lebesgue measure on $\Reals$, the change of variables theorem and Fubini's theorem give
\begin{align*}
\int_{B^{n}} f(x)\, dV
  &= \int_{[0,1]\times S^{n-1}} f\bigl(P(r, e)\bigr) |\det DP(r, e)|\, dS\, dr \\
  &= \int_{[0,1]\times S^{n-1}} f(re)r^{n-1}\, dS\, dr \\
  &= \int_{0}^{1} \int_{S^{n-1}} f(re)r^{n-1}\, dS\, dr.
\end{align*}
Geometric argument: The sphere
$$
S^{n-1}(r) = \{x \in \Reals^{n} : \|x\|^{2} = r\}
$$
is a "scaled copy" of the unit sphere $S^{n-1}$; since the dimension is $(n - 1)$ and the scaling is uniform, volumes scale by $r^{n-1}$. In physics notation, $dS(r) = r^{n-1}\, dS$.
The "cylinder" $[r, r + dr] \times S^{n-1}(r)$ therefore has volume $r^{n-1}$ times the $(n - 1)$-volume of $S^{n-1}(1)$. Moreover, a thin shell
$$
\{x \in \Reals^{n} : r \leq \|x\|^{2} = r + dr\} \subset B^{n}
$$
may be identified with the cylinder, in the sense that if $f$ is an integrable function of $n$ variables, then the integrals of $f$ over these two domains differ by terms of order $dr^{2}$. (In the plane $\Reals^{2}$, for example, the shell has area $\pi\bigl[(r + dr)^{2} - r^{2}\bigr] = 2\pi r\, dr + dr^{2}$ and the cylinder has area $2\pi r\, dr$.) Consequently, the difference vanishes when you pass to an integral (i.e., in the limit as $dr \to 0$).
Since the ball $B^{n}$ is a union of thin shells, the integral of $f$ over $B^{n}$ is equal to the integral of $f$ over the corresponding cylinders, i.e.,
$$
\int_{B^{n}} f(x)\, dV
  = \int_{[0, 1] \times S^{n-1}} f(re) r^{n-1}\, dS\, dr
  = \int_{0}^{1} \int_{S^{n-1}} f(re) r^{n-1}\, dS\, dr.
$$
