How should I prove $\operatorname{vol}_{n+1}B_{n+1}=\int_0^1 \operatorname{vol}_n S^n(r)dr$ without using spherical coordinates? 
Let $B_n:=\{x\in{\Bbb R}^n:|x|\leq 1\}$ and $S^n(r):=\{x\in{\Bbb R}^{n+1}:|x|=r\}$. Then we have the following formula
  $$
\operatorname{vol}_{n+1}B_{n+1}=\int_0^1 \operatorname{vol}_n S^n(r)dr.
$$

Geometrically this is clear: one can sum up the shells to get the volume of the unit ball. I'm looking for a proof of this formula. 
Using the generalization of spherical coordinates is quite cumbersome. (It is definitely feasible though.) I would like to do it in an alternative way. For any $x\in B_{n+1}$, one can write it as
$$
x=(r,\xi),\quad r\in[0,1], \ \xi\in S^{n}(1).
$$ 
I think this might be useful but I don't see how to go on. 

I'm also interested in related references about this formula, which have detailed discussion instead of informal casual explanation. 

Your idea is a good one; here's a detailed sketch.
The map $\sigma:[0, 1]\times S^n(1) \to B_{n+1}(1)$ defined by $\sigma(r, \xi) = r\xi$ (scalar multiplication of $\xi$ by $r$) is smooth, and bijective away from $\{r= 0\}$. At each point of the slice $\{r\} \times S^n(1)$, the Jacobian $\det(D\sigma)$ is $r^n$. Pulling back the Euclidean volume form $\omega$ of $\mathbf{R}^{n+1}$ to $[0, 1] \times S^n(1)$ gives
$$
\sigma^*\omega = r^n\, dr\, d\operatorname{vol}_{S^n(1)}
  = dr\, d\operatorname{vol}_{S^n(r)}.
$$
(The second equality holds because the $n$-volume of $S^n(r)$ is $r^n$ times the $n$-volume of $S^n(1)$.)
Applying change of variables and Fubini's theorem gives the desired claim.
