Evans 's PDE proof Again, I got stuck. Please help me to understand the following:

What is the meaning when you change from integration over the Ball B(x,r) to the surface integration dB(x,s), with another integral of s from 0 to r outside.
How can you get the $n\cdot\alpha(n)\cdot s^{n-1}$ ? is that the surface area of the n-ball ?
 A: In this Theorem 2 (Mean-value formulas for Laplace's equation), p. 25 in Evans, it is proven that for $C^2$ harmonic function $u$ it holds
$$
u(x) = \frac{1}{n \alpha(n) r^{n-1}} \int_{\partial B(x,r)} u dS,
$$
where $n \alpha(n) r^{n-1}$ is, yes, the surface area of the sphere $\partial B(x, r)$ (see the definition (vi) on p.616).
Therefore, 
$$
\int_{\partial B(x,r)} u dS = n \alpha(n) r^{n-1} \, u(x).
$$
Hence, substitution of this equality in your expression leads to
\begin{align}
\int_{B(x,r)} u dy &= \int_0^r \left( \int_{\partial B(x,s)} u dS \right) ds \\
&= \int_0^r \left( n \alpha(n) s^{n-1} \, u(x) \right) ds \\
&= n \alpha(n) \, u(x) \int_0^r s^{n-1} ds = \alpha(n) r^{n} \, u(x).
\end{align}
A: Actually, you don't have a single integral over the ball. You have $n$-th multiple integral: 
$$
\int\limits_{B(x,r)} u(y) \, dy = \idotsint \limits_{B(x,r)} u(y_1,\dots,y_n) \,dy_1 \dots dy_n,
$$
since $y \in \mathbb{R}^n$.
And Theorem 4, p.628 in Evans says that you can find the intergal in the following way:
$$
\idotsint \limits_{B(x,r)} u(y_1,\dots,y_n) \,dy_1 \dots dy_n 
= \int\limits_0^r \left( \idotsint \limits_{\partial B(x,r)} u(S_1,\dots,S_{n-1}) \,dS_1 \dots dS_{n-1} \right) \, dr.
$$
Note that into the brackets you have $(n-1)$-th multiple integral (since the surface of the ball is $(n-1)$-dimensional). Therefore, the number of integrals on the left-hand side and right-hand side is equal.
Just to feel what happens: On the left-hand side you calculate the integral like that:

and on the right-hand side like that:

