In the following example p.260 Evans. 
I think I understand everything except for one calculus fact in the second last equation: $$ \int_{\partial B(0,\varepsilon)}\varepsilon^{-\alpha}\,dS=\varepsilon^{n-1-\alpha}. $$
Could anyone show me how the integral computed explicitly?
On
Here is a possible explicit calculation:
Let $A$ be the area of the unit sphere $S(0,1)=\{x\in\mathbb{R}^n;\ |x|=1\}$. Employing polar coordinates (see Theorem 2.49 in the Folland's book) we conclude that
$$\int_{B(0,\varepsilon)}1\ dx=\int_0^\varepsilon\int_{\partial B(0,1)} r^{n-1}\ dS\ dr=A\int_0^\varepsilon r^{n-1}\ dr=A\frac{\varepsilon^n}{n}.$$
Thus, by item (ii) of Theorem 4 in appendix C.3 of the Evans book,
$$\int_{\partial B(0,\varepsilon)}\varepsilon^{-\alpha}\;dS=\varepsilon^{-\alpha}\int_{\partial B(0,\varepsilon)}1\ dS=\varepsilon^{-\alpha}\left(\frac{d}{d\varepsilon}\int_{B(0,\varepsilon)}1\ dx\right)=A\varepsilon^{n-1-\alpha}.$$
It is not true that they are equal, instead the left hand side is bounded by the right hand side multiplied by some constant that is independant of $\epsilon$, it seems to be part of the the $C$ in the example. This fact arises because $\partial B(0,\epsilon)$ is an $n-1$ dimensional object, so $$\int_{\partial B(0,\epsilon)}\,dS=|\partial B(0,\epsilon)|\le C\epsilon^{n-1}.$$ Now multiply by $\epsilon^{-\alpha}$ and you get result.