Problem with a $N$-dimensional integral In a proof that I'm studying there is this equality:
$$c \int_{\partial B(0,\epsilon)}\frac{1}{\lvert x \rvert^\alpha}d\sigma_x=c_1\frac{1}{\epsilon^\alpha}\epsilon^{N-1},$$
where $B(0,\epsilon)$ is the $N$-dimensional ball centred in $0$ with radius $\epsilon$ and $c,c_1$ costants.
A friend of mine tries to explain it to me but i think it's wrong (for me it doesn't make sense). 
His work:
$$c \int_{\partial B(0,\epsilon)}\frac{1}{\lvert x \rvert^\alpha}d\sigma_x=\frac c {\epsilon^\alpha}\int_{\partial B(0,\epsilon)} d\sigma_x=c\frac{1}{\epsilon^\alpha}\epsilon^{N-1}$$
Someone could help me, please? Thank you, and sorry for my bad English.
 A: The boundary of the ball consists precisely of those $x$ with distance $\epsilon$ to the origin: that is, $|x|=\epsilon$. So you get 
$$
c \int_{\partial B(0,\epsilon)}\frac{1}{\lvert x \rvert^\alpha}d\sigma_x=\frac c {\epsilon^\alpha}\int_{\partial B(0,\epsilon)} d\sigma_x.
$$
This last integral is precisely the area of the ball, which is of the form $c_2\epsilon^{N-1}$ for some constant $c_2$ that depends on $N$ (but not on $\epsilon$). Then 
$$
c \int_{\partial B(0,\epsilon)}\frac{1}{\lvert x \rvert^\alpha}d\sigma_x=c\frac{1}{\epsilon^\alpha}\,c_2\epsilon^{N-1}=c_1\frac{1}{\epsilon^\alpha}\epsilon^{N-1},
$$
where $c_1=c\,c_2$. 
A: Your friend is right when he says that, if $x\in\partial B(0,\epsilon)$ then $|x|=\epsilon$.
After accepting this, you'll need to compute the surface of a $N$-dimensional ball with radius $\epsilon$. If you know how much is the volume of $B(0,\epsilon)$ (if you don't, you may want to see this link), deriving it with respect to $\epsilon$ will yield the surface formula.
