A integral in Evans PDE chapter 2 problem 2 (straight calculus problem) Here I am attempting problem 3 in Chapter 2 of the book PDE by Evans.
Thankfully I have found the solution on the internet
enter link description here
However, there is one line confuses me I was starring it for 3 long hours. (See page 3)

I think this must got to do with the polar coordinate formula in appendix, i.e.

I am really struggling seeing this, I just cannot see why everything including $\frac{1}{t^{n-2}}$ is in term of $x$. Could anyone help explaining the equality of the two integrals?

I think there is an error, as the LHS depends on $\epsilon$ and the RHS does not.  On that same page, he calls
$$
I=\int_\epsilon^r\int_{\partial B(0,t)}\frac{1}{t^{n-2}}f\,dSdt\,\,\,\,\,\text{ and }\,\,\,\,\,J=\frac{1}{\epsilon^{n-2}}\int_{B(0,\epsilon)}f\,dy.
$$
Then
$$
I=\int_\epsilon^r\int_{\partial B(0,t)}\frac{1}{|y|^{n-2}}f\,dS(y)dt=\int_{\epsilon\leq|y|\leq r}\frac{1}{|y|^{n-2}}f\,dy=\int_{B(0,r)}\frac{1}{|y|^{n-2}}f\,dy-\int_{B(0,\epsilon)}\frac{1}{|y|^{n-2}}f\,dy
$$
since $t=|y|$ when $y\in\partial B(0,t)$ and $\bigcup_{t\in[\epsilon,r]}\partial B(0,t)=B(0,r)\backslash B(0,\epsilon)$.  For $n\geq 2$ we have 
$$
\left|\int_{B(0,\epsilon)}\frac{1}{|y|^{n-2}}f\,dy\right|\leq\max\,|f|\int_{0}^\epsilon\frac{1}{t^{n-2}}\int_{\partial B(0,t)}\,dS(y)dt=\max\,|f|\int_0^\epsilon n\alpha_nt\,dt\to 0
$$
as $\epsilon\to 0$.  He earlier states $J\to 0$ as $\epsilon\to 0$.  Hence,
$$
I+J\to \int_{B(0,r)}\frac{1}{|y|^{n-2}}f\,dy\,\,\text{ as }\,\,\epsilon\to 0
$$
and not as he states.
