$ \int_{B(0,R)}|\nabla \chi_R^r|^n=\frac{\omega_n}{\log(R/r)^{n-1}} $ Consider the function on $\mathbb{R}^n$ given by $\chi_R^r(y)=\frac{\log(|y|/R)}{\log(r/R)}$ if $r\leq |y|\leq R$, $1$ if $|y|\leq r$ and $0$ if $|y|\geq R$. I want to show that
$$
\int_{B(0,R)}|\nabla \chi_R^r|^n=\frac{\omega_n}{\log(R/r)^{n-1}}
$$
where $\omega_n$ is the measure of unit sphere in $\mathbb{R}^n$.
I have $\partial_i (\chi_R^r)=\frac{y_i}{R\cdot 
 \log(r/R)|y|^2} $ hence $$|\nabla \chi_R^r|^n= \left(\frac{1}{R^n\cdot \log(r/R)^n}\frac{1}{|y|^n} \right) $$
and then
$$
\int_{B(0,R)}|\nabla \chi_R^r|^ndy=\frac{1}{R^n\cdot \log(r/R)^n}\int_{B(0,R)}\frac{1}{|y|^n}dy
$$
I got stuck here. Is it right ultil here and how can I go further? Can someone help?
 A: For $r < |y| < R$ and $1 \leq i \leq n$,
$$\frac{\partial \chi_{R}^{r}}{\partial y_{i}} = \frac{1}{\text{log}(R/r)}\frac{y_{i}/(R|y|)}{|y|/R} = \frac{y_{i}}{\text{log}(r/R)|y|^{2}} $$
$$\Rightarrow |\nabla \chi_{R}^{r}| = \frac{|y|}{\text{log}(R/r)|y|^{2}} = \frac{1}{\text{log}(R/r)|y|} \Rightarrow |\nabla\chi_{R}^{r}|^{n} = \frac{1}{\text{log}(R/r)^{n}|y|^{n}} $$
Note that in the above line, we have $\text{log}\frac{R}{r}$ instead since $R > r$ and $\text{log}\frac{r}{R} = -\text{log}\frac{R}{r}$. Also note that $\nabla \chi_{R}^{r} = 0$ for $|y| < r$. Our integral is:
$$\int_{\mathbb{R}^{n}}|\nabla \chi_{R}^{r}(y)|^{n}dy = \frac{1}{\text{log}(R/r)^{n}}\int_{\{r \leq |y| \leq R \}}\frac{dy}{|y|^{n}} $$
$$ = \frac{1}{\text{log}(R/r)^{n}}\int_{r}^{R}\{ \int_{\partial B(0, t)}\frac{1}{|y|^{n}}dS(y) \}dt\text{, by polar coordinates} $$
$$ = \frac{1}{\text{log}(R/r)^{n}}\int_{r}^{R}\omega_{n}t^{n - 1}(\frac{1}{t^{n}})dt = \frac{\omega_{n}}{\text{log}(R/r)^{n}}\int_{r}^{R}\frac{dt}{t} = \frac{\omega_{n}}{\text{log}(R/r)^{n - 1}} $$
as desired. See Appendix C.3 of Evans for a reference on polar coordinates.
