The following is part of a proof that I am studying from the book Partial Differential Equations by Lawrence Evans.


Given that $\displaystyle u \in C^{1}(\mathbb{R}^{n})$ with $t \in (0,s)$ and $\displaystyle w \in \partial B(0,1)$.


Consider the integral $\displaystyle \int_{0}^{s}\int_{\partial B(0,1)} |Du(x+tw)|dSdt = \int_{0}^{s}\int_{\partial B(0,1)}|Du(x+tw)|\frac{t^{n-1}}{t^{n-1}}dSdt$.


How does it follow then that if we let $\displaystyle y=x+tw$ so that $\displaystyle t = |x-y|$. Then converting from polar coordinates, we have the inequality:


$\displaystyle \int_{\partial B(0,1)}|u(x+sw) - u(x)|dS \leq \int_{B(x,s)}\frac{|Du(y)|}{|x-y|^{n-1}}dy$


How does this final inequality follow? How is it a result of polar coordinate conversion?
Let me know if anything is unclear. Thanks.