Integral of a radially increasing function Fix a radius $R > 0$, a positive integer $n$ and suppose $B(s, R) \subset \mathbb{R}^{d}$. Let $f(s) := \int_{B(s, R)}|x|^{n}\, dx$. Is $f$ monotonically increasing in $|s|$? That is, suppose we are given 2 points $s$ and $t$ such that $|s| \leq |t|$, then is $f(s) \leq f(t)$?
One way to do this is to try to compute $f(s)$. In the case of $s = 0$, $$f(0) = \int_{B(0, R)}|x|^{n}\, dx = \int_{S^{d - 1}}\int_{0}^{R}r^{n + d - 1}\, dr\, d\sigma = \frac{1}{n + d}R^{n + d}\int_{S^{d - 1}}\, d\sigma$$ but what about general $s$?
 A: This is true for every radially increasing function $w(x)=\phi(|x|)$, in particular for $w(x)=|x|^n$. Indeed, the layer cake representation allows us to write $w$ as an integral of functions of the form $\chi_{\{|x|\ge \rho\}}$. So, it remains to show that for each fixed $\rho$, the measure of the intersection of $B(s,R)$ with $\{|x|\ge \rho\}$ is an increasing function of $s$. 
The problem reduces to the statement that

The measure of the intersection of two balls of fixed radii is a decreasing function of the distance between their centers.

First, one deals with the one-dimensional case: the length of the intersection $[-R,R]\cap [d-r,d+r]$ is a decreasing function of $r$ for $r>0$. 
The $n$-dimensional case is then handled with Fubini's theorem.  Let $d$ be the distance between the centers.  Take any line $L$  parallel to the line connecting the centers. The intersection of each ball with $L$ is a line segment whose length is independent of $d$. The distance between the midpoints of these segments is $d$. Therefore,  thus, the length of the intersection is a decreasing function of $d$. 
More details here.
