Frequency distribution of volumes of spheres is positively skew if the radii are symmetrically distributed 
Suppose the frequency distribution of radii of $n$ spheres is symmetric. Show that the frequency distribution of the volumes of these spheres is positively skew.

I am trying to show this using Bowley's measure of skewness.
Denoting the radius by $r(>0)$, we have $Q_3(r)-Q_2(r)=Q_2(r)-Q_1(r)$ where $Q_i$ denotes the $i$th quartile. Then I need to prove that $Q_3(f(r))-Q_2(f(r))>Q_2(f(r))-Q_1(f(r))$ where $f(r)=\frac{4}{3}\pi r^3$. Is there a neat argument to prove the result without much calculation? My guess is that the median is getting pushed towards the right in the case of volume, but I cannot justify that.
 A: This follows from the fact that the local effect of the function $R\to R^3$ is to expand values more and more as $R$ increases.  We don't actually need any ideas of Calculus to demonstrate this, though: by formulating the situation in the right way, it comes down to demonstrating a simple inequality.  Just don't lose site of the meaning of that inequality.

To simplify the exposition, without any loss of generality, choose a unit of radial measure for which the median radius $R$ is $1$.
The implications of symmetry
The symmetry of the distribution of $R$ implies its quantiles are equidistant from the median.  Specifically, for any $\delta$ with $0 \lt \delta\le 1$,
$$\Pr(R \le 1-\delta) = \Pr(R \ge 1 + \delta) = \alpha,$$
say. This makes $1-\delta$ an $\alpha$ quantile of $R$ and $1+\delta$ a $1-\alpha$ quantile, its "complementary quantile."  (Since $\delta\gt 0$, note that $\alpha\le 1/2$.)  Geometrically, this asserts that all pairs of complementary quantiles are equally spaced around the median.
The effect of cubing
Cubing of positive numbers is a monotonically, strictly increasing operation, whence cubing the radius preserves these inequalities.  Consequently, writing $C$ for the constant factor $4\pi/3$ and $V=CR^3$ for the volume, we can obtain useful information about the distribution of $V:$
$$\eqalign{\Pr(V \le C(1-\delta)^3) &= \Pr(R^3 \le (1-\delta)^3) &\\&= \Pr(R^3 \ge (1+\delta)^3) = \Pr(V \ge C(1+\delta)^3).\tag{1}
}$$
The assumption $0\lt \delta \le 1$ when multiplied by the inequality $2\delta^2 \le 6 \delta^2$ shows $2\delta^2 \lt 6\delta^3$, which is equivalent to
$$(1+\delta)^3 - 1 \gt 1 - (1-\delta)^3.$$
The consequences
In light of $(1)$, this shows that

for all possible quantiles $\alpha \le 1/2$, a $1-\alpha$ quantile of $V$--namely, $C(1+\delta)^3$--is further from its median $C$ than is the complementary $\alpha$ quantile--namely, $C(1-\delta)^3$.

Geometrically, it says that in any pair of complementary quantiles, the upper quantile lies further from the median than the lower one.
That is a very strong statement of positive skewness.

Ramifications
A comparable argument holds for any power $p \gt 1$ in place of $p=3$ and any positive constant $C$.  With a minor modification it will show that for $p \lt 1$, $p\ne 0$, the distribution of $CR^p / p$ is negatively skewed.  This provides insight into what the Box-Cox transformations $R \to (R^p-1)/p$ accomplish in terms of making distributions more or less skewed.
