The bases of a general trapezoid are $a$ and $b$, and its height is $m$. The distance of the trapezoid's centroid from one of the parallel sides needs to be determined.
In this case

$$
V=\frac{a+b}{2} m \cdot 2 d \pi
$$

if $d$ is the distance of the center of gravity from $b$. But if the trapezoid rotates around $b$ $(b>a)$, then:

$$
V=a m^{2} \pi+\frac{b-a}{3} m^{2} \pi-m^{2} \pi \frac{b+2 a}{3}
$$

and thus

$$
\frac{a+b}{2} m \cdot 2 d \pi=m^{2} \pi \frac{b+2 a}{3}
$$

from which

$$
d=\frac{m(b+2 a)}{3(a+b)}
$$

(Lázár Lajos, Budapest.)

The problem was also solved by: Bayer B., Bogdán G., König D., Lukhaub Gy., Póka Gy., Scharff J., Sasvári J.