1154. A smaller parallel side of an isosceles trapezoid is \(a\) and its height is \(m\). What are the surface areas and volumes of the solids that are formed when the trapezoid (1) rotates around one of its parallel sides, (2) rotates around the other parallel side, (3) rotates around one of its non-parallel sides? The non-parallel sides form a \(45^{\circ}\) angle with the base.
Let the base of the trapezoid be $AB$, the side parallel to it be $CD$, and $CD=a$. Then $AB=2m+a$ and $AC=m\sqrt{2}$.

$1^{\circ}$ The trapezoid rotates around $CD$. The surface of the resulting solid is formed by the lateral surface of a cylinder and the lateral surfaces of two cones; thus,

$$
F_{1}=2m\pi(2m+a)+2m^{2}\pi\sqrt{2}=2m\pi[a+(2+\sqrt{2})m]
$$

The volume is obtained by subtracting the volumes of two cones from the volume of a cylinder; that is,

$$
K_{1}=m^{2}\pi(2m+a)-2\cdot\frac{\pi}{3}m^{3}=m^{2}\pi\left(a+\frac{4m}{3}\right)
$$

$2^{\circ}$ The trapezoid rotates around $AB$. The surface of the solid of revolution is formed by the lateral surface of a cylinder and the lateral surfaces of two cones; thus,

$$
F_{2}=2m\pi a+2m^{2}\pi\sqrt{2}=2m\pi(a+m\sqrt{2})
$$

The volume is obtained by adding the volumes of two cones to the volume of a cylinder; that is,

$$
K_{2}=m^{2}\pi a+2\cdot\frac{\pi}{3}m^{3}=m^{2}\pi\left(a+\frac{2m}{3}\right).
$$

$3^{\circ}$ The rotation occurs around $BD$. Let the intersection point of sides $AC$ and $BD$ be $E$, and $AE=R$ and $CE=r$. The surface of the solid of revolution is composed of the lateral surfaces of two cones and the area of an annulus. Thus,

$$
\begin{aligned}
F_{3} & =R\pi(a+2m)+r\pi a+R^{2}\pi-r^{2}\pi= \\
& =\pi\left[R(a+2m)+ra+R^{2}\pi-r^{2}\right]
\end{aligned}
$$

where

$$
R=\frac{1}{2}(a+2m)\sqrt{2} \quad \text{and} \quad r=\frac{a}{2}\sqrt{2}
$$

Therefore,

$$
F_{3}=2m\pi[(m+a)(1+\sqrt{2})]+a^{2}\pi\sqrt{2}
$$

The volume is equal to the difference in the volumes of two cones; that is,

$$
K_{3}=\frac{\pi}{3}\left(R^{3}-r^{3}\right)=\frac{m\pi\sqrt{2}}{6}\left(4m^{2}+6ma+3a^{2}\right)
$$

(Jánosy Gyula, Budapest.)