Consider an equilateral triangle with one side $a$. We draw a line $T_{1} T_{2}$ through one of the vertices of the triangle, which forms a $30^{\circ}$ angle with the opposite side. Calculate the surface area and volume of the solid of revolution that is formed when the triangle rotates around the $T_{1} T_{2}$ axis.
First solution. Let $ABC$ be the given triangle, through whose vertex $C$ we draw the axis $T_{1} T_{2}$. We draw the perpendicular $BE$ from the vertex $B$ of the triangle to $T_{1} T_{2}$. The solid of revolution is obtained by subtracting the cone formed by the revolution of the triangle $CBE$ from the frustum of a cone formed by the revolution of the trapezoid $CABE$. Since

$$
CD=\frac{a}{2} \sqrt{3} \text { and } BE=\frac{a}{2}
$$

the sought volume is:

$$
\begin{gathered}
V=\frac{a \sqrt{3} \pi}{2 \cdot 3}\left(a^{2}+\frac{a^{2}}{4}+\frac{a^{2}}{2}\right)-\frac{a^{2}}{4} \pi \cdot \frac{a}{2 \cdot 3} \sqrt{3}= \\
=\frac{7}{24} a^{3} \sqrt{3} \pi-\frac{a^{3}}{24} \sqrt{3} \pi=\frac{a^{3}}{4} \sqrt{3} \pi
\end{gathered}
$$

The sought surface area is:

$$
F=a^{2} \pi+\left(a+\frac{a}{2}\right) a \pi+\frac{a^{2}}{2} \pi=3 a^{2} \pi
$$

(Fried Ernö, Budapest.)

Second solution. According to Guldin's rule (K. M. L. VIII. 137.), we solve the problem as follows:

The surface area of the solid of revolution is obtained by multiplying the length of the rotating lines by the distance traveled by their center of mass during the rotation.

Thus:

$$
F=a \cdot 2 \cdot \frac{a}{2} \cdot \pi+a \cdot 2 \cdot \frac{a}{4} \cdot \pi+a \cdot 2 \cdot \frac{3 a}{4} \cdot \pi=3 a^{2} \pi
$$

The volume is obtained by multiplying the distance traveled by the center of mass by the area of the rotating figure. Therefore

$$
K=2 \cdot \frac{a}{2} \cdot \pi \cdot \frac{a^{2}}{4} \sqrt{3}=\frac{a^{3}}{4} \sqrt{3} \pi
$$

(Szende György, Budapest.)