2.24. An isosceles trapezoid with bases 2 and 3 cm and an acute angle of $60^{\circ}$ rotates around the smaller base. Calculate the surface area and volume of the resulting solid of revolution.
2.24. The desired volume $V=V_{\text{cyl}}-2 V_{\text{cone}}$, and the surface area $S=S_{\text{cyl}}+2 S_{\text{cone}}$, where $V_{\text{cyl}}, V_{\text{cone}}$ and $S_{\text{cyl}}, S_{\text{cone}}$ are the volumes and lateral surfaces of the cylinder and cone, respectively (Fig. 2.20). We have $V_{\text{cyl}}=\pi r^{2} h$, where $r=CK, h=AD=3 \text{ cm}$. Since $DK=\frac{1}{2}(AD-BC)=\frac{1}{2} \text{ cm}$, and $CD=2 DK=1 \text{ cm}$, then

$$
CK=\sqrt{CD^{2}-DK^{2}}=\sqrt{1-\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{3}}{2} \text{ cm}
$$

Then

$$
V_{\text{cyl}}=\pi \cdot \frac{3}{4} \cdot 3=\frac{9 \pi}{4} \text{ cm}^{3}, V_{\text{cone}}=\frac{1}{3} \pi \cdot \frac{3}{4} \cdot \frac{1}{2}=\frac{\pi}{8} \text{ cm}^{3}
$$

from which $V=\frac{9 \pi}{4}-\frac{\pi}{4}=2 \pi \text{ cm}^{3}$. Finally, we find

$$
\begin{gathered}
S_{\text{cyl}}=2 \pi r h=2 \pi \cdot \frac{\sqrt{3}}{2} \cdot 3=3 \pi \sqrt{3} \text{ cm}^{2}, \\
S_{\text{cone}}=\pi r l=\pi \cdot \frac{\sqrt{3}}{2} \cdot 1=\frac{\pi \sqrt{3}}{2} \text{ cm}^{2},
\end{gathered}
$$

from which $S=3 \pi \sqrt{3}+\pi \sqrt{3}=4 \pi \sqrt{3} \text{ cm}^{2}$.

Answer: $4 \pi \sqrt{3} \text{ cm}^{2} ; 2 \pi \text{ cm}^{3}$.