11.099. 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.
Solution.

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.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-710.jpg?height=432&width=310&top_left_y=98&top_left_x=849)

Fig. 11.86. We have $V_{\text {cyl }}=\pi r^{2} h$, where $r=CK, \quad h=AD=3$ cm. Since $\quad DK=\frac{1}{2}(AD-BC)=\frac{1}{2}$ (cm), and $CD=2 DK=1$ (cm), then $CK=\sqrt{CD^{2}-DK^{2}}=\sqrt{1-\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{3}}{2}$ (cm). Therefore, $V_{\text {cyl }}=\pi \cdot \frac{3}{4} \cdot 3=\frac{9 \pi}{4}\left(\mathrm{~cm}^{3}\right), V_{\text {cone }}=\frac{1}{3} \pi \cdot \frac{3}{4} \cdot \frac{1}{2}=\frac{\pi}{8}\left(\mathrm{~cm}^{3}\right)$, from which $V=\frac{9 \pi}{4}-\frac{\pi}{4}=2 \pi\left(\right.$ cm $\left.^{3}\right)$. Finally, we find $S_{\text {cyl }}=2 \pi r h=2 \pi \cdot \frac{\sqrt{3}}{2} \cdot 3=3 \pi \sqrt{3}\left(\mathrm{cm}^{2}\right), S_{\text {cone }}=\pi r l=\pi \cdot \frac{\sqrt{3}}{2} \cdot 1=\frac{\pi \sqrt{3}}{2}\left(\mathrm{~cm}^{2}\right)$, from which $S=3 \pi \sqrt{3}+\pi \sqrt{3}=4 \pi \sqrt{3}\left(\mathrm{~cm}^{2}\right)$.

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