11 Solution: (1) Let the volume of the solid obtained by rotating triangle $\mathrm{BCO}$ around $\mathrm{CD}$ once be $\mathrm{s}, \mathrm{S}$ is equal to the volume of a cone with a height of 10 cm and a base radius of 6 cm, minus the volumes of 2 cones with heights of 5 cm and base radii of 3 cm.
(2) That is:
$$
\begin{array}{l}
S=\frac{1}{3} \times 6^{2} \times 10 \times \pi-2 \times \frac{1}{3} \times 3^{2} \times 5 \times \pi=90 \pi \\
2 S=180 \pi=565.2 \text { (cubic centimeters) }
\end{array}
$$

Answer: The volume is 565.2 cubic centimeters.
11. As shown in the figure, $ABCD$ is a rectangle, $BC=6 \text{ cm}, AB=10 \text{ cm}, AC$ and $BD$ are diagonals. If the shaded area in the figure rotates one full turn around point $C$, what is the volume of the solid swept by the shaded area in cubic centimeters? $(\pi$ is taken as 3.14)