10. As shown in Figure 4, $AB$ is the diameter of $\odot O_{1}$, $AO_{1}$ is the diameter of $\odot O_{2}$, chord $MN \parallel AB$, and $MN$ is tangent to $\odot O_{2}$ at point $C$. If the radius of $\odot O_{1}$ is 2, then the area of the figure enclosed by $O_{1}B$, $\overparen{BN}$, $NC$, and $\overparen{CO_{1}}$ is equal to
10. $\frac{\pi}{12}+\frac{\sqrt{3}}{2}+1$.

As shown in Figure 8, draw $O_{1} D \perp M N$, with the foot of the perpendicular at $D$, and connect $C O_{2}$. Then $O_{1} D=1, O_{1} N=2$.

Therefore, $\angle O_{1} N C^{-\prime}=30^{\circ}$, $\angle D O_{1} N=60^{\circ}$, $\angle N O_{1} B=30^{\circ}$.
Thus, $S_{\text {white shape } O_{1} B}=\frac{4}{12} \pi=\frac{\pi}{3}$, $S_{\triangle N N_{1} D}=\frac{1}{2} \cdot O_{1} D \cdot D N=\frac{1}{2} \times 1 \times \sqrt{3}=\frac{\sqrt{3}}{2}$. The area enclosed by $O_{1} D$, $C D$, and $\overparen{C O_{1}}$ is
$$
S=S_{\text {square } C 00_{1} o_{2}}-S_{\text {sector } 00_{1} O_{2}}=1-\frac{\pi}{4} \text {. }
$$

Therefore, the area of the required shape is
$$
S=\frac{\pi}{3}+\frac{\sqrt{3}}{2}+1-\frac{\pi}{4}=\frac{\pi}{12}+\frac{\sqrt{3}}{2}+1 \text {. }
$$