 Circular-cap carrier count in a square Brillouin zone

For $s=Q/(2R)$, each face cap has area $C=R^2[\arccos s-s\sqrt{1-s^2}]$. The total second-band electron area is $4C$; the first-band hole area is $Q^2-\pi R^2+4C$. Multiplying either area by the spin degeneracy two and dividing by $(2\pi {)}^2$ gives the carrier densities. This is an infinitesimal-potential count, not an exact finite-gap formula.