= Solution
The length, area and volume of a first <Brillouin zone> are $(2\pi/a)^d$. The corresponding occupied $k$-space measures are
$$
2k_F^{(1)}=\frac\pi a=\frac12\frac{2\pi}a,
$$
$$
\pi(k_F^{(2)})^2=\frac{2\pi^2}{a^2}=\frac12\left(\frac{2\pi}a\right)^2,
$$
$$
\frac{4\pi}3(k_F^{(3)})^3=\frac{4\pi^3}{a^3}=\frac12\left(\frac{2\pi}a\right)^3.
$$
Hence
$$
\boxed{\frac{\text{occupied }k\text{-space measure}}{\text{one Brillouin-zone measure}}=\frac12\quad\text{in all three dimensions}.}
$$
This is state counting, independent of the interval/disk/ball geometry. One <Brillouin zone> contains one spatial state per <unit cell> in a single <energy band>, and that band can hold two <Electrons> per cell because of <spin>. One <Electron> per cell therefore occupies half of it. The “volume enclosed” is the occupied <Fermi sea>, not the lower-dimensional area of the <Fermi surface> itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.
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