= Solution
For $k=1$, put $r=\sin\chi$. The spatial <metric tensor> becomes
$$
a^2\left[d\chi^2+\sin^2\chi(d\theta^2+\sin^2\theta\,d\phi^2)\right].
$$
The global closed spatial slice is a <3-sphere> with $0\leq\chi\leq\pi$; the original $r$ chart covers only a hemisphere, so two charts are needed. A <circle> at fixed $\chi$ in an equatorial angular plane has circumference $\boxed{2\pi a\sin\chi}$. The full spatial volume is
$$
\boxed{4\pi a^3\int_0^\pi\sin^2\chi\,d\chi=2\pi^2a^3.}
$$
Integrating a single $r\in[0,1]$ chart without its second copy would give only half this volume.
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