Solution
= Solution
Let $a=h/2$ and place the hole axis along $z$. At the rim, $r_0^2=R^2-a^2$. A horizontal slice at $|z|\le a$ is an annulus of area $\pi[(R^2-z^2)-r_0^2]=\pi(a^2-z^2)$. Thus the <napkin ring problem> gives
$$
\boxed{V=\pi\int_{-a}^{a}(a^2-z^2)\,dz
=\frac{4\pi a^3}{3}=\frac{\pi h^3}{6}.}
$$
Each slice depends only on $a$ and $z$, explaining the independence from the original sphere radius, rather than obtaining it merely by cancellation after integration.