. A cylindrical hole $6 \mathrm{~cm}$ long is drilled through a sphere, the axis of the cylinder passing through the center of the sphere. What is the remaining volume?

(It is recalled that the volume of a spherical cap is $\pi h^{2}(R-h / 3)$, where $R$ is the radius of the sphere and $h$ is the height of the cap.)
. Let $R$ be the radius of the sphere. By the Pythagorean theorem, the radius of the cylinder is $\sqrt{R^{2}-9}$. The remaining volume is equal to the volume of the sphere, which is $\frac{4}{3} \pi R^{3}$, minus the volume of the cylinder, which is $\pi\left(R^{2}-9\right) \times 6$, and the volume of the two spherical caps, which is $2 \times \pi(R-3)^{2}\left(R-\frac{R-3}{3}\right)=\frac{2}{3} \pi(R-3)^{2}(2 R+3)$. The calculation gives

$$
V=\frac{4}{3} \pi R^{3}-\pi\left(R^{2}-9\right) \times 6-\frac{2}{3} \pi(R-3)^{2}(2 R+3)=36 \pi \mathrm{cm}^{3}
$$

and we observe in particular that the result does not depend on $R$!