A hole through a ball

Idea. Every horizontal slice of the ring is an annulus whose area does not depend on the radius of the ball. So the volume does not either, and the shortcut of a hole of zero width (a ball of diameter 6) gives it.

Step 1: a slice. Let the ball have radius $R$ and the hole height $h=6$, so the hole has radius $a$ with $a^2=R^2-9$. At height $z$, with $|z|\le 3$, the slice is an annulus between radius $\sqrt{R^2-z^2}$ and $a$, of area $$\pi (R^2-z^2)-\pi (R^2-9)=\pi (9-z^2),$$ with no $R$ left.

Step 2: integrate. $$\begin{array}{rl}V& ={\int }_{-3}^3\pi (9-z^2)dz=\pi \left(54-18\right)\\ & =36\pi \approx \mathrm{113.1.}\end{array}$$

The shortcut. As the hole gets thinner, the ball shrinks towards a ball of diameter 6, whose volume is $\frac{4}{3}\pi \cdot 3^3=36\pi$. In general the ring has volume $\frac{\pi h^3}{6}$, whether it was cut from an orange or from the Earth.