Compute the volume of a sphere with a cylindrical hole of length 2h removed

Given a solid sphere, drill a cylindrical hole of length $2h$ through the center of the sphere. Prove that the volume of the resulting ring is $\pi a h^3$ , where $a \in \mathbb{Q}$ .

Proof. Let $r_s$ denote the radius of the sphere, and $r_c$ denote the radius of the cylindrical hole. Then the volume of the ring is the volume of the solid of revolution formed by rotating the area between the functions

$f(x) = \sqrt{r_s^2 - x^2}, \qquad g(x) = r_c, \qquad -h \leq x \leq h$

about the $x$ -axis. Since the length of the hole is $2h$ , we know $f(h) = g(h)$ ; thus, $r_s^2 - h^2 = r_c^2 \implies r_s^2 - r_c^2 = h^2$ . So, we have,

$\begin{align*} V &= \pi \int_{-h}^h \left( (\sqrt{r_s^2 - x^2})^2 - r_c^2 \right) \, dx \\ &= \pi \int_{-h}^h (r_s^2 - r_c^2 - x^2) \, dx \\ &= \pi h^2 (2h) - pi \left. \frac{x^3}{3} \right|_{-h}^h \\ &= 2 \pi h^3 - \frac{2}{3} \pi h^3 \\ &= \frac{4}{3} \pi h^3 \\ &= \pi a h^3 & \text{where } a = \frac{4}{3} \in \mathbb{Q}. \qquad \blacksquare \end{align*}$

Stumbling Robot

Compute the volume of a sphere with a cylindrical hole of length 2h removed

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