Area and Volume Revisited — Question 5

Question 5 – Solution

Strategy. Use cylindrical shells with $a\le r\le R$; the sphere supplies the upper and lower $z$-bounds.

Step 1: Bounds

The sphere is $r^2+z^2\le R^2$. Hence $$0\le\theta\le 2\pi,\qquad a\le r\le R,\qquad -\sqrt{R^2-r^2}\le z\le\sqrt{R^2-r^2}.$$

Step 2: Evaluate $$\begin{align*} V&=\int_0^{2\pi}\int_a^R 2\sqrt{R^2-r^2}\,r\,dr\,d\theta\\ &=4\pi\left[-\frac{(R^2-r^2)^{3/2}}{3}\right]_a^R =\frac{4\pi}{3}(R^2-a^2)^{3/2}. \end{align*}$$ Since $R^2-a^2=h^2/4$, $$\boxed{V=\frac{\pi h^3}{6}}.$$

Verification The final formula contains only the remaining height. Thus every drilled sphere with the same height $h$ leaves the same volume, regardless of its original $R$ and hole radius $a$.