Hemisphere and cylinder via Cavalieri

Problem

A hemisphere and a cylinder both have base radius $R$ and height $R$. Show with Cavalieri that the volume of the hemisphere equals $2/3$ of the volume of the cylinder.

Solution

We compare the hemisphere with the solid “cylinder minus cone” (a cylinder of radius $R$ and height $R$ from which a cone of radius $R$ and height $R$ is removed, with its apex at the bottom). At height $z\in[0,R]$ from the base:

• Cross-section of the hemisphere: disc of radius $\sqrt{R^2-z^2}$, area $\pi(R^2-z^2)$.
• Cross-section of the cylinder minus cone: annulus of outer radius $R$ and inner radius $z$, area $\pi R^2-\pi z^2 = \pi(R^2-z^2)$.

The areas coincide at every height, so by Cavalieri the two solids have the same volume: $V_{\text{semisfera}} = V_{\text{cil}}-V_{\text{cono}} = \pi R^3 - \tfrac13\pi R^3 = \tfrac{2}{3}\pi R^3.$ Since the volume of the cylinder is $V_{\text{cil}} = \pi R^2\cdot R = \pi R^3$, we obtain $\boxed{\dfrac{V_{\text{semisfera}}}{V_{\text{cil}}} = \dfrac{2}{3}.}$

Links

Topics: Integral
Concepts: Cavalieri’s principle · Volume by cross-sections
Methods: Cavalieri’s principle
Skills: Prove · Synthetic geometry
People: Archimedes
Exercise types: Proof