A Generalized Cavalieri-Zu Principle
Sidney Kung

Volume of a Sphere and Volume of an Ellipsoid

We give two ways to find the volume of a sphere. The first is to take a rectangular prism \(P\) of height equal to the radius \(R\) of a hemisphere \(S\).

Volume of sphere by the Cavalier-Zu generalized principle

A 4-pyramid \(Q\) is inscribed into \(P\). A plane cuts the solids at a distance \(d\) from the base. Denote the cross-section areas of \(S\), \(P\), and \(Q\) by \(A\), \(A_1\), and \(A_2\), respectively. Since \(\frac{A_2}{A_1}=\frac{d^2}{R^{2}}\),

\(A=\pi (R^{2}-d^{2})=\pi R^{2}-\pi A_{2}R^{2}/A_{1}=A_{1}-A{2},\)

so, from (**), the volume of a sphere is


\(V_{S}=2\times (\pi R^{3}- \frac{\pi}{3}R^{2}R)=\frac{4}{3}\pi R^{3}.\)

The second way is to construct a wedge \(W\) whose dimensions are shown below

Volume of sphere via a wedge by the Cavalier-Zu generalized principle

We find the cross-section area of \(W\) a distance \(d\) from the base as follows. Let \(ET\perp ABCD\) with foot \(T\), and let \(TH\perp AD\). Then we have

\(\displaystyle \frac{MK}{AD}=\frac{EI}{EH}=\frac{ES}{ET}=\frac{R-d}{R}\),

or \(MK=R-d\). It's easy to see that \(MN=\pi R +\pi d\). So, \((MK)(MN)=\pi (R^{2}-d^{2})=A\). We know that the volume of a wedge is given by the formula



where \(a\) and \(b\) are base lengths, \(l\) is the top edge length, \(h\) is the height of the wedge. Thus, by letting \(\pi R\), \(b=R\), \(l=2\pi R\), and \(h=R\) in (4') we obtain the volume of a sphere

\(V=2\times \left[\frac{1}{6}R\cdot R(2\pi R+2\pi R)\right]=\frac{4}{3}R^{3}.\)

To find the volume of an ellipsoid, we use a hemisphere instead of a cone [Needham] as a companion solid to a semi-ellipsoid.

Volume of ellipsoid by the Cavalier-Zu generalized principle

\(a_{1}\) and \(c_{1}\) are calculated based on the traces (semi-ellipses) on the \(by\)- and \(Hz\)-planes, respectively. \(a_{1}=\frac{a\sqrt{a^{2}-h^{2}}}{b}\) and \(c_{1}=\frac{ c\sqrt{b^{2}-h^{2}}}{b^{2}}\), \(A=\pi a_{1}c_{1}=\frac{\pi a}{b^{2}}(b^{2}-h^{2}) =\frac{a}{b^{2}}\pi r^{2}=\frac{a}{b^{2}}A'\). Hence, by (**), the volume of an ellipsoid is


\(\displaystyle 2\frac{ac}{b^{2}}(\frac{2}{3}\pi b^{3})=\frac{4}{3}\pi abc.\)


  1. Joseph Needham, Science and Civilization of China, V.3, 1959, Caves Nooks, Taipei, 143.

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