Cavalieri's Principle

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Theorem

Let two solid figures $S_1$ and $S_2$ have equal height.

Let sections made by planes parallel to their bases and at equal distances from the bases always have equal area.


Then the volumes of $S_1$ and $S_2$ are equal.


Extension

An extension of Cavalieri's Principle is as follows:


Let two solid figures $S_1$ and $S_2$ have equal height.

Let the areas of the sections made by planes parallel to their bases and at equal distances from the bases always have the same ratio.


Then the volumes of $S_1$ and $S_2$ are in that same ratio.


Proof

Let $H$ be the common height of the two figures.

The volume of a solid figure is its Lebesgue Measure in $\R^3$.


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Therefore:

\(\ds \map V {S_1}\) \(=\) \(\ds \map {\lambda^3} {S_1}\)
\(\ds \) \(=\) \(\ds \int_{\R^3} \chi_{S_1} \rd \lambda^3\) Definition of Integral of Positive Simple Function
\(\ds \) \(=\) \(\ds \int_{\R \times \R^2} \chi_{S_1} \rd \paren {\lambda \times \lambda^2}\)
\(\ds \) \(=\) \(\ds \int_\R \int_{\R^2} \paren {\chi_{S_1} }_x \rd \lambda^2 \rd \lambda\) Fubini's Theorem
\(\ds \) \(=\) \(\ds \int_{\closedint 0 H} \int_{\R^2} \paren {\chi_{S_1} }_x \rd \lambda^2 \rd \lambda\) Figure is bounded between planes $x = 0$ and $x = H$


This article, or a section of it, needs explaining.
In particular: Lebesgue measure isn't a product measure. We need a variant of Fubini that works on completions of measure spaces.
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In the same manner:

$\ds \map V {S_2} = \int_{\closedint 0 H} \int_{\R^2} \paren {\chi_{S_2} }_x \rd \lambda^2 \rd \lambda$


The area of a plane figure is its Lebesgue Measure in $\R^2$.

The area of a section made in $S_1$ by the plane $x = x_0$ is:

\(\ds \map A {S_1 \cap P_{x \mathop = x_0} }\) \(=\) \(\ds \map {\lambda^2} {S_1 \cap P {x = x_0} }\)
\(\ds \) \(=\) \(\ds \int_{\R^2} \paren {\chi_{S_1} }_{x \mathop = x_0} \rd \lambda^2\) Definition of Integral of Positive Simple Function

Likewise:

$\ds \map A {S_2 \cap P_{x \mathop = x_0} } = \int_{\R^2} \paren {\chi_{S_2} }_{x \mathop = x_0} \rd \lambda^2$


But by assumption:

$\map A {S_1 \cap P_{x \mathop = x_0} } = \map A {S_2 \cap P_{x \mathop = x_0} }$

for every $x_0 \in \closedint 0 H$.

Therefore:

\(\ds \leadsto \ \ \) \(\ds \int_{\R^2} \paren {\chi_{S_1} }_x \rd \lambda^2\) \(=\) \(\ds \int_{\R^2} \paren {\chi_{S_2} }_x \rd \lambda^2\)
\(\ds \int_{\closedint 0 H} \int_{\R^2} \paren {\chi_{S_1} }_x \rd \lambda^2 \rd \lambda\) \(=\) \(\ds \int_{\closedint 0 H} \int_{\R^2} \paren {\chi_{S_2} }_x \rd \lambda^2 \rd \lambda\)

Thus we conclude that $\map V {S_1} = \map V {S_2}$

$\blacksquare$


Source of Name

This entry was named for Bonaventura Francesco Cavalieri.


Sources

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