Cavalieri's Principle

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:

Proof
Let $H$ be the shared height of the two figures.

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

Therefore:

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:

And 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:

And thus, we may conclude that $\map V {S_1} = \map V {S_2}$