$\lambda\epsilon$, and one-half the spheroid is twice as great as the same cone. In the parallelogram $\lambda\zeta$ draw the straight lines $\phi\chi$ and $\psi\omega \| \alpha\gamma$ through the points $\beta$ and $\delta$ and imagine a cylinder whose bases are the circles on the diameters $\phi\psi$ and $\chi\omega$, and whose axis is $\alpha\gamma$. Now since the cylinder whose axes form the parallelogram $\phi\omega$ is twice as great as the cylinder whose axes form the parallelogram $\phi\delta$ because their bases are equal but the axis of the first is twice as great as the axis of the second, and since the cylinder whose axes form the parallelogram $\phi\delta$ is three times as great as the cone whose vertex is at $\alpha$ and whose base is the circle on the diameter $\beta\delta$ perpendicular to $\alpha\gamma$, then the cylinder whose axes form the parallelogram $\phi\omega$ is six times as great as the aforesaid cone. But it has been shown that the spheroid is four times as great as the same cone, hence the cylinder is one and one half times as great as the spheroid. Q. E. D. \section*{Proposition IV} That a segment of a right conoid cut by a plane perpendicular to its axis is one and one half times as great as the cone having the same base and axis as the segment, can be proved by the same method in the following way: Let a right conoid be cut through its axis by a plane intersecting the surface in a parabola $\alpha\beta\gamma$ [Fig.~4]; let it be also cut by another plane perpendicular to the axis, and let their common line of intersection be $\beta\gamma$. Let the axis of the segment be $\delta\alpha$ and let it be produced to $\theta$ so that $\theta\alpha = \alpha\delta$. Now imagine $\delta\theta$ to be a scale-beam with its center at $\alpha$; let the base of the segment be the circle on the diameter $\beta\gamma$ perpendicular to $\alpha\delta$; imagine a cone whose base is the circle on the diameter $\beta\gamma$, and whose vertex is at $\alpha$. Imagine also a cylinder whose base is the circle on the diameter $\beta\gamma$ and its axis $\alpha\delta$, and in the parallelogram let a straight line $\mu\nu$ be drawn $\| \beta\gamma$ and on $\mu\nu$ construct a plane perpendicular to $\alpha\delta$; it will intersect the cylinder in a circle whose diameter is $\mu\nu$, and the segment of the right conoid in a circle whose diameter is $\xi o$. Now since $\beta\alpha\gamma$ is a parabola, $\alpha\delta$ its diameter and $\xi\sigma$ and $\beta\delta$ its ordinates, then [\emph{Quadr. parab.} 3] $\delta\alpha : \alpha\sigma = \beta\delta^2 : \xi\sigma^2$. But $\delta\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma^2 : \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2$ = the circle in the cylinder whose diameter is $\mu\nu$ : the circle in the segment of the right conoid whose diameter is $\xi o$, hence $\theta\alpha : \alpha\sigma$ = the circle with the diameter $\mu\nu$