and those [in the cone] will in their present position be in equilibrium [with all circles] in the cone, if these are transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is their center of gravity; [therefore the hemisphere and cone also] are in their position [in equilibrium at the point $\alpha$] with the cone if it is transferred and so arranged [on the scale-beam at $\theta$] that $\theta$ is its center of gravity. \section*{Proposition VII} By [this method] it may also be perceived that [any segment whatever] of a sphere bears the same ratio to a cone having the same [base] and axis [that the radius of the sphere + the axis of the opposite segment : the axis of the opposite segment] \dotfill and [Fig.~7] on $\mu\nu$ construct a plane perpendicular to $\alpha\gamma$; it will intersect the cylinder in a circle whose diameter is $\mu\nu$, the segment of the sphere in a circle whose diameter % \begin{wrapfigure}{r}{0.6\textwidth} \includegraphics[width=0.6\textwidth]{fig07.png} \begin{center} {\small Fig.~7.} \end{center} \end{wrapfigure} % is $\xi o$ and the cone whose base is the circle on the diameter $\epsilon\zeta$ and whose vertex is at $\alpha$ in a circle whose diameter is $\pi\rho$. In the same way as before it may be shown that a circle whose diameter is $\mu\nu$ is in its present position in equilibrium at $\alpha$ with the two circles [whose diameters are $\xi o$ and $\pi\rho$ if they are so arranged on the scale-beam that $\theta$ is their center of gravity. [And the same can be proved of all corresponding circles.] Now since cylinder, cone, and spherical segment are filled up with such circles, the cylinder in its present position [will be in equilibrium at $\alpha$] with the cone + the spherical segment if they are transferred and attached to the scale-beam at $\theta$. Divide $\alpha\eta$ at $\phi$ and $\chi$ so that $\alpha\chi = \chi\eta$ and $\eta\phi = \frac{1}{3}\alpha\phi$; then $\chi$ will be the center of gravity of the cylinder because it is the center of the axis $\alpha\eta$. Now because the above mentioned bodies are in equilibrium at $\alpha$, cylinder : cone with the diameter of its base $\epsilon\zeta$ + the spherical segment $\beta\alpha\delta = \theta\alpha : \alpha\chi$. And because $\eta\alpha = 3\eta\phi$ then [$\gamma\eta \times \eta\phi$] = $\frac{1}{3}\alpha\eta \times \eta\gamma$. Therefore also $\gamma\eta \times \eta\phi = \linebreak\frac{1}{3}\beta\eta^2$. \dotfill \section*{Proposition VIIa} In the same way it may be perceived that any segment of an ellipsoid cut off by a perpendicular plane, bears the same ratio to a cone having the same base and the same axis, as half of the axis of the ellipsoid + the axis of the opposite segment bears to the axis of the opposite segment. \dotfill \section*{Proposition VIII}