$\mu$ = $\theta\alpha : \alpha\phi = \gamma\alpha : \alpha\phi$. But cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$; hence by subtraction, cylinder $\mu$ : cylinder $\nu$ = $\alpha\phi : \gamma\phi$. And cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$; hence cone $\epsilon\alpha\zeta$ : cylinder $\nu$ = $\gamma\alpha : \gamma\phi = \theta\alpha : \gamma\phi$. But it was also demonstrated that segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi : \chi\alpha$; hence {\selectlanguage{greek}di' \~isou} segment $\beta\alpha\delta$ : cylinder $\nu$ = $\zeta\alpha : \alpha\chi$. And it was demonstrated that segment $\beta\alpha\delta$ is in equilibrium at $\alpha$ with the cylinder $\nu$ and $\theta$ is the center of gravity of the cylinder $\nu$, consequently the point $\chi$ is also the center of gravity of the segment $\beta\alpha\delta$. \section*{Proposition IX} In a similar way it can also be perceived that the center of gravity of any segment of an ellipsoid lies on the straight line which is the axis of the segment so divided that the portion at the vertex of the segment bears the same ratio to the remaining portion as the axis of the segment + 4 times the axis of the opposite segment bears to the axis of the segment + twice the axis of the opposite segment. \section*{Proposition X} It can also be seen by this method that [a segment of a hyperboloid] bears the same ratio to a cone having the same base and axis as the segment, that the axis of the segment + 3 times the addition to the axis bears to the axis of the segment of the hyperboloid + twice its addition [\emph{De Conoid.} 25]; and that the center of gravity of the hyperboloid so divides the axis that the part at the vertex bears the same ratio to the rest that three times the axis + eight times the addition to the axis bears to the axis of the hyperboloid + 4 times the addition to the axis, and many other points which I will leave aside since the method has been made clear by the examples already given and only the demonstrations of the above given theorems remain to be stated. \section*{Proposition XI} When in a perpendicular prism with square bases a cylinder is inscribed whose bases lie in opposite squares and whose curved surface touches the four other parallelograms, and when a plane is passed through the center of the circle which is the base of the cylinder and one side of the opposite square, then the body which is cut off by this plane [from the cylinder] will be $\frac{1}{6}$ of the entire prism. This can be perceived through the present method and when it is so warranted we will pass over to the geometrical proof of it. \begin{figure}[t] \begin{minipage}[5]{0.45\textwidth} \includegraphics[width=\textwidth]{fig09.png} \begin{center} {\small Fig.~9.} \end{center} \end{minipage} \hfill \begin{minipage}[5]{0.45\textwidth} \includegraphics[width=\textwidth]{fig10.png} \begin{center} {\small Fig.~10.} \end{center} \end{minipage} \end{figure}