Part 3

Let a sphere be cut by a plane through its center intersecting the surface in the circle $\alpha\beta\gamma\delta$ [Fig.6], $\alpha\gamma$ and $\beta\delta$ being two diameters of the circle perpendicular to each other. Let a plane be constructed on $\beta\delta$ perpendicular to $\alpha\gamma$. Then imagine a cone whose base is the circle with the diameter $\beta\delta$, whose vertex is at $\alpha$ and its lateral boundaries are $\beta\alpha$ and $\alpha\delta$; let $\gamma\alpha$ be produced so that $\alpha\theta = \gamma\alpha$, imagine the straight line $\theta\gamma$ to be a scale-beam with its center at $\alpha$ and in the semi-circle $\beta\alpha\delta$ draw a straight line $\xi o \| \beta\delta$; let it cut the circumference of the semicircle in $\xi$ and $o$, the lateral boundaries of the cone in $\pi$ and $\rho$, and $\alpha\gamma$ in $\epsilon$. On $\xi o$ construct a plane perpendicular to $\alpha\epsilon$; it will intersect the hemisphere in a circle with the diameter $\xi o$, and the cone in a circle with the diameter $\pi\rho$. Now because $\alpha\gamma : \alpha\epsilon = \xi\alpha^2 : \alpha\epsilon^2$ and $\xi\alpha^2 = \alpha\epsilon^2 + \epsilon\xi^2$ and $\alpha\epsilon = \epsilon\pi$, therefore $\alpha\gamma : \alpha\epsilon = \xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$. But $\xi\epsilon^2 + \epsilon\pi^2 : \epsilon\pi^2$ = the circle with the diameter $\xi o$ + the circle with the diameter $\pi\rho$ : the circle with the diameter $\pi\rho$, and $\gamma\alpha = \alpha\theta$, hence $\theta\alpha : \alpha\epsilon$ = the circle with the diameter $\xi o$ + the circle with % \begin{wrapfigure}{r}{0.33\textwidth} \includegraphics[width=0.33\textwidth]{fig06.png} \begin{center} {\small Fig.~6.} \end{center} \end{wrapfigure} % the diameter $\pi\rho$ : circle with the diameter $\pi\rho$. Therefore the two circles whose diameters are $\xi o$ and $\pi\rho$ in their present position are in equilibrium at the point $\alpha$ with the circle whose diameter is $\pi\rho$ if it is transferred and so arranged at $\theta$ that $\theta$ is its center of gravity. Now since the center of gravity of the two circles whose diameters are $\xi o$ and $\pi\rho$ in their present position [is the point $\epsilon$, but of the circle whose diameter is $\pi\rho$ when its position is changed is the point $\theta$, then $\theta\alpha : \alpha\epsilon$ = the circles whose diameters are] $\xi o$ [,$\pi\rho$ : the circle whose diameter is $\pi\rho$. In the same way if another straight line in the] hemisphere $\beta\alpha\delta$ [is drawn $\| \beta\delta$ and a plane is constructed] perpendicular to [$\alpha\gamma$ the] two [circles produced in the cone and in the hemisphere are in their position] in equilibrium at $\alpha$ [with the circle which is produced in the cone] if it is transferred and arranged on the scale at $\theta$. [Now if] the hemisphere and the cone [are filled up with circles then all circles in the] hemisphere 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}

\dotfill\linebreak produce $\alpha\gamma$ [Fig.~8] making $\alpha\theta = \alpha\gamma$ and $\gamma\xi$ = the radius of the sphere; imagine $\gamma\theta$ to be a scale-beam with a center at $\alpha$, and in the plane cutting % \begin{wrapfigure}{r}{0.4\textwidth} \includegraphics[width=0.4\textwidth]{fig08.png} \begin{center} {\small Fig.~8.} \end{center} \end{wrapfigure} % off the segment inscribe a circle with its center at $\eta$ and its radius = $\alpha\eta$; on this circle construct a cone with its vertex at $\alpha$ and its lateral boundaries $\alpha\epsilon$ and $\alpha\zeta$. Then draw a straight line $\kappa\lambda \| \epsilon\zeta$; let it cut the circumference of the segment at $\kappa$ and $\lambda$, the lateral boundaries of the cone $\alpha\epsilon\zeta$ at $\rho$ and $o$ and $\alpha\gamma$ at $\pi$. Now because $\alpha\gamma : \alpha\pi = \alpha\kappa^2 : \alpha\pi^2$ and $\kappa\alpha^2 = \alpha\pi^2 + \pi\kappa^2$ and $\alpha\pi^2 = \pi o^2$ (since also $\alpha\eta^2 = \epsilon\eta^2$), then $\gamma\alpha : \alpha\pi = \kappa\pi^2 + \pi o^2 : o\pi^2$. But $\kappa\pi^2 + \pi o^2 : \pi o^2$ = the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle with the diameter $o\rho$ and $\gamma\alpha = \alpha\theta$; therefore $\theta\alpha : \alpha\pi$ = the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle with the diameter $o\rho$. Now since the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ : the circle with the diameter $o\rho$ = $\alpha\theta : \pi\alpha$, let the circle with the diameter $o\rho$ be transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity; then $\theta\alpha : \alpha\pi$ = the circle with the diameter $\kappa\lambda$ + the circle with the diameter $o\rho$ in their present positions : the circle with the diameter $o\rho$ if it is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. Therefore the circles in the segment $\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ are in equilibrium at $\alpha$ with that in the cone $\alpha\epsilon\zeta$. And in the same way all circles in the segment $\beta\alpha\delta$ and in the cone $\alpha\epsilon\zeta$ in their present positions are in equilibrium at the point $\alpha$ with all circles in the cone $\alpha\epsilon\zeta$ if they are transferred and so arranged on the scate-beam at $\theta$ that $\theta$ is their center of gravity; then also the spherical segment $\alpha\beta\delta$ and the cone $\alpha\epsilon\zeta$ in their present positions are in equilibrium at the point $\alpha$ with the cone $\epsilon\alpha\zeta$ if it is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. Let the cyIinder $\mu\nu$ equal the cone whose base is the circle with the diameter $\epsilon\zeta$ and whose vertex is at $\alpha$ and let $\alpha\eta$ be so divided at $\phi$ that $\alpha\eta = 4\phi\eta$; then $\phi$ is the center of gravity of the cone $\epsilon\alpha\zeta$ as has been previously proved. Moreover let the cylinder $\mu\nu$ be so cut by a perpendicularly intersecting plane that the cylinder $\mu$ is in equilibrium with the cone $\epsilon\alpha\zeta$. Now since the segment $\alpha\beta\delta$ + the cone $\epsilon\alpha\zeta$ in their present positions are in equilibrium at $\alpha$ with the cone $\epsilon\alpha\zeta$ if it is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity, and cylinder $\mu\nu$ = cone $\epsilon\alpha\zeta$ and the two cylinders $\mu + \nu$ are moved to $\theta$ and $\mu\nu$ is in equilibrium with both bodies, then will also the cylinder $\nu$ be in equilibrium with the segment of the sphere at the point $\alpha$. And since the spherical segment $\beta\alpha\delta$ : the cone whose base is the circle with the diameter $\beta\delta$, and whose vertex is at $\alpha = \xi\eta : \eta\gamma$ (for this has previously been proved [\emph{De sph. et cyl.} II, 2 Coroll.]) and cone $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = the circle with the diameter $\beta\delta$ : the circle with the diameter $\epsilon\zeta = \beta\eta^2 : \eta\epsilon^2$, and $\beta\eta^2 = \gamma\eta \times \eta\alpha$, $\eta\epsilon^2 = \eta\alpha^2$, and $\gamma\eta \times \eta\alpha : \eta\alpha^2 = \gamma\eta : \eta\alpha$, therefore cone $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta = \gamma\eta : \eta\alpha$. But we have shown that cone $\beta\alpha\delta$ : segment $\beta\alpha\delta$ = $\gamma\eta : \eta\xi$, hence {\selectlanguage{greek}di' \~isou} segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\xi\eta : \eta\alpha$. And because $\alpha\chi : \chi\eta = \eta\alpha + 4\eta\gamma : \alpha\eta + 2\eta\gamma$ so inversely $\eta\chi : \chi\alpha = 2\gamma\eta + \eta\alpha : 4\gamma\eta + \eta\alpha$ and by addition $\eta\alpha : \alpha\chi = 6\gamma\eta + 2\eta\alpha : \eta\alpha + 4\eta\gamma$. But $\eta\xi = \frac{1}{4} (6\eta\gamma + 2\eta\alpha)$ and $\gamma\phi = \frac{1}{4} (4\eta\gamma + \eta\alpha)$; for that is evident. Hence $\eta\alpha : \alpha\chi = \xi\eta : \gamma\phi$, consequently also $\xi\eta : \eta\alpha = \gamma\phi : \chi\alpha$. But it was also demonstrated that $\xi\eta : \eta\alpha$ = the segment whose vertex is at $\alpha$ and whose base is the circle with the diameter $\beta\delta$ : the cone whose vertex is at $\alpha$ and whose base is the circle with the diameter $\epsilon\zeta$; hence segment $\beta\alpha\delta$ : cone $\epsilon\alpha\zeta$ = $\gamma\phi : \chi\alpha$. And since the cylinder $\mu$ is in equilibrium with the cone $\epsilon\alpha\zeta$ at $\alpha$, and $\theta$ is the center of gravity of the cylinder while $\phi$ is that of the cone $\epsilon\alpha\zeta$, then cone $\epsilon\alpha\zeta$ : cylinder $\mu$ = $\theta\alpha : \alpha\phi = \gamma\alpha : \alpha\phi$. 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}

Imagine a perpendicular prism with square bases and a cylinder inscribed in the prism in the way we have described. Let the prism be cut through the axis by a plane perpendicular to the plane which cuts off the section of the cylinder; this will intersect the prism containing the cylinder in the parallelogram $\alpha\beta$ [Fig.~9] and the common intersecting line of the plane which cuts off the section of the cylinder and the plane lying through the axis perpendicular to the one cutting off the section of the cylinder will be $\beta\gamma$; let the axis of the cylinder and the prism be $\gamma\delta$ which is bisected at right angles by $\epsilon\zeta$ and on $\epsilon\zeta$ let a plane be constructed perpendicular to $\gamma\delta$. This will intersect the prism in a square and the cylinder in a circle.

Now let the intersection of the prism be the square $\mu\nu$ [Fig.~10], that of the cylinder, the circle $\xi o\pi\rho$ and let the circle touch the sides of the square at the points $\xi$, $o$, $\pi$ and $\rho$; let the common line of intersection of the plane cutting off the cylinder-section and that passing through $\epsilon\zeta$ perpendicular to the axis of the cylinder, be $\kappa\lambda$; this line is bisected by $\pi\theta\xi$. In the semicircle $o\pi\rho$ draw a straight line $\sigma\tau$ perpendicular to $\pi\chi$, on $\sigma\tau$ construct a plane perpendicular to $\xi\pi$ and produce it to both sides of the plane enclosing the circle $\xi o\pi\rho$; this will intersect the half-cylinder whose base is the semicircle $o\pi\rho$ and whose altitude is the axis of the prism, in a parallelogram one side of which = $\sigma\tau$ and the other = the vertical boundary of the cylinder, and it will intersect the cylinder-section likewise in a parallelogram of which one side is $\sigma\tau$ and the other $\mu\nu$ [Fig.~9]; and accordingly $\mu\nu$ will be drawn in the parallelogram $\delta\epsilon \| \beta\omega$ and will cut off $\epsilon\iota = \pi\chi$. Now because $\epsilon\gamma$ is a parallelogram and $\nu\iota \| \theta\gamma$, and $\epsilon\theta$ and $\beta\gamma$ cut the parallels, therefore $\epsilon\theta : \theta\iota = \omega\gamma : \gamma\nu = \beta\omega : \upsilon\nu$. But $\beta\omega : \upsilon\nu$ = parallelogram in the half-cylinder : parallelogram in the cylinder-section, therefore both parallelograms have the same side $\sigma\tau$; and $\epsilon\theta = \theta\pi$, $\iota\theta = \chi\theta$; and since $\pi\theta = \theta\xi$ therefore $\theta\xi : \theta\chi$ = parallelogram in half-cylinder : parallelogram in the cylinder-section. lmagine the parallelogram in the cylinder-section transferred and so brought to $\xi$ that $\xi$ is its center of gravity, and further imagine $\pi\xi$ to be a scale-beam with its center at $\theta$; then the parallelogram in the half-cylinder in its present position is in equilibrium at the point $\theta$ with the parallelogram in the cylinder-section when it is transferred and so arranged on the scale-beam at $\xi$ that $\xi$ is its center of gravity. And since $\chi$ is the center of gravity in the parallelogram in the half-cylinder, and $\xi$ that of the parallelogram in the cylinder-section when its position is changed, and $\xi\theta : \theta\chi$ = the parallelogram whose center of gravity is $\chi$ : the parallelogram whose center of gravity is $\xi$, then the parallelogram whose center of gravity is $\chi$ will be in equilibrium at $\theta$ with the parallelogram whose center of gravity is $\xi$. In this way it can be proved that if another straight line is drawn in the semicircle $o\pi\rho$ perpendicular to $\pi\theta$ and on this straight line a plane is constructed perpendicular to $\pi\theta$ and is produced towards both sides of the plane in which the circle $\xi o\pi\rho$ lies, then the parallelogram formed in the half-cylinder in its present position will be in equilibrium at the point $\theta$ with the parallelogram formed in the cylinder-section if this is transferred and so arranged on the scale-beam at $\xi$ that $\xi$ is its center of-gravity; therefore also all parallelograms in the half-cylinder in their present positions will be in equilibrium at the point $\theta$ with all parallelograms of the cylinder-section if they are transferred and attached to the scale-beam at the point $\xi$; consequently also the half-cylinder in its present position will be in equilibrium at the point $\theta$ with the cylinder-section if it is transferred and so arranged on the scale-beam at $\xi$ that $\xi$ is its center of gravity.

\section*{Proposition XII}

Let the parallelogram $\mu\nu$ be perpendicular to the axis [of the circle] $\xi o$ [$\pi\rho$] [Fig.~11]. Draw $\theta\mu$ and $\theta\eta$ and erect upon them two planes perpendicular to the plane in which the semicircle $o\pi\rho$ lies and extend these planes on both sides. The result is a prism whose base is a triangle similar to $\theta\mu\eta$ and whose altitude is equal to the axis of the cylinder, and this prism is $\frac{1}{4}$ of the entire prism which contains the cylinder. In the semicircle $o\pi\rho$ and in the square $\mu\nu$ draw two straight lines $\kappa\lambda$ and $\tau\upsilon$ at equal distances from $\pi\xi$; these will cut % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig11.png} \begin{center} {\small Fig.~11.} \end{center} \end{wrapfigure} % the circumference of the semicircle $o\pi\rho$ at the points $\kappa$ and $\tau$, the diameter $o\rho$ at $\sigma$ and $\zeta$ and the straight lines $\theta\eta$ and $\theta\mu$ at $\phi$ and $\chi$. Upon $\kappa\lambda$ and $\tau\upsilon$ construct two planes perpendicular to $o\rho$ and extend them towards both sides of the plane in which lies the circle $\xi o\pi\rho$; they will intersect the half-cylinder whose base is the semicircle $o\pi\rho$ and whose altitude is that of the cylinder, in a parallelogram one side of which = $\kappa\sigma$ and the other = the axis of the cylinder; and they will intersect the prism $\theta\eta\mu$ likewise in a parallelogram one side of which is equal to $\lambda\chi$ and the other equal to the axis, and in the same way the half-cylinder in a parallelogram one side of which = $\tau\zeta$ and the other = the axis of the cylinder, and the prism in a parallelogram one side of which = $\nu\phi$ and the other = the axis of the cylinder.\dotfill

\section*{Proposition XIII}