Part 2

That a sphere is four times as large as a cone whose base is equal to the largest circle of the sphere and whose altitude is equal to the radius of the sphere, and that a cylinder whose base is equal to the largest circle of the sphere and whose altitude is equal to the diameter of the circle is one and a half times as large as the sphere, may be seen by the present method in the following way:

\begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig02.png} \begin{center} {\small Fig.~2.} \end{center} \end{wrapfigure}

Let $\alpha\beta\gamma\delta$ [Fig.~2] be the largest circle of a sphere and $\alpha\gamma$ and $\beta\delta$ its diameters perpendicular to each other; let there be in the sphere a circle on the diameter $\beta\delta$ perpendicular to the circle $\alpha\beta\gamma\delta$, and on this perpendicular circle let there be a cone erected with its vertex at $\alpha$; producing the convex surface of the cone, let it be cut through $\gamma$ by a plane parallel to its base; the result will be the circle perpendicular to $\alpha\gamma$ whose diameter will be $\epsilon\zeta$. On this circle erect a cylinder whose axis $= \alpha\gamma$ and whose vertical boundaries are $\epsilon\lambda$ and $\zeta\eta$. Produce $\gamma\alpha$ making $\alpha\theta = \gamma\alpha$ and think of $\gamma\theta$ as a scale-beam with its center at $\alpha$. Then let $\mu\nu$ be any straight line whatever drawn $\| \beta\delta$ intersecting the circle $\alpha\beta\gamma\delta$ in $\xi$ and $o$, the diameter $\alpha\gamma$ in $\sigma$, the straight line $\alpha\epsilon$ in $\pi$ and $\alpha\zeta$ in $\rho$, and on the straight line $\mu\nu$ construct a plane perpendicular to $\alpha\gamma$; it will intersect the cylinder in a circle on the diameter $\mu\nu$; the sphere $\alpha\beta\gamma\delta$, in a circle on the diameter $\xi o$; the cone $\alpha\epsilon\zeta$ in a circle on the diameter $\pi\rho$. Now because $\gamma\alpha \times \alpha\sigma = \mu\sigma \times \sigma\pi$ ( for $\alpha\gamma = \sigma\mu$, $\alpha\sigma = \pi\sigma$), and $\gamma\alpha \times \alpha\sigma = \alpha\xi^2 = \xi\sigma^2 + \alpha\pi^2$ then $\mu\sigma \times \sigma\pi = \xi\sigma^2 + \sigma\pi^2$. Moreover, because $\gamma\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$ and $\gamma\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$. But it has been proved that $\xi\sigma^2 + \sigma\pi^2 = \mu\sigma \times \sigma\pi$; hence $\alpha\theta : \alpha\sigma = \mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2$. But it is true that $\mu\sigma^2 : \xi\sigma^2 + \sigma\pi^2 = \mu\nu^2 : \xi\alpha^2 + \pi\rho^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$ the circle in the cone whose diameter is $\pi\rho$ + the circle in the sphere whose diameter is $\xi o$; hence $\theta\alpha : \alpha\sigma =$ the circle in the cylinder $:$ the circle in the sphere $+$ the circle in the cone. Therefore the circle in the cylinder in its present position will be in equilibrium at the point $\alpha$ with the two circles whose diameters are $\xi o$ and $\pi\rho$, if they are so transferred to $\theta$ that $\theta$ is the center of gravity of both. In the same way it can be shown that when another straight line is drawn in the parallelogram $\xi\lambda \| \epsilon\zeta$, and upon it a plane is erected perpendicular to $\alpha\gamma$, the circle produced in the cylinder in its present position will be in equilibrium at the point $\alpha$ with the two circles produced in the sphere and the cone when they are transferred and so arranged on the scale-beam at the point $\theta$ that $\theta$ is the center of gravity of both. Therefore if cylinder, sphere and cone are filled up with such circles then the cylinder in its present position will be in equilibrium at the point $\alpha$ with the sphere and the cone together, if they are transferred and so arranged on the scale-beam at the point $\theta$ that $\theta$ is the center of gravity of both. Now since the bodies we have mentioned are in equilibrium, the cylinder with $\kappa$ as its center of gravity, the sphere and the cone transferred as we have said so that they have $\theta$ as center of gravity, then $\theta\alpha : \alpha\kappa =$ cylinder $:$ sphere $+$ cone. But $\theta\alpha = 2\alpha\kappa$, and hence also the cylinder $= 2 \times$ (sphere $+$ cone). But it is also true that the cylinder = 3 cones [Euclid, \emph{Elem.} XII, 10], hence 3 cones = 2 cones + 2 spheres. If 2 cones be subtracted from both sides, then the cone whose axes form the triangle $\alpha\epsilon\zeta =$ 2 spheres. But the cone whose axes form the triangle $\alpha\epsilon\zeta =$ 8 cones whose axes form the triangle $\alpha\beta\delta$ because $\epsilon\zeta = 2\beta\delta$, hence the aforesaid 8 cones = 2 spheres. Consequently the sphere whose greatest circle is $\alpha\beta\gamma\delta$ is four times as large as the cone with its vertex at $\alpha$, and whose base is the circle on the diatneter $\beta\delta$ perpendicular to $\alpha\gamma$.

Draw the straight lines $\phi\beta\chi$ and $\psi\delta\omega \| \alpha\gamma$ through $\beta$ and $\delta$ in the parallelogram $\lambda\zeta$ 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 large as the cylinder whose axes form the parallelogram $\phi\delta$ and the latter is three times as large as the cone the triangle of whose axes is $\alpha\beta\delta$, as is shown in the Elements [Euclid, \emph{Elem.} XII, 10], the cylinder whose axes form the parallelogram $\phi\omega$ is six times as large as the cone whose axes form the triangle $\alpha\beta\delta$. But it was shown that the sphere whose largest circle is $\alpha\beta\gamma\delta$ is four times as large as the same cone, consequently the cylinder is one and one half times as large as the sphere, Q. E. D.

After I had thus perceived that a sphere is four times as large as the cone whose base is the largest circle of the sphere and whose altitude is equal to its radius, it occurred to me that the surface of a sphere is four times as great as its largest circle, in which I proceeded from the idea that just as a circle is equal to a triangle whose base is the periphery of the circle and whose altitude is equal to its radius, so a sphere is equal to a cone whose base is the same as the surface of the sphere and whose altitude is equal to the radius of the sphere.

\section*{Proposition III}

By this method it may also be seen that a cylinder whose base is equal to the largest circle of a spheroid and whose altitude is equal to the axis of the spheroid, is one and one half times as large as the spheroid, and when this is recognized it becomes clear that if a spheroid is cut through its center by a plane perpendicular to its axis, one-half of the spheroid is twice as great as the cone whose base is that of the segment and its axis the same.

For let a spheroid be cut by a plane through its axis and let there be in its surface an ellipse $\alpha\beta\gamma\delta$ [Fig.~3] whose diameters are $\alpha\gamma$ and $\beta\delta$ and whose center is $\kappa$ and let there be a circle in the spheroid on the diameter $\beta\delta$ perpendicular to $\alpha\gamma$; then imagine a cone whose base is the same circle but whose vertex is at $\alpha$, and producing its surface, let the cone be cut by a plane through $\gamma$ parallel to the base; the intersection will be a circle perpendicular to $\alpha\gamma$ with $\epsilon\zeta$ as its diameter. Now imagine a cylinder whose base is the same circle with the diameter $\epsilon\zeta$ and whose axis is $\alpha\gamma$; let $\gamma\alpha$ be produced so that $\alpha\theta = \gamma\alpha$; think of $\theta\gamma$ as a scale-beam with its center at $\alpha$ and in the parallelogram $\lambda\theta$ draw a straight line $\mu\nu \| \epsilon\zeta$, and on $\mu\nu$ construct a plane perpendicular to $\alpha\gamma$; this will intersect the cylinder in a circle whose diameter is $\mu\nu$, the spheroid in a circle whose diameter is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Because $\gamma\alpha : \alpha\sigma = \epsilon\alpha : \alpha\pi = \mu\sigma : \sigma\pi$, and $\gamma\alpha = \alpha\theta$, therefore $\theta\alpha : \alpha\sigma = \mu\sigma : \sigma\pi$. But $\mu\sigma : \sigma\pi = \mu\sigma^2 : \mu\sigma \times \sigma\pi$ and $\mu\sigma \times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$, for $ \alpha\sigma \times \sigma\gamma : \sigma\xi^2 = \alpha\kappa \times \kappa\gamma : \kappa\beta^2 = \alpha\kappa^2 : \kappa\beta^2$ (for both ratios are equal to the ratio between the diameter and the parameter [Apollonius, \emph{Con.} I, 21]) $ = \alpha\sigma^2 : \sigma\pi^2$ therefore $\alpha\sigma^2 : \alpha\sigma \times \sigma\gamma = \pi\sigma^2 : \sigma\xi^2 = \sigma\pi^2 : \sigma\pi \times \pi\mu$, consequently $\mu\pi \times \pi\sigma = \sigma\xi^2$. If $\pi\sigma^2$ is added to both sides then $\mu\sigma \times \sigma\pi = \pi\sigma^2 + \sigma\xi^2$. Therefore $\theta\alpha : \alpha\sigma = \mu\sigma^2 : \pi\sigma^2 + \sigma\xi^2$. But $\mu\sigma^2 : \sigma\xi^2 + \sigma\pi^2 =$ the circle in the cylinder whose diameter is $\mu\nu :$ the circle with the diameter $\xi o$ + the circle with the diameter $\pi\rho$; hence the circle whose diameter is $\mu\nu$ will in its present position be in equilibrium at the point $\alpha$ with the two circles whose diameters are $\xi o$ and $\pi\rho$ when they are transferred and so arranged on the scale-beam at the point $\alpha$ that $\theta$ is the center of gravity of both; and $\theta$ is the center of gravity of the two circles combined whose diameters are $\xi o$ and $\pi\rho$ when their position is changed, % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig03.png} \begin{center} {\small Fig.~3.} \end{center} \end{wrapfigure} % hence $\theta\alpha : \alpha\sigma =$ the circle with the diameter $\mu\nu :$ the two circles whose diameters are $\xi o$ and $\pi\rho$. In the same way it can be shown that if another straight line is drawn in the parallelogram $\lambda\zeta \| \epsilon\zeta$ and on this line last drawn a plane is constructed perpendicular to $\alpha\gamma$, then likewise the circle produced in the cylinder will in its present position be in equilibrium at the point $\alpha$ with the two circles combined which have been produced in the spheroid and in the cone respectively when they are so transferred to the point $\theta$ on the scale-beam that $\theta$ is the center of gravity of both. Then if cylinder, spheroid and cone are filled with such circles, the cylinder in its present position will be in equilibrium at the point $\alpha$ with the spheroid $+$ the cone if they are transferred and so arranged on the scale-beam at the point $\alpha$ that $\theta$ is the center of gravity of both. Now $\kappa$ is the center of gravity of the cylinder, but $\theta$, as has been said, is the center of gravity of the spheroid and cone together. Therefore $\theta\alpha : \alpha\kappa =$ cylinder $:$ spheroid $+$ cone. But $\alpha\theta = 2\alpha\kappa$, hence also the cylinder = 2 $\times$ (spheroid $+$ cone) = 2 $\times$ spheroid + 2 $\times$ cone. But the cylinder = 3 $\times$ cone, hence 3 $\times$ cone = 2 $\times$ cone + 2 $\times$ spheroid. Subtract 2 $\times$ cone from both sides; then a cone whose axes form the triangle $\alpha\epsilon\zeta$ = 2 $\times$ spheroid. But the same cone = 8 cones whose axes form the $\Delta\alpha\beta\delta$; hence 8 such cones = 2 $\times$ spheroid, 4 $\times$ cone = spheroid; whence it follows that a spheroid is four times as great as a cone whose vertex is at $\alpha$, and whose base is the circle on the diameter $\beta\delta$ perpendicular to $\lambda\epsilon$, 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 $\mu\nu$ : the circle with the diameter $\xi o$; therefore the circle in the cylinder whose % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig04.png} \begin{center} {\small Fig.~4.} \end{center} \end{wrapfigure} % diameter is $\mu\nu$ is in its present position, in equilibrium at the point $\alpha$ with the circle whose diameter is $\xi o$ if this be transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. And the center of gravity of the circle whose diameter is $\mu\nu$ is at $\sigma$, that of the circle whose diameter is $\xi o$ when its position is changed, is $\theta$, and we have the inverse proportion, $\theta\alpha : \alpha\sigma$ = the circle with the diameter $\mu\nu$ : the circle with the diameter $\xi o$. In the same way it can be shown that if another straight line be drawn in the parallelogram $\epsilon\gamma \| \beta\gamma$ the circle formed in the cylinder, will in its present position be in equilibrium at the point $\alpha$ with that formed in the segment of the right conoid if the latter is so transferred to $\theta$ on the scale-beam that $\theta$ is its center of gravity. Therefore if the cylinder and the segment of the right conoid are filled up then the cylinder in its present position will be in equilibrium at the point $\alpha$ with the segment of the right conoid if the latter is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. And since these magnitudes are in equilibrium at $\alpha$, and $\kappa$ is the center of gravity of the cylinder, if $\alpha\delta$ is bisected at $\kappa$ and $\theta$ is the center of gravity of the segment transferred to that point, then we have the inverse proportion $\theta\alpha : \alpha\kappa$ = cylinder : segment. But $\theta\alpha = 2\alpha\kappa$ and also the cylinder = 2 $\times$ segment. But the same cylinder is 3 times as great as the cone whose base is the circle on the diameter $\beta\gamma$ and whose vertex is at $\alpha$; therefore it is clear that the segment is one and one half times as great as the same cone.

\section*{Proposition V}

That the center of gravity of a segment of a right conoid which is cut off by a plane perpendicular to the axis, lies on the straight line which is the axis of the segment divided in such a way that the portion at the vertex is twice as great as the remainder, may be perceived by our method in the following way:

Let a segment of a right conoid cut off by a plane perpendicular to the axis be cut by another plane through the axis, and let the intersection in its surface be the parabola $\alpha\beta\gamma$ [Fig.~5] and let the common line of intersection of the plane which cut off the segment and of the intersecting plane be $\beta\gamma$; let the axis of the segment and the diameter of the parabola $\alpha\beta\gamma$ be $\alpha\delta$; produce $\delta\alpha$ so that $\alpha\theta = \alpha\delta$ and imagine $\delta\theta$ to be a scale-beam with its center at $\alpha$; % \begin{wrapfigure}{r}{0.5\textwidth} \includegraphics[width=0.5\textwidth]{fig05.png} \begin{center} {\small Fig.~5.} \end{center} \end{wrapfigure} % then inscribe a cone in the segment with the lateral boundaries $\beta\alpha$ and $\alpha\gamma$ and in the parabola draw a straight line $\xi o \| \beta\gamma$ and let it cut the parabola in $\xi$ and $o$ and the lateral boundaries of the cone in $\pi$ and $\rho$. Now because $\xi\sigma$ and $\beta\delta$ are drawn perpendicular to the diameter of the parabola, $\delta\alpha : \alpha\sigma = \beta\delta^2 : \xi\sigma^2$ [\emph{Quadr. parab.} 3]. But $\delta\alpha : \alpha\sigma = \beta\delta : \pi\sigma = \beta\delta^2 : \beta\delta \times \pi\sigma$, therefore also $\beta\delta^2 : \xi\sigma^2 = \beta\delta^2 : \beta\delta \times \pi\sigma$. Consequently $\xi\sigma^2 = \beta\delta \times \pi\sigma$ and $\beta\delta : \xi\sigma = \xi\sigma : \pi\sigma$, therefore $\beta\delta : \pi\sigma = \xi\sigma^2 : \sigma\pi^2$. But $\beta\delta : \pi\sigma = \delta\alpha : \alpha\sigma = \theta\alpha : \alpha\sigma$, therefore also $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$. On $\xi o$ construct a plane perpendicular to $\alpha\delta$; this will intersect the segment of the right conoid in a circle whose diameter is $\xi o$ and the cone in a circle whose diameter is $\pi\rho$. Now because $\theta\alpha : \alpha\sigma = \xi\sigma^2 : \sigma\pi^2$ and $\xi\sigma^2 : \sigma\pi^2$ = the circle with the diameter $\xi o$ : the circle with the diameter $\pi\rho$, therefore $\theta\alpha : \alpha\sigma$ = the circle whose diameter is $\xi o$ : the circle whose diameter is $\pi\rho$. Therefore the circle whose diameter is $\xi o$ will in its present position be in equilibrium at the point $\alpha$ with the circle whose diameter is $\pi\rho$ when this is so transferred to $\theta$ on the scale-beam that $\theta$ is its center of gravity. Now since $\sigma$ is the center of gravity of the circle whose diameter is $\xi o$ in its present position, and $\theta$ is the center of gravity of the circle whose diameter is $\pi\rho$ if its position is changed as we have said, and inversely $\theta\alpha : \alpha\sigma$ = the circle with the diameter $\xi o$ : the circle with the diameter $\pi\rho$, then the circles are in equilibrium at the point $\alpha$. In the same way it can be shown that if another straight line is drawn in the parabola $\| \beta\gamma$ and on this line last drawn a plane is constructed perpendicular to $\alpha\delta$, the circle formed in the segment of the right conoid will in its present position be in equilibrium at the point $\alpha$ with the circle formed in the cone, if the latter is transferred and so arranged on the scale-beam at $\theta$ that $\theta$ is its center of gravity. Therefore if the segment and the cone are filled up with circles, all circles in the segment will be in their present positions in equilibrium at the point $\alpha$ with all circles of the cone if the latter are transferred and so arranged on the scale-beam at the point $\theta$ that $\theta$ is their center of gravity. Therefore also the segment of the right conoid in its present position will be 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. Now because the center of gravity of both magnitudes taken together is $\alpha$, but that of the cone alone when its position is changed is $\theta$, then the center of gravity of the remaining magnitude lies on $\alpha\theta$ extended towards $\alpha$ if $\alpha\kappa$ is cut off in such a way that $\alpha\theta : \alpha\kappa$ = segment : cone. But the segment is one and one half the size of the cone, consequently $\alpha\theta = \frac{3}{2}\alpha\kappa$ and $\kappa$, the center of gravity of the right conoid, so divides $\alpha\delta$ that the portion at the vertex of the segment is twice as large as the remainder.

\section*{Proposition VI}

[The center of gravity of a hemisphere is so divided on its axis] that the portion near the surface of the hemisphere is in the ratio of $5 : 3$ to the remaining portion.