Proposition II

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:

Let αβγδ [Fig. 2] be the Fig. 2. largest circle of a sphere and αγ and βδ its diameters perpendicular to each other; let there be in the sphere a circle on the diameter βδ perpendicular to the circle αβγδ, and on this perpendicular circle let there be a cone erected with its vertex at α; producing the convex surface of the cone, let it be cut through γ by a plane parallel to its base; the result will be the circle perpendicular to αγ whose diameter will be ζ. On this circle erect a cylinder whose axis = αγ and whose vertical boundaries are λ and ζη. Produce γα making αθ = γα and think of γθ as a scale-beam with its center at α. Then let μν be any straight line whatever drawn ∥βδ intersecting the circle αβγδ in ξ and o, the diameter αγ in σ, the straight line α in π and αζ in ρ, and on the straight line μν construct a plane perpendicular to αγ; it will intersect the cylinder in a circle on the diameter μν; the sphere αβγδ, in a circle on the diameter ξo; the cone α ζ in a circle on the diameter πρ. Now because γα × ασ = μσ × σπ ( for αγ = σμ, ασ = πσ), and γα×ασ = αξ² = ξσ² + απ² then μσ × σπ = ξσ² + σπ². Moreover, because γα: ασ = μσ: σπ and γα = αθ, therefore θα: ασ = μσ: σπ = μσ²: μσ × σπ. But it has been proved that ξσ² + σπ² = μσ × σπ; hence αθ: ασ = μσ²: ξσ² + σπ². But it is true that μσ²: ξσ² + σπ² = μν²: ξα² + πρ² = the circle in the cylinder whose diameter is μν: the circle in the cone whose diameter is πρ + the circle in the sphere whose diameter is ξo; hence θα: ασ = 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 α with the two circles whose diameters are ξo and πρ, if they are so transferred to θ that θ 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 ξλ ζ, and upon it a plane is erected perpendicular to αγ, the circle produced in the cylinder in its present position will be in equilibrium at the point α 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 θ that θ 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 α with the sphere and the cone together, if they are transferred and so arranged on the scale-beam at the point θ that θ is the center of gravity of both. Now since the bodies we have mentioned are in equilibrium, the cylinder with κ as its center of gravity, the sphere and the cone transferred as we have said so that they have θ as center of gravity, then θα: ακ = cylinder: sphere + cone. But θα = 2ακ, and hence also the cylinder = 2 × (sphere + cone). But it is also true that the cylinder = 3 cones [Euclid, 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 α ζ = 2 spheres. But the cone whose axes form the triangle α ζ = 8 cones whose axes form the triangle αβδ because ζ = 2βδ, hence the aforesaid 8 cones = 2 spheres. Consequently the sphere whose greatest circle is αβγδ is four times as large as the cone with its vertex at α, and whose base is the circle on the diatneter βδ perpendicular to αγ. Draw the straight lines φβχ and ψδω αγ through β and δ in the parallelogram λζ and imagine a cylinder whose bases are the circles on the diameters φψ and χω and whose axis is αγ. Now since the cylinder whose axes form the parallelogram φω is twice as large as the cylinder whose axes form the parallelogram φδ and the latter is three times as large as the cone the triangle of whose axes is αβδ, as is shown in the Elements [Euclid, Elem. XII, 10], the cylinder whose axes form the parallelogram φω is six times as large as the cone whose axes form the triangle αβδ. But it was shown that the sphere whose largest circle is αβγδ 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.