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 αβγδ [Fig. 3] whose diameters are αγ and βδ and whose center is κ and let there be a circle in the spheroid on the diameter βδ perpendicular to αγ; then imagine a cone whose base is the same circle but whose vertex is at α, and producing its surface, let the cone be cut by a plane through γ parallel to the base; the intersection will be a circle perpendicular to αγ with ζ as its diameter. Now imagine a cylinder whose base is the same circle with the diameter ζ and whose axis is αγ; let γα be produced so that αθ = γα; think of θγ as a scale-beam with its center at α and in the parallelogram λθ draw a straight line μν ζ, and on μν construct a plane perpendicular to αγ; this will intersect the cylinder in a circle whose diameter is μν, the spheroid in a circle whose diameter is ξo and the cone in a circle whose diameter is πρ. Because γα: ασ = α: απ = μσ: σπ, and γα = αθ, therefore θα: ασ = μσ: σπ. But μσ: σπ = μσ²: μσ × σπ and μσ × σπ = πσ² + σξ², for ασ × σγ: σξ² = ακ × κγ: κβ² = ακ²: κβ² (for both ratios are equal to the ratio between the diameter and the parameter [Apollonius, Con. I, 21]) = ασ²: σπ² therefore ασ²: ασ × σγ = πσ²: σξ² = σπ²: σπ × πμ, consequently μπ × πσ = σξ². If πσ² is added to both sides then μσ × σπ = πσ² + σξ². Therefore θα: ασ = μσ²: πσ² + σξ². But μσ²: σξ² + σπ² = the circle in the cylinder whose diameter is μν: the circle with the diameter ξo + the circle with the diameter πρ; hence the circle whose diameter is μν will in its present position be in equilibrium at the point α with the two circles whose diameters are ξo and πρ when they are transferred and so arranged on the scale-beam at the point α that θ is the center of gravity of both; and θ is the center of gravity of the two circles combined whose diameters are ξo and πρ when their position is changed, hence θα: ασ = the circle with the diameter μν: the two circles whose diameters are ξo and πρ. In the same way it can be shown that if another straight line is drawn in the parallelogram λζ ζ and on this line last drawn a plane is constructed perpendicular to αγ, then likewise the circle produced in the cylinder will in its present position be in equilibrium at the point α 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 θ on the scale-beam that θ is the center of gravity of both. Then if Fig. 3. cylinder, spheroid and cone are filled with such circles, the cylinder in its present position will be in equilibrium at the point α with the spheroid + the cone if they are transferred and so arranged on the scale-beam at the point α that θ is the center of gravity of both. Now κ is the center of gravity of the cylinder, but θ, as has been said, is the center of gravity of the spheroid and cone together. Therefore θα: ακ = cylinder: spheroid + cone. But αθ = 2ακ, hence also the cylinder = 2 × (spheroid + cone) = 2 × spheroid + 2 × cone. But the cylinder = 3 × cone, hence 3 × cone = 2 × cone + 2 × spheroid. Subtract 2 × cone from both sides; then a cone whose axes form the triangle α ζ = 2 × spheroid.

But the same cone = 8 cones whose axes form the ∆αβδ; hence 8 such cones = 2 × spheroid, 4 × cone = spheroid; whence it follows that a spheroid is four times as great as a cone whose vertex is at α, and whose base is the circle on the diameter βδ perpendicular to λ, and one-half the spheroid is twice as great as the same cone.

In the parallelogram λζ draw the straight lines φχ and ψω αγ through the points β and δ 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 great as the cylinder whose axes form the parallelogram φδ 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 φδ is three times as great as the cone whose vertex is at α and whose base is the circle on the diameter βδ perpendicular to αγ, then the cylinder whose axes form the parallelogram φω 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.