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 αβγ [Fig. 4]; let it be also cut by another plane perpendicular to the axis, and let their common line of intersection be βγ. Let the axis of the segment be δα and let it be produced to θ so that θα = αδ. Now imagine δθ to be a scale-beam with its center at α; let the base of the segment be the circle on the diameter βγ perpendicular to αδ; imagine a cone whose base is the circle on the diameter βγ, and whose vertex is at α. Imagine also a cylinder whose base is the circle on the diameter βγ and its axis αδ, and in the parallelogram let a straight line μν be drawn βγ and on μν construct a plane perpendicular to αδ; it will intersect the cylinder in a circle whose diameter is μν, and the segment of the right conoid in a circle whose diameter is ξo. Now since βαγ is a parabola, αδ its diameter and ξσ and βδ its ordinates, then [Quadr. parab. 3] δα: ασ = βδ²: ξσ². But δα = αθ, therefore θα: ασ = μσ²: σξ². But μσ²: σξ² = the circle in the cylinder whose diameter is μν: the circle in the segment of the right conoid whose diameter is ξo, hence θα: ασ = the circle with the diameter μν: the circle with the diameter ξo; therefore the circle in the cylinder whose diameter is μν is in its present position, in equilibrium at the point α with the circle whose diameter is ξo if this be transferred and so arranged on the scale-beam at θ that θ is its center of gravity. And the center of gravity of the circle whose diameter is μν is at σ, that of the circle whose diameter is ξo when its position is changed, is θ, and we have the inverse proportion, θα: ασ = the circle with the diameter μν: the circle with the diameter ξo. In the same way it can be shown that if another straight line be drawn in the Fig. 4. parallelogram γ βγ the circle formed in the cylinder, will in its present position be in equilibrium at the point α with that formed in the segment of the right conoid if the latter is so transferred to θ on the scale-beam that θ 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 α with the segment of the right conoid if the latter is transferred and so arranged on the scale-beam at θ that θ is its center of gravity. And since these magnitudes are in equilibrium at α, and κ is the center of gravity of the cylinder, if αδ is bisected at κ and θ is the center of gravity of the segment transferred to that point, then we have the inverse proportion θα: ακ = cylinder: segment. But θα = 2ακ and also the cylinder = 2 × segment. But the same cylinder is 3 times as great as the cone whose base is the circle on the diameter βγ and whose vertex is at α; therefore it is clear that the segment is one and one half times as great as the same cone.