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]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. and [Fig. 7] on μν construct a plane perpendicular to αγ; it will intersect the cylinder in a circle whose diameter is μν, the segment of the sphere in a circle whose diameter is ξo and the cone whose base is the circle on the diameter ζ and whose vertex is at α in a circle whose diameter is πρ. In the same way as before it may be shown that a circle whose diameter is μν is in its present position in equilibrium at α with the two circles [whose diameters are ξo and πρ if they are so arranged on the scale-beam that θ 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 Fig. 7. cylinder in its present position [will be in equilibrium at α] with the cone + the spherical segment if they are transferred and attached to the scale-beam at θ. Divide αη at φ and χ so that αχ = χη and ηφ = 1/3 αφ; then χ will be the center of gravity of the cylinder because it is the center of the axis αη. Now because the above mentioned bodies are in equilibrium at α, cylinder: cone with the diameter of its base ζ + the spherical segment βαδ = θα: αχ. And because ηα = 3ηφ then [γη × ηφ] = 1/3 αη × ηγ. Therefore also γη × ηφ = 1/3 βη². . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .