Proposition VIII

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . produce αγ [Fig. 8] making αθ = αγ and γξ = the radius of the sphere; imagine γθ to be a scale-beam with a center at α, and in the plane cutting off the segment inscribe a circle with its center at η and its radius = αη; on this circle construct a cone with its vertex at α and its lateral boundaries α and αζ. Then draw a straight line κλ ζ; let it cut the circumference of the segment at κ and λ, the lateral boundaries of the cone α ζ at ρ and o and αγ at π. Now because αγ: απ = ακ²: απ² and κα² = απ² + πκ² and απ² = πo² (since also αη² = η² ), then γα: απ = κπ² + πo²: oπ². But κπ² + πo²: πo² = the circle with the diameter κλ + the circle with the diameter oρ: the circle with the diameter oρ and γα = αθ; therefore θα: απ = the circle with the diameter κλ + the circle with the diameter oρ: the circle with the diameter oρ. Now since the circle with the diameter κλ + the circle with the diameter oρ: the circle with the diameter oρ = αθ: πα, let the circle with the diameter oρ be transferred and Fig. 8. so arranged on the scale-beam at θ that θ is its center of gravity; then θα: απ = the circle with the diameter κλ + the circle with the diameter oρ in their present positions: the circle with the diameter oρ if it is transferred and so arranged on the scale-beam at θ that θ is its center of gravity. Therefore the circles in the segment βαδ and in the cone α ζ are in equilibrium at α with that in the cone α ζ. And in the same way all circles in the segment βαδ and in the cone α ζ in their present positions are in equilibrium at the point α with all circles in the cone α ζ if they are transferred and so arranged on the scate-beam at θ that θ is their center of gravity; then also the spherical segment αβδ and the cone α ζ in their present positions are in equilibrium at the point α with the cone αζ if it is transferred and so arranged on the scale-beam at θ that θ is its center of gravity. Let the cyIinder μν equal the cone whose base is the circle with the diameter ζ and whose vertex is at α and let αη be so divided at φ that αη = 4φη; then φ is the center of gravity of the cone αζ as has been previously proved. Moreover let the cylinder μν be so cut by a perpendicularly intersecting plane that the cylinder μ is in equilibrium with the cone αζ. Now since the segment αβδ + the cone αζ in their present positions are in equilibrium at α with the cone αζ if it is transferred and so arranged on the scale-beam at θ that θ is its center of gravity, and cylinder μν = cone αζ and the two cylinders μ + ν are moved to θ and μν is in equilibrium with both bodies, then will also the cylinder ν be in equilibrium with the segment of the sphere at the point α. And since the spherical segment βαδ: the cone whose base is the circle with the diameter βδ, and whose vertex is at α = ξη: ηγ (for this has previously been proved [De sph. et cyl. II, 2 Coroll.]) and cone βαδ: cone αζ = the circle with the diameter βδ: the circle with the diameter ζ = βη²: η², and βη² = γη × ηα, η² = ηα², and γη × ηα: ηα² = γη: ηα, therefore cone βαδ: cone αζ = γη: ηα. But we have shown that cone βαδ: segment βαδ = γη: ηξ, hence δι' ισου segment βαδ: cone αζ = ξη: ηα. And because αχ: χη = ηα + 4ηγ: αη + 2ηγ so inversely ηχ: χα = 2γη + ηα: 4γη + ηα and by addition ηα: αχ = 6γη + 2ηα: ηα + 4ηγ. But ηξ = 1/4 (6ηγ + 2ηα) and γφ = 1/4 (4ηγ +ηα); for that is evident. Hence ηα: αχ = ξη: γφ, consequently also ξη: ηα = γφ: χα. But it was also demonstrated that ξη: ηα = the segment whose vertex is at α and whose base is the circle with the diameter βδ: the cone whose vertex is at α and whose base is the circle with the diameter ζ; hence segment βαδ: cone αζ = γφ: χα. And since the cylinder μ is in equilibrium with the cone αζ at α, and θ is the center of gravity of the cylinder while φ is that of the cone αζ, then cone αζ: cylinder μ = θα: αφ = γα: αφ. But cylinder μν = cone αζ; hence by subtraction, cylinder μ: cylinder ν = αφ: γφ. And cylinder μν = cone αζ; hence cone αζ: cylinder ν = γα: γφ = θα: γφ. But it was also demonstrated that segment βαδ: cone αζ = γφ: χα; hence δι' ισου segment βαδ: cylinder ν = ζα: αχ. And it was demonstrated that segment βαδ is in equilibrium at α with the cylinder ν and θ is the center of gravity of the cylinder ν, consequently the point χ is also the center of gravity of the segment βαδ.