Cor. 5.—The volume of a sphere is two-thirds of the volume of a circumscribed cylinder.
Dem.—Let AB be the diameter of the semicircle which describes the sphere; ABCD the rectangle which describes the cylinder. Take two points E, F indefinitely near each other in the semicircle. Join EF, which will be a tangent, and produce it to meet the diameter PQ perpendicular to AB in N. Let R be the centre. Join RE; draw EG, FH, NL parallel to AB; and EI, FK parallel to PQ; and produce to meet LN in M and L; and let O, O′ be the middle points of the rectangles EH, EK.
Now the rectangle NG.GR = PG.GQ, because each is equal to GE2. Hence NG : GP :: GQ : GR, or ME : IE :: RP + RG : RG. Now, denoting the radii of the circles described by the points O, O′ by ρ, ρ′ respectively, we have ultimately ρ = GR and ρ′ =
(RP + RG). Hence ME : IE :: 2ρ′ : ρ; but ME : IE :: rectangle EL : rectangle EK :: [I. xliii.] EH : EK;
| ∴ EH : EK :: 2ρ′ : ρ; | |||
| ∴ 2πρ.EH = 2(2πρ′.EK). |
Hence the solid described by EH equal twice the solid described by EK. Therefore we infer, as in the last Cor., that the whole volume of the sphere is equal to twice the difference between the cylinder and sphere. Therefore the sphere is two-thirds of the cylinder.
Cor. 6.—If r be the radius of a sphere,