Dem.—Let ABCD be a rectangle whose diagonal is AC. The triangle ABC will describe a cone, and the rectangle a cylinder by revolving round AB. Take two points E, F infinitely near each other in AC, and form two rectangles, EH, EK, by drawing lines parallel to AD, AB. Now if O, O′ be the middle points of these rectangles, it is evident that, when the whole figure revolves round AB, the circumference of the circle described by O′ will ultimately be twice the circumference of the circle described by O; and since the parallelogram EK is equal to EH [I. xliii.], the solid described by EK (Cor. 1) will be equal to twice the solid described by EH. Hence, if AC be divided into an indefinite number of equal parts, and rectangles corresponding to EH, EK be inscribed in the triangles ABC, ADC, the sum of the solids described by the rectangles in the triangle ADC is equal to twice the sum of the solids described by the rectangles in the triangle ABC—that is, the difference between the cylinder and cone is equal to twice the cone. Hence the cylinder is equal to three times the cone. Or thus: We may regard the cone and the cylinder as limiting cases of a pyramid and prism having the same base and altitude; and since (v. Cor. 2) the volume of a pyramid is one-third of the volume of a prism, having the same base and altitude, the volume of the cone is one-third of the volume of the cylinder. Cor. 4.—If r be the radius of the base of a cone, and h its height, 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, PROP. VII.—Theorem. The surface of a sphere is equal to the convex surface of the circumscribed cylinder.