PROP. VII.—Theorem.
The surface of a sphere is equal to the convex surface of the circumscribed cylinder.
Dem.—Let AB be the diameter of the semicircle which describes the sphere. Take two points, E, F, indefinitely near each other in the semicircle. Join EF, and produce to meet the tangent CD parallel to AB in N. Draw EI, FK parallel to PQ. Produce EI to meet AB in G. Let O be the centre. Join OE.
| Now we have | FE : KI :: EN : | IN [VI. ii.]; | |||||||||
| but | EN : IN :: | OE : EG, |
because the triangles ENI and OEG are similar.
| Hence | FE : KI :: | OE : EG; | |||||||||
| but | OE = | IG. |
Hence EF : IK :: IG : EG; and IG : EG :: circumference of circle described by the point I : circumference of circle described by the point E. Hence the rectangle contained by EF, and circumference of circle described by E is equal to the rectangle contained by IK, and circumference of circle described by I—that is, the portion of the spherical surface described by EF is equal to the portion of the cylindrical surface described by IK. Hence it is evident, if planes be drawn perpendicular to the diameter AB—that the portions of cylindrical and spherical surface between any two of them are equal. Hence the whole spherical surface is equal to the cylindrical surface described by CD.
Or thus: Conceive the whole surface of the sphere divided into an indefinitely great number of equal parts, then it is evident that each of these may be regarded as the base of a pyramid having the centre of the sphere as a common vertex. Therefore the volume of the sphere is equal to the whole area of the surface multiplied by one-third of the radius. Hence if S denote the surface, we have
| S × | = [vi., Cor. 6]; | ||||||||||
| therefore | S | = 4πr2. |