962. The lateral area of a frustum of a cone of revolu tion is equal to half the sum of the circumferences of its bases multiplied by its slant height.

Denote the circumferences of its bases by C and c, its slant height by L, and its lateral area by S.

Circumscribe about the frustum a frustum of a regular pyramid.

Denote the perimeters of its bases by C' and c', and its lateral area by S'. Its slant height will also be L.

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Now let the number of lateral faces of the frustum of a regular pyramid be indefinitely increased.

Then

S' approaches S as a limit,

C'+' approaches C+c as a limit,

(C'+c)× L approaches (C+c) x L as a limit.

and

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§ 189

186 Q. E. D.

963. COR. I. If R and r are the radii of the bases, L the slant height, and S the lateral area of a frustum of a cone of revolution,

S=(2πR+2πr) x L =π(R+r) x L.

964. COR. II. The last formula may be written

If K is the radius of a section half-way between the bases of the frustum,

That is, the lateral area of a frustum of a cone of revolu tion is equal to the circumference of a section half-way between its bases multiplied by its slant height.

965. The volume of a cone is equal to one-third the product of its base and altitude.

GIVEN any cone, of which B is the base, H the altitude, and V the volume.

Circumscribe about the cone a pyramid. Denote its base by B', and its volume by V'. Its altitude is H.

966. COR. I. If the base of the cone is a circle of radius R,

V=TRH.

967. COR. II. The volumes of two similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases.

Hint. The method of proof is the same as that followed in § 942.

968. Def.-The altitude of a frustum of a cone is the perpendicular distance between its bases.

PROPOSITION VIII. THEOREM

969. A frustum of a cone is equivalent to the sum of three cones whose common altitude is the altitude of the frustum and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum.

GIVEN-a frustum of a cone. Denote its bases by B and b, its altitude by h, and its volume by V.

TO PROVE—V=}h(B+b+√Bxb), which is the algebraic statement of the theorem.

Circumscribe about the frustum of a cone a frustum of a pyramid. Denote its bases by B' and b', and its volume by V'.

Its altitude will be h.

Hence

V' = {h(B' + b' + √B' x b').

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Now let the number of lateral faces of the frustum of a

and (B'+b'+VB'x b') approaches (B+b+ √Bxb). Therefore V=3h(B+b+ √Bxb).

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Q. E. D.

970. COR. If the frustum is the frustum of a circular

cone, let R and r be the radii of its bases.

THE SPHERE

971. Defs.-A zone is a portion of the surface of a sphere bounded by the circumferences of two circles whose planes are parallel.

The bounding circumferences are called the bases, and the perpendicular distance between their planes the altitude of the zone.

972. Def.--If the plane of one bounding circumference is tangent to the sphere, the zone is called a zone of one base.

973. Defs.-A spherical segment is a portion of a sphere contained between two parallel planes.

The bounding circles are called the bases, and the perpendicular distance between their planes the altitude of the segment.

974. Def.-A spherical segment of one base is a spherical segment one of whose bounding planes is tangent to the sphere.

The curved surface of a spherical segment is a zone.