Proposition 5. Theorem.

773. The lateral area of a cone of revolution is equal to the product of the circumference of its base by half its slant height.

Hyp. Let S, C, L denote the lateral area of the cone, the Oce of its base, and its slant height, respectively.

To prove

S = CXL.

S

Proof. Inscribe in the cone a pyramid S-ABCD, having a regular base of any number of sides; and let S' denote its lateral area, C' the perimeter of its base,

B

and L' its slant height.

Then, since the edges of the pyramid are elements of the cone (769), the pyramid is regular.

Now let the number of lateral faces of the inscribed pyramid be indefinitely increased.

C' will approach C as its limit.

(430)

... L' and S' will approach Land S respectively, as their

limits.

Because, however great the number of lateral faces,

S' = C' × L',

... SC XL.

Q.E.D.

774. COR. 1. If R is the radius of the base of a cone of revolution, and T is the total area, we have

S = 2πR × L

=

πRL.

T = 7RL + 7R2 = πR (L+ R).

(436)

775. COR. 2. By the process that was employed in (756) we may show that the lateral areas, or the total areas, of two similar cones of revolution are to each other as the squares of their radii, or of their slant heights, or of their altitudes.

Proposition 6. Theorem.

776. 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.

Hyp. Let S, C, c, L denote the lateral area of the frustum, the Oces of its bases, and its slant height, respectively.

Proof. Inscribe in the frustum of the cone the frustum of a regular pyramid; and let S' denote its lateral area, C' and c' the

perimeters of its lower and upper bases, respectively, and L' its slant height.

Then,

S′ = {(C′ + c')L'.

(629)

Now let the number of lateral faces of the inscribed frustum be indefinitely increased.

C' and c' will approach C and c respectively, as their limits.

(430)

... L' and S' will approach L and S respectively, as their limits.

... S(C + c) L.

Q.E.D.

777. COR. The lateral area of a frustum of a cone of revolution is equal to the circumference of a section equidistant from its bases* multiplied by its slant height.

* Called the mid-section.

Proposition 7. Theorem.

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

Hyp. Let V, B, H denote the volume of

the cone, the area of its base, and its altitude, respectively.

Proof. Inscribe in the cone a pyramid, and let V' and B' denote its volume and the area of its base.

Then,

V'B' x H.

(632)

B

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

B' will approach B as its limit.

.. V' will approach V as its limit.

(430)

Q. E.D.

779. COR. 1. For the volume of a cone of revolution, whose altitude is H and radius of the base is R, we have

780. COR. 2. The volumes of similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases.

(759)

EXERCISES.

1. Required the lateral area and volume of a right circular cone whose altitude is 24 inches and radius of the base 10 inches. Ans. 816.82 sq. ins.; 2513.3 cu. ins. 2. Required the entire surface of a right circular cone whose altitude is 16 inches and radius of the base 12 inches. Ans. 1206.37 sq. ins.

Proposition 8. Theorem.

781. The volume of a frustum of any cone is equal to the sum of the volumes 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.

Hyp. Let V, B, b, H denote the volume

of the frustum, its bases, and its altitude, respectively.

Proof. Inscribe in the frustum of the

cone the frustum of a pyramid, and let

V', B', b' denote its volume and the areas of its bases.

Then,

V' = H(B′ + b′ + √ B′ʊ').

3

(635)

Now let the number of lateral faces of the inscribed frustum be indefinitely increased.

B' and b' will approach B and b respectively, as their limits.

... V' will approach V as its limit.

(430)

Q. E. D.

782. COR. 1. If the frustum is that of a right circular cone, and the radii of its bases are R and r, we have B = πR2, b = πг2.

· · . V = {7H(R2 + y2 + Rr).

783. COR. 2. This formula may be put into the form

NOTE.-The volume of a cask may be found approximately by this formula, in which H = total height of cask, R = radius of mid-section, and r = radius of end.

The nearest approximation in the case of most casks is given by the formula

V = TH[2R2 + r2 — }(R2 — 1·2)].*

*See Rouché et Comberousse, p. 155.

THE SPHERE.

DEFINITIONS.

784. A zone is a portion of the surface of a sphere included between two parallel planes.

The altitude of the zone is the perpendicular distance between the parallel planes.

The bases of the zone are the circumferences of the circles which bound the zone.

If one of the parallel planes touches the sphere, the zone is called a zone of one base.

785. A spherical segment is a portion of the volume of a sphere included between two parallel planes.

The altitude of the segment is the perpendicular distance between the parallel planes.

The bases of the segment are the sections of the sphere made by the parallel planes.

A segment of one base is a segment one of whose bounding planes touches the sphere.

786. A spherical sector is a portion of the volume of a sphere generated by the revolution of a circular sector about a diameter of the circle.

G

D

787. Let the sphere be generated by the revolution of the semicircle ACDEFB about its diameter AB as an axis; and let CG and DH be drawn perpendicular to the axis. The arc CD generates a zone whose altitude is GH, and the figure CDIIG generates a spherical segment whose altitude is GH. The circumferences generated by the points C and

E

B

D are the bases of the zone, and the circles generated by CG and DH are the bases of the segment.