that is, volume generated by the polygon approaches volume spherical sector AOD;

and OA is constant.

Therefore vol. sph. sect.A OD=area zone AD × OA. § 186

987. COR. I. Let H denote the altitude of the zone which forms the base of the spherical sector.

Then

vol. sph. sector=2πRH × R

=TRH.

988. COR. II. The volume of a sphere is equal to the area of its surface multiplied by one-third of its radius.

Hint.-A sphere may be regarded as a spherical sector whose base is the surface of the sphere.

989. COR. III. If V is the volume of a sphere, R its radius, and D its diameter,

V=4πR2 × }R={πR® = {πD3.

990. COR. IV. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters.

991. COR. V. The volume of a spherical pyramid is equal to the area of its base multiplied by one-third the radius of

the sphere.

Hint.-Let v be the volume of the spherical pyramid, s the area of its base, and R the radius of the sphere.

Also let V be the volume of the sphere and S the area of its surface.

992. The volume of the solid generated by a circular seg ment revolving about a diameter exterior to it is equal to onesixth the area of the circle whose radius is the chord of the segment multiplied by the projection of that chord upon the axis.

GIVEN―a circular segment ACB revolving about the diameter HS. Let A'B' be the projection of AB upon HS.

TO PROVE

Draw the radii OA, OB, and draw OM perpendicular to AB.

Then vol. ACB=vol. sector AOB-vol. triangle AOB.

PROPOSITION XIV. THEOREM

993. The volume of a spherical segment is equal to half the sum of its bases multiplied by its altitude increased by the volume of a sphere whose diameter is equal to that altitude.

GIVEN a spherical segment, generated by the revolution of the figure ACBB'A' about the diameter HS of the semicircle HBS, the lines AA' and BB' generating the bases, and the arc ACB generating the curved surface of the segment. Denote BB' by r, AA' by r', A'B' by h, and the volume of the spherical segment by V.

The volume of the spherical segment is the sum of the volume generated by the circular segment ACB and the volume of the frustum of a cone generated by the trapezoid ABB'A'.

Hence VAB2 xh+r+r22+rr'). (1) $$ 992,970 Draw AK perpendicular to BB'.

Then

Hence

Now

BK=r-r'.

BKr2+r"-2rr'.

AB2 = AK2 + BK2 = h2+r2+r'2 - 2rr'.

Substituting this value for AB in (1), we get

V = 1⁄2 (πr2 + πr13)h+}πh3.

8317

Q. E. D.

994. COR. The formula for the volume of a spherical segment of one base is

Hint. This is obtained from the preceding formula by making the radius r' of one base equal to zero.

PROBLEMS OF DEMONSTRATION

995. Exercise.-The lateral area of a cylinder of revolution is equal to the area of a circle the radius of which is a mean proportional between the altitude of the cylinder and the diameter of its base.

996. Exercise.-The volume of a cylinder is equal to the product of the area of a right section by an element.

997. Exercise.-The area of a sphere is equal to the lateral area of a circumscribed cylinder of revolution.

998. Exercise. The volume of a sphere is two-thirds the volume of a circumscribed cylinder of revolution.

999. Exercise.-If a cylinder of revolution of which the altitude is equal to the diameter of the base, and a cone of

revolution of which the slant height is equal to the diameter of the base, be inscribed in a sphere; the total area of the cylinder is a mean proportional between the area of the sphere and the total area of the cone, and the volume of the cylinder is a mean proportional between the volume of the sphere and the volume of the cone.

1000. Exercise.-If a cylinder of revolution of which the altitude is equal to the diameter of the base, and a cone of revolution of which the slant height is equal to the diameter of the base, be circumscribed about a sphere; the total area of the cylinder is a mean proportional between the area of the sphere and the total area of the cone, and the volume of the cylinder is a mean proportional between the volume of the sphere and the volume of the cone.

1001. Exercise.-Show that two cylinders of revolution, whose lateral areas are equal, are to each other as their radii, or inversely as their altitudes.

PROBLEMS FOR COMPUTATION

1002. (1.) A right section of a cylinder is a circle whose radius is 3 ft.; an element of the cylinder is 13 ft. Find the lateral area.

(2.) A cylindrical boiler is 12 ft. long and 6 ft. in diameter. Find its surface, and the number of gallons of water it will hold.

(3.) A cylindrical pail is 6 in. deep and 7 in. in diameter. Find its contents and the amount of tin required for its construction.

(4.) Find the volume generated by a rectangle 9 dcm. long and 4 dcm. broad (a) in revolving about its longer side; (b) in revolving about its shorter side.