III... 205.2406 cu. in the volume of the segment. NOTE. From the formula_V=πa(3r+a2), we have V=1⁄2ñar}+а3 2 1, and altitude la, and .. The volume of a is the base of the seg But ar is the volume of a cylinder whose radius is Ta3 is the volume of a sphere whose diameter is a segment of a sphere is equal to a cylinder whose base ment and altitude half the altitude of the segment, plus a sphere whose diameter is the altitude of the segment. Prob. CVIII. To find the volume of a frustum of a sphere,or the portion included between two parallel planes. * Formula.-V={ña[3(r2+r2)+a2]={a(πr2+ar2)+ a3, in which r1 is the radius of the lower base, r2 dius of the upper base. the ra Rule. To three times the sum of the squared radii of the two ends, add the square of the altitude; multiply this sum by .5235987 times the altitude. I. What is the volume of the frustum of a sphere, the radius of whose upper base is 2 feet and lower base 3 feet and altitude foot? By formula, V=}πa[3(r2+r2)+a2]=}π׆[3(9+4)+4]= 8.03839 cu. ft. II. 1. 3 ft. the radius of the lower base. 3. 39 sq. ft.=3(32+22)=three times the sum of the squares 4. 5. sq. ft. the square of the altitude. XX3918.03839 cu. ft. the volume of the frustum. III. ... 8.03839 cu. ft. the volume of the frustum. Prob. CIX. To find the volume of spherical sector. A Spherical Sector is the volume generated by any sector of a semi-circle which is revolved about its diameter. Formula.-VaR2, where a is the altitude of the zone of the sector. Rule.-Multiply its zone by one-third the radius. 2 *NOTE.-α(Tri+r)=the volume of two cylinders whose bases are the upper and lower bases of the segment and whose altitude is half the altitude of the segment. 3 is the volume of a sphere whose diameter is the altitude of the segment. Hence the volume of a segment of a sphere of two bases is equivalent to the volume of two cylinders whose bases are the upper and lower bases respectively of the segment and whose common altitude is the altitude of the segment, plus the volume of a sphere whose diameter is the altitude of the segment. For a demonstration of this and the preceding formula, see Wentworth's Plane and Solid Geometry, Bk. IX., Prob. XXXII. I. What is the volume of a spherical sector the altitude of whose zone is 2 meters and the radius of the sphere 6 meters? By formula, V=} nаR2={π ×2×6o= 150.7964m3. II.. 1. 2m. the altitude BD of the zone gener- F cle is revolved about AB. 2. 6m. the radius EC of the sphere. 3. 276m.=37.699104 m =the circumference 4. 276X2-75.398208 m2. the area of the spherical sector. FIG. 47. the volume of the III .. The volume of the spherical sector is 150.796416 m3. I. Find the diameter of a sphere of which a sector contains 7853.98 cu. ft., when the altitude of its zone is 6 feet. By formula, V=πar2=π×6×r2. ..πX6Xr2= 7853.98 cu. ft., or 4r2-2500 sq. ft., whence 2r=50 feet, the diameter of the sphere. II.< III. 1. 6 ft. the altitude of the zone. 2... X6Xr2=the volume of the sector. But 3. 7853.98 cu. ft. the volume. 4... X6Xr2-7853.98 cu. ft. 5. 2-625 sq. ft, by dividing by 47. 6. .. 2r=50 ft., the diameter of the sphere. .. The diameter of the sphere is 50 feet. Prob. CX. To find the area of a lune. A Lune is that portion of a sphere comprised between two great semi-circles. A =47 R2u, where u is Formula.—5—47 R2 (360°)=4π the quotient of the angle of the lune divided by 360°. Rule.-Multiply the surface of the sphere by the quotient of the angle of the lune divided by 360° I. Given the radius of a sphere 10 inches; find the area of a lune whose angle is 30°. By formula,S=4π R2 u=4×π X102 X (30°÷360°)= 102-104.7197 sq. in. ·II. 1. 10 in. the radius of the sphere. 2. 30° the angle of the lune. 3. -30°÷360° the quotient of the angle of the lune divided by 360°. 4. 4 π 102-400 π=1256.6368 sq. in. the surface of the sphere. 5.1256.6368 sq. in. 104.7198 sq. in. lune. III. .. The area of the lune is 104.7198 sq. in. the area of the Wentworth's New Plane and Solid Geometry, p. 371, Ex. 585. Prob. CXI. To find the volume of a spherical ungula. A Spherical Ungula is a portion of a sphere bunded by a lune and two great semi-circles. Formula.-V=π R3u, where u is the same as in the last problem. Rule.-Multiply the area of the lune by one-third the radius; or, multiply the volume of the sphere by the quotient of the angle of the lune divided by 360°. I. What is the volume of a spherical ungula the angle of whose lune is 20°, if the radius of the sphere is 3 feet? By formula, V=‡ñR3u=‡π ×33×(20°÷360°) = 6.283184 cu. ft. II. 1. 3 ft. the radius of the sphere. 2. 4732X (20°÷360°)=6.283184 sq. ft. the area of the lune, by Prob CX 3... ×3×6.283184—6.283184 cu. ft. the volume of the ungula. III. .. 6.283184 cu. ft. is the volume of the ungula. Formula.-S=2π R2 × (A+B+C—180°)÷360°, in which A, B, and C are the angles of the spherical triangle. Rule.-Multiply the area of the hemisphere in which the triangle is situated by the quotient of the spherical excess (the excess of the sum of the spherical angles over 180°) divided by 360°. I. What is the area of a spherical triangle on a sphere whose diameter is 12, the angles of the triangle being 820, 98°, and 100°? By formula, S=2πR2 × (A+B+C—180°)÷÷÷360°-2π 62 X (820+980+100°-180°)÷360°-2762×5-62.83184-area. II. 1. 6 the radius of the sphere. 2. 2762-72π the area of the hemisphere. 5 3. (820+980+100°-180°)=100° the spherical excess. (5... 1×72π=62.83184-the area of the spherical triangle. III. .. The area of the spherical triangle is 62.83184. (Olney's Geometry and Trigonometry, Un. Ed., p.238,Ex. 8.) Prob. CXIII. To find the volume of a spherical pyramid. A Spherical Pyramid is the portion of a sphere bounded by a spherical polygon and the planes of its sides. Formula.— V={π R3 × ( E÷÷360°), where E is the spherical excess. Rule.-Multiply the area of the base by one-third of the radius of the sphere I. The angles of a triangle, on a sphere whose radius is 9 feet, are 100°, 115°, and 120°; find the area of the triangle and the volume of the corresponding spherical pyramid. о By formula, V={π R3×(E÷360°)=π R3 × (A+B+C— 180°)÷360°-793 x (100° +115° +120°-180°)÷360°= 793=657.377126 cu. ft. = [1. 9 ft. the radius of the sphere. 2. 2792 the area of the hemisphere in which the pyramid is situated. 3 (100+115°+120°-180°)=155° the sperical ex II.4. III. cu. ft. 5. .. cess. 1=155°÷360°—the quotient of the spherical excess divided by 360°. mid. X27927 92 the area of the base of the pyra6...X9××2792-657.377126 cu. ft. the volume of the pyramid. .. The volume of the spherical pyramid is 657.377126 (Van Amringe's Davies' Geometry and Trigonometry, p. 278, Ex, 15. I. Find the area of a spherical hexagon whose angles are 96°, 110°, 128°, 136°, 140°, and 150°, if the circumference of a great circle of the sphere is 10 inches. Formula.—S—2πR2 [ T—(n—2)180° ] where T is 360° the sum of the angles of the polygon and n the number of sides. (96°+110° +128°+136°+140°+150°—(6—2)×180°)÷ II. π 50 x(760°-720°)÷÷360°==1.7684 sq. in. π (1. 5÷7=the radius of the sphere, since 2πR=10 in.. 4. 3. 760°-(6-2) × 180°-40° the spherical excess. 5. 2л π 2 the area of the hemisphere on which the polygon is situated. 6. .. ƒ×27(+-)2 =¡×50÷7=1.7684 sq. in. π III. .. The area of the polygon is 1.7684 sq. in. XIII, SPHEROID. 1. A Spheroid is a solid formed by revolving an ellipse about one of its diameters as an axis of revolution. 1. THE PROLATE SPHEROID. 1. The Prolate Spheroid is the spheroid formed by revolving an ellipse about its transverse diameter as an axis of revolution. Prob. CXIV. To find the surface of a prolate spheroid. Formulae.—(a)S=2 ƒ2xy ds=2 ɲ2ny√1+dy® dx= ́a + y2 + b2x2 + Απ fab α a4y2 2 2 ) * dx,= 47 fo[a2 (a2b2 —b2x2 )+b^x2]*dx— (a2—e2x2)dx=2nb2+27absin ̄'e,—276(b+sine), where |