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. D A 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 CDHG 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. The arc AC generates a zone of one base, and the figure ACG a spherical segment of one base. The circular sector OEF generates a spherical sector whose base is the zone generated by the arc EF; the other bounding surfaces are the conical surfaces generated by the radii OE and OF. If OF coincides with OB, the spherical sector is bounded by a conical surface and a zone of one base. If OE is perpendicular to OB, the spherical sector is bounded by a plane surface, a conical surface, and a zone. Proposition 9. Theorem. 788. The area generated by a straight line revolving about an axis in its plane, is equal to the product of the projection of the line on the axis by the circumference whose radius is the perpendicular erected at the middle point of the line and terminated by the axis. Hyp. Let AB be the revolving line, CD its projection on the axis GO, and EO the at the mid. pt. of AB and termi nating in the axis. To prove area ABCD × 2πЕО. Proof. Draw EFL, and AH || to GO. The area generated by AB is the lateral B area of a frustum of a cone of revolution, whose slant height is AB and axis CD. The ... area AB = AB × 2πEF. AS ABH and EOF are similar. .·. AB × EF = AH × EO = CD × EO. ... area AB = CD × 2πEO. (777) (316) If AB meets the axis, or is || to it, thus generating a conical, or a cylindrical surface, the result is the same from (773) and (753). Q.E.D. Proposition 10. Theorem. 789. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. Hyp. Let the sphere be generated by the revolution of the semicircle ABDF about the diameter AF, let O be the centre, R the radius, and denote the surface of the sphere by S. Proof. Inscribe in the semicircle a regular semi-polygon ABCDEF, of any number of sides. Draw Bb, Cc, etc., 1 to AF and OHL to AB. Then Similarly, area AB Ab × 2πОН. (201) (788) area BC = bc × 27ОH, and so on. In equal os equal chords are equally distant from the centre (206). Now the sum of the projections Ab, bc, etc., of all the sides of the semi-polygon make up the diameter AF. .. the entire surface generated by the revolving semipolygon = AF X 2πΟΗ. Now let the number of sides of the inscribed semi-polygon be indefinitely increased. The semi-perimeter will approach the semi-circumference as its limit, and OH will approach the radius R as its limit. (430) ... the surface of revolution will approach the surface of the sphere as its limit. ... S = : AF × 2πR. Q.E.D. 790. COR. 1. Since AF = 2R, S = 2R × 2πR = 4πR2. (789) Therefore, the area of the surface of a sphere is equal to the area of four great circles. 791. COR. 2. The areas of the surfaces of two spheres are to each other as the squares of their radii, or as the squares of their diameters. 792. COR. 3. The area of a zone is equal to the product of its altitude by the circumference of a great circle. For, the area of the zone generated by the revolution of the arc BD = bd × 2πR. 793. COR. 4. Zones on the same sphere, or on equal spheres, are to each other as their altitudes. 794. COR. 5. Since the arc AB generates a zone of one base whose area is Ab × 2πR = πАь × АF = ΠΑΒ, (325) therefore, the area of a zone of one base is equal to the area of the circle whose radius is the chord of the zone. 795. COR. 6. If a cylinder is circumscribed about a sphere, the total area of the cylinder = 6πR2. (755) Therefore, the area of the surface of a sphere is equal to two-thirds the total area of the circumscribing cylinder. EXERCISES. 1. Find the area of the surface of a sphere whose radius is 4 inches. 2. Prove that two zones on different spheres are to each other as the products of their altitudes by the radii of the spheres. Proposition 11. Theorem. 796. The volume of a sphere is equal to the area of its surface multiplied by one-third of its radius. Hyp. Let V denote the volume of a sphere, S the area of its surface, and R its radius. To prove V = S × }R. Proof. Conceive the whole surface of the sphere to be divided into a great number of equal spherical polygons. Form pyramids by joining the vertices of these polygons together successively and to the centre of the sphere. It is evident that these pyramids will have a common altitude. The volume of each pyramid = base altitude. (631) .. the sum of all the pyramids = sum of bases titude. al Now let the number of spherical polygons be indefinitely increased. The sum of the bases of all the pyramids will approach the surface of the sphere as its limit; and the altitude of each pyramid will approach the radius of the sphere as its limit. .. the sum of the volumes of all the pyramids will approach the volume of the sphere as its limit. NOTE. This result might be obtained by regarding the sphere as the limit of a circumscribed polyedron, the number of whose faces was indefinitely increased. For, if pyramids are formed having the faces of the polyedron as their bases, and the centre of the sphere as their common vertex, these pyramids will have a common altitude equal to the radius of the sphere. Then each pyramid = face X altitude. .. sum of pyramids = sum of faces altitude. But in the limit sum of faces of polyedron = surface of sphere: and sum of volumes of pyramids = volume of sphere. .. V = SX } R. |