and by the portion of the surface of the sphere which they intercept. This piece will resemble a triangular pyramid; so that we may readily admit that the volume of the piece is equal to one third of the product of the radius of the sphere into the intercepted portion of the surface of the sphere. The whole sphere may be supposed to be cut up into a very large number of very small pieces like that just considered; and thus we are easily led to the result given in Art. 349. The student will observe the resemblance of these remarks to those in Art. 177. 351. The sphere has the following remarkable property: of all solids of a given volume the sphere is that which has the least surface, and of all solids of a given surface the sphere is that which has the greatest volume. The student may verify this statement by such examples as 16...20 at the end of the present Chapter. 352. We will now solve some exercises. (1) The area of the surface of a sphere is 200 square inches: find the diameter, and the volume of the sphere. The product of the square of the diameter into 3.1416 is equal to 200; therefore the square of the diameter = 63 6618: the square root of this number will be found to be 7.9789. Thus the diameter is 7.98 inches very nearly. = 200 3.1416 Then, by Art. 349, the volume of the sphere in cubic feet (2) The volume of a sphere is 1000 cubic inches: find the area of its surface. By Art. 291 the cube of the diameter of the sphere 1000 = 1909.8548; therefore the diameter of the sphere *5236 in inches is the cube root of this number: it will be found that this is 12:407. Then by Art. 345 we shall obtain for the area of the surface 483'6 square inches very nearly. EXAMPLES. XXXVIII. Find the areas of the surfaces of spheres having the following dimensions: 1. Radius 5 inches. 3. Radius 2.2 feet. 5. Circumference 4 feet. 2. Radius 15 inches. 4. Circumference 20 inches. 6. Circumference 6'4 feet. Find the diameters of the spheres having the following superficial areas: 7. 400 square inches. 8. 64 square feet. 9. 75 square feet. Find the volumes of the spheres having the following superficial areas: 10. 20 square feet. 11. 50 square feet. 12. 100 square feet. 13. Find the volume of a sphere when its surface is equal to that of a circle 4 feet in diameter. 14. Find the volume of a sphere when its surface is equal to that of a circle 9 feet in diameter. 15. A cylinder 5 feet long and 3 feet in diameter is closed by a hemisphere at each end: find the area of the whole surface. 16. The radius of the base of a right circular cylinder is 10 inches, and the height is 10 inches; the surface of a sphere is equal to the whole surface of this cylinder: find the volume of each. 17. The surface of a sphere is equal to that of a cube the length of which is one foot: find the volume of each. 18. The surface of a sphere is equal to that of a right circular cylinder the radius of the base of which is one foot, and the height two feet: find the volume of each. The following examples involve the extraction of the cube root: 19. The volume of a sphere is equal to that of a cube the length of which is one foot: find the surface of each. 20. The volume of a sphere is equal to that of a right circular cylinder the radius of the base of which is one foot, and the height two feet: find the surface of each. XXXIX. ZONE OF A SPHERE. SEGMENT OF A SPHERE. 353. The surface of a zone of a sphere consists of two circular ends and another portion which we shall call the curved surface. The surface of a segment of a sphere consists of a circular base and another portion which we shall call the curved surface. 354. To find the area of the curved surface of a zone of a sphere or of a segment of a sphere. RULE. Multiply the circumference of the sphere by the height of the zone or segment. 355. Examples. (1) The height of a segment of a sphere is 6 inches, and the diameter of the sphere is 18 inches: find the area of the curved surface. 6 x 18 x 3'1416339°2928. Thus the area of the curved surface is 339'3 square inches nearly. (2) The ends of a zone of a sphere are distant 2 feet and 4 feet respectively from the centre of a sphere, and are on the same side of the centre; the diameter of the sphere is 14 feet: find the area of the whole surface of the zone. The area of the curved surface in square feet By Art. 89 the square of the radius of one end and the square of the radius of the other end =11x3=33: thus the area of the two ends in square feet = 78 × 3·1416. Therefore the area of the whole surface in square feet = 106 × 3·1416=333 0096. 356. It appears from Arts. 320 and 354 that the curved surface of a zone of a sphere or of a segment of a sphere is equal to the area of the curved surface of a right circular cylinder which has its height equal to that of the zone or segment, and the diameter of its ends equal to the diameter of the sphere. This remarkable result holds also for the surface of a sphere, if by the height of the sphere we understand the diameter of the sphere: see Art. 348. 357. We will now solve some exercises. (1) The height of a segment of a sphere is 7 inches, and the circumference of the sphere is 64 inches: find the area of the whole surface of the segment. The area of the curved surface in square inches = 7 × 64 = 448. The diameter of the sphere is 64 3.1416 inches; therefore, by Art. 79, the square of the radius of the base of the segment is obtained by subtracting 7 from multiplying the remainder by 7; so that it is 64 3.1416' 7 × 64 3.1416 and - 49. The area of the base of the segment in square inches is the product of this result into 31416; therefore it is 7 × 64-49 x 3'1416, that is 448-49 × 31416. Thus the area of the whole surface in square inches is The method which we have adopted in solving this exercise may appear rather artificial and difficult to a beginner; but it deserves attention. It will be seen that in effect we establish the following Rule: the whole surface of a segment of a sphere is equal to twice the excess of the curved surface above a circle which has the height of the segment for its radius. (2) A zone of a sphere is the difference of two segments of the heights 13 inches and 9 inches respectively, and the circumference of the sphere is 82 inches: find the arca of the whole surface of the zone. The area of the curved surface in square inches By Art. 79 the square of the radius of one end of the zone is obtained by subtracting 9 from 82 3.1416' tiplying the remainder by 9; so that it is and mul The area of this end of the zone in square inches is the product of this result into 3 1416; therefore it is 9 × 82-81 × 3·1416. Similarly the area of the other end of the zone in square inches will be found to be 13 x 82-169 × 31416. Hence the area of the whole surface of the zone is equal to the sum of 4 × 82, 9 × 82, and 13 × 82, diminished by the sum of 81 × 31416 and 169 × 31416, that is, to 2 × 13 × 82-250 × 3·1416. Thus the area of the whole surface is 1346'6 square inches. It will be seen that in effect we establish the following Rule for finding the area of the whole surface of a zone of a sphere, regarded as the difference of two segments of the sphere: from twice the curved surface of the large segment subtract the areas of two circles having for their radii respectively the heights of the segments. (3) The radius of a sphere is 12 feet; from a point which is at the distance of 15 feet from the centre of the sphere straight lines are drawn to touch the sphere, thus determining a segment of the sphere: find the area of the curved surface of this segment. |