342. Examples. (1) The radius of one end of a frustum of a right circular cone is 10 inches, and the radius of the other end is 15 inches; the slant height is 16 inches: find the area of the curved surface. The sum of the circumferences in inches is the product of 3.1416 into the sum of 20 and 30, that is, into 50; thus the sum of the circumferences is 50 x 3'1416 inches. x 16 x 50 x 314168 × 50 × 3·1416=400 × 3·1416=1256.64. Thus the area of the curved surface is 1256 64 square inches. (2) The radius of one end of a frustum of a right circular cone is 5 feet, and the radius of the other end is 8 feet; the slant height is 8 feet: find the area of the whole surface. The area of the curved surface in square feet = 13 × 8 × 31416=104 × 3·1416. = The area of one end in square feet 25 × 31416, and the area of the other end in square feet = 64 × 3°1416. Hence the area of the whole surface is the product of 31416 into the sum of 104, 25, and 64, that is, into 193: thus the area of the whole surface in square feet 343. It may be inferred from the Rule of Art. 341 that the slant height of a frustum of a right circular cone bears the same proportion to the difference of the radii of the ends as the area of the curved surface bears to the difference of the areas of the ends. 344. We will now solve some exercises. (1) The radii of the ends of a frustum of a right circular cone are 7 inches and 10 inches, and the height of the frustum is 4 inches: find the area of the curved surface. is the height of the frustum, and the other side is the difference of the radii of the ends. In the present case the height of the frustum is 4 inches, and the difference of the radii is 3 inches; therefore, by Art. 55, the slant height is 5 inches. Therefore the area of the curved surface in square inches = 5 x 17 × 3·1416=267'036. (2) The diameter of the ends of a frustum of a right circular cone are 16 feet and 24 feet respectively; the height of the frustum is equal to the product of these diameters divided by their sum: find the area of the curved surface, and the area of the two ends. 16 × 24 4 × 24 10 The height of the frustum in feet: =9'6. We must determine the slant height. The square of 969216; the difference of the radii of the ends is 4 feet; the square of 4=16; 92:16+16=108'16; the square root of 108.16=10'4. Thus the slant height is 104 feet. The area of the curved surface in square feet = 10′4 × 20 × 3·1416 = 208 × 3·1416653'4528. The area of the two ends is the product of 3.1416 into the sum of the squares of 8 and 12, that is, the product of 31416 into the sum of 64 and 144, that is, the product of 3.1416 into 208. Hence the area of the two ends is equal to the area of the curved surface. It will be found on trial that this is always the case if the height of the frustum of a right circular cone is equal to the product of the diameters of the ends divided by their sum. T. M.. 14 (3) The radii of the ends of a frustum of a right circular cone are 8 inches and 10 inches respectively; the slant height is 6 inches: if the frustum be divided into two of equal curved surfaces, find the slant height of each. We must determine the slant height of the whole cone. By a method like that of the fourth Exercise of Art. 77, we find that the slant height of the whole cone is 30 inches; and therefore the slant height of the smaller cone is 24 inches. Then we proceed after the manner of Art. 210 to determine the slant distance from the vertex of the plane which divides the curved surface of the frustum into two equal parts. The square of 24 is 576; the square of 30 is 900; half the sum of these squares is 738: the square root of 738 will be found to be 27.166 very nearly. Subtract 24 from this, and the remainder is 3.166. Thus the slant height of one part is very nearly 3.166 inches; and therefore the slant height of the other part is rather less than 2.834 inches. EXAMPLES. XXXVII. Find the area of the curved surface of frustums of right circular cones, having the following dimensions : 1. Circumference of ends 15 inches and 17 inches, slant height 11 inches. 2. Circumference of ends 19 inches and 23 inches, slant height 13 inches. 3. Radii of ends 7 inches and 9 inches, slant height 5 inches. 4. Radii of ends 2.6 feet and 34 feet, slant height 5 feet. 5. Radii of ends 11 and 16 inches, height 12 inches. 6. Radii of ends 4 feet and 3 feet, height 2 feet 11 inches. 7. Radii of ends 4 feet and 5 feet, height 3 feet. 8. Radii of ends 5 feet and 6 feet, height 2 feet. Find the area of the whole surface of frustums of right cones having the following dimensions : 9. Circumference of ends 14 and 16 inches, slant height 10 inches. 10. Circumference of ends 17 and 21 inches, slant height 9 inches. 11. Radii of ends 2 feet and 3 feet, slant height 2 feet. 12. Radii of ends 34 and 4.2 feet, slant height 2 feet. 13. Radii of ends 12 and 18 inches, height 8 inches. 14. Radii of ends 12 and 20 inches, height 15 inches. Find in square feet the area of the curved surface and the area of the two ends, supposing the height of the frustum equal to the product of the diameters of the ends divided by their sum, for frustums with the following dimensions: 15. Diameter of ends 6 feet and 4 feet. 19. The radii of the ends of a frustum are 5 feet and 8 feet, and the slant height is 4 feet: if the frustum be divided into two of equal curved surface, find the slant height of each part. 20. A tent is made in the form of a frustum of a right circular cone surmounted by a cone: find the number of square yards of canvass required for the tent, supposing the diameters of the ends of the frustum to be 28 feet and 16 feet respectively, the height of the frustum 8 feet, and of the conical part 6 feet. XXXVIII. SPHERE. 345. To find the area of the surface of a sphere. RULE. Multiply the square of the diameter by 3·1416. 346. Examples. (1) The diameter of a sphere is 9 inches, 9 × 9 × 3 1416=2544696. Thus the area of the surface is 254 47 square inches nearly. (2) The diameter of a sphere is 3 feet. 3.5 x 35 x 3.1416-38 4846. Thus the area of the surface is about 38-4846 square feet. 347. Other modes of expressing the Rule in Art. 345 may be given: multiply the diameter of the sphere by its circumference; or, divide the square of the circumference by 31416. By the circumference of the sphere is meant the circumference of the circle which will produce the sphere in the manner explained by Art. 223, that is, the circumference of a great circle of the sphere. 348. It follows from Arts. 320 and 345, that the area of the surface of a sphere is equal to the area of the curved surface of a right circular cylinder which has its height and the diameter of its ends equal to the diameter of the sphere. 349. From Arts. 291 and 345 we may deduce the following important result: the volume of a sphere is equal to one third of the product of the area of the surface into the radius. 350. The result just given may be easily remembered by its resemblance to the Rule for finding the volume of a pyramid or a cone. Let O denote the centre of a sphere; and let P, Q, R denote three points on the surface of the sphere very near to each other. Suppose we cut from the sphere the piece bounded by the planes POQ, QOR, ROP, |