Alfo by example 4 prob. 10, the height is 2062.2955. Then (R+ + +b2) × 1·5708h 17249136 X 2062 2955 X 15708 = 55877778668, the folidity of each temperate zone. Otherwife. Since the radii of the ends of this zone, are the fines of 23 and 66 degrees, which are complements the one of the other, the fum of the fquares of those radii will be equal to the fquare of the radius of the fphere; and therefore (dd+bh) × 1.5708h = 17249136 × 2062·2955 × 1·5708 = 55877778668, the content, as before. PROBLEM XVIII. To find the Solidity of the Middle Zone of a Sphere. Multiply, either the fum of the fquare of the diameter of the end and of the fquare of the height, or the difference between the fquare of the diameter of the fphere and of the fquare of the height, by the height, and the product by 7854 for the con Required the folidity of the middle zone of a fphere whofe top and bottoin diameters are each 3 feet, and height 4 feet. Here (DD+bb) × 7854b = (32 + 3×42)× 4 ×·7854 = 59 × 4 × •2618 = 61.7848, the content required. EXAMPLE II. What is the folidity of the torrid zone of the earth, which extends to 23 degrees on each fide of the equator; the diameter of the earth being 7957 miles? By the example to prob. 16, the fine of 23 degrees, or the height of the zone, is 1586.57282526, and the whole height is 3173.14565052 = b. Then (ddbb) x 7854b = × (7957 752-3173 145650522) × 3173*14565052 X7854149455081137, the content. SCHOLIUM. From the last three problems we find that The two frigid zones = 2647359420 The two temperate zones 111755557336 The torrid zone 149455081137 whofe fum 263857997893 is the whole sphere, nearly the fame as found in prob. 15. To find the Solidity of the Second Segment of a Sphere. DEFINITION. A fecond fegment is a part cut off a fegment by a plane perpendicular to the base. 3 RULE. A C H E K 0 By prob. 14 find the curve furface of the fecond fegment AECE, which being drawn into of the radius of the fphere, will produce the content of the fpheric fector OEACE; from which if there be taken the pyramid OE Ae, whofe bafe is the fegment AEe, and height OH; and from the remainder be taken the pyramid Oɛce, whofe bafe is the fegment cɛe, and height Ko, or HI; it is evident that the laft remainder will be the fecond fegment AEce: for thefe two pyramids and the fecond fegment compose the spheric pyramid. PROBLEM XX. To find the Surface of a Circular Spindle, or of any Segment or Fruftum of it. From the product of the height of the folid and radius of the revolving arc, fubtract the product of the faid arc and central distance; multiply the remainder by 3.1416; and double the product will be the furface defcribed by that arc; whether it be the whole or any part of the fpindle.* That - Put the arc CP, its fine RE, r and c for the radius and central distance, and p = 3.14159. Then the fluxion of the furfaces is === 2px X RP = 2px (~/TT — xxcx); but, by the property of the circle, rrx is; therefore s= 2p × (r cz), and s = 2px (rx- cx) = 2px (r X RE - CX PC). Corol Required the furface of a circular spindle whofe greatest diameter is 30, and its length 40 inches. Here, by the property of the circle, EK AE÷ EC 400 15 26; and 263 + 15 = 41 is the diameter CK, or 20% the radius of the circle. P 3 And Corol. 1. When RE - AE, the rule becomes 2p x (r X AE - cx Ac) for the furface of half the fpindle, or 2px (rx A B -CX ACB) for that of the whole. Corol. 2. If from the furface of the femi-fpindle be taken that of the fruitum, there will remain 2p × (r × ARC X AP) for that of the fegment APQ: fo that the rule is general. Corol. 3. When E coincides with F, c vanishes, and the spindle becomes a fphere; and then the theorem becomes barely 2prx, the fame with that before found for the sphere. Corol. 4. From cor. 1, it appears that the radius is to the cofine of an arc, always in a greater proportion than that of the arc to its fine, or of the double arc to its chord. But those ratios approximate to an equality as the are diminishes, till, when the arc vanishes, they become accurately equal, and the arc and chord vanish alfo in a ratio of equality.-Confequently, in fmall arcs, the arc is a fourth proportional to the cofine, fine, and radius, nearly. Thus, the fine and cofine of an arc of 1 degree are 0174524 and 9998477 to the radius 1; then '9998477: 017452410174550 the arc nearly, the true figures being 0174533. PROBLEM XXII. To find the Content of the Middle Fruftum or Zone of a Circular Spindle. From the fquare of half the length of the fpindle, take of the fquare of half the length of the middle zone; and multiply the remainder by the faid half length of the zone; from the product fubtract the product of the generating circular area, and central diftance; then the remainder drawn into 2 times 3.14159 will be the content of the middle zone. That is, putting IER half the length of the zone, a the generating area R P Y W. Then [(LL11)l — ac] × 2p the zone PYZQ EXAMPLE. If a cafk, in the form of the middle fruftum of a circular fpindle, have its head diameter 24, bung diameter 32, and length 40 inches; how many ale gallons will it hold? Here CE PR 16 hence 100 12 = 4 TC, and PT2 TC 400 ÷ 4 = 100 = TK; +4=104 the diameter, and 52 of the generating circle. the radius FC CE = 52 16 36 EF the central diftance; and AFFE✓ 522 — 362 2 41392 = 4√88 = 8 √22 AE half the length of the fpindle. 480 the area RPTY W. PRX RW 12 X 40 The |