Ex. 2. Required the folidity of a zone, whofe greater diameter is 2 feet, the lefs 1 foot 4 inches, and the height 1 foot & inches. Anf. 10723.328 inches. Ex. 3. What is the folid content of a zone, whofe height is 30, and end diameters 60 and 40 inches? Anf. 75398.4 cubic inches. Ex. 4. What is the folidity of a zone, whofe height is 8 inches, and diameter of the ends 12 inches? Anf. 1172 864 cubic inches. PROBLEM XVI. Fig. 96. To find the area of a circular spindle. RULE. Multiply the length of the spindle by the radius of the revolving arch; again multiply the distance between the centre of the revolving arch and the centre of the spindle by the length of 1 of the revolving arch; fubtract this laft product from the former, and multiply the remainder by 6.2832 for the fuperficies. EXAMPLE. Required the area of a circular spindle, whose length is 40 and thickness 30 inches. AD2 BD 400 15 AD2+BD AB the chord of the arch ABC; that is, = = 400+225 25 =DH and DH+BD=FB rad. alfo FB-BD=DF cent. dift. 2 =26.6 and 26.6+16=20.83 rad. alfo 20.83-15=5.83=DF Then fay, As 360°: 147° 2 :: 3.1416×41.6: 53.58 leng. of arch, Or thus: 25 200 40 3 160 53-3 the arch nearly. Ex. 2. Required the number of fquare inches which will cover a circular spindle, whofe length is 80 and thickness 16 inches? Anf. 2747.3166336. Ex. 3. Required the area of a circular fpindle, whofe length is 12, and thicknefs y inches. Anf. 294.3621 fq. inches. PROBLEM XVII. To find the folidity of a circular spindle. Multiply the area of the revolving fegment by the distance between the centres of the arch and fpindle, fubtract the prcduct from the cube of half the length of the spindle, then multiply the remainder by 4, and this product again by 3.1416 for the folidity. See the last figure. EXAMPLE I. Required the area of a circular spindle, whofe length is 60 and diameter 45. D d AD |