EXAMPLES. 1. The diameters of the two concentric circles being AB 10 and DG 6; required the area of the ring contained between their circumferences AEBA and DFGD. 2. The diameters of two concentric circles being 20 and 10; required the area of the ring between their circumfer ences. Ans. 235'62. 3. What is the area of the ring, the diameters of whose bounding circles are 6 and 4? Ans. 15'708. COR. If DI be a perpendicular at the point D, then will the area of the ring be equal to that of a circle, whose radius is DI. RULE 2. Multiply half the sum of the circumferences by half the difference of the diameters, and the product will be the area. PROBLEM XVI. To measure long irregular figures. RULE. Take the breadth in several places at equal distances. Add all the breadths together, and divide the sum by the number of them, for the mean breadth, which multiply by the length, for the area. EXAMPLES. 1. The breadths of an irregular figure, at five equidistant places being AD 8'1, mp 7'4, nq 9'2, or 10'1, BC 8'6; and the length AB 39; required the area. 2. The length of an irregular figure being 84, and the breadths at 6 places 17'4, 20'6, 14′2, 16'5, 20'1, 24'3 what is the area? Ans. 1583'4.. PROBLEM XVII. To find the circumference of an ellipse.* RULET. Add the two axes together, and multiply the sum by 1'5708, for the circumference nearly. EXAMPLES. 1. Required the circumference of an ellipse, whose two axes are 70 and 50. For definitions of ellipse, parabola, and hyperbola, see CONIC SECTIONS. It will be evident, that this rule is very near the truth, if it be considered, that this arithmetical mean between the axes exceeds their geometrical mean; and, that the geometrical mean is the diameter of a circle, equal in area to the ellipse; which circle is of less ambit than the ellipse, or any other figure of the same area. 2. What is the periphery of an ellipse, whose two axes are *24 and 20? Ans. 69 1152. PROBLEM XVIII. To find the area of an ellipse RULE.* Multiply the transverse by the conjugate, and the product, multiplied by '7854, will be the area. Or, Multiply '7854 by one axe, and the product by the ❤ther. EXAMPLES. 1. To find the area of an ellipse, whose two axes are 70 and 50. 7854 39°2700 70 2748*9000 answer. 2. What is the area of an ellipse, whose two axes are 24 and 18? Ans. 339 2928. The demonstration of this rule is contained in that of the Bext problem. PROBLEM XIX. To find the area of an elliptic segment. RULE.* Divide the height of the segment by the axis of the ellipse, of which it is a part; and find, in the table of circular segments, a circular segment having the same versed sine as this quotient. Then multiply continually together this seg ment and the two axes, for the area required. EXAMPLES. 1. What is the area of an elliptic segment EAF, whose height AP is 20; the transverse AB being 70, and the conjugate CD 50? 70)20(28′′ the tab. vers. DEMONSLRATION. Let the transverse diameter AB=a, the conjugate CD=c, AG=x, and EG=y; then, by the prop erty of the curve, we shall have y=—✔ax- 2, and the flux c ion of the area EAF=(yx)== | EAF=(yx)= — × x√ux—x3. But x√ax-x2 is known to express the fluxion of the corresponding circular seg ment, whose versed sine is x, and diameter a. Let the fluent of this expression, therefore, be denoted by A, and then the fluent of × √3 will be = A, whence the rule is formed. Q. E. I. COR. The ellipse is equal to a circle, whose diameter is at mean proportional between the two axes, and hence the rule is formed for the whole ellipse. |