by 5, is equal to 10: this being multiplied by the whole length, viz. 24, gives 240 instead of 256, the exact area of the curve found by the last Example; the difference of this Example being too small by part of the true area, which is very considerable. PROBLEM II. To find the superficial content of a mixed figure, partly a curve, and partly right lined. Find the area of the curve part of the figure by the last Problem, by dividing it into equidistant ordinates; divide the right-lined parts of the figure by ordinates drawn through every angle, which will divide the right-lined part of the figure into trapezoids and triangles; find the area of each part, according to their respective rules; add the areas of all the parts together, and the sum will give the area of the whole figure. EXAMPLE I. Plate 54, Fig. 5, let ABKNORS be the figure proposed, to find its area. As the end AB turns round very perpendicular to the base AS, draw the ordinate CB in such a manner, as it may cut off the most perpendicular part of the curve AB at the end, and divide it by ordinates, which are respectively 1, 2, 1, 1, 0, at the distance of 3 from each other; the part CBKL of the curvilineal space is also divided into four equal parts, between the first and last ordinates BC, KL, by the ordinates ED, GF, HI, and LK, which are respectively 12, 13, 12, 10, and 9, and their common distance 4; the other Gg 2 parts parts of the figure are divided into three trapezoids, KLMN, MNOP, OPQR, and the triangle QRS, by ordinates from the angles at K, N, O, and R; the whole figure being thus prepared, by dividing it into curvilineal spaces, trapezoids and a triangle, each part will be measured according to their respective rules. The measures or dimensions are marked on their respective places on the figure; the contents of each part is computed separately, as is shown in the following operation. 368.1 sum of the areas, or contents of the whole figure. PROBLEM III. To measure any irregular figure, bounded wholly by right lines. Divide the whole figure into trapeziums and triangles, divide each trapezium into two triangles, by means of diagonals from the other angles, let fall perpendiculars to the diagonals; then to find the area of any of the trapeziums, multiply the half sum of the two perpendiculars in that trapezium by its diagonal; find the area or content of all the trapeziums in this manner, and of the triangles, if any; add their several areas together, and the sum will give the area of the whole figure. EXAMPLE. Plate 54, Fig. 6, let ABCDEFG be the figure proposed, which is divided into two trapeziums ABFG, BCEF, and a triangle CDE; the diagonal BG of the trapezium ABFG is 16, and the two perpendiculars AH and IF are respectively 4 and 8; the diagonal BE of the trapezium BCEF is 10, and the perpendiculars KF and CL are respectively 14 and 6; the base CE of the triangle CDE is 8, and its perpendicular 5, required the area of the whole figure ABCDEFG. Then 96, the area of the trapezium ABFG. 100, the area of the trapezium BCDF. And 216, the area of the whole figure ABCDEFG. PROBLEM PROBLEM IV. To find the solidity of a solid, by means of equidistant sections or planes. Divide the length of the solid into any even number of equal parts; find the area of all the parallel sections passing through these parts perpendicular to the axis of the solid; then to four times the sum of the areas of all the even planes, add twice the sum of the areas of all the odd planes, except the first and last, and the areas of the two ends; divide the sum by 3, multiply the quotient by the common distance, and the product will give the solidity. N. B. If the sections are circular, the rule may be as follows to four times the area of the even planes, add twice the sum of the areas of the odd planes, excepting the first and last, and the sum of the areas of the ends, or the first and last planes; then multiply the sum by 26187854, and that product by the common distance, and it will give the solidity sought. SCHOLIUM. This Problem is accurately true for parabolic curves, or solids, generated from the revolution of conic sections, or right lines. For all kinds of pyramids, or frustums of pyramids, or any other kind of areas and solidities, it is a very near approximation. It is evident, that the greater the number of ordinates. or sections are used, the more accurate will the area or |