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ties of the ordinates are joined by straight lines, the area so found will be exactly true: but the following is a method of approximation still nearer to the truth, whether the curve be concave or convex to the axis.

Method 2. Divide the given curve, by ordinates, into any even number of equal parts ; then add into one sum four times the sum of all the even ordinates ; twice the sum of all the odd ordinates, except the first and last, and also the first and last ordinates; and if onethird of that sum is multiplied by the common distance between any two ordinates, the product will be the answer.

Er. 1. Let fig. 1, be a curve of any kind, whose equidistant ordinates, AB, CD, EF, GH, IK, LM, and NO, are respectively 5ft. 5ft. 6in. Oft. 7ft. Oft. 10ft. and 8ft. and the distance between the ordinates is 3fi., required the area of the curve.

CD, GH, and LM, will be the even ordinates ; that is, the second, ourth, and sixth ; EF, and IK, the odd ordinates, that is, the third and fifth ; AB, and NO, the fust and last.

f. in.

f.
CD= 5 6

EE= 6
GH= 7 0

IK= 9
LM310 0

15
22 6

2
4

30
90 O four times the sum of the even ordinates.
30 0 twice the sum of the odd ordinates.
5 O first ordinate.
8 o last ordinate.

3)133 O sum.
44 4

3

133 0 the area, or superficial contento

Now by coni aring this area, viz. 133 feet, with the area found in Method 1, Example viz. 132 feet, there appears to be a differeuce of 1 foot; but this last method is the most correct.

Er. 2. In fig. 3, let ACEGILN be a concave curve, whose equidistant ordinates, A, BC, DE, FG, HI, KL, and MN, are respectively i to 3. 6, 10, 15, 21, and the common distance 2; required the

a.

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90 the area, very near the truth. Es. 3. Let AHI, fig. 4, be a parabola, whose ordinates, A, BC, DE, FG, and HI, are respectively 0, 7, 12, 15, and 16, and their common distance 6; reouired the area of the curve. 7

12 15

2

22

24
4
88 four times the sum of the even ordinater,
24 twice the sum of the odd ordinates.

16 sum of the end
3)128
423

6
256 the true aroa.

T2

There is ancther method for finding the areas of curvilinear spaces, besides what has already been shown, which is as follows : divide the sum of all the ordinates by the number of them, for a mean breadth, which is to be multiplied by the length for the content; but this rule is a very false one ; it gives the area by far too great when the curve is concave, and by far too small when it is convex, and will not give the true area in any case whatever, except the curve become a straight line ; in which case, all the other rules will coincide with it. But in order to show the falsity of this rule, suppose it were required to find the contents of the same figure, as in the last example, then the sum of all the ordinates, viz. 0+7+12+15+16=50, their number is 5; and 50 divided 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 1•16th part of the true area, which is very considerable.

Prob. 2. To find the superficial content of a mixed figure, partly a curve, and partly right-lined.

Find the area of the curved 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 die vide 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.

Ex. 1. 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, 14, 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 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.

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23

2)25 4

12
92

4
24
12

50 area of the trapezoid 9

KLMN 3)16

3) 137
54
3

45.6

4
16 area of the
part ABC.

182:6 area of the part CBKL.
16
13

2)14
13
14

7
16
2)29
2)27

7 182.6 50 14.5 131

49 area of the tri. 43.5

3
2

angle QRS.
27.0
49:0

43:5

area of 27 area of OPQR.

MNOP. 368:1 sum of the areas, or contents of the whole figure.

Prob. 3. 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 the 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.

Er. Let ABCDEFG, fig. 6, be the figure proposed, which is die vided 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 BĖ 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.

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а

And 216, the area of the whole figure ABCDEFG. Prob. 4. To find the solidity of a solid, by means of equidistant rections 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 - 2618 equal .? , and that pro Juct 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 otber kind of areas and solidities, it is a very near approximation.

It is evident, that the greater the number of ordinates or sections used, the more accurate will the area or solidity be determined : but_in practice a few sections will be found sufficieut to answer the purpose. This is the best method that possibly van be devised for the practice of gauging, or for measuring curved timber trees, or the like unequally thick ; or any kind of carved solids whatever, generated about an axis ; and when all other methodo fail, this is the only one that can be depended upon for its accuracy.

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Prob. 1. To measure timber scantling.

Find the area at either end, and multiply it by the lengtu, will give the solidity.

Er. I. Suppose a joist is ein, by gin. and 8ft. long, what is the solidity ?

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