Page:EB1911 - Volume 18.djvu/160

The volume of a briquette can be found in this way if the area of the section by any principal plane can be expressed in terms of the distance of this plane from a fixed plane of the same set. The result of treating this area as if it were the ordinate of a trapezette leads to special formulae, when the data are of the kind mentioned in § 44.

49. (B) Mensuration of Graphs of Algebraical Functions.—The first class of cases to be considered comprises those cases in which u is an algebraical function (i.e. a rational integral algebraical function) of x, or of x and y, of a degree which is known.

50. The simplest case is that in which u is constant or is a linear function of x, i.e. is of the form px + q. The trapezette is then a right trapezium, and its area, if m＝l, is 1/2h(u₀, + u₁) or hu_1/2.

51. The next case is that in which u is a quadratic function of x, i.e. is of the form px² + qx + r. The top is then a parabola whose axis is at right angles to the base; and the area can therefore (§ 34) be expressed in terms of the two bounding ordinates and the mid-ordinate. If we take these to be u₀ and u₀, and u₁, so that m＝2, we have

area＝1/6h(u₀ + 4u₁ + u₂)＝1/3h(u₀ + 4u₁ + u₂).

This is Simpson’s formula.

If instead of u₀, u₁, and u₂, we have four ordinates u₀, u₁, u₂ and u₃, so that m＝3, it can be shown that

area＝3/8h(u₀ + 3u₁ + 3u₂ + u₃).

This is Simpson’s second formula. It may be deduced from the formula given above. Denoting the areas of the three strips by A, B, and C, and introducing the middle ordinate u_3/2, we can express A + B; B + C; A + B + C; and B in terms of u₀, u₁, u₂; u₁, u₂, u₃; u₀, u_3/2, u₃; u₁, u_3/2, u₂ respectively. Thus we get two expressions for A + B + C, from which we can eliminate u_3/2.

A trapezette of this kind will be called a parabolic trapezette.

52. Simpson’s two formulae also apply if u is of the form px³ + qx² + rx + s. Generally, if the area of a trapezette for which u is an algebraical function of x of degree 2n is given correctly by an expression which is a linear function of values of u representing ordinates placed symmetrically about the mid-ordinate of the trapezette (with or without this mid-ordinate), the same expression will give the area of a trapezette for which u is an algebraical function of x of degree 2n + 1. This will be seen by taking the mid-ordinate as the ordinate for which x＝0, and noticing that the odd powers of x introduce positive and negative terms which balance one another when the whole area is taken into account.

53. When u is of degree 4 or 5 in x, we require at least five ordinates. If m＝4, and the data are u₀, u₁, u₂, u₃, u₄, we have

area＝2/45h(7u₀ + 32u₁ + 12u₂, + 32u₃, 7u₄).

For functions of higher degrees in x the formulae become more complicated.

54. The general method of constructing formulae of this kind involves the use of the integral calculus and of the calculus of finite differences. The breadth of the trapezette being mh, it may be shown that its area is

mh ${\Big \{}$u_1/2m + 1/24m²h²u″_1/2m + 1/1920m⁴h⁴u^iv_1/2m + 1/1920m⁶h⁶u^vi_1/2m + 1/92897280m⁸h⁸u^viii_1/2m + . . . ${\Big \}}$,

where u_1/2m, u″_1/2m, u ‴_1/2m , . . . denote the values for x＝x_1/2m of the successive differential coefficients of u with regard to x; the series continuing until the differential coefficients vanish. There are two classes of cases, according as m is even or odd; it will be convenient to consider them first for those cases in which the data are the bounding ordinates of the strips.

(i) If m is even, u_1/2m, will be one of the given ordinates, and we can express h²u″_1/2m, h⁴u^iv_1/2m in terms of u_1/2m and its even central differences (see Differences, Calculus of). Writing m＝2p, and grouping the coefficients of the successive differences, we shall find

area＝2ph $\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$u_p + p²/6δ²u_p + 3p⁴ −5p²/360δ⁴u_p + 3p⁶ −21p⁴ + 28p²/15120δ⁴u_p + . . . $\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$ .

If u is of degree 2f or 2f + 1 in x, we require to go up to δ^2fu_p, so that m must be not less than 2f Simpson’s (first) formula, for instance, holds for f＝1, and is obtained by taking p＝1 and ignoring differences after δ²u_p.

(ii) If m is odd, the given ordinates are u₀, . . . u_1/2m−1/2, u_1/2m+1/2 . . . u_m. We then have

area＝mh $\scriptstyle {\left\{{\begin{matrix}\ \\\ \end{matrix}}\right.}$ μu_1/2m + m² − 3/24μδ²u_1/2m + 3m⁴ − 50m²+135/5760μδ⁴u_1/2m + 3m⁶ − 147m⁴+1813m² − 4725/967680μδ⁶u_1/2m + . . . $\scriptstyle {\left.{\begin{matrix}\ \\\ \end{matrix}}\right\}\,}$,

where μu_1/2m, μδ²u_1/2m, . . . denote 1/2(u_1/2m−1/2 + u_1/2m+1/2), 1/2(δ²u_1/2m−1/2 + δ²u_1/2m +1/2), . . . Simpson’s second formula is obtained by taking m＝3 and ignoring differences after μδ²u_1/2m.

55. The general formulae of § 54 (p being replaced in (i) by 1/2m) may in the same way be applied to obtain formulae giving the area of the trapezette in terms of the mid-ordinates of the strips, the series being taken up to δ^2fu_1/2m or μδ^2fu_1/2m at least, where u is of degree 2f or 2f + 1 in x. Thus we find from (i) that Simpson’s second formula, for the case where the to is a parabola (with axis, as before, at right angles to the base) and) there are three strips of breadth h, may be replaced by

area＝3/8h(3u_1/2 + 2u_3/2 + 3u_5/2).

This might have been deduced directly from Simpson’s first formula, by a series of eliminations.

56. Hence, for the case of a parabola, we can express the area in terms of the bounding ordinates of two strips, but, if we use mid-ordinates, we require three strips; so that, in each case, three ordinates are required. The question then arises whether, by removing the limitation as to the position of the ordinates, we can reduce their number.

Fig. 8.

Suppose that in fig. 6 (§ 34) we draw ordinates QD midway between KA and MC, and RE midway between MC and LB, meeting the top in D and E (fig. 8), and join DE, meeting KA, LB, and MC in H, J, and W. Then it may be shown that DE is parallel to AB, and that the area of the figure between chord DE and arc DE is half the sum of the areas DHA and EJB. Hence the area of the right trapezium KHJL is greater than the area of the trapezette KACBL.

If we were to take QD and RE closer to MC, the former area would be still greater. If, on the other hand, we were to take them very close to KA and LB respectively, the area of the trapezette would be the greater. There is therefore some intermediate position such that the two areas are equal; i.e. such that the area of the trapezette is represented by KL . 1/2(QD + RE).

To find this position, let us write QM＝MR＝θ . KM. Then

WC＝θ².VC, VW＝(1 − θ²) VC;

curved area ACB＝2/3 of parallelogram AFGB＝2/3KL . VC;

parallelogram AHJB＝KL . VW＝(1 − θ²) KL . VC.

Hence the areas of the trapezette and of the trapezium will be equal if

1 − θ²＝2/3, θ＝1/√3.

This value of θ is the same for all parabolas which pass through D and E and have their axes at right angles to KL. It follows that, by taking two ordinates in a certain position with regard to the bounding ordinates, the area of any parabolic trapezette whose top passes through their extremities can be expressed in terms of these ordinates and of the breadth of the trapezette.

The same formula will also hold (§ 52) for any cubic trapezette through the points.

57. This is a particular case of a general theorem, due to Gauss, that, if u is an algebraical function of x of degree 2p or 2p + 1, the area can be expressed in terms of p + 1 ordinates taken in suitable positions.

58. The Prismoidal Formula.—It follows from §§ 48 and 51 that, if V is a solid figure extending from a plane K to a parallel plane L, and if the area of every cross-section parallel to these planes is a quadratic function of the distance of the section from a fixed plane parallel to them, Simpson’s formula may be applied to find the volume of the solid. If the areas of the two ends in the planes K and L are S₀ and S₂, and the area of the mid-section (i.e. the section by a plane parallel to these planes and midway between them) is S₁, the volume is 1/6H(S₀ + 4S₁ + S₂), where H is the total breadth.

This formula applies to such figures as the cone, the sphere, the ellipsoid and the prismoid. In the case of the sphere, for instance, whose radius is R, the area of the section at distance x from the centre is π(R²−x²), which is a quadratic function of x; the values of S₀, S₁, and S₂ are respectively 0, πR², and 0, and the volume is therefore 1/6 . 2R . 4πR²＝4/3πR³.

Fig. 9.

To show that the area of a cross-section of a prismoid is of the form ax² + bx + c, where x is the distance of the section from one end, we may proceed as in § 27. In the case of a pyramid, of height h, the area of the section by a plane parallel to the base and at distance x from the vertex is clearly x²/h² × area of base. In the case of a wedge with parallel ends the ratio x²/h² is replaced by x/h. For a tetrahedron, two of whose opposite edges are AB and CD, we require the area of the section by a plane parallel to AB and CD. Let the distance between the parallel planes through AB and CD be h, and let a

plane at distance x from the plane through AB cut the edges AC,