Page:Encyclopædia Britannica, Ninth Edition, v. 16.djvu/32

Rh 22 MENSURATION equation y= A + Ex + Cx&quot;+ .... + Kx n for n + 1 points between P! and P n +i, then the area of the curvilinear figure bounded by the straight lines A. 1 P l, AjAn+j, and A,,+iP rt+ i and the curve PjPH+i will agree very nearly with the curvilinear figure bounded by the same straight lines and the curve whose equation is y = A + Bx + Cx*+ .... +Kx n, and the greater the number of common points the closer will be the agreement. Let AjAn-t-i be divided into n equal parts, each equal to h, then Aj A n _|-i = nh. Now whenx=0, y = y l = A; .... + K7i&quot; ; +K(ph) From these ?i + l equations the ?i + l quantities A, B ..... K can be determined as functions of 2/j, 2/2, l/n+i, and h. Next let AjAn+i be divided into m equal parts each equal to h. Thus mh = nh and hence 7i = m Now the area of the rectangle A P A P+1 P P R = A p A p +i x Ap? p. But AP = ?/ 7* = ; therefore area of A P R ?i7i I + Enh ( m m* m 6 Hence the area of the whole figure A + B ^ m m* L ^ where S,, = l n + 2 + 3&quot; + . . +m n . Now if we take the limit of each of the terms S Sj S 2 S n TO in? m 3 m n + 1 we obtain area of curvilinear figure + jiA-}- ?i 2 A 2 + ^ 3 - +1 From this general result we can deduce &quot; Simpson s Rule &quot;and also another rule called &quot; Weddle s Rule.&quot; Thus let 7i = 2 ; that is, assume that the curve under consideration has three points in common with the curve whose equation is c 2, i.e., with a parabola, then 2/i = A, Now the area is approximately = 2/i { A + B27i + JC 2 2 7i 2 } = ^{ 2/i + 4y. 2 f7/ 3 } , Simpson s Rule. If we now put n = 6, we have area of curvilinear figure Now G(Qk) 6 . From this system of equations we can determine A, B, C, . . . G, and substituting the values so obtained in the above expression we obtain the following remarkable formula for the approximate area : This formula, called Weddle s Rule, gives the closest approximation to the curvilinear area that can be obtained by any simple rule. We are now in a position to find the approximate area of any irregular plane figure. For the given figure can be divided into plane rectilinear and cur vilinear figures, the areas of which can be separately determined by the rules already given. For example, APQRS (fig. 38) = ABC + APD + BRC - DQB - ASC. PART II. SOLIDS. SECTION I. SOLIDS CONTAINED BY PLANES. A. Prisms, Pyramids, and Prismatoids. 66. Volume of a Right Prism. First let the prism be a rect angular parallelepiped (fig. 39), and let the side AB contain a units of length, BC b units of length, and CD c units of length. If we divide AB into a equal parts, BC into b equal parts, and CD into c equal parts, and if, through the points of division we draw planes parallel to the sides of the parallelepiped, these planes will divide it into a series of parallelepipeds, whose edges are each equal to the unit of length. Each horizontal layer contains ab of these cubes, and since there are c layers the whole number of cubes will be abc. volume But each of these is the unit of volume, and therefore 6 x c = area of base ABC x altitude c. In the above demonstration we have assumed the edges to be commensurable, but from 2 it follows that the proof will hold also when the edges are incommensurable. If the parallelepiped be cut by a plane BGE it will be divided into two equal triangular right prisms, and hence volumeof right triangular prism = ^ab x c = area of its base x altitude. Since every prism can be divided into triangular prisms as in fig. 40, we have at once volume of right prism A ABCDE = A ABC + A ACD + A ADE = ABC x BB + ACD x CC + ADE x DD = ( ABC + ACD + ADE) x altitude (since BB = CC = DD = altitude) = area of base ABCDE x altitude. Fig. 40. Fig. 41. 67. Volume of an Oblique Prism. Draw the right section A&quot;B&quot;C&quot;D&quot;E&quot; (fig. 41), and let A denote its area and A the area of the base A B C D E. Let I denote the length of the prism, h its altitude, and a the angle between the planes A B C D E and A&quot;B&quot;C&quot;D&quot;E&quot;. Conceive the part above the right section placed at the other extremity of the prism. Then we have a right prism, whose volume = A x I ( 66); but A = Acos o, since A is the projection of A ( 51), and Z= - ; hence cos a volume = A x I = A cos a x cos a A x h or the volume of any prism is equal to the area of its base multi plied by its altitude.
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