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Hence the area of the cycloid is trisected by the base circle on its axis, and the companion to the cycloid ; see fig. 20.

Ex. 11. The whole area of the loop of the curve whose equation is ay2 = 22 (a? — x2), included between x = a and x = 0,

4a2

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Ex. 12. The area included between the axis of x, two ordinates, and the logarithmic curve y = a“, is

Qin – Quito

log a' and that included between the curve, the asymptote, and the axis

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Ex. 13. The area included between the axis of x and the curve y = a sin, for the limits x = 0 and x = ma, is 2 a?.

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Ex. 14. The area oaPm of the catenary in fig. 7 is equal to the rectangle contained by oa and the arc AP, and therefore is equаl tо twice the triangle PHм.

Ex. 15. If [*' dy dx=ay; it is required to find the equation of the curve.

ay = f*Lay de

= /*y dx ; is a dy = y dx ;

y = be. Ex. 16. If ["L"ay do = minm ay, shew that (3) * = (%)".

221.] In all the preceding examples, the y-integration has preceded the x-integration, and we have by this process first determined the general value of a differential slice of infinitesimal breadth dx, contained between parallel ordinates, and by the summation of these determined the required area. The order of integration however, as we remarked in the preceding chapter, is indifferent; though if the order is changed, of course the limits must be changed: this we shall exemplify in a few cases.

Ex. 1. In determining the area OAB, fig. 21, where on = a, AB=b, and the equation to the bounding curve is ayo=b2x; if we first effect the x-integration, y being the same for all, we sum the elements along the line pk, that is, from x = a to X = a; and thereby obtain the area of the slice PQLK contained between two parallel abscissæ separated by the distance dy; which slices must again be summed with respect to y, the limits of integration being b and 0. Hence

a

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the same result as that of Ex. 1, 'Art. 220.

Ex. 2. The equation to the equitangential curve being

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and x being the general value of the abscissa to the curve, the area included between the curve, the axis of x, and the axis of y

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the same result as that of Ex. 6, Art. 220.

222.] If it is required to determine the area contained between two ordinates corresponding to Xn, Xo, and between two curves whose equations are y=f(x), y = $(x), the former being the equation to the upper, and the latter that to the lower curve; then, as is evident from fig. 22, the y-integration must be first effected, and for the limits f (x) and $ (r); the result of which will give the area of the slice pp'Q'Q; and the subsequent definite x-integration will give the sum of all such slices between the assigned limits; and this will be the required area. Thus

the area = /**/** dy dx.

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If however the superficies, whose area is required, is of a form such as that delineated in fig. 23, it is more convenient to resolve it into slices whose bounding lines are parallel to the axis of x, that is, first to effect the x-integration, for in such a case the equation to the curves will give the limits of integration : the equation to the curve AQPB giving the superior and that to AQ'P'B giving the inferior limit; which manifestly they do not, if the y-integration is first performed; in this case, if the equations to the curves are

x = f(y), X = $(y),
and if the ordinates to A and B are yn and yo,

syn p(y)
the area = | 7 dx dy.

yo (y) Sometimes also it is necessary to divide a problem of quadrature into two or more parts, and to integrate each of the double integrals in the order which is most convenient for its form and limits : such division however must be left to the ingenuity of the student, for no general rules can be given, but the principles of the calculus are of sufficient breadth to include all such cases.

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223.] Examples illustrative of the preceding principles.

Ex. 1. To determine the area included between the parabola whose equation is = 4ax, and the straight line whose equation is y = Bx; see fig. 24. The coordinates to the point B, determined by elimination be

4a 4a tween the given equations, are oa=22, AB = t; therefore

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2 (ar)

the area OPB =

= )" {2(x) @e} die
= [40*7 ) Sep

Ex. 2. To find the area contained between an hyperbola, its transverse axis, and a central radius vector; fig. 25.

Let P, to which the radius vector is drawn be (@ne Yn); and let the limits of x be denoted by X, and xo; then the equation to OP is

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The order in which the integrations have been effected, in relation to the limits, deserves attention; as the superficies P.,OA admits of being resolved into slices by lines parallel to the axis of X, the limits of which are given by the equations to the straight line and the curve, we have effected first the x-integration, and subsequently the y-integration ; but the order could have been reversed, only subject to other conditions : viz. if we had integrated first with respect to y, the limits would have been the ordinate to the straight line and zero, for all values of x from o to A, but at A, and thence on to Pm, the superior and inferior limits would have been respectively the ordinate to the straight line and the ordinate to the hyperbola; and the definite integral must have been divided into two parts corresponding to these limits. Ex. 3. AP being the catenary, in fig. 7, whose equation is

y = {ea + e-á}, and oa = a, on = 0, the area apr = {5 log 2–3}.

Ex. 4. The area included between a parabola whose equation is y2 = 4ax, and a straight line through the focus inclined at 45° to the axis of x, is = a 21.

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Ex. 5. The equation to a curve being (y x)2 = a— xa, the area = nao.

224.] The quadrature of an area is frequently facilitated by a substitution, and chiefly by putting the equation to the bounding curve into simultaneous equations by the introduction of a subsidiary angle, according to the method of Art. 193, Vol. I. The following are examples of the process.

Ex. 1. The equations to an ellipse are x = a cos 0, y = b sino; find the whole area.

Let b sin , which is the ordinate of the ellipse, be denoted by Y: then

the area = L * S*dydx = 2 so v sin $ dx. Now dx = -a sin o do; and when x = a, 0 = 0; when x = -a, = . Consequently

the area = 2ab "(sin op)? = ab.

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Ex. 2. The equations to the cycloid, fig. 6, are x = a(0-sine), y = a (1-сos 6). Then, using the notation of the last example,

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