No constant is added because o vanishes with 0. Thus and since c is less than unity, 4 is greater than c (1 + c)2. 228. We will give one more proposition in this subject, by establishing a relation among Elliptic Functions of the second order, analogous to that proved in Art. 225 for functions of the first order. If cos e cos - sine sin √(1-c2 sin2 μ) = cos μ, then will E (c, 0) + E (c, d) – E (c, μ) = c2 sin 0 sin & sin μ. By virtue of the given equation connecting the amplitudes, 4 is a function of 0; thus we may assume E (c, 0)+E (c, d) — E (c, μ) =ƒ(0). Differentiate; thus аф ƒ′ (0) = √/(1 — c2 sin2 0) + √/ (1 — c2 sin2 4) do But sin2 + sin2 + 2 cos 0 cos & cos μ 1 2 sine sin o sinμ No constant is added, because f(0) obviously vanishes with 0. 180 CHAPTER XI. CHANGE OF THE VARIABLES IN A MULTIPLE INTEGRAL. 229. WE have seen in Art. 62 that the double integral B rb [*[* (x, y) dx dy is equal to f(x, y) dy do when the ανα a limits are constant, that is, a change in the order of integration produces no change in the limits for the two integrations. But when the limits of the first integration are functions of the other variable, this statement no longer holds, as we have seen in several examples in the seventh and eighth chapters. We add here a few additional examples. The limits of the integration with respect to y here are y=0 and y=√(a2x2); that is, we may consider the integral extending from the axis of x to the boundary of a circle, having its centre at the origin, and radius equal to a. Then the integration with respect to x extends from the axis of y to the extreme point A of the quadrant. Thus if we consider z=(x, y) as the equation to a surface, the above double integral represents the volume of that solid which is contained between the surface, the plane of (x, y), and a line moving perpendicularly to this plane round the boundary ОАРВО. It is then obvious from the figure that if the integration with respect to x is performed first, the limits will be x=0 and x=√(ay), and then the limits for y will be y=0 and y=a. Thus the transformed integral is faf √ia2-39$ (x, y) dy dx. 231. Change the order of integration in OA as dia Thus the Let OA=2a, and describe a semicircle on meter. Let POX = 0, then OP=2a cos 0. double integral may be considered as the limit of a summation of values of † (r, 0) r ▲ ▲r over all the area of the semicircle. Hence when the order is changed we must inte grate for from 0 to cos1 and for r from 0 to 2a. Դ The integration for y is taken from y = x2 4a The equation y = belongs to a parabola OLD, and y=3a-x to a straight line BLC, which passes through L, the extremity of the latus rectum of the parabola. Thus the integration may be considered as extending over the area OLBSO. Now let the order of integration be changed; we shall have to consider separately the spaces OLS and BLS. For the space OLS we must integrate from x=0 to x=2(ay), and then from y = 0 to y = a; and for the space BLS we must integrate from x = 0 to x=3a-y, and then from y = a to y = 3a. Thus the transformed integral is (a (2√(ay) 00 3a (3a-y 4 (x, y) dy dx + ***a ̄*4 (x, y) dy dx. a o |