. Differential and integral calculus, an introductory course for colleges and engineering schools. §215 MULTIPLE INTEGRALS 327 Moreover, the equations of Si and S2 are z = c + Vr2 - (x - af - (y - 6)2 = fx{x, y), and z = c - Vr2 - (x - a)2 - (y - b)2 = f2 (x, y).Hence ° x = a+r y=S(x) Vx=j f[c + ^r*-(x-ay-(y-by]dydx, a a+r x—a— r y=y(x) x=a+r y=d(x and v2 -s Jl- Hence x=a—r y=y{x) dx. x=a-\-r y=S(x) V=V1-V2 = ff(fl-f2)dydx = 2J f Vr2 - (x - a)2 -(y- 6)2 dy dx. R x = a—r y=y{x) Integrating by the table of integrals, we have y=8(x) f Vr2-(x-aY-(y-b)2dy = | (y -b)Vr2-(x-a y-{y-by y=y{x) |2/=6+V/r
. Differential and integral calculus, an introductory course for colleges and engineering schools. §215 MULTIPLE INTEGRALS 327 Moreover, the equations of Si and S2 are z = c + Vr2 - (x - af - (y - 6)2 = fx{x, y), and z = c - Vr2 - (x - a)2 - (y - b)2 = f2 (x, y).Hence ° x = a+r y=S(x) Vx=j f[c + ^r*-(x-ay-(y-by]dydx, a a+r x—a— r y=y(x) x=a+r y=d(x and v2 -s Jl- Hence x=a—r y=y{x) dx. x=a-\-r y=S(x) V=V1-V2 = ff(fl-f2)dydx = 2J f Vr2 - (x - a)2 -(y- 6)2 dy dx. R x = a—r y=y{x) Integrating by the table of integrals, we have y=8(x) f Vr2-(x-aY-(y-b)2dy = | (y -b)Vr2-(x-a y-{y-by y=y{x) |2/=6+V/r2-(x-a)2 + ir2_(a;_a)2(sin-l y ° = [r2_(a:_a)2]. Vr2 —(a: —a)2Jy=6-Vr2-(x-a)2 ^Therefore x=a-\-r a+r V = 7T Mr2 — (a; — a)2] dx = v \r2x — - (x — a)3 = -ttt3. Aws. 215. A Plane Area Expressed as a Double Integral. In formula (B) let /(«, ?/) = 1. Then the double integral takes thez. special form J> dx and is rep- resented geometrically by the vol-ume of a right cylinder of base Rand of altitude 1. But becausethe volume is in this case expressedby the same number as the area ofthe base, we may regard this doubleintegral as giving the area of this point of view the area of 328 INTEGRAL CALCULUS 216 R is the limit of a sum of rectangles of dimensions Ay and have then the general formula for a plane area, x=b y=8(x) R = f I dy dx = I j dy dx. Now and R S(x) I dy - 8(x)y(x) x=a y=y(x) y(x) = AiAi, nb nb R = / 8(x) dx — I y(x) dx = aAA2Bb — aAAxBb. U a J a From this it appears that this process amounts after all to anapplication of our familiar formula for a plane area, A = I y dx. 0a In fact, there is seldom any advantage in using a double integralrather than a single integral to calculate a plane area. Problem. Express as a double integral in polar coordinates, p, 6, thearea of a sector bounded by a curve and two radii vectores. Suggestion. Divide
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