. Differential and integral calculus. ;72. Ans. 7Ttf3. Ans. # 7r«^r. 340 Integral Calculus CHAPTER VIII. GEOMETRIC APPLICATIONS. 215. Quadrature of Plane Areas. From § 206 (a), we have dA = this equation with respect to y we have d2A = , A = J j dydx. (1) From § 206, (b), we have dA = ±r*d0; hence d2A = rdrdO; »rdrdO (2) ?—/./ The order of the differentials in (1) and (2) are obviouslyimmaterial. Formulas (1) and (2) enable us to determine the areasbounded by curves by double integration. EXAMPLES. 1. To find the area of the circle x2 -f-72 = a1 by double inte-gra


. Differential and integral calculus. ;72. Ans. 7Ttf3. Ans. # 7r«^r. 340 Integral Calculus CHAPTER VIII. GEOMETRIC APPLICATIONS. 215. Quadrature of Plane Areas. From § 206 (a), we have dA = this equation with respect to y we have d2A = , A = J j dydx. (1) From § 206, (b), we have dA = ±r*d0; hence d2A = rdrdO; »rdrdO (2) ?—/./ The order of the differentials in (1) and (2) are obviouslyimmaterial. Formulas (1) and (2) enable us to determine the areasbounded by curves by double integration. EXAMPLES. 1. To find the area of the circle x2 -f-72 = a1 by double inte-gration. In this case (1) § 215 becomes for the area of the first quadrant Geometric Applications 341 /•a /Vis si A= I I axdxdyJo Jo and it will be noted that the limits of y are from y = o (jt~-axis) to _y = Vtf — #2 (any point on the curve). Thus the first integration (x and its differential being constant) would give the area contained in the strip MNPQ, which, itself, would be the differential area of the portion Fig. 58. Hence the limits of x are selected as insingle integration from the origin to the limits of the curve on thejf-axis, , from x = o to x = a, andThe entire area is then found from A = *XX dxdy = 4 J V#2 — x1 dx id*{ — I 2 7T 1 \ 2 J Ex. 1, p. 322. 2. Find the area of the circle using a polar equation. Let OXbe the polar axis and O the pole, then r = 2 a cos 6will be its polar equation. Hence, (2) § 215, we have for theupper half


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