. Differential and integral calculus. Geometric Applications 321 dA = \ = \r2dQ\ , (b) .-. a=* r w* (3) where <£ = ^tfJT and ^ = POX. Equation (3), it will be ob-served, gives the area bounded by the curve and terminal radii-vectores. For example, let us find the area of one loop of the lemniscata, r2 = a2 cos 2 Fig. 54- From tne equation we observe that the limiting values ofare 450 and - 450; Hence, •• = 45° and i// = - 450. =£ r w* jr w : I / *rV0 = % I Vcos 2 0//0 U IT *J IT 14 = \ a2 sin 2 0 ; hence, A = - , the area of the loop is \ the square constructed on a. 322 Integr
. Differential and integral calculus. Geometric Applications 321 dA = \ = \r2dQ\ , (b) .-. a=* r w* (3) where <£ = ^tfJT and ^ = POX. Equation (3), it will be ob-served, gives the area bounded by the curve and terminal radii-vectores. For example, let us find the area of one loop of the lemniscata, r2 = a2 cos 2 Fig. 54- From tne equation we observe that the limiting values ofare 450 and - 450; Hence, •• = 45° and i// = - 450. =£ r w* jr w : I / *rV0 = % I Vcos 2 0//0 U IT *J IT 14 = \ a2 sin 2 0 ; hence, A = - , the area of the loop is \ the square constructed on a. 322 Integral Calculus Find area of the circle from its rectangular equation, J fa f*a ydx — 4 I y/a2—.x2dxo Jo = 4j-V^^+-sin-1-((2 2 a) See Ex. 3, p. 285. 4 Ksin-Vj + C- ^sin^o - C j «2 7T 4- 2 2,2 2. Find the area of the circle from its polar equationr — 2 a cos 6, the left-hand extremity of the horizontal diameterbeing the pole and the diameter being the initial line. Here A = 1 | r2d6 Jty 2 ( 6 sin 2 0 ) I*= 2 at See Ex. 2, p. 300. <4 4 V 4 4 /) l2 cos2 2 7T 7T 2 ar < - -t- -4 4 = 7T^. 1374 58 Geometric Applications 323 3. Find the area in Ex. (2), (a), when the center is the poleand any diameter is the initial line; (b), when any point on thecurve is the pole and the
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