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SalasSV_09_05 - 548 CHAPTER 9 THE CONIC SECTIONS POLAR...

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9.5 AREA IN POLAR COORDINATES Here we develop a technique for calculating the area of a region the boundary of which is given in polar coordinates. As a start, we suppose that α and β are two real numbers with α < β α + 2 π . We take ρ as a function that is continuous on [ α , β ] and keeps a constant sign on that interval. We want the area of the polar region generated by the curve r = ρ ( θ ), α θ β . Such a region is portrayed in Figure 9.5.1. In the figure ρ ( θ ) remains nonnegative. If ρ ( θ ) were negative, the region would appear on the opposite side of the pole. In either case, the area of is given by the formula = θ β polar axis = θ α r = ρ θ ( ) Γ O Figure 9.5.1 (9.5.1) A = β α 1 2 [ ρ ( θ )] 2 d θ . PROOF We consider the case where ρ ( θ ) 0. We take P = { θ 0 , θ 1 , . . . , θ n } as a partition of [ α , β ] and direct our attention to what happens from θ i 1 to θ i . We set r i = min value of ρ on [ θ i 1 , θ i ] and R i = max value of ρ on [ θ i 1 , θ i ]. The part of that lies between θ i 1 and θ i contains a circular sector of radius r i and central angle θ i = θ i θ i 1 and is contained in a circular sector of radius R i with the same central angle θ i . (See Figure 9.5.2.) Its area A i must therefore satisfy the inequality polar axis R i r i i 1 θ i θ = α θ o = β θ n O Figure 9.5.2 1 2 r 2 i θ i A i 1 2 R 2 i θ i . By summing these inequalities from i = 1 to i = n , we can see that the total area A must satisfy the inequality (1) L f ( P ) A U f ( P ) † The area of a circular sector of radius r and central angle α is 1 2 r 2 α .
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9.5 AREA IN POLAR COORDINATES 549 where f ( θ ) = 1 2 [ ρ ( θ )] 2 . Since f is continuous and (1) holds for every partition P of [ α , β ], we can conclude that A = β α f ( θ ) d θ = β α 1 2 [ ρ ( θ )] 2 d θ . Example 1 Calculate the area enclosed by the cardioid r = 1 cos θ (Figure 9.5.3) SOLUTION The entire curve is traced out as θ increases from 0 to 2 π , and 1 cos θ 0 on [0, 2 π ]. Therefore polar axis [2, ] π r = 1 cos θ 1, π 2 1, π 2 3 [ ] [ ] Figure 9.5.3 A = 2 π 0 1 2 (1 cos θ ) 2 d θ = 1 2 2 π 0 (1 2 cos θ + cos 2 θ ) d θ = 1 2 2 π 0 ( 3 2 2 cos θ + 1 2 cos 2 θ ) d θ . half-angle formula: cos 2 θ = 1 2 + 1 2 cos 2 θ Since 2 π 0 cos θ d θ = sin θ 2 π 0 = 0 and 2 π 0 cos 2 θ d θ = 1 2 sin 2 θ 2 π 0 = 0, we have A = 1 2 2 π 0 3 2 d θ = 3 4 2 π 0 d θ = 3 2 π .
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