SalasSV_09_05

# 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 ±gure 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 = θβ polar axis = θα r = ρθ ( ) Γ O Figure 9.5.1 (9.5.1) A = ³ 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 ²θ i = θ i θ i 1 and is contained in a circular sector of radius R i with the same central angle ²θ 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 = αθ o = βθ n O Figure 9.5.2 1 2 r 2 i ²θ i A i 1 2 R 2 i ²θ i . † By summing these inequalities from i = 1to 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, 2 1, 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 ² ² 2 π 0 = 0 and ± 2 π 0 cos 2 θ d θ = 1 2 sin 2 θ ² ² ² 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 . ± A slightly more complicated type of region is pictured in Figure 9.5.4. We approach the problem of calculating the area of the region ± in the same way that we calculated the area between two curves in Section 5.5; that is, we calculate the area out to
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SalasSV_09_05 - 548 CHAPTER 9 THE CONIC SECTIONS POLAR...

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