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**Unformatted text preview: **ries of the region and the intervals of θ values over which these boundaries are traced out. Since the polar coordinates of a point are not unique, extra care is needed to determine these intervals of θ values.
Example 2
SOLUTION r=1 r = 2 cos θ O polar axis Find the area of the region that consists of all points that lie within the circle r = 2 cos θ but outside the circle r = 1.
θ=
5π 3 The region is shown in Figure 9.5.5. Our ﬁrst step is to ﬁnd values of θ for the two points where the circles intersect: 2 cos θ = 1, cos θ = 1 , 2 θ = 1π, 5π. 3 3 Figure 9.5.5 550 CHAPTER 9 THE CONIC SECTIONS; POLAR COORDINATES; PARAMETRIC EQUATIONS Since the region is symmetric about the polar axis, the area below the polar axis equals the area above the polar axis. Thus A=2
0 π /3 1 ([2 cos θ ]2 2 − [1]2 ) d θ .
1 2 If you carry out the integration, you will see that A = 1 π + 3
θ= π
r = 1 – 2 cosθ
1 3 √ 3 ∼ 1. 91. = Example 3 limaçon Find the area A of the region between the inner and outer loops of the r = 1 − 2 cos θ .
(Figure 9.5.6) SOLUTION You can verify that r = 0 at θ = π/3 and at θ = 5π/3. The outer loop is formed as θ increases from π/3 to 5π/3. Thus
θ= π
5 3 area within outer loop = A1 = 5π/3 π/3 1 [1 2 − 2 cos θ ]2 d θ . Figure 9.5.6 The lower half of the inner loop is formed as θ increases from 0 to π/3, and the upper half as θ increases from 5π/3 to 2π (verify this). Therefore area within inner loop = A2 =
0 π /3 1 [1 2 − 2 cos θ ]2 d θ + 2π 5π/3 1 [1 2 − 2 cos θ ]2 d θ . Now 1 [1 2 − 2 cos θ ]2 d θ = = = 1 2 1 2 1 2 [1 − 4 cos θ + 4 cos2 θ ] d θ [1 − 4 cos θ + 2(1 + cos 2θ )] d θ [3 − 4 cos θ + 2 cos 2θ ] d θ = 1 [3θ − 4 sin θ + sin 2θ ] + C . 2
5π/3 π/3 Therefore A1 = 1 2 3θ − 4 sin θ + sin 2θ = 2π + 3 2 √ 3 and A2 = = 1 2 3θ − 4 sin θ + sin 2θ −
3 4 π /3 0 + 1 2 3θ − 4 sin θ + sin 2θ 2π 5π/3 1 π 2 √ 3+ 1 π 2 √ √ − 3 = π − 3 3. 2
3 4 Thus, A = A 1 − A 2 = 2π + 3 2 √ 3 − (π − 3 2 √ √ 3...

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