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**Unformatted text preview: **[θi−1 , θi ]. The part of that lies between θi−1 and θi contains a circular sector of radius ri and central angle θi = θi − θi−1 and is contained in a circular sector of radius Ri with the same central angle θi . (See Figure 9.5.2.) Its area Ai must therefore satisfy the inequality
12 r 2i θi ≤ Ai ≤ 1 R2 2i θi . † O Figure 9.5.2 By summing these inequalities from i = 1 to i = n, we can see that the total area A must satisfy the inequality (1) Lf (P ) ≤ A ≤ Uf (P ) † The area of a circular sector of radius r and central angle α is 1 r 2 α . 2 9.5 AREA IN POLAR COORDINATES 549 where f (θ ) = 1 [ρ (θ )]2 . 2 Since f is continuous and (1) holds for every partition P of [α , β ], we can conclude that A=
Example 1
β α [1, ]
r = 1 – cos θ [2, π ] polar axis π 2 f (θ ) d θ = β α 1 [ρ (θ )]2 d θ . 2 Calculate the area enclosed by the cardioid r = 1 − cos θ
(Figure 9.5.3) [1, ]
Figure 9.5.3 3π 2 SOLUTION The entire curve is traced out as θ increases from 0 to 2π , and 1 − cos θ ≥ 0 on [0, 2π ]. Therefore A=
0 2π 1 (1 2 − cos θ )2 d θ = = ↑ 1 2 1 2 2π 0 2π 0 (1 − 2 cos θ + cos2 θ ) d θ ( 3 − 2 cos θ + 1 cos 2θ ) d θ . 2 2
half-angle formula: cos2 θ =
1 2 + 1 2 cos 2θ 2π Since
0 cos θ d θ = sin θ 2π 0 =0
1 2 2π 0 3 2 2π and
0 cos 2θ d θ = 1 2 sin 2θ 2π 0 = 0, we have A= dθ = 3 4 2π 0 dθ = 3 π . 2
θ=β 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 r = p2 (θ ) and subtract from it the area out to r = p1 (θ ). This gives area of which can be written as
(9.5.2) Ω r = ρ1(θ ) r = ρ2(θ ) θ=α = β α 1 [ρ (θ )]2 d θ 22 − β α 1 [ρ (θ )]2 d θ , 21 O polar axis Figure 9.5.4 area of = β α 1 ([ρ2 (θ )]2 2 − [ρ1 (θ )]2 ) d θ . θ= π 3 To ﬁnd the area between two polar curves, we ﬁrst determine the curves that serve as outer and inner bounda...

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