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# Find the area between the inner and outer loops of the curve r=2cos(theta)-1.?

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- Mike GLv 71 year agoFavorite Answer
Because of symmetry I just considered the top part and will multiply by 2

Area of smaller semi-loop =

∫(1/2)r^2 dθ [0,π/3] =

∫(1/2)[2cosθ-1]^2 dθ [0,π/3] =

(1/2)∫[4cos^2(θ)-4cosθ+1] dθ [0,π/3] =

∫cos(2θ)-2cosθ+3/2 dθ [0,π/3] =

[sin(2θ)/2-2sinθ+3θ/2] [0,π/3] =

sin(2π/3)/2-2sin(π/3)+π/2 =

(√3)/4 - √3 + π/2

≈ 0.2718

Area of the larger semi-loop has the same integral but different bounds

The values of θ forming the upper loop are in Q3 and Q4

A = [sin(2θ)/2-2sinθ+3θ/2] [π, 5π/3] ≈ 4.4406

4.4406 - 0.2718 = 4.1688

Required area = 4.1688*2

= 8.3376

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