Anonymous
Anonymous asked in Science & MathematicsMathematics · 7 months ago

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

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  • Mike G
    Lv 7
    7 months ago
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    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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