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Math Help - Double Integration in Polar Coordinates

  1. #1
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    Double Integration in Polar Coordinates

    Ok, so I have tried this problem like 4 times and have not gotten the correct answer yet. The instructions say, "Use a double integral to find the area of the shaded region."

    11. r=2(1-cosθ)
    Then there is a graph with the entire limacon filled in.

    This is what I have done but I haven't gotten the correct answer.

    ∫∫rdrdθ

    ∫ (1/2 r^2 |(0 to 2(1-cosθ)) )dθ

    ∫ (1/2 *4 (sin^2(θ)))dθ

    ∫ (2 sin^2(θ))dθ

    ∫ (2*-1/2 (cos(2θ)-1))dθ

    ∫ (-cos(2θ) + 1)dθ

    u=2θ
    du=2dθ

    -1/2 ∫ (cosu)du + ∫1dθ

    -1/2 (sin2θ) | (0 to 2π) + θ | (0 to 2π)

    0 - 0 + 2π - 0 =

    Why is this not correct? It is supposed to be 6π. Where have I made a mistake?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by dude15129 View Post
    Ok, so I have tried this problem like 4 times and have not gotten the correct answer yet. The instructions say, "Use a double integral to find the area of the shaded region."

    11. r=2(1-cosθ)
    Then there is a graph with the entire limacon filled in.

    This is what I have done but I haven't gotten the correct answer.

    ∫∫rdrdθ

    ∫ (1/2 r^2 |(0 to 2(1-cosθ)) )dθ

    ∫ (1/2 *4 (sin^2(θ)))dθ This is where your mistake is: \color{red}(1-\cos\vartheta)^2\neq \sin^2\vartheta!!!

    ∫ (2 sin^2(θ))dθ

    ∫ (2*-1/2 (cos(2θ)-1))dθ

    ∫ (-cos(2θ) + 1)dθ

    u=2θ
    du=2dθ

    -1/2 ∫ (cosu)du + ∫1dθ

    -1/2 (sin2θ) | (0 to 2π) + θ | (0 to 2π)

    0 - 0 + 2π - 0 =

    Why is this not correct? It is supposed to be 6π. Where have I made a mistake?
    --Chris
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