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Math Help - line integral of a shifted semi-circle

  1. #1
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    line integral of a shifted semi-circle

    This problem is to find the line integral of a semi-circle, from (3,0) to (1,0), centered at (2,0) above the x-axis, y > 0. The vector field is (x i + y j)

    I have calculated that the range for t is: π < t < 0,

    Parameterizing of a unit circle:
    r(t) = x(t)i + y(t)j = cos(t)i + sin(t)j
    x=cos(t)
    y=sin(t)

    This is what Ive done, but the answer is obviously incorrect:
    The circles equation is (x-2)^2 + (y-0)^2 = 1
    x-2=cos(t), so that:
    x=cos(t)+2
    y=sin(t)

    F = [(cos(t)+2] i + [sin(t)] j
    r(t) = dr/dt = x(t) i + y(t) j = { -sin(t) i + cos(t) j }
    dr = { -sin(t) i + cos(t) j } dt
    F dr = ∫(from 0 to π) { [(cos(t)+2] i + [sin(t)] j -sin(t) i + cos(t) j } dt

    = ∫(from 0 to π) { -sin(t)cos(t)-2sin(t)] +cos(t)sin(t) } dt
    = { -sin(π)cos(π) - 2sin(π) + cos(π)sin(π) } - { -sin(0)cos(0) - 2sin(0) + cos(0)sin(0) }
    = { 0 - 0 + 0 } - { -0 - 0 + 0 } (all the sins equal zero)
    = 0

    (As a side note, how does π < t < 0 differ from π ≤ t ≤ 0 in calculating the answer?)

    Thanks in advance.
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  2. #2
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    Re: line integral of a shifted semi-circle

    Before I go any further, are you only travelling along the curved surface of the semicircle, or are you also travelling along the x axis in order to close the semicircle?
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  3. #3
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    Re: line integral of a shifted semi-circle

    only along the semicircle, not including the x-axis. also , the area of the field is y>0, so the endpoints of the semicircle are not included.
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  4. #4
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    Re: line integral of a shifted semi-circle

    I did more work on this problem. My answer is -4.
    Could you please see if Ive solved it correctly?

    (Previously, I miswrote the range. It should be 0 < t < π )

    F dr = ∫ (from 0 to π) -sin(t)cos(t)dt - 2sin(t)dt + sin(t)cos(t)

    Integrating this, I broke it into 3 separate parts:

    1: ∫ (from 0 to π) -sin(t)cos(t) dt
    2: ∫ (from 0 to π) 2sin(t) dt
    3: ∫ (from 0 to π) sin(t)cos(t) dt


    1: using substitution:
    u = cos(t)
    du/dt = 1sin(t)
    du = -sin(t)dt
    dt = du/-sin(t)

    ∫ (from 0 to π) -sin(t)cos(t)(du/-sin(t))
    = ∫ (from 0 to π) cos(t)(du)
    = ∫ (from 0 to π) u du
    = (1/2)u^2, evaluated from π to 0
    = (1/2)cos^2(t), evaluated from π to 0
    = (1/2)cos^2(π) - (1/2)cos^2(0)
    using the trig identity cos^2(x) = { [1 + cos(2x)] / 2 }
    = (1/2) { ( [1+cos(2π)] / 2) - ( [1+cos(2(0))] / 2) }
    = 1/2(1/2 +1/2) - 1/2(1/2 +1/2)
    = (1/2)(1) - (1/2)(1)
    = 0


    2:
    - ∫ (from 0 to π) 2sin(t) dt
    = -2 ∫ (from 0 to π) sin(t) dt
    = -2[-cos(t)] evaluated from π to 0
    = 2[cos(t)] evaluated from π to 0
    = 2[cos(π) - cos(0)]
    = 2[(-1) - (1)]
    = -4

    3:
    u = cos(t)
    du/dt = 1sin(t)
    du = -sin(t)dt
    dt = du/-sin(t)

    ∫ (from 0 to π) -sin(t)cos(t)(du/-sin(t))
    = ∫ (from 0 to π) - u du
    = (-1/2)u^2, evaluated from π to 0
    = (-1/2)cos^2(t), evaluated from π to 0
    = (-1/2) [cos^2(π) - cos^2(0)]
    = (-1/2) { [1+cos(2π)] / 2) - ( [1+cos(2(0)] / 2) }
    = (-1/2) { [(1/2) + (1/2)] - [(1/2) + (1/2)] }
    = 0

    (0) + (-4) + (0) = -4
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