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Math Help - Line Integral for a sawtooth pattern

  1. #1
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    Line Integral for a sawtooth pattern

    I have to evaluate the line integral:
    \int_Z(sin y + 2x) dx + (2y + x cos y)dy

    from the point (0,0) to (1,1) to (3,-1) to (4,0).

    I evaluated the integrals for each of the 3 separate lines such that:
    \int_Z = \int_{C_1} + \int_{C_2} + \int_{C_3}

    I got the answer of 2cos(1) - 3sin(1) -7 but I am pretty sure this is wrong. Can someone help?
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  2. #2
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    Re: Line Integral for a sawtooth pattern

    Quote Originally Posted by tammyl View Post
    I have to evaluate the line integral:
    \int_Z(sin y + 2x) dx + (2y + x cos y)dy

    from the point (0,0) to (1,1) to (3,-1) to (4,0).

    I evaluated the integrals for each of the 3 separate lines such that:
    \int_Z = \int_{C_1} + \int_{C_2} + \int_{C_3}

    I got the answer of 2cos(1) - 3sin(1) -7 but I am pretty sure this is wrong. Can someone help?

    Show your work on \int_{C_1}, so that we can see what you are doing.
    Thanks from topsquark
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  3. #3
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    Re: Line Integral for a sawtooth pattern

    Thanks for the response.

    So for C_1,

    Parameterisation of straight line from (0,0) to (1,1)
    x = a(1-t) + ct = t where a=0, c=1
    y = b(1-t) + dt = t where b=0, d=1
    Thus
    dx=dt and dy=dt where 0\leq t \leq1

    So,
    \int_{C_1} (sin y + 2x)dx + (2y + x cos y)dt
    = \int_0^1 (sin t + 2t)dt + (2t + t cos t)dt
    = \int_0^1 [sin t + 4t + t cos t] dt
    = [-cost + 2t^2 + t sin t + cos t]_0^1
    = [2t^2 + t sin t]_0^1
    = (2.(1)^2 - 2.(0)^2) + (1.sin (1) - 0.sin(0))
    = 2 + sin (1)

    Repeating the procedure for the other two straight lines, I got
    \int_{Z} = \int_{C_1} + \int_{C_2} + \int_{C_3} = (2 + sin (1)) + (-16 - 4sin (1))+(7+2cos(1))= 2 cos(1) - 3 sin(1) -7

    I really am hoping my technique is right and its my integration that sucks.

    Any help would be great.
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  4. #4
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    Re: Line Integral for a sawtooth pattern

    Quote Originally Posted by tammyl View Post
    Thanks for the response.
    So for C_1,
    Parameterisation of straight line from (0,0) to (1,1)
    x = a(1-t) + ct = t where a=0, c=1
    y = b(1-t) + dt = t where b=0, d=1
    Thus
    dx=dt and dy=dt where 0\leq t \leq1

    So,
    \int_{C_1} (sin y + 2x)dx + (2y + x cos y)dt
    = \int_0^1 (sin t + 2t)dt + (2t + t cos t)dt
    = \int_0^1 [sin t + 4t + t cos t] dt
    = [-cost + 2t^2 + t sin t + cos t]_0^1
    = [2t^2 + t sin t]_0^1
    = (2.(1)^2 - 2.(0)^2) + (1.sin (1) - 0.sin(0))
    = 2 + sin (1)

    Repeating the procedure for the other two straight lines, I got
    \int_{Z} = \int_{C_1} + \int_{C_2} + \int_{C_3} = (2 + sin (1)) + (-16 - 4sin (1))+(7+2cos(1))= 2 cos(1) - 3 sin(1) -7
    Good, the above is correct.

    If you are doing the other two carefully then the result should be correct.

    The difficulty that I see is that both contours are more complicated.
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