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Thread: 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:
    $\displaystyle \int_Z(sin y + 2x) dx + (2y + x cos y)dy$

    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:
    $\displaystyle \int_Z = \int_{C_1} + \int_{C_2} + \int_{C_3}$

    I got the answer of $\displaystyle 2cos(1) - 3sin(1) -7$ 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:
    $\displaystyle \int_Z(sin y + 2x) dx + (2y + x cos y)dy$

    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:
    $\displaystyle \int_Z = \int_{C_1} + \int_{C_2} + \int_{C_3}$

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

    Show your work on $\displaystyle \int_{C_1}$, 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 $\displaystyle C_1$,

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

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

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

    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 $\displaystyle C_1$,
    Parameterisation of straight line from (0,0) to (1,1)
    $\displaystyle x = a(1-t) + ct = t$ where $\displaystyle a=0, c=1$
    $\displaystyle y = b(1-t) + dt = t$ where $\displaystyle b=0, d=1$
    Thus
    $\displaystyle dx=dt$ and $\displaystyle dy=dt$ where $\displaystyle 0\leq t \leq1$

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

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