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Evaluate the line integral

  1. Dec 16, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex] \int xydx+ 4ydy[/tex]
    where C is the curve from (1,2) to (3,5) made up of the twoline segments parallel to the coordinate axes.
    [tex]c_1:(1,2)\rightarrow(3,2)[/tex]
    [tex]c_2:(3,2)\rightarrow(3,5)[/tex]

    2. Relevant equations


    3. The attempt at a solution
    i got c2 correct, y=2+3t, and x = 0, for t goes from 0 to 1.
    but i got c1 wrong, for c1, i see only x is changing, x=1+2t. so x'(t)= 2. if y value is not changing, it means that dy=0, my professor had y=1, by setting 2xy=2x, i guess 2xy is the first half of the initial integral by replacing dx with 2, but i dont understand what is 2x on the right side of the equation.
     
  2. jcsd
  3. Dec 16, 2015 #2
    Hi qq:

    I don't understand why you introduce the t variable unless it is that you are used to doing that in general for arbitrary curves. I also don't understand why you use t=0 and t= 1 as the limits. For this particular problem I suggest you write down the integrals with the limits specified for both c1 and c2. Since y is constant for c1 and x is constant for c2, these two integrals should be easy to integrate.

    Hope t his helps.

    Regards,
    Buzz
     
  4. Dec 16, 2015 #3

    BvU

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    ##\int xy\, dx## with y constant is ## y \int x\, dx##. For the first leg that means the same as xy = 2x and you calculate ##2\int_1^3 x\, dx##.

    If you want to end up with an ##\int_0^1## ( a sort of parametrization that isn't really necessary here, as BB explains), you substitute t = (x-1)/2 so dt = dx/2 to get ##2 \int_0^1 2(1+2t) \, dt ##

    same result :rolleyes:

    --
     
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