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Line Integral Evaluation

  1. Dec 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Evaluate the line integral [itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
    (a) a straight line from (0,1) to (1,2);
    (b) the parabola x=t, y=t2 + 1;
    (c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

    2. Relevant equations
    Equation of line: y = mx + c

    3. The attempt at a solution
    (a)
    To convert given integral into one variable I used Equation of line
    y=mx + c
    where slope m = [itex]\frac {y_2 - y_1}{x_2 - x_1} = \frac{2-1}{1-0}=1[/itex]
    and y intercept c = 1
    which gives us y = x+1
    dy = dx

    Thus given integral

    [itex]∫^{(1,0)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]

    becomes one variable x dependent

    [itex] = ∫^{1}_{0} (x^2-x-1)dx + (({x+1})^2+x)dx[/itex]
    [itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+2x+1+x)dx[/itex]
    [itex] = ∫^{1}_{0} (x^2-x-1)dx + (x^2+3x+1)dx[/itex]
    [itex] = ∫^{1}_{0} (x^2-x-1+x^2+3x+1)dx[/itex]
    [itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex] ------------------ (1)

    solving
    [itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]
    applying limits gives
    [itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]

    (b) need some hints for this part

    (c) straight lines from (0,1) to (1,1) and then from (1,1) to (1,2).

    from equation (1) above
    [itex] = ∫^{1}_{0} (2x^2+2x)dx[/itex]

    solving
    [itex] \frac{2x^3}{3} +\frac{2x^2}{2}[/itex]

    from (0,1) to (1,1)
    we use only x: 0 to 1
    we get [itex]\frac{2}{3} + 1 = \frac{2+3}{3} = \frac{5}{3}[/itex]

    from (1,1) to (1,2)
    we use only x: 1 to 1
    we get [itex]\frac{2}{3} - \frac{2}{3} + 1 - 1 = 0 [/itex]

    hence adding both parts gives [itex] \frac{5}{3} [/itex]


    Please tell me am I on the right track?
     
    Last edited: Dec 25, 2014
  2. jcsd
  3. Dec 25, 2014 #2

    ehild

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    Check the upper boundary, it must be (1,2). With that boundary, your answer to (a) is correct.

    What is t at the lower boundary x=0, y=1, and what is it at x=1, y=2?
    Write x, dx, and y, dy in the integrand in terms of t, and integrate with respect to t.

    You can not use eq (1). Separate the integral from(0,1) to (1,1) along the straight line connecting these points (a horizontal path) and then from (1,1) to (1,2) (a vertical path).
     
  4. Dec 27, 2014 #3
    a)
    A vector parallel to the line segment from (0,1) to (1,2) is <1,1>. Therefore the vector equation for the line segment is r(t) = <0,1> + t<1,1> = <t, t +1>. So the parameter of the line integral is :
    x =t
    y = t+1
    0 <= t <= 1.
    dy/dt = 1
    dx/dt = 1

    so the integral becomes: ∫(t^2 - t - 1) + (t+1)^2 +t dt from 0 to 1
    = ∫2t^2 + 2t dt from 0 to to 1 which evaluates to 5/3.

    follow the same procedure for the rest of the section.
    1)parameter
    2)find dy/dt and dx/dt
    3)plug in the x's and y's in term of t
    4)plug in dy = dy/dt * dt and dx = dx/dt *dt
    5) integrate from 0 to 1

    note: You will have to divide C into to problems identical to A.
     
  5. Dec 27, 2014 #4
    yes, you are right. It's a typo, upper bound should be (1,2).
     
  6. Dec 27, 2014 #5
    The line integral is evaluated along a line segment C. It makes no sense to have limits in the integral.
     
  7. Dec 27, 2014 #6
    The problem is from Mathematical Methods for Physicists by Tai L. Chow, see the attached file.
     

    Attached Files:

  8. Dec 27, 2014 #7
    Yes, those limits given are the exact same limits given by the line segments. My guess is that those limits are given for problem b. For problems A and C, ignore the given limits and focus on the limits given by the line segment
     
  9. Dec 27, 2014 #8

    haruspex

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    No, it makes sense, but it in (a) and (c) it's redundant because the bounds are also given separately. You could equally well complain, instead, that the curve specifications should be:
    (a) a straight line segment;
    (b) a segment of the parabola x=t, y=t2 + 1;
    (c) two straight line segments meeting at (1,1)
    Does it really matter?
     
  10. Dec 27, 2014 #9
    So for part (b)
    Evaluate the line integral [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
    (b) the parabola x=t, y=t2 + 1

    x= t
    dx = dt

    [itex]y= t^2 +1[/itex]
    dy = 2tdt

    when x=0 then t=0
    when x=1 then t=1
    when y=1 then t = imaginary
    when y=2 then t= 1

    Hence given integral
    [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]
    in terms of t is
    [itex](t^2 -t^2 -1)dt + [(t+1)^2 + t]2tdt[/itex]
    [itex](-1)dt + [(t^2 + 2t +1) + t]2tdt[/itex]
    [itex]∫^{1}_{0} [-1 + 2t^3 + 6t^2 +2t]dt[/itex]
    [itex]-t + \frac{2}{4}t^4 + \frac{6}{3}t^3 + \frac{2}{2}t [/itex]
    after applying lower and upper bound we get
    [itex]-1 +\frac{1}{2} + 2 + 1[/itex]
    [itex] \frac{1}{2} + 2 = \frac{5}{2}[/itex]

    Which doesn't seem right.
     
  11. Dec 27, 2014 #10

    haruspex

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    No. Try that again.
    You dropped an exponent.
     
  12. Dec 27, 2014 #11
    When y=1 then t = 0

    thx! yes and it should be
    [itex] (t^2 -t^2 -1)dt + [(t^2 + 1)^2 +1]2dt[/itex]

    After solving and applying bounds we get [itex] \frac{5}{3}[/itex]
     
  13. Dec 27, 2014 #12

    haruspex

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    I get a slightly higher value. Please post your working.
     
  14. Dec 27, 2014 #13
    Well technically it does not; however a line integral is defined as an integral which is taken around a curve. By definition line integrals must be calculated along a certain path, so it does not make sense to give coordinates in the integral itself.
     
  15. Dec 28, 2014 #14

    Ok! solving again

    Evaluate the line integral [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex] along
    (b) the parabola x=t, y=t2 + 1

    x= t
    dx = dt

    [itex]y= t^2 +1[/itex]
    dy = 2tdt

    when x=0 then t=0
    when x=1 then t=1
    when y=1 then t = 0
    when y=2 then t= 1

    Hence given integral
    [itex]∫^{(1,2)}_{(0,1)} (x^2-y)dx + (y^2+x)dy[/itex]
    in terms of t is
    [itex](t^2 -t^2 -1)dt + [(t^2+1)^2 + t]2tdt[/itex]
    [itex](-1)dt + [(t^4 + 2t^2 +1) + t]2tdt[/itex]
    [itex][-1 + t^4 + 2t^2 +1 + t]2tdt[/itex]
    [itex]∫^{1}_{0} [2t^5 + 4t^3 +2t^2]dt[/itex]
    [itex] \frac{2}{6}t^6 + \frac{4}{4}t^4 + \frac{2}{3}t^3 [/itex]
    after applying lower and upper bound we get
    [itex]\frac{1}{3} + 1 + \frac{2}{3} [/itex]
    [itex] \frac{1+3+2}{3} = \frac{5}{3}[/itex]
     
  16. Dec 28, 2014 #15

    ehild

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    1+3+2=6 :D
     
  17. Dec 28, 2014 #16
    :oops: now I know my problem, I don't know simple arithmetic :).

    Jokes apart. Integral suppose to be path independent so why two different answers for part (a) and (b)?
     
  18. Dec 28, 2014 #17

    haruspex

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    You also went wrong in this step
    but it turned out to make no numerical difference.
     
  19. Dec 28, 2014 #18

    haruspex

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    Why should it be?
     
  20. Dec 28, 2014 #19

    ehild

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    No, it is only path-independent if you integrate an exact differential. http://mathworld.wolfram.com/ExactDifferential.html
     
  21. Dec 28, 2014 #20
    Part(b) Correction

    [itex]=\left(-1+2t^{5}+4t^{3}+2t^{2}+2t\right)dt[/itex]

    [itex]=-t+\frac{2t^{6}}{6}+\frac{4t^{4}}{4}+\frac{2t^{3}}{3}+\frac{2t^{2}}{2}[/itex]

    [itex]=-t+\frac{t^{6}}{3}+t^{4}+\frac{2t^{3}}{3}+t^{2}[/itex]

    applying bounds from 0 to 1

    [itex]=-1+\frac{1}{3}+1+\frac{2}{3}+1[/itex]

    [itex]=\frac{1}{3}+\frac{2}{3}+1=\frac{1+2+3}{3}=\frac{6}{3}=2[/itex]
     
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