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Line integral over a vector field

  1. Apr 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Evaluate ∫C < −y, x − 1 > dr where C is the closed piecewise continuous curve formed by the line segment joining the point A(− √ 2, √ 2) to the point B( √ 2, − √ 2) followed by the arch of the circle of radius 2, centered at the origin, from B to A.
    2. The attempt at a solution
    Vector field, F = < −y, x − 1 > = -y i + (x-1) j
    F
    (r(t)) = (-sin(t))i + (cos(t)-1)j
    r
    (t) = cos(t)i+sin(t)j
    r
    '(t) = -sin(t)i+cos(t)j
    F
    (r(t))⋅r'(t)=sin2(t)+cos2(t)-1
    Now I need to evaluate that between the regions 0≤t≤π
    I just wanted to check if I've set this up correctly. The arc from point B to A is = 180°= π, so that's why I've defined the region of t as going from 0≤t≤π.
     
  2. jcsd
  3. Apr 25, 2016 #2

    LCKurtz

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    Your parameterization is not a circle of radius ##2##. And your ##t## range is incorrect. It needs to start at the right place when ##t=0##.
     
  4. Apr 25, 2016 #3
    Oh, I forgot the radius = 2. I think (below) is better!
    r(t) = 2cos(t)i+2sin(t)j
    r'(t) = -2sin(t)i+2cos(t)j
    As for the range.
    ( √ 2, − √ 2)≤t≤(− √ 2, √ 2)
    This is the same as the range 0≤t≤π.
     
  5. Apr 25, 2016 #4

    LCKurtz

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    No it isn't. ##\vec r(0) =\langle 2,0\rangle##.
     
  6. Apr 25, 2016 #5
    The same radial distance*

    edit: not sure if you were correcting my last statement or my r(t)?

    7π/4≤t≤3π/4
     
    Last edited: Apr 25, 2016
  7. Apr 25, 2016 #6

    Mark44

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    This makes no sense. t is a real number that you're comparing to two points. In addition there is no ordering principle for points, so you can't say ( √ 2, − √ 2) ≤ (− √ 2, √ 2)
    This is saying, among other things, that 7π/4 ≤ 3π/4, which is clearly not true.
     
  8. Apr 25, 2016 #7
    Ok. I see what you mean. I'm not sure where to go from here though with defining the range...
     
  9. Apr 25, 2016 #8

    LCKurtz

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    Your original post did not tell you which arch to take getting back from B to A. You have to decide that first and make sure your parameterization goes in the correct direction, either clockwise or counterclockwise. Then you need to think about what geometric angle ##t## represents in your parameterization. Make sure it starts at B and traverses the arch to A in the direction you want.
     
  10. Apr 25, 2016 #9
    I'm having trouble understanding what you mean, sorry. I'm assuming we are calculating the integral from B to A, so it's counterclockwise. It start at an angle of 315 degrees, and finishes at an angle of 135 degrees, a difference of 180 degrees.

    I understood the question to mean that we have a vector field, < −y, x − 1 >, and a path from point B to point A, which followed the arc of a circle that was centred at the origin and had a radius 2. i.e. I pictured it as a particle moving in a semi-circle from point B to point A. The particle's path is determined by the field vector, F, and it's own movement, r(t). (Sorry if I don't sound to mathematically rigorous here, I'm still trying to get the intuition of line integrals over vector fields.)
     
  11. Apr 25, 2016 #10

    LCKurtz

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    While there is nothing wrong with assuming it is counterclockwise, your original problem doesn't say that. It's OK with me to assume that, but understand that you could just as well go clockwise on the circle from B to A, and the answers are probably different. The direction should have been stated in the problem, or maybe you just left it out. Now, assuming you want to go counterclockwise and start at 315 degrees, you won't end at 135 degrees. Counterclockwise motion adds to the angle.

    Yes. You just didn't specify which semi-circle. See above.

    The particles path is determined by ##\vec r(t)##, not by the field vector ##\vec F##. The vector field affects how much work is done in the process.
     
  12. Apr 26, 2016 #11
    Ok, I'm trying to understand that comment a bit more with a much simpler example. I jump from point B to point A. I only got to point A because I put in positive work, while the vector field (gravity) did negative work, which made me land at point A. The amount of work I did in getting from B to A is equal to the ∫F(r(t))⋅r'(t).

    But back to the question:
    If I start at 315 degrees, i.e. 315 degrees = 0, and I end at 135 degrees, that would mean I start at 0 degrees and end at 180 degrees. So 0≤t≤180 or 0≤t≤π. I calculated the integral with this range and got -π as a result.
     
  13. Apr 26, 2016 #12

    LCKurtz

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    That integral calculates the work done by the vector field.

    ##315\ne 0##. And it doesn't agree with ##t=0## either. See below

    You need to show your work. Are you still using ##\vec r(t) = 2\cos t i + 2\sin(t) j##? If so, I have pointed out to you before that when ##t=0## it gives (2,0). And at ##t=\pi## it gives ##(-2,0)##. That does not describe the correct semicircle.
     
    Last edited: Apr 26, 2016
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