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Potential function of a gradient field.

  1. Apr 1, 2008 #1
    1. The problem statement, all variables and given/known data
    For the vector field = -yi + xj, find the line integral along the curve C from the origin along the x-axis to the point (6, 0) and then counterclockwise around the circumference of the circle x2 + y2 = 36 to the point (6/sqrt(2), 6/sqrt(2)) . Give an exact answer.

    2. Relevant equations

    3. The attempt at a solution

    I got a potential function f = xy. I tried to consider it as a path independent potential function, and plugged in F(6/sqrt(2), 6/sqrt(2))-F(0,0). The answer I got was 18. Which webassign calls wrong. Then again, webassign incorrect on about 20% of the problems, which makes learning math very frustrating.
  2. jcsd
  3. Apr 1, 2008 #2


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    grad(f) is yi+xj. That's not what you want. The vector field doesn't have a gradient function, it has nonzero curl. You really have to do the path integral.
  4. Apr 1, 2008 #3
    Thanks. I just read a few chapters ahead into the curl stuff.
  5. Apr 8, 2009 #4
    sorry to bring this up again but I got the same question with the OP and I am just confused what the real answer should be if the vector field doesn't have a grad function
  6. Apr 8, 2009 #5


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    Normally one does path integrals before "potential functions" so this is a peculiar question.

    Do it as two separate integrals. On the x-axis use x= t, y= 0. On the circle [itex]x^2+ y^2= 36[/itex] use x= [itex]6cos(\theta)[/itex], [itex]y= 6 sin(\theta)[/itex].
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