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Line integrals

  1. May 13, 2009 #1
    1. The problem statement, all variables and given/known data

    F = (3x2 + 2y cos(xy))i + (2y + 2x cos(xy))j

    a - show that F is a gradient field

    b - calculate the integral of F dot dr where c includes the points -2,0 and 2,0

    c - determine the value of the integral of F dot dr where c is any curve joining -2,0 and 2,0

    2. Relevant equations



    3. The attempt at a solution


    a..

    grad f = F,

    I found f = x3 + y2 - 2sin(xy)


    b.....

    curlf F = 0, therefore the integral F dot dr = curl F dot dA = 0

    c.......

    wouldnt that be the same as the above b
     
  2. jcsd
  3. May 13, 2009 #2

    gabbagabbahey

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    looks like you're off by a negative sign; [itex]\frac{d}{dx}\sin(xy)=+y\cos(xy)[/itex]


    You must have a closed path to use Stoke's theorem. Is the curve in (b) closed? It is not clear from your description of the problem...

    It sounds like you have an open curve from (-2,0) to (2,0) and so you can't use stokes theorem (an open curve does not bound a surface) try using the fundamental theorem of gradients instead...
     
  4. May 13, 2009 #3

    dx

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    Should be f = x3 + y2 + 2sin(xy) (you got the sign wrong).

    For b, the integral will be zero only if C is a closed curve. Does it say that in the question?

    For c, no it won't. If F = ∇f, then [itex] \int_a^b \nabla f \cdot dr = f(b) - f(a) [/itex].

    EDIT: Oops, gabba beat me to it.
     
  5. May 13, 2009 #4
    b is an open curve


    c is a closed curve
     
  6. May 13, 2009 #5
    so c.... should be -8 - 8 = -16 what about b
     
  7. May 13, 2009 #6

    dx

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    The curve in c is not a closed curve. It's a curve joining (-2,0) and (2,0).
     
  8. May 14, 2009 #7
    Ok i will explain this a little better

    Part B Calculate integral F dot dr where C is the picture included

    Part C Determine the value of integral F dot dr where C is anycurve joining -2,0 to 2,0. Explain Reasoning


    Whats the difference in the question being asked
     

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