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Fundamental theorem for line integrals

  1. Sep 21, 2012 #1
    Hi, I have a question. In my calculus book, I always see the fundamental theorem for line integrals used for line integrals of vector fields, where f=M(x,y)i + N(x,y)j is a vector field.

    The fundamental theorem tells me that if a vector field f is a gradient field for some function F, then f is path independent and then given a curve paramaterized by r(t) from a <= t <=b, ∫f°dr = F(r(a)) - F(r(b))

    However, what is the analogue of this for line integrals of scalar fields over the arc length of a curve? You know, the ol' ∫f(x,y)ds type of line integrals where f(x,y) is just a regular function of two variables? I know the gradient is a vector field, so clearly f here isn't a gradient, but it could still be exact (meaning there's a function F such that ∂F/∂x = ∂F/∂y). If that's the case, is F a potential function for f, and can I apply the fundamental theorem?

    Another issue I have is visualizing the problem. It's simple to visualize the line integral of a scalar field using arc length: it's just the "Area" of the "curtain" of f above the curve. But I can't for the life of me visualize the vector field version ∫Mdx +Ndy. But I know that for the vector field version, ∫f°dr = ∫f°Tds where T is the unit tangent vector so I know they're related by the tangent vector. How can I use the idea of the unit tangent vector to help visualize/understand the line integral of a vector field? The problem is my book just gives examples of physics and work, and to be honest, I just don't know anything about physics. How can I visualize it from a purely geometric perspective?
    Last edited: Sep 21, 2012
  2. jcsd
  3. Sep 21, 2012 #2


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    If you have "ds" then you have "s" as a parameter for the curve: x= g(s), y= h(s). So
    [itex]\int f(x,y)ds= \int f(x(s), y(s))ds[/itex] which is just an ordinary integral.
  4. Sep 21, 2012 #3
    [tex]\oint f(r) \; dr = -\int \nabla f(r) \times dA[/tex]

    (for scalar [itex]f(r)[/itex], vector [itex]r[/itex], [itex]dr[/itex], [itex]dA[/itex].

    All such integrals follow from the generalized fundamental theorem of calculus.
  5. Sep 21, 2012 #4

    I'm confused on what you're saying. When I was talking about "ds" I was talking about the change in arc length. In all the examples in my book, they end up paramaterizing the curve in terms of a variable t, so I end up with:

    ∫f(x,y)ds = ∫f(x(t),y(t))ds(dt/dt) = ∫f(x(t),y(t))(ds/dt)dt = ∫f(x(t),y(t))√(dx/dt)^2 + (dy/dt)^2) dt

    (t is paramaterizing the curve)
  6. Sep 21, 2012 #5

    So when you say dr here, is that a vector r(t) which is paramaterizing the curve? (for example, if our curve was the portion of a circle of radius 1 in the first quadrant, we could have r(t)=<cost,sint> 0<=t<=∏/2

    Then wouldn't we end up having:

    ∫f(r)dr = ∫f(r(t))dr*(dt/dt) = ∫f(r(t))(dr/dt)dt = ∫f(r(t))<-sint,cost>dt ? (I mean isn't this what the left hand side would be)

    And as for dA on the right hand side, this is the change in area? I'm confused, the area of what?
  7. Sep 21, 2012 #6
    Yeah, [itex]dr[/itex] could be taken equal to [itex](dr/dt) \; dt[/itex] for a scalar parameter [itex]t[/itex]. The same thing is done in vector line integrals; it's just that either you have a dot or a cross product between the vector field and [itex]dr/dt[/itex] in practice.

    [itex]dA[/itex] is an area element of a surface whose boundary is the closed path of the line integral. If the path of the line integral is not closed, then there is no area integral on the right to compare against.
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