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Line/Curve Integral Question

  1. May 2, 2012 #1
    Hey guys. I've just had a few lectures on line integrals. My lecturer has told me the following:

    [itex]\int_{C}^{} f(s) \text{d}s = \int_{a}^{b} f(\gamma(t))\|\gamma'(t)\| \text{d}t[/itex]

    Unfortunately he hasn't explained this topic very well. I understand what's going on with a line integral but have these few questions:

    - What is the 'ds' on the left hand side?
    - How do we determine what gamma is?
  2. jcsd
  3. May 2, 2012 #2
    The "ds" is just a notation to indicate the variable towards we integrate. You can change it by dt or du or whatever (although ds seems like the canonical choice).

    The [itex]\gamma[/itex] is a parametrization of the curve. This is usually given, although not always.

    For example, if C is a circle, then

    [itex]\gamma:[0,2\pi[\rightarrow \mathbb{R}^2:t\rightarrow (\cos(t),\sin(t))[/itex]

    This is one of the many possible parametrizations of the circle.
    Note that the integral does depend on the parametrization used: for example, if we decide to run C in the other direction, then the integral can turn up to be negative. Thus C should always come with some kind of orientation if we want to find a right parametrization.
  4. May 2, 2012 #3
    Firstly, thanks for the reply.

    Can you describe what you mean by a parametrization? In your example you've used a range, but in an example we did in class the lecturer used the function of the curve.

    I'm confused :(
  5. May 2, 2012 #4


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    "ds" is the "differential of arc length". A curve, being one dimensional, can be expressed in terms of one variable. For example, a curve in 3 dimensional space, given an ordinary xyz-coordinate system, can be written x= f(t), y= g(t), z= h(t). If you like, you think of t as "time" and those three equations give the position of an object moving along that curve.

    If we have two points, say [itex](x_0, y_0, z_0)[/itex] and [itex](x_1, y_1, z_1)[/itex], then the straight line distance between the points is [itex]\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2+ (z_1-z_0)^2}[/itex], from the Pythagorean theorem. For a curve, do the usual reduction to "differentials" or derivatives that you have surely seen in Calculus to get the "differential of arc length", [itex]ds= \sqrt{dx^2+ dy^2+ dx^2}[/itex] or, in terms of the parameter, t, [itex]dx= \sqrt{(dx/dt)^2+ (dy/dt)^2+ (dz/dt)^2}dt[/itex].

    IF the curve, in two dimensions, is given as a function, y= f(x), you can use x itself as parameter. Formally, that would be "x= t, y= f(t)" so that the differential of arc length would be [itex]ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt= \sqrt{1+ (df/dt)^2}dt[/itex] or, less formally, [itex]ds= \sqrt{(1+ (dy/dx)^2}dx[/itex].
    Last edited by a moderator: May 2, 2012
  6. May 2, 2012 #5
    Not sure about the second part but the "ds" basically means the derivative of variable "s." This vairable could have been any other letters, such as x - in that case, the derivate of "x" would be stated as "dx."
  7. May 2, 2012 #6
    You mean differential, not derivative. Derivatives are things like dy/dx, and dx and dy are called differentials.
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