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Surface integral

  1. Mar 5, 2014 #1
    The line integral can be expressed, at least, in this three different ways: [tex]\int \vec{f} \cdot \hat{t} ds = \int \vec{f} \cdot d\vec{s} = \int \vec{f} \cdot d\vec{r}[/tex] The surface integral too (except by least expression above): [tex]\iint \vec{f} \cdot \hat{n} d^2S = \iint \vec{f} \cdot d^2\vec{S}[/tex] My ask is: exist some equivalent/analogous expression to ∫ f·dr for surface integral? In other words, is possible to write the surface integral in terms of the position vector r? Maybe: ##\iint \vec{f} \cdot d^2\vec{r}## ?
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  3. Mar 9, 2014 #2
    I don't really agree with the third formula you used for the line integral. The second one is definitely correct, the line integral takes the component of the vector field parallel to the line itself. The third one is not always valid. It is true in the case of radial motion, when you path is actually parallel to the position vector itself.
    Last edited: Mar 9, 2014
  4. Mar 9, 2014 #3
    Great observation! But, are you sure???? I already saw the 3rd definition a lot times.


  5. Mar 10, 2014 #4
    If you look at the Wikipedia page you linked it says that r(t) is a bijective parametrization of the curve, not the position vector. It's just a confusing notation, that why one usually uses s(t).

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  6. Mar 10, 2014 #5


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    r(t) doesn't have to be the position vector, the Wikipedia page is saying it can be more general, but it CAN be the position vector. For a line integral, you want the differential to be a vector parallel to the curve whose magnitude is the line element, and that's exactly what dr is. Equivalently, one can write dr = (dr/dt) dt.

    For the surface integral, you want the differential to be a vector orthogonal to the surface whose magnitude is equal to the element of surface area. Suppose the surface is described by giving the position vector as a function of two variables, r(s, t). Then the differential is the vector d2S = (∂r/∂s) x (∂r/∂t) ds dt.
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