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I have curve $C$ and I don't have its parametric equations. I want to evaluate the line integral along C.

$\oint_C F \ ds $

How do I that?

Imagine we don't have parametric equation for the circle; how do we evaluate the line integral along that circle? When we have the parametric equations, it's easy.

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    One problem with sticking to an implicitly-defined curve: how do you indicate direction? The sign of a line integral about a circular contour depends on whether you're traversing the circle clockwise or anticlockwise...2011-12-24

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What you probably need to do is to express your vector field $\vec F$ as the gradient of some function $f, \nabla f = \vec F$. Then you can easily evaluate all line integrals of $\vec F$: they will be equal to $f(b) -f(a)$ where $a, b$ are the endpoints of your curve $C$.

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    $F$ is scalar in this case.2011-12-24
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When a curve $\gamma\subset {\mathbb R}^2$ is not given in the form $y=f(x)$ $\ ( a\leq x\leq b)$ or more generally in the form $t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)$ $\ (a\leq t\leq b)$ but implicitly as the zero set of some function $F\colon\ {\mathbb R}^2\to{\mathbb R}$ then the computation of a line integral $\int_\gamma \Phi({\bf z})\ d{\bf z}$ (or similar) is not easy.

Example: Let $\gamma$ be given implicitly by the simple condition $\gamma:=\{(x,y)\ |\ x^2+y^2=1\}$. Then $\int_\gamma 1\ |d{\bf z}| =2\pi$. Where would the transcendental number $2\pi$ come from if the computation starting with the equation $x^2+y^2=1$ would be an easy matter?

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    It could come from $\int e^{-\lambda x^2}dx=\sqrt{\pi/\lambda}$.2012-04-23