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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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    What do you mean by a "5 dimensional curve"? Do you mean a (1-dimensional) curve that exists in 5-dimensional space? And what do you mean by $C(x_1,\ldots,x_5)$? If not the parametric equations, what information about the curve do you have?2011-12-24
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    I mean if I want to specify a line I will write $C(x_1,x_2) \equiv x_1 + x_2 = 10$. Should I remove that 5 dimensional thing?2011-12-24
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    A single equation will not in general yield a line; it will most often give you something $(n-1)$-dimensional (possibly with folds and self-intersections), where $n$ is the total number of variables/unknowns. So the solutions to a single equation with 3 unknowns will be something in 3-dimensional space that looks like a surface, but might have singularities.2011-12-24
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    @Arthur Just assume that C is nice.2011-12-24
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    @PratikDeoghare My quarrel was actually more about the dimensions than the singularities, but yeah. Anyways, the singularities (at least of polynomials) tend to be negligible with respect to integrals.2011-12-24
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    To paraphrase Arthur: a single equation $C(x_1,\ldots, x_5) = \text{something}$ will (generally) give you a 4-dimensional shape (which is definitely not a curve) in 5-dimensional space. If you would like to express a curve in 5-dimensional space, then you will (generally) need a system of 4 equations in 5 variables.2011-12-24
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    Ok. Got it. Imagine we don't have parametric equation for circle then how do we evaluate line integral along that circle?2011-12-24
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    I would like users to explain their downvotes. I find this to be an interesting question: how can one calculate a line integral when given only an implicit equation of the curve? (Pratik, I think this is what you mean to ask.)2011-12-24
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    Yes. That is exactly what I mean. Thanks @Jesse!2011-12-24
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    I think the question is too general to have a precise and satisfactory answer. With $F$ and $C$ unknown, there are many different ways to answer the question. For example, if $F$ is holomorphic and $C$ is a closed curve, then $\int_C F=0$ by Cauchy theorem; if $F$ has a simple pole $z_0$ insider a closed curve $C$, then $\int_C F=2\pi iRes(F,z_0)$, etc etc. There are many possible answers depending on the situation.2011-12-24
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    @Paul $0$C$ is not a closed curve. – 2011-12-24
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    @Pratik: Then it would be better for you to mention it when you posted your question. If you look at http://math.stackexchange.com/faq, you can see that we are encouraged to ask question with precise detail.2011-12-24
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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