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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? – Jesse Madnick Dec 24 '11 at 6:25
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? – Pratik Deoghare Dec 24 '11 at 6:28
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. – Arthur Dec 24 '11 at 6:34
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.) – Jesse Madnick Dec 24 '11 at 6:56
@Pratik: Then it would be better for you to mention it when you posted your question. If you look at, you can see that we are encouraged to ask question with precise detail. – Paul Dec 24 '11 at 7:29

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. – Pratik Deoghare Dec 24 '11 at 6:41

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 $\int e^{-\lambda x^2}dx=\sqrt{\pi/\lambda}$. – George Lowther Apr 23 '12 at 23:33

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