# How to evaluate line integral along a curve C without parametrizing it

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 math.stackexchange.com/faq, 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 $C$.
$F$ 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?
It could come from $\int e^{-\lambda x^2}dx=\sqrt{\pi/\lambda}$. – George Lowther Apr 23 '12 at 23:33