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I am very interested in a certain problem and I am wondering what methods currently exist to solve it. Given a curve defined by a function which maps any given arc length, s, from an arbitrarily chosen original point on the curve (so s designates the arc length from the original point to another point on the curve), to a certain tangential angle, theta, what is the area enclosed by this curve if said curve is a Jordan curve?

Edit: More specifically, if a curve, A, is defined by the aforementioned function, and this curve is closed (it returns to its original point after some finite arc length), what is the area within the entire curve. For instance, with this construction, we have a Jordan curve. As the Jordan Curve Theorem states that a Jordan curve divides the plane into two sections, an inner section and an outer section, what is the area of the inner section define by the previous curve?

Edit 2: I understand that you can use Green's Theorem for this solution, but as for many different curves described in the aforementioned way, this would be a difficult, messy, and not very elegant way to solve the problem. Is there a simpler solution?

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What's the significance of the title "Research Advice"? –  Srivatsan Nov 12 '11 at 1:11
    
@Srivatsan I am researching this problem and I need guidance. –  analysisj Nov 12 '11 at 1:22
    
This strikes me as something one could compute using line integrals and standard multivariable calculus techniques. However, I am unclear as to your exact meaning. –  Bill Cook Nov 12 '11 at 1:48
    
Do you mean to measure the area bounded by a portion of a curve and a secant line (a line from your initial point to a point $s$ arc length units down the curve)? –  Bill Cook Nov 12 '11 at 1:50
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Could you give an example? –  Potato Nov 12 '11 at 1:50
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1 Answer

When the boundary $\partial A$ of a domain $A\subset{\mathbb R}^2$ is given in parametric form $$\partial A:\quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)$$ then the area $|A|$ can only be obtained via Green's theorem, i.e., by means of a formula of the type $$|A|=\int_{\partial A} x\ dy=\int_a^b x(t)\ y'(t)\ dt\qquad(*)$$ (there are several versions). Make sure that $A$ is to the left of $\partial A$, or you will get the wrong sign.

$\bigl({\it Remark}:$ For the proof of $(*)$ you don't need the full strength of Green's theorem. Assume $A$ is a convex domain in the right half-plane. Then the integral $\int_{\partial A} x\ dy$ can be interpreted as (and approximated by) an algebraic sum of rectangles of width $x$ and signed height $dy\ .\bigr)$

In your case you are not yet there. You are only given a function $$s\mapsto\theta(s)\qquad(0\leq s\leq L)$$ where $\theta$ denotes the argument of the unit tangent vector $(\dot x,\dot y)$ along $\partial A$. This amounts to $$\dot x(s)=\cos\theta(s)\ ,\quad \dot y(s)=\sin\theta(s)\qquad(0\leq s\leq L)\ .$$ If you want to apply the version $(*)$ of the area formula (with $s$ as parameter) you first have to compute explicitly the function $x(s):=\int_0^s \cos\theta(s')\ ds'$, and then the area $|A|$ is obtained as $$|A|\ =\ \int_0^L x(s)\ \dot y(s)\ ds\ .$$

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Thank you. I wonder, though, if there is a method which does not require this transformation? Is there a method which only uses the function theta=f(s)? And not via some parametrization? –  analysisj Nov 12 '11 at 14:50
    
and without Green's theorem... –  analysisj Nov 12 '11 at 15:21
    
@analysisj: There is no "transformation" involved. The variable $s$ remains variable of the parametrization. $-$ Concerning Green's theorem see my edit. –  Christian Blatter Nov 12 '11 at 18:39
    
thank you, but I mean without any sort of parametrization. –  analysisj Nov 12 '11 at 20:02
    
and I am voting this response up –  analysisj Nov 12 '11 at 20:03
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