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I'm trying to find the area of an irregular domain that is bounded by $x = c$, $y = c$, and $c = -A\sin(x/2)\sin(y/2)+\cos(x/2)\cos(y/2)$, where A can vary in the range [-1,1], and x and y are only defined over $x\in[0,\pi]$ and $y\in[0,\pi]$. I need to know the area as a function of the parameter A. I was thinking of trying to use green's theorem and just integrating around the border of the region, but in order to do so I need a smart parameterization of $c = -A\sin(x)\sin(y)+\cos(x)\cos(y)$. Can anyone think of a good way to parameterize this function or the boundary in general? Or any better ideas of how to get the area of this region?

In practice I express c as $c = \cos(z/2)$ so that the solution to the equation $\cos(z/2) = -A\sin(x/2)\sin(y/2)+\cos(x/2)\cos(y/2)$ when c is a constant are level sets. The equations $x = c$, $y = c$, and $c = -A\sin(x/2)\sin(y/2)+\cos(x/2)\cos(y/2)$, as well as the domain boundaries, $x = 0$, $x = \pi$, $y = 0$, and $y = \pi$, subdivide the domain into a number of regions and I'm actually interested in the area of each of the regions as a function of the parameters A and c, an example of a representative situation is given here, where you can see 7 different regions whose area I am interested in.

Thank you so much for your help.

if the "here" link doesn't work:
http://www3.wolframalpha.com/input/?i=ImplicitPlot[{0.8+*+Sin[x%2F2]+*+Sin[y%2F2]%2BCos[x%2F2]+*+Cos+[y%2F2]%3DCos[pi%2F6]%2Cx%3DCos[pi%2F6]%2Cy%3DCos[pi%2F6]}%2C{x%2C0%2Cpi}%2C{y%2C0%2Cpi}]

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    $\begingroup$ Did I misread the question? If $A=1,c=1$, then the equation reduces to $\cos(x-y)=1\Rightarrow x-y=0$. The three lines $x=c,y=c, x=y$ does not give me a well-defined bounded domain. $\endgroup$
    – GWu
    Commented Apr 12, 2011 at 1:27
  • $\begingroup$ @GWu: You mean $A=-1$? $\endgroup$
    – joriki
    Commented Apr 12, 2011 at 4:07
  • $\begingroup$ @okj: I think you need to specify more clearly which area you're interested in. For instance, in this case: tiny.cc/h0btm, do you want the area with $0<x,y<0.8$ or $0.8<x,y<\pi$? $\endgroup$
    – joriki
    Commented Apr 12, 2011 at 5:26
  • $\begingroup$ @okj: I've merged your two duplicate accounts. Please consider registering: it will help the software keep track of your account and allow you to edit your own questions without approval vote from a higher reputation user. $\endgroup$
    – Willie Wong
    Commented Apr 12, 2011 at 16:09
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    $\begingroup$ @Christian Blatter: Thank you, your suggestion was very helpful! When I expand your alternate form of the equation I see the connection, but I'm quite impressed at the cleverness of how you got there in the reverse direction. Is this a common technique to rewrite equations (multiply by a scalar and then add terms that sum to zero), or how did you think to manipulate it the way you did? $\endgroup$
    – okj
    Commented Apr 12, 2011 at 16:19

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I suggest you show us a figure of what you have in mind; maybe this helps to clarify ideas. A parametrization of the boundary arc in question you can obtain in the following way:

The given equation is equivalent to $$2c=(1+A)\cos(x+y)+(1-A)\cos(y-x)\ .$$

Putting $x+y=:u$ and $y-x=:v$ you have $2c=(1+A)\cos u+(1-A)\cos v$ which can be parametrized by $$u=\arccos{c+t\over 1+A}\ ,\qquad v=\arccos{c-t\over 1-A}\ .$$ The required $t$-interval, which depends on $c$ and $A$, has to be determined by looking at the figure. When everything is set you can compute your area using Green's theorem.

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