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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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    @okj: The symmetry of the given equation suggests to "linearize" it by introducing $x+y$ and $y-x$ as new variables. So we arrive at $2c=(1+A)\cos u+(1-A) \cos v$. But this means that $(1+A)\cos u$ has to be $c+t$ when at the same time $(1-A)\cos v$ is $c-t$. - As for your second question, $\cos z\equiv\cos (-z)$, so it doesn't make a difference, apart from the fact that now the $(u,v)$ coordinate system is again right-handed.2011-04-12

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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.