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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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    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.2011-04-12
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    @GWu: You mean $A=-1$?2011-04-12
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    @okj: I think you need to specify more clearly which area you're interested in. For instance, in this case: http://tiny.cc/h0btm, do you want the area with $0 or $0.8?2011-04-12
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    @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.2011-04-12
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    @GWu: Thanks for your clarification, the function is actually defined in terms of half angles over the domain I gave (I had posted the the full angles just to simplify notation, but forgot to adjust the domain appropriately), I've made the change above. Also, you are correct that the region changes depending on the value of A. When the three equations given do not define a closed region with the given 3 equations, the boundary is given by including the limits of the domain (i.e. $x = \pi$ and $y = \pi$) in addition to the three equations already given.2011-04-12
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    @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?2011-04-12
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    @Christian Blatter: Am I mistaken or should the equation you gave have the x and y switched in the second term on the right hand side? Like: $2c = (1+A)\cos(x+y)+(1-A)\cos(x-y)$. Since the identity is: $\cos(x-y) = \cos(x)\cos(y)+\sin(x)\sin(y)$ ( See [Mathworks](http://mathworld.wolfram.com/TrigonometricAdditionFormulas.html)).2011-04-12
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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.