[Computational Geometry]How to find the area between intersecting circles I have a bunch of circles that intersect. For instance, in the diagram below, there are four points $A, B, C$ and $D$ located at (0,0), (0,1), (1,1) and (1,0). There are a total of 6 edges, one between each pair of points. There are 6 circles, one for each edge, with the center at the middle of the edge and radius is half the edge length.

I need to find the area of each of the regions demarcated by the circle arcs -- for instance, the area demarcated by circle $c$ lying between point $B$ and $C$ right on the top of the image.
In one of the references, a method is provided for calculating these areas by application of the Green's theorem. I provide the relevant snapshot below:

I am unable to clearly understand how Green's theorem can be used to evaluate the various areas. Any help is appreciated. For instance, why do the authors take $P(x,y)=-y$ and why is $Q(x,y)=x$. Can one of the area's calculation be provided in an easy-to-understand manner that expands on the summation that the authors have finally specified?
 A: Big hint:
Call $A, C, L$ the resp. areas of the axe-shape, crescent-shape, (convex) lens-shape.
The 4 little circles have common radius $r=\tfrac12$, therefore area $\pi r^2 =\tfrac{\pi}{4}$.
The big circle has (using Pythagoras) radius $R=\tfrac{1}{\sqrt{2}}$, therefore area $\pi R^2=\tfrac{\pi}{2}$.
Consequently:
$$\begin{cases}
A+C+2L&=&\tfrac{\pi}{4}\\
4A \ \ \ \ +4L&=&\tfrac{\pi}{2}
\end{cases}\tag{1}$$
Besides, the lens-like area $L$ can be computed separately as $L=2(\pi/16-1/8)$ by reasoning on isosceles rectangle triangle GID (and half of the lens' area).  Do you see why ?
Therefore, using (1), it shouldn't be difficult to find areas $A$ and $C$.
A: You ask:
I am unable to clearly understand how Green's theorem can be used to evaluate the various areas. Any help is appreciated. For instance, why do the authors take $(,)=−$ and why is $(,)=$. Can one of the area's calculation be provided in an easy-to-understand manner that expands on the summation that the authors have finally specified?
The functions $(,)=−$ and $(,)=$ are chosen because in this case the integral
$$
\iint_S\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy 
$$
gives exactly the doubled area ($2S$) over which the integral is taken. Therefore by the Green's theorem the line integral over a closed curve $C$:
$$
\int_C (-ydx+x dy)\tag1
$$
will give the doubled area surrounded by the curve. To facilitate the integration it remains to express $x,y$ via a parameter $t$ so that $(x(t),y(t))$ give a point on the curve.
For an arc of a circle with radius $R_i$ centered at $(x_i,y_i)$ the suitable parameter would be the angle $\phi$:
$$
x=x_i+R_i\cos\phi,\quad y=y_i+R_i\sin\phi\tag2
$$
so that the equation $(1)$ can be rewritten as:
$$
\int_{\phi_1}^{\phi_2}[R_i^2+R_i(x_i\cos\phi+y_i\sin\phi)]d\phi\tag3
$$
which can be directly integrated. The angles $\phi_1,\phi_2$ can be found substituting in $(2)$ the coordinates of the intersection points.
