<s>[INST] Finding the area of a region of circle Consider the following shape. It is requested to find the area of the shaded region. The radius of the circles and the overlapping half circles is 1 unit.
The region is irregular and I couldn't find a way to find it.

 [/INST] Rotate the image, and place the origin of the Cartesian system at the center of the main circle. The other two circles will be centered at $(\pm 1/ \sqrt{2},-1/ \sqrt{2})$.

Obviously, it's enough to find half of the area for $x>0$.
For the green circle for $y>0$ we have:
$$y=\sqrt{1-x^2}$$
While for the right blue circle:
$$y =\sqrt{1-(x-1/ \sqrt{2})^2}-1/ \sqrt{2}$$
While polar coordinates are a good way to go, the problem can be solved as well in Cartesian.
We need to find the intersection point for $y>0$ of the two circles. Let's work with the equations:
$$y+1/ \sqrt{2} =\sqrt{1-x^2+\sqrt{2} x-1/2}=\sqrt{y^2+\sqrt{2} x-1/2}$$
Squaring:
$$y^2+\sqrt{2}y+1/2 =y^2+\sqrt{2} x-1/2$$
$$\sqrt{2}y =\sqrt{2} x-1$$
$$y =x-1/\sqrt{2}$$
From the second equation again:
$$y+1/ \sqrt{2} =\sqrt{1-y^2}$$
$$y^2+\sqrt{2}y+1/2 =1-y^2$$
$$2y^2+\sqrt{2}y-1/2 =0$$
$$y_0=\frac{\sqrt{3}-1}{2 \sqrt{2}}$$

$$x_0=\frac{\sqrt{3}+1}{2 \sqrt{2}}$$

Now all we need to do is integrate from $x=0$ to $x_0$ (subtracting the result for the blue circle from the green one):
$$S/2=\int_0^{x_0} \sqrt{1-x^2} dx-\int_0^{x_0}(\sqrt{1-(x-1/ \sqrt{2})^2}-1/ \sqrt{2}) dx=$$
$$=x_0/ \sqrt{2}+(x_0 \sqrt{1-x_0^2}+\arcsin x_0 )/2-(y_0 \sqrt{1-y_0^2}+\arcsin y_0 )/2+$$
$$+(-1/ \sqrt{2} \sqrt{1-1/2}+\arcsin (-1/ \sqrt{2}) )/2$$
$$S=\frac{\sqrt{3}+1}{2}+\frac{1}{4}+\frac{5 \pi}{12}-\frac{1}{4}-\frac{ \pi}{12}-\frac{1}{2}-\frac{ \pi}{4}$$

$$S=\frac{6 \sqrt{3}+\pi}{12}$$

Same result as Allawonder obtained.
This method is tedious (I have made liberal use of Wolfram Alpha), but we didn't need to think much.
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