Integrate area of the shadow? Today I found an interesting article here. It computes (approximately) area of the shadow.

I was wondering what is exact value of the area. My first thought was to use integrals but it doesn't seem to be easy. How to compute this using integrals?
Only thing I know is that $$S\approx 2.92$$
 A: There is of course a much easier approach that does not require any integrals whatsoever.  Let the points of intersection of the circles be $A$ and $B$, the center of the small circle be $C=(1,1)$ and the center of the large circle, i.e., the lower left vertex of the square be $O=(0,0)$.  (We will scale the resulting area to the original area of the square; in this case, we will multiply our result by $5$.)

The shaded area $K$ is equal to $K = K_1-K_2 + 2 K_3$, where $K_1$ is the area of the sector $CAB$, $K_2$ is the area of the sector $OAB$, and $K_3$ is the area of the triangle $\Delta OAC$.  We have that
$$K_1 = \frac12 \cdot 1^2 \cdot \theta$$
$$K_2 = \frac12 \cdot 2^2 \cdot \phi$$
$$K_3 = \sqrt{s (s-2)(s-1)(s-\sqrt{2})} $$
where $\theta$ is the angle subtended by the sector $CAB$, $\phi$ is the angle subtended by the sector $OAB$, and $s$ is the semiperimeter of triangle $\Delta OAC$. It should be clear, then, that $K_3 = \sqrt{7}/4$.
We just need to find the angles $\theta$ and $\phi$.  We do this by finding the points of intersection $A$ and $B$ and applying simple trigonometry.  The points of intersection are found by solving the equations
$$(x-1)^2+(y-1)^2=1$$
$$x^2+y^2=4$$
The result is that $A = \left (\frac{5-\sqrt{7}}{4},\frac{5+\sqrt{7}}{4} \right )$ and $B = \left (\frac{5+\sqrt{7}}{4},\frac{5-\sqrt{7}}{4} \right )$.  (The symmetry is expected.)  Finding the angles is a relatively simple affair, so long as one has the proper respect for the range of our inverse trig functions.  The result is that
$$\theta = \frac{\pi}{2}+2 \arctan{\left (\frac{\sqrt{7}-1}{\sqrt{7}+1} \right )} $$
$$\phi = \frac{\pi}{2}-2 \arctan{\left (\frac{5-\sqrt{7}}{5+\sqrt{7}} \right )} $$
The final result for the area is then

$$K = 4 \arctan{\left (\frac{5-\sqrt{7}}{5+\sqrt{7}} \right )} + \arctan{\left (\frac{\sqrt{7}-1}{\sqrt{7}+1} \right )} + \frac{\sqrt{7}}{2} - \frac{3 \pi}{4} $$

To compare with the OP's monte carlo result, I get from WA that $5 K \approx 2.927625 \dots$
ADDENDUM
Actually, there is a far more elegant and simpler method of attack.  We don't need to find the points $A$ and $B$.  Rather, we just find the angles $\theta$ and $\phi$ from the law of cosines and sines, respectively, using $\Delta OAC$.  Using the law of cosines, I find that $\cos{(\angle OCA)} = -1/(2 \sqrt{2})$, so that $ \sin{(\angle OCA)} = \sqrt{7}/(2 \sqrt{2}) $.  Thus, $\angle OCA = \pi - \arcsin{(\sqrt{7}/(2 \sqrt{2}))}$ and therefore $\theta = 2 \arcsin{(\sqrt{7}/(2 \sqrt{2}))}$.  Using the law of sines, we find that $\phi = 2 \arcsin{(\sqrt{7}/(4 \sqrt{2}))}$. Thus,

$$K = \arcsin{\left (\sqrt{\frac{7}{8}} \right )} - 4 \arcsin{\left (\frac12 \sqrt{\frac{7}{8}} \right )}+ \frac{\sqrt{7}}{2}$$

This produces the exact same result as above, and much more elegantly.
