How to find the shaded area How to find the shaded area crossed by semi-circle of radius 2 and quarter-circle of radius 4?

 A: The shaded area is shaped from two intersecting circles like an asymmetric lens! in order to find the area of this lens we simply split it in two parts:

A: Hint.

(This space intentionally left blank.)

Solution.

$$\begin{align}
\frac12\cdot\text{target area} &= A + B \\
&= \left( \frac12 \cdot (2s)^2\cdot\alpha - 4 C\right) + \left(\frac12 \cdot s^2 \cdot \beta - C \right) \\[4pt]
&= \frac12 s^2\left( 4 \alpha + \beta \right) - 5 C \\[4pt]
&=\frac12 s^2 \left( \frac\pi2 + 3\alpha \right)-5 \cdot \frac{1}{20}(2s)^2 \\[4pt]
&= \frac12 s^2 \left( \frac\pi2 + 3\operatorname{atan}\frac{1}{2} - 2 \right)
\end{align}$$
So, the target area, with $s = 2$, is

$$2 \pi + 12 \operatorname{atan}\frac{1}{2} - 8 $$

A: This problem is so funny. @Robert Z's first answer of
$$4\sin^{-1}\left(\frac2{\sqrt5}\right)+8\sin^{-1}\left(\frac45\right)-8$$
was correct as is his latest answer of
$$4\left(\pi-2-\tan^{-1}\left(\frac2{11}\right)\right)$$
Integrating in polar coordinates I got
$$12\tan^{-1}\left(\frac12\right)-8+2\pi$$
Which was also correct. Taking the difference between sectors of circles and triangles, I get
$$16\sin^{-1}\left(\frac1{\sqrt5}\right)-8+4\sin^{-1}\left(\frac2{\sqrt5}\right)$$
Amazing that all $4$ of these expressions are the same! Even
$$8\pi-12\tan^{-1}2-8$$
As was seen in two nearly simultaneous answers was OK. How about sticking to Pythagorean triples with
$$6\sin^{-1}\left(\frac45\right)+2\pi-8$$
A: Let $ABCD$ be our square, where $B$ is a center of the circle with the radius $4$ and $AD$ is a diameter of the second circle. 
Let these circles be intersected in points $A$ and $E$ and $G$ be the center of the second circle.
Thus, the needed area it's:
$$\frac{1}{2}\cdot4^2\cdot2\arctan\frac{1}{2}+\frac{1}{2}\cdot2^2\cdot2\arctan2-S_{ABEG}=$$
$$=16\arctan\frac{1}{2}+4\arctan2-4\cdot2=$$
$$=16\left(\frac{\pi}{2}-\arctan2\right)+4\arctan2-4\cdot2=8\pi-12\arctan2-8.$$
