How calculate the shaded area in this picture? Let the centers of four circles with the radius  $R=a$ be on 4 vertexs a square with edge size $a$. How calculate the shaded area in this picture?
 

Let $(APD)$ be the area of the figure $APD$.
And let $x,y,z$ be $(KEPM),(PAD),(MPD)$ respectively.
First, we have
$$(\text{square}\ ABCD)=a^2=x+4y+4z.\tag1$$
Second, we have
$$(\text{sector}\ BDA)=\frac{\pi a^2}{4}=x+2y+3z.\tag2$$
Third, note that $KA=KD=a$ and that $(\triangle KAD)=\frac{\sqrt 3}{4}a^2$ since $\triangle KAD$ is a equilateral triangle. 
So, since we have
$$\begin{align}(K(E)AD(M))&=(\text{sector}\ AKD)+(\text{sector}\ DKA)-(\triangle KAD)\\&=\frac{\pi}{6}a^2+\frac{\pi}{6}a^2-\frac{\sqrt 3}{4}a^2\\&=\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2,\end{align}$$
we have
$$\frac{\pi}{3}a^2-\frac{\sqrt 3}{4}a^2=x+y+2z.\tag3$$
Solving $(1),(2),(3)$ gives us
$$(KEPM)=x=\left(1+\frac{\pi}{3}-\sqrt 3\right)a^2.$$
