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?
 
 A: 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
$$(KEPM)=x=\left(1+\frac{\pi}{3}-\sqrt 3\right)a^2.$$
A: Here is a geometric solution.
Let's find the angle $EDK$. Since $KA=KD=a=AD$ hence the triangle $AKD$ is equiliteral and the angle $KDA$ is $\pi/3$. Therefore angle $KDC$ is $\pi/6$. In the same way the angle $EDA$ is also $\pi/6$ and we get that the angle $EDK$ is $\pi/6$.
Now the shaded area 
$$(EPMK) = ({\rm square\,} EPMK) + 4\times ({\rm segment\,}EKE).$$
The edge of the square $EPMK$ (from the triangle $EDK$) is $$EK=2a\sin\frac{\pi}{12}=a\frac{\sqrt{3}-1}{\sqrt{2}},$$
hence the area
$$({\rm square\,} EPMK) =(2-\sqrt{3})a^2.$$
And the area
$$({\rm segment\,}EKE) = \frac{1}{2}\left(\frac{\pi}{6}-\sin\frac{\pi}{6}\right)a^2=\frac{1}{4}\left(\frac{\pi}{3}-1 \right)a^2.$$
So
$$(EPMK) = (2-\sqrt{3})a^2 + \left(\frac{\pi}{3}-1 \right)a^2=\left(\frac{\pi}{3}+1-\sqrt{3}\right)a^2.$$
A: Well this one is very easy to solve. We put a coordinate system directly in the middle of your square. The upper arc from the top left to bottom right is a quater circle and can be considered the graph of a function $g$. Now, since it is on a circle that must have its middle point at $(-a/2,-a/2)$, $g$ satisfies
$$ (g(x)+a/2)^2 + (x + a/2)^2 = a^2,$$
because $a$ is its radius. Now we solve this equation for $g(x)$, keeping in mind, that it is the upper half of the circe and get
$$ g(x) = \sqrt{a^2 - (x+a/2)^2} - a/2. $$
Due to the symmetry along the line $x=y$, it must be $g(g(0)) = 0$. Due to further symmetries, we get the area as
$$ A = 4\int_0^{g(0)} g(x)\,dx. $$
We can easily see, by standard integration techniques, that
$$ G(x) = \frac{1}{8} \left(4 a^2 \tan ^{-1}\left(\frac{a+2 x}{\sqrt{(a-2 x) (3 a+2 x)}}\right)-4 a x+(a+2 x) \sqrt{(a-2 x) (3 a+2 x)}\right) $$
is an antiderivative. Now we use the fundamental theorem of calculus and finally get
$$ A = 4(G(g(0)) - G(0)) = 4 \frac1{12} a^2 (3 - 3\sqrt 3 + \pi) = a^2(1 - \sqrt 3 + \pi/3).$$
