Find area of striped region enclosed by a quarter and two semi circles along with two lines inside a square The figure is a square of side 2$a$. It is requested to find the striped area, but no information is given of the position of the lines that intersect the sides and that makes the area vary greatly. There is a special case for the two lines bisecting sides, but I I imagine that the problem points to a general solution with those mobile sides which complicates me just enough to do it.
For conventional geometry if there is a way to do it I do not see it, for integration I imagine defining possible points that I will call
on the left side and y on the right x where $0 <x <= 2a$ e $0 <y <=2a$
Any idea how to raise it?

 A: 
Assume that A and D are midpoints. Let $(x,y)$ be the location of the point Y, which is the intersection of the circle $x^2+y^2 = 4a^2$ and the line $y=2(x-a)$. Solve to get $x=\frac85a$ and $y=\frac65a$.
Let [.] denote areas. Then the striped area is 
$$A = 2[OAY]+[OYZ]- [OAXD] - 2[DXC]$$
$$=2\left(\frac12  \frac65a^2\right)+\frac12 (2a)^2\alpha
 -a^2 -2\left( \frac12 a^2\beta\right) = \left( \frac15+2\alpha-\beta\right)a^2$$
$\beta$ satisfies $\tan\beta = \frac12$ and $\alpha$ is derived from $\tan\theta = \frac yx = \frac34$ as follows,
$$\tan\alpha = \tan(90-2\theta) = \cot(2\theta) 
= \frac{1-\tan^2\theta}{2\tan\theta} = \frac7{24}$$
Thus, the area is
$$A= \left( \frac15+2\tan^{-1}\frac7{24}-\tan^{-1}\frac12\right)a^2$$
or about $0.3a^2$, which is verified from numerical integration.
A: 
The area of the region $OMNDP$
is found by subtracting 
the sum of the areas of semicircle $AOPD$
and half of the area of the square $AEOF$
from the sum of 
$45^\circ$ sector of the circle ($AMND$)
and the area of the sector $EOA$:
\begin{align}
[OMNDP]&=
\tfrac12\,\pi a^2
+
\tfrac14\,\pi a^2
-\tfrac12\,\pi a^2
-\tfrac12\, a^2
=\tfrac14\,a^2\,(\pi-2)
\tag{1}\label{1}
.
\end{align} 
The area of the half of the region in question
can be found by subtracting 
the area $[DPN]$ from \eqref{1}.
Using the coordinate system with the origin $F$,
$x$-axis 
emanating from it along the direction $QD$
and
$y$-axis along $FC$,
\begin{align}
[DPN]&=\int_0^{|DQ|} \int_{f_1(x)}^{f_2(x)}\,dy\,dx
\tag{2}\label{2}
,\\
f_1(x)&=\sqrt{a^2-x^2}
,\\
f_2(x)&=A_y+\sqrt{4a^2-(x-A_x)^2}
.
\end{align}
We can easily find that 
\begin{align}
|DQ|&=\tfrac{2\sqrt5}5\,a
,\\
A_x&=-D_x
=-\tfrac{2\sqrt5}5\,a
,\\
A_y&=-D_y
=-\tfrac{\sqrt5}5\,a
\end{align}
and \eqref{2} becomes
\begin{align} 
[DPN]&=
\tfrac1{10}\,a^2(35\arctan(2)-6-10\pi)
\end{align}
and 
\begin{align} 
[OMNP]&=[OMNDP]-[DPN]
=
\tfrac1{20}\,a^2(25\pi+2-70\arctan(2))
,
\end{align}
and 
\begin{align} 
[OKLMNP]=
2[OMNP]&=
\tfrac1{10}\,a^2(25\pi+2-70\arctan(2))
\approx .30394\,a^2
.
\end{align}
Note, that despite it looks slightly different,
the result is the same as in
the @Quanto's answer. 
Another way could be a standard integration 
using coordinate system at the origin $O$
and $OC$ and $OD$ as $x$ and $y$ axes respectively.
