Is there a way to calculate the area of this intersection of four disks without using an integral?

Is there anyway to calculate this area without using integral ?

Assume that the side of the square $\overline{AB}=1$. Consider the diagram

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By symmetry, $\overline{EC}=\overline{CD}$; therefore, $\overline{CD}=1/2$. Since $\overline{AC}=1$ and $\overline{AD}\perp\overline{CD}$, we have that $\angle CAD=\pi/6$ ($30$-$60$-$90$ triangle). Similarly, $\angle GAF=\pi/6$, leaving $\angle CAG=\pi/6$.

Since base $\overline{AB}=1$ and altitude $\overline{CD}=1/2$, $\triangle ABC$ has area $1/4$.

Since $\angle CAB=\pi/6$, the circular sector $CAB$ has area $\pi/12$.

Therefore, the area of the purple half-wedge between $B$ and $C$ is $\color{#A050A0}{\dfrac{\pi-3}{12}}$.

Furthermore, $\overline{CG}^2=\overline{BC}^2=\overline{CD}^2+\overline{DB}^2=\left(\frac12\right)^2+\left(1-\frac{\sqrt3}{2}\right)^2=\color{#50B070}{2-\sqrt3}$.

Therefore, the area requested is $\color{#50B070}{2-\sqrt3}+4\left(\color{#A050A0}{\dfrac{\pi-3}{12}}\right)=1+\dfrac\pi3-\sqrt3$

Your "curvilinear square" just cuts the quarter-circles in thirds, so the distance between two adjacent vertices is $2l\sin\frac{\pi}{12}=\frac{\sqrt{3}-1}{\sqrt{2}}l$, given that $l$ is the length of the side of the original square. So the area of the "circular square" is given by $(2-\sqrt{3})l^2$ plus four times the area of a circular segment.

The area of such a circular segment is the difference between the area of a circular sector and the area of an isosceles triangle having base length $l\frac{\sqrt{3}-1}{\sqrt{2}}$ and height $l\cos\frac{\pi}{12}=\frac{\sqrt{3}+1}{2\sqrt{2}}l$, hence: $$ S = \left(\frac{\pi}{12}-\frac{1}{4}\right)l^2 $$ and the area of the "circular square" is just: $$ Q = \left(1-\sqrt{3}+\frac{\pi}{3}\right)l^2.$$

With integrals, by following Shabbeh's suggestion, we have: $$ Q = 4l^2\int_{1/2}^{\sqrt{3}/2}\left(\sqrt{1-x^2}-\frac{1}{2}\right)dx = 2l^2\left.\left(x\sqrt{1-x^2}-x+\arcsin x\right)\right|_{1/2}^{\sqrt{3}/2}$$ that obviously leads to the same result. Just a matter of taste, as usual.