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 ?

 A: Assume that the side of the square $\overline{AB}=1$. Consider the diagram
$\hspace{3cm}$
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 $\color{#50B070}{2-\sqrt3}+4\left(\color{#A050A0}{\dfrac{\pi-3}{12}}\right)=1+\dfrac\pi3-\sqrt3$
A: Consider the quarter-circle of radius $r$ and a $\pi\over 2$ rotation of it where the two arcs share two common corners and each shares three corners with the square of side length $r$.  Call the area not covered by quarter-circle pieces a "counter-arc" (a triangle with missing arc pieces).  The area covered by quarter-circles can be broken up into intersecting and non-intersecting sections; note that we can cut in straight lines and get two $\pi\over 12$ arc sections and one equilateral triangle, so we have counter-arc area 
$$c=r^2-\frac 12 (r)\frac {\sqrt 3}2r-2\cdot \frac 1{12}\pi r^2=r^2\left(1-\frac {\sqrt 3}4-\frac {\pi}6\right)$$
Now we have the counter-arc, we can get the arrow-arc (the area outside one quarter-circle and between two counter-arcs), which is $$a=r^2-{\pi r^2\over 4}-2c=r^2\left(1-\frac {\pi}4-2+\frac{\sqrt 3}2+\frac {\pi}3\right)$$
The arrow-arcs and counter-arcs together make up the outside perimeter of the indicated shape, so we have shaded area
$$r^2-4a-4c=r^2\left(1-4+\sqrt 3+\frac {2\pi}3-4+\pi+8-2\sqrt 3-\frac {4\pi}3\right)=r^2\left(1-\sqrt 3+\frac {\pi}3\right)$$
A: 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.
A: Let $R$ be its radius and $D$ its diameter: $R = 5$, $D = 10$.
$$\begin{align}
\text{Area of big square} &= D^2 = 100 \\
\text{Area of circle} &= \frac{\pi D^2}{4} \approx 78.54 \\
\text{Area outside circle} &= 100 - 78.54 = 21.46 \\
\text{Area of 4 petals} &= 78.54 - 21.46 = 57.08 \\
\text{Area of single petal} &= \frac{57.08}{4} = 14.27 \\
\text{Area of small square} &= R^2 = 25
\end{align}$$
Let $x$ denote the area of the portion selected.
$$\begin{align}
\text{Area of 2 petals} &= 2 \cdot 14.27 = 28.54 \\
0 &= 25 - 28.54 + x  \\
\text{Area of 8 petals} &= 2 \cdot 57.08 = 114.16 \\
\text{OR}\\
100 - 114.16 + 4 x &= 0
\end{align}$$
$$ x = 3.54 $$
Here we have a small area which needs to be added,
that I found out by modeling to be $4.34$
$$\begin{align}
\text{This gives us the desired area of}\\ 
\text{4.34+3.54} &\approx 7.88\\
\text{& a percentage of} &\approx 4 \times 7.88\\
\text{that is} &\approx 31.52\%
\end{align}$$
A: $$\begin{align}
\text{Let $r$ be radius of circle in square, $if$}\\
\text{ $r$} &= 10\\
\text{by coordinate geometry marked Area}
&= \left( \frac {\pi + 3 - 3 \sqrt 3}{3} \right) \cdot r^2\\
&\approx 31.51
\end{align}$$
