How to compute the area of the shadow? 
If we can not use the integral, then how to compute the area of the shadow?
It seems easy, but actually not?
Thanks!
 A: First find the intersection points of the circles.  If you make the square of side $2$ with the center at the origin, they are $(\frac 14(1-\sqrt 7),\frac 14(1+\sqrt 7))$ and $(\frac 14(1+\sqrt 7),\frac 14(1-\sqrt 7))$.  The line segment between them has length $\frac {\sqrt {14}}2$ which gives the angle of the circular sector of each circle.  Then the shaded area is the difference in area of two circular segments, one of radius $2$ and one of radius $1$
A: Let $a$ be a side of the square. Consider the following diagram

The area we need to calculate is as follows.
$$\begin{eqnarray} \color{Black}{\text{Black}}=(\color{blue}{\text{Blue}}+\color{black}{\text{Black}})-\color{blue}{\text{Blue}}. \end{eqnarray}$$
Note that the blue area can be calculated as
$$\begin{eqnarray}\color{blue}{\text{Blue}}=\frac14a^2\pi-2\cdot\left(\color{orange}{\text{Yellow}}+\color{red}{\text{Red}}\right).\end{eqnarray}$$
We already know most of the lengths. What's stopping us from calculating the black area is lack of known angles. Because of symmetry, almost any angle would do the trick.
It's fairly easy to calculate angles of triangle $\begin{eqnarray}\color{orange}{\triangle POA}\end{eqnarray}$, if we use cosine rule.
$$\begin{eqnarray}
|PA|^2&=&|AO|^2+|PO|^2-2\cdot|AO|\cdot|PO|\cos\angle POA\\
a^2&=&\frac{a^2}{4}+\frac{2a^2}{4}-2\cdot\frac a2\cdot\frac{a\sqrt2}{2}\cdot\cos\angle POA\\
4a^2&=&3a^2-2a^2\sqrt2\cos\angle POA\\
1&=&-2\sqrt2\cos\angle POA\\
\cos\angle POA&=&-\frac{1}{2\sqrt2}=-\frac{\sqrt2}{4}.
\end{eqnarray}$$
Now, because of symmetry, we have $\angle POA=\angle POB$, so $\angle AOB=360^\circ-2\angle POA$. So the cosine of angle $\angle AOB$ can be calculated as follows:
$$\begin{eqnarray}
\cos\angle AOB&=&\cos(360^\circ-2\angle POA)=\cos(2\pi-2\angle POA)\\
\cos\angle AOB&=&\cos(-2\angle POA)=\cos(2\angle POA)\\
\cos\angle AOB&=&\cos^2(\angle POA)-\sin^2(\angle POA)\\
\cos\angle AOB&=&\cos^2(\angle POA)-(1-\cos^2(\angle POA))\\
\cos\angle AOB&=&2\cos^2(\angle POA)-1\\
\cos\angle AOB&=&2\cdot\left(-\frac{\sqrt2}{4}\right)^2-1=-\frac34\\
\end{eqnarray}$$
From this, we can easily calculate the sine of angle $\angle AOB$, using Pythagorean identity.
$$ \sin\angle AOB=\sqrt{1-\frac9{16}}=\sqrt\frac{16-9}{16}=\frac{\sqrt7}4 $$
Going this way, I believe it's not hard to calculate other angles and use known trigonometry-like formulas for area. Then you can easily pack it together using the first equation with colors.
A: Assume a square of side length $2$ for simplicity.
Denote the various sub-areas as $A, B, \ldots, H$ from top to bottom, left to right. So, the shaded area is $C$, the small disk is $C+D=\pi$, the big quarter disk is $\pi = A+E+G+H$.
Note that $A+B=F+H=D=G=1-\frac\pi 4$.
We locate the right vertex of $A$, measured from the center by solving $x^2+y^2=1$ and $(1+y)^2+(1+x)^2=4$, which implies 
$$x+y=\frac{(1+y^2)+(1+x)^2-(x^2+y^2)-2}2=\frac12$$
and then $x^2+(\frac12-x)^2=1$, hence $x=\frac{1\pm\sqrt 7}4$ and by sign consideration, 
$$ x=\frac{1-\sqrt 7}4, \quad y=\frac{1+\sqrt 7}4.$$
Now $B$ can be computet as the difference of a rectangle $1\times (1-y) =\frac{3-\sqrt 7}4$ and thow halves of circular segments.
But for these segments (using e.g. the formula $ r^2\arccos(1-\frac hr)-(r-h)\sqrt{(2r-h)h}$ for the area of a segment), we need to compute some angles!
I doubt that anything better than the $\arccos$ expression can be found
for these angles (i.e. they are not a rational multiple of $\pi$).
Having thus obtained an expression 
$$B=a+b\sqrt 7+c\arccos\frac{1+\sqrt 7}4+d\arccos\frac{5+\sqrt 7}8$$ 
with $a,b,c,d$ rational, we finally find
$$C = A+G+H = 3G-2B = 3-\frac34\pi -2B. $$
A: EDIT: Sorry this has an error in it... I solved with a $2E$ rather than a $3E$.
Denote the corners of this unit square by $X,Y,Z,W$ and draw the circles of unit radii with centres at $X$, $Y$, $Z$ and $W$.
If you do this your square is divided up into 21 regions. Denote the central region by $C$. Denote the arrow heads by $E$. Denote the 'other bits' at the edge by $D$, the things that look like annulli by $A$ and the other bits --- the round hats by $B$:

Now the total area is
$$4A+4B+C+8D+4E\overset{!}{=}1,$$
the area of the little circle is
$$4A+4B+C\overset{!}{=}\pi\left(\frac{1}{2}\right)^2,$$
and the area of a quarter of one of those unit circles is
$$3A+2B+C+4D+3E\overset{!}{=}\frac{1}{4}\pi(1)^2.$$
Now solve these equations for $A$ and $B$ in terms of $C$ which should be possible. Now the area you are looking for is $A+2B$... it turns out this EDIT DOES HAVE $E$ dependence and you are left with
$$\frac{4-\pi}{8}+E.$$
