Area of the intersection of four circles of equal radius 
This picture basically shows a rearrangement of four quarters of a circle of radius 1. It asks for the shaded area.
I got the answer to be $\frac{2\pi + 6}{13}$. But then it is incorrect. 
The way I did it is by rearranging the four pieces into respectively a circle a square and a larger square with a inscribed circle. And naming each different pieces in the picture as x y and z.
 A: 
To figure out the area of the shape you described, we can split it into two parts: the area in the square, and the part just outside it. For both, we need to find out the angle between which each circle intersects. We can find this easily by using the equations of circle. To find the top intersection, consider $x^2+y^2=1$ and $(x-1)^2+y^2=1$. These circles intersect at $(\frac12,\frac{\pm\sqrt3}{2})$, which is at an angle of $\frac\pi3$ with the x-axis. Similarily, the coordinates of the right point make an angle of $\frac\pi6$ with the x-axis. Hence, each circle intersects in an angle of $\frac\pi6$.
Now to find the areas. To find the areas of the four smaller sections between the square and circles, it is the area of the sector minus the triangle in the sector. Hence the area of each is 
$$\frac12\cdot\frac\pi6-\frac12\sin\left(\frac\pi6\right)=\frac\pi{12}-\frac14$$.
Now for the square. If the side length of the the square is $s$, then by connecting the top and right points with the bottom left point of the big square, we create a triangle with sides $1,1$ and $s$, and the angle between the 1s is $\frac\pi6$. By the cosine rule, the area of the square is
$$s^2=1^2+1^2-2\cos\left(\frac\pi6\right)=2-\sqrt3$$
Hence, the area of the shape between the four circles is
$$4\left(\frac\pi{12}-\frac14\right)+(2-\sqrt3)=\frac\pi3+1-\sqrt3$$
No calculus required.
Alternatively:

Consider the set of parametric equations $x(t)=\cos(t)-\frac12,y(t)=\sin(t)-\frac12$. This is the equation of the circle centred at the bottom left point. Point $D$ is a $t=0$, E at $t=\frac\pi6$, and $F$ at $t=\frac\pi3$. Hence, the centre area is 4 times the area between $F$ and $E$, so its area is 
$$4\int_{t=\frac\pi3}^{t=\frac\pi6} y \,dx=-4\int_{\frac\pi6}^{\frac\pi3}y(t)x'(t)\,dt=4\int_{\frac\pi6}^{\frac\pi3}\left(\sin(t)-\frac12\right)\sin(t)\,dt=\frac\pi3+1-\sqrt3$$
A: Consider; What about trying to fit a rectangle in a quarter area?
As a rectangle is 2 x triangles, using the Pythagorean Theorem is handy. $1^2 = x^2 + {\frac{1}{2}}^2$
Thus we can figure out that $x = \sqrt{\frac{3}{2}}$, then integrate to get the area under the curve where there would be no rectangle.
Let $A$ denote the area, 
$$A = \int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{1}{2}(1-\sin(2u))\,du = \frac{\pi}{12}$$
and let $R$ denote the area of the rectangle, and $S$ denote the shaded region,
$$R=\frac{1}{2}(\sqrt{\frac{3}{2}} - \frac{1}{2}) = \frac{\sqrt{3} -1}{4}$$
Which leaves us with $$\frac{S}{4} = A-R \implies\\ S = \frac{\pi}{3} - \sqrt{3} + 1$$

Also, googling "quarter circles" gives a pretty few good results...
A: Note that we could use the first quadrant and then get the final answer by symmetry. We have three quarter-circles in this region with the equations$$\begin{array}{l}{C_1}:{\left( {x - \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\\{C_2}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = 1\\{C_3}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\end{array}$$we want to obtain the area beneath ${C_3}$ (to the $x$-axis) but to obtain the endpoint, we have to find the cross point of ${C_2}$ and ${C_3}$, which is $$\begin{array}{l}x = \frac{{\sqrt 3  - 1}}{2}\\y = 0\end{array}$$now, the final answer is $$I = 4\int\limits_0^{\frac{{\sqrt 3  - 1}}{2}} {\left( { - \frac{1}{2} + \sqrt {1 - {{\left( {x + \frac{1}{2}} \right)}^2}} } \right)dx}  = 1 - \sqrt 3  + \frac{\pi }{3}$$
A: To solve this problem, the only tool that you need is:-
(P ∩ Q) = (P) + (Q) – (P ∪ Q)
The following is an example.

