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.

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...
