Let y = y(x) be the solution of the differential equation $x\tan \left( {{y \over x}} \right)dy = \left( {y\tan \left( {{y \over x}} \right) - x} \right)dx$, $ - 1 \le x \le 1$, $y\left( {{1 \over 2}} \right) = {\pi \over 6}$. Then the area of the region bounded by the curves x = 0, $x = {1 \over {\sqrt 2 }}$ and y = y(x) in the upper half plane is :