Lil Bit Complex 2

$\begin{cases} |x+y| \le 1 & \Rightarrow \begin{cases} y \ge -x-1 \\ y \le -x + 1 \end{cases} & \text{within a pair of parallel lines } -45^\circ \text{ with x-axis} \\ |y-x| \le 1 & \Rightarrow \begin{cases} y \ge x-1 \\ y \le x + 1 \end{cases} & \text{within a pair of parallel lines } 45^\circ \text{ with x-axis} \\ 2x^2 + 2y^2 = 1 & \Rightarrow x^2+y^2 = \frac{1}{2} & \text{a circle centered at origin with radius } \frac{1}{\sqrt{2}} \end{cases}$

The four parallel lines and the circle are as in the figure below. Therefore, the bounded area $A =$ the area of square with $\sqrt{2}$ side length $-$ the area of the circle with $\frac{1}{\sqrt{2}}$ radius.

$\begin{aligned} A & = \left(\sqrt{2}\right)^2 - \pi \left(\frac{1}{\sqrt{2}} \right)^2 = 2 - \frac{\pi}{2} \approx \boxed{0.4292} \end{aligned}$

Same as Chew-Seong Cheong ,( for whom I have up voted,)put differently.
The four lines define a square with diagonal= 2, and along the x-axis and y-axis, intersecting at the origin. So its area is $\frac 1 2*2*2=2.$ . Area of the circle fully inside it with radius $\sqrt{\frac{1} 2} ,~~ is ~~\dfrac \pi 2$ . So the difference of the two areas= $2-\dfrac \pi 2 = .4292.$