What Does It Look Like?

First we consider the case where $x,y\geq 0$ to eliminate the absolute value signs. We can manipulate equation to rewrite it in the standard form of a circle :

$x^2\textcolor{#3D99F6}{-x}+y^2\textcolor{#3D99F6}{-y}=0$

$x^2-x \textcolor{#3D99F6}{+\dfrac{1}{4}}+y^2-y \textcolor{#3D99F6}{+\dfrac{1}{4}}=\textcolor{#3D99F6}{\dfrac{1}{2}}$

$\left(x-\dfrac{1}{2}\right) ^2 + \left(y-\dfrac{1}{2}\right) ^2 = \dfrac{1}{2} = \left(\dfrac{\sqrt{2}}{2}\right) ^2$

So, we know the circle has center $\left(\dfrac{1}{2} , \dfrac{1}{2}\right)$ and radius $\dfrac{\sqrt{2}}{2}$ . We also know that it crosses the axes at $(1,0)$ and $(0,1)$ . By symmetry, we can tell that there are like circles in all four quadrants for each of the cases of $x$ and $y$ being positive/negative. Thus, the graph of the equation is as shown:

Note that the region is made up of 4 semicircles lying on sides of a square with side length $\sqrt{2}$ . The semicircles each have area $\dfrac{1}{2}\pi \left(\dfrac{\sqrt{2}}{2}\right) ^2=\dfrac{1}{4} \pi$ . Therefore, the combined area is $\boxed{\pi+2}$ .

The polar equation of the portion of the bounding curve in the first quadrant is $r \; = \; \cos\theta + \sin\theta \qquad \qquad 0 \le \theta \le \tfrac12\pi$ and so the area enclosed (by symmetry) is $4 \times \tfrac12\int_0^{\frac12\pi}(\cos\theta + \sin\theta)^2\,d\theta \; = \; 2\int_0^{\frac12\pi} (1 + \sin2\theta)\,d\theta \; = \; \pi+2$

the edge of the square = the diameter of the half circles= √ $(1^2 + 1^2)$ = √2

total area = area of square + area of 4 half circles = $(√2)^2 + (\frac{4×π×(√2)^2}{4×2}$ ) = $π + 2$

N.B. the sides of the square is the diameter of the half circles because it contains the centers of the circles: $x^2 + y^2 ± x ± y = 0$

As shown by Mohtasim Nakib, by graphing the function and realizing that it is a square and four semicircles, I got the area.