What is the graph?

The circle $(x-1)^2 + (y-1)^2 = 2$ has polar equation $r = 2(\sin\theta + \cos\theta)$ , and so the area of the portion of that circle within the first quadrant is $\tfrac12\int_0^{\frac12\pi} 4(\sin\theta + \cos\theta)^2\,d\theta \; = \; 2\int_0^{\frac12\pi}(1 + \sin2\theta)\,d\theta \; = \; \pi + 2$ Thus the area of the desired region defined by $x^2 + y^2 \le 2|x| + 2|y|$ is $4(\pi + 2)$ . Taking away $4\pi$ , the area of the circle $x^2 + y^2 \le 4$ , we obtain the answer $\boxed{8}$ .

Given $4\leq x^2+y^2\leq 2|x|+2|y|$

So $x^2+y^2\geq4$

This is the region outside and on the circle of radius $2$ and centre at origin.

Now, $x^2+y^2\leq 2|x|+2|y| \implies (|x|-1)^2+(|y|-1)^2\leq2$

This represent the regions inside the circles of radius $\sqrt2$ and centres $(1,1) , (-1,1) , (-1,-1) , (1,-1)$

So the required region is out side the circle(1) $x^2+y^2=4$ and inside the circles(2) of centre $(\pm1,\pm1)$ and radius $\sqrt2$

Consider any one circle of centre $(1,1)$ of family of circles(2). Then the intersecting points of both the circles are $(0,2),(2,0)$ And every four circles of family of circles(2) passes through origin. Here we can see that the line joining points $(1,1)$ and $(2,0)$ and $(1,1)$ and $(0,2)$ have same slope so they represent a single line. Also this line passes through centre of circle(2) so this is a diameter for circle(2).

Hence outside this on R.H.S the value of area is $\pi$ . But this is not the required answer.We have to minus the are of circle (1) from this. Now consider the right angled triangle having vertices $(0,2),(2,0), (0,0)$ So it's area will be $2$ .Now area of circle(1) in 1st quadrant is $\pi$ . So if we remove the area of triangle from area of circle(1) then we will get that area which has to be removed from our outside region in order to get our answer. So $\pi-2$ has to be removed from $\pi$ . So we got $2$

Now there are four such circles. So $2\times4 = \boxed{8}$ which is our required area.