Given region,

$$ S=\left\{(x, y): 2 y-y^2 \leq x^2 \leq 2 y, x \geq y\right\} $$

Here, we have three curves

$$ \begin{aligned} &2 y-y^2 =x^2 ..........(i)\\\\ &x^2 =2 y ..........(2)\\\\ &\text {and}~~ x = y ...........(3) \end{aligned} $$



From Eq. (i),

$$ x^2+y^2-2 y=0 $$

$\Rightarrow x^2+(y-1)^2=1$ is a circle with centre $(0,1)$ and radius is 1.

Intersection points of Equations (i) and (iii) are,

$$ \begin{array}{rlrl} & 2 x-x^2 =x^2 \\\\ &\Rightarrow 2 x^2 =2 x \\\\ &\Rightarrow x(x-1) =0 \\\\ &\Rightarrow x =0,1 \\\\ & \text { When, } x=0 \text {, then } y=0 \\\\ & \text { and when } x=1 \text {, then } y=1 \end{array} $$

$\therefore$ Required points are $(0,0)$ and $(1,1)$.

Also, intersection points of (ii) and (iii)

$$ \begin{aligned} & \quad x^2=2 x \\\\ & \Rightarrow x(x-2)=0 \\\\ & \Rightarrow x=0,2 \\\\ & \text { When, } x=0 \text {, then } y=0 \\\\ & \text { and when } x=2 \text {, then } y=2 \\\\ & \therefore \text { Required points are }(0,0) \text { and }(2,2) \end{aligned} $$

$\therefore$ Required area $=$ Area of triangle - Area of part I - Area of part II

$$ =\frac{1}{2} \times 2 \times 2-\int\limits_0^2 \frac{x^2}{2}-\left(\frac{\pi}{4}-\frac{1}{2}\right) $$

$$ \begin{aligned} & =2-\frac{1}{2}\left(\frac{x^3}{3}\right)_0^2-\frac{\pi}{4}+\frac{1}{2} \\\\ & =2-\frac{1}{2} \times \frac{8}{3}-\frac{\pi}{4}+\frac{1}{2} \\\\ & =\frac{7}{6}-\frac{\pi}{4} \\\\ & =\frac{5+2}{5+1}-\frac{\pi}{5-1} \\\\ & \therefore n =5 \end{aligned} $$