QJEEM23Mae20f: JEE Main 2023 Mathematics - Class 12 JEEM23Mae20f Solution: If the area of the region is equal to , then the natural number …

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

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

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