If the area of the region S=(x,y):2y-y^2 x^2 2y,x y is equal to (n+2)/(n+1)-( )/(n-1), the…

Question

If the area of the region $S={(x,y):2y-{y}^{2}\le {x}^{2}\le 2y,x\ge y}$ is equal to $\frac{n+2}{n+1}-\frac{\pi }{n-1}$, then the natural number $n$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Given region,

$S={(x,y):2y-{y}^{2}\le {x}^{2}\le 2y,x\ge y}$

Here, we have three curves

$\begin{gathered} 2y-{y}^{2}={x}^{2}..........(i) \ {x}^{2}=2y..........(2) \ x=y...........(3) \end{gathered}$

From Eq. (i),

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

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

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

$\begin{gathered} 2x-{x}^{2}={x}^{2} \ \Rightarrow 2{x}^{2}=2x \ \Rightarrow x(x-1)=0 \ \Rightarrow x=0,1 \ x=0y=0 \ x=1y=1 \end{gathered}$

$	herefore $ Required points are $(0,0)$ and $(1,1)$.

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

$\begin{gathered} \, {x}^{2}=2x \ \Rightarrow x(x-2)=0 \ \Rightarrow x=0,2 \ x=0y=0 \ x=2y=2 \ 	herefore (0,0)(2,2) \end{gathered}$

$	herefore $ Required area $=$ Area of triangle - Area of part I - Area of part II

$=\frac{1}{2}	imes 2	imes 2-\int _{0}^{2}\frac{{x}^{2}}{2}-(\frac{\pi }{4}-\frac{1}{2})$

$\begin{gathered} =2-\frac{1}{2}{(\frac{{x}^{3}}{3})}_{2}^{0}-\frac{\pi }{4}+\frac{1}{2} \ =2-\frac{1}{2}	imes \frac{8}{3}-\frac{\pi }{4}+\frac{1}{2} \ =\frac{7}{6}-\frac{\pi }{4} \ =\frac{5+2}{5+1}-\frac{\pi }{5-1} \ 	herefore n=5 \end{gathered}$

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