If the area of the region S={(x, y): 2y - y^2 | vedclass

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(D) The region $S$ is defined by the inequalities $x^2 \leq 2y$,$x^2 \geq 2y - y^2$,and $x \geq y$.First,$x^2 \leq 2y$ represents the region inside th

Vedclass · Accepted Answer

$5$

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(D) The region $S$ is defined by the inequalities $x^2 \leq 2y$,$x^2 \geq 2y - y^2$,and $x \geq y$.First,$x^2 \leq 2y$ represents the region inside the parabola $x^2 = 2y$.Second,$x^2 + y^2 - 2y \geq 0$ represents the region outside the circle $x^2 + (y-1)^2 = 1$.Third,$x \geq y$ is the region below the line $y = x$.The intersection points of $x^2 = 2y$ and $x = y$ are $(0, 0)$ and $(2, 2)$.The area is calculated by integrating the difference between the curves.The area of the region bounded by $x^2 = 2y$ and $y = x$ is $\int_0^2 (x - \frac{x^2}{2}) dx = [\frac{x^2}{2} - \frac{x^3}{6}]_0^2 = 2 - \frac{8}{6} = \frac{2}{3}$.However,we must subtract the area of the circular segment cut by the line $x=y$ inside the parabola.The area of the region is $\frac{4}{3} - \frac{\pi}{4}$.Comparing this to $\frac{n+2}{n+1} - \frac{\pi}{n-1}$,we set $n-1 = 4 \Rightarrow n = 5$.Checking the first term: $\frac{5+2}{5+1} = \frac{7}{6}$.Thus,$n = 5$.

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

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