Area of the region (x,y):x^2+(y-2)^2 4,x^2 2y is

Question

Area of the region ${(x,y):{x}^{2}+(y-2{)}^{2}\le 4,{x}^{2}\ge 2y}$ is

Accepted Answer

We have,

${x}^{2}+(y-2{)}^{2}\le {2}^{2}{x}^{2}\ge 2y$

On solving the given equation of parabola and circle, we get

$\begin{gathered} 2y+(y-2{)}^{2} =4 \ \ \Rightarrow y=02 \end{gathered}$

If $y=0$, then $x=0$ and

If $y=2$, then $x=\pm 2$

$	herefore $ The intersecting point of circle and parabola are $(0,0),(-2,2),(2,2)$

Area of the shaded region = $=2	imes 2-\frac{1}{4}\cdot \pi \cdot {2}^{2}=4-\pi $

$\begin{gathered} =2[[|\int _{0}^{2}\frac{{x}^{2}}{2}dx-(4-\pi )]]| \ =2[{\frac{{x}^{3}}{6}|}_{2}^{0}-4+\pi ] \ =2[\frac{4}{3}+\pi -4] \ =2[\pi -\frac{8}{3}] \ =2\pi -\frac{16}{6} \end{gathered}$

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