Area of the region { ( x , y ) : x 2 + ( y - 2 ) 2 4 , x 2 2 y } is [2023]

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Solution

Q.
Area of the region ${(x, y) : x^{2} + {(y - 2)}^{2} \leq 4, x^{2} \geq 2 y}$ is [2023]

1 $2 π - \frac{16}{3}$

2 $π - \frac{8}{3}$

3 $π + \frac{8}{3}$

4 $2 π + \frac{16}{3}$

Ans.
(1)

$Given that x^{2} + {(y - 2)}^{2} \leq 2^{2} and x^{2} \geq 2 y$

$Convert the given inequalities into equations:$

$x^{2} + {(y - 2)}^{2} = 4 and x^{2} = 2 y$

$Solving circle and parabola simultaneously:$

$2 y + y^{2} - 4 y + 4 = 4,$ $y^{2} - 2 y = 0,$ $y = 0, 2$

$Put y = 2 in x^{2} = 2 y \Rightarrow x = \pm 2 \Rightarrow (2, 2) and (- 2, 2)$

$A_{1} = 2 \times 2 - \frac{1}{4} \cdot π \cdot 2^{2} = 4 - π$

$Required area = 2 [\int_{0}^{2} \frac{x^{2}}{2} d x - (4 - π)]$

$= 2 [\frac{x^{3}}{6} |_{0}^{2} - 4 + π] = 2 [\frac{4}{3} + π - 4] = 2 [π - \frac{8}{3}] = 2 π - \frac{16}{3} sq. units$

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