The area of the region { ( x , y ) : x 2 y 8 - x 2 , y 7 } is [2023]

Question

The area of the region ${(x, y) : x^{2} \leq y \leq 8 - x^{2}, y \leq 7}$ is [2023]

Smart Achievers · Accepted Answer

(2)

$We have,$

$x^{2} \leq y$

$8 - x^{2} \geq y$

$y \leq 7$

$Converting the given inequations into equations, we get x^{2} = y and y = 8 - x^{2}$

$Solving these equations to find their point of intersection$

i.e., $x^{2} = 8 - x^{2}$

$\Rightarrow 2 x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$

$∴$ $y = 4$

$Required area = \int_{- 2}^{2} (8 - x^{2} - x^{2}) d x - \int_{- 1}^{1} (8 - x^{2} - 7) d x$

$= \int_{- 2}^{2} (8 - 2 x^{2}) d x - \int_{- 1}^{1} (1 - x^{2}) d x$ $= {[8 x - \frac{2}{3} x^{3}]}_{- 2}^{2} - {[x - \frac{x^{3}}{3}]}_{- 1}^{1}$

$= [(16 - \frac{16}{3}) - (- 16 + \frac{16}{3})] - [\frac{2}{3} + \frac{2}{3}]$

$= 16 - \frac{16}{3} + 16 - \frac{16}{3} - \frac{4}{3}$ $= 20 sq. units$

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