The area of the region in the first quadrant inside the circle x 2 + y 2 = 8 and outside the parabola y 2 = 2 x is equal to: [2024]

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Solution

Q.
The area of the region in the first quadrant inside the circle $x^{2} + y^{2} = 8$ and outside the parabola $y^{2} = 2 x$ is equal to: [2024]

1 $\frac{π}{2} - \frac{2}{3}$

2 $\frac{π}{2} - \frac{1}{3}$

3 $π - \frac{1}{3}$

4 $π - \frac{2}{3}$

Ans.
(4)

We have, $x^{2} + y^{2} = 8 and y^{2} = 2 x$

Required area = area of the shaded region

$= \int_{0}^{2} \frac{y^{2}}{2} d y + \int_{2}^{2 \sqrt{2}} \sqrt{8 - y^{2}} d y$

$= \frac{1}{6} {[y^{3}]}_{0}^{2} + {[\frac{y}{2} \sqrt{8 - y^{2}} + 4 \sin^{- 1} (\frac{y}{2 \sqrt{2}})]}_{2}^{2 \sqrt{2}}$

$= \frac{4}{3} + [2 π - 2 - π] = π - \frac{2}{3}$

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