JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 4)
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 :
$$\frac{\pi}{2}-\frac{1}{3}$$
$$\pi-\frac{1}{3}$$
$$\pi-\frac{2}{3}$$
$$\frac{\pi}{2}-\frac{2}{3}$$
Explanation
We have, $$x^2+y^2=8$$ and $$y^2=2 x$$
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Area of shaded region
$$\begin{aligned} & =\int_\limits0^2 \frac{y^2}{2} d y+\int_\limits2^{2 \sqrt{2}} \sqrt{8-y^2} d y \\ & =\frac{1}{6}\left[y^3\right]_0^2+\left[\frac{y}{2} \sqrt{8-y^2}+4 \sin ^{-1}\left(\frac{y}{2 \sqrt{2}}\right)\right]_2^{2 \sqrt{2}} \\ & =\frac{4}{3}+\pi-2=\pi-\frac{2}{3} \end{aligned}$$
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