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 :

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Aicharya · Accepted Answer

Required area $=\operatorname{Ar}($ circle from 0 to 2$)-$ $\operatorname{ar}($ para from 0 to 2$)$ $\begin{aligned} & =\int_0^2 \sqrt{8-x^2} d x-\int_0^2 \sqrt{2 x} d x \\ & =\left[\frac{x}{2} \sqrt{8-x^2}+\frac{8}{2} \sin ^{-1} \frac{x}{2 \sqrt{2}}\right]_0^2-\sqrt{2}\left[\frac{x \sqrt{x}}{3 / 2}\right]_0^2 \\ & =\frac{2}{2} \sqrt{8-4}+\frac{8}{2} \sin ^{-1} \frac{2}{2 \sqrt{2}}-\frac{2 \sqrt{2}}{3}(2 \sqrt{2}-0) \\ & \Rightarrow 2+4 \cdot \frac{\pi}{4}-\frac{8}{3}=\pi-\frac{2}{3}\end{aligned}$

The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2…

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