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

Question

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 :

Accepted Answer

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

Area of shaded region

$$\begin{aligned}
& =\int_\limits0^2 \frac{y^2}{2} d y+\int_\limits2^{2 \sqrt{2}} \sqrt{8-y^2} d y \
& =\frac{1}{6}\left[y^3\right]_0^2+\left[\frac{y}{2} \sqrt{8-y^2}+4 \sin ^{-1}\left(\frac{y}{2 \sqrt{2}}\right)\right]_2^{2 \sqrt{2}} \
& =\frac{4}{3}+\pi-2=\pi-\frac{2}{3}
\end{aligned}$$

Solutions · 1