The area of the region \(x,y):x^2+y^2≤ 8, y^2≥ 2x\ is

The circle has area $8\pi$. The parabola $y^2=2x$ meets the circle where $x^2+2x=8$, i.e. at $x=2$, $y=\pm2$. The part of the disc lying inside the parabola ($y^2\le2x$) is bounded by the parabola for $0\le x\le2$ and by the circle for $2\le x\le2\sqrt2$. Its area is $2\left[\int_0^2\sqrt{2x}\,dx+\int_2^{2\sqrt2}\sqrt{8-x^2}\,dx\right]$. The first integral equals $8/3$. The second equals $\left[\tfrac x2\sqrt{8-x^2}+4\sin^{-1}\tfrac{x}{2\sqrt2}\right]_2^{2\sqrt2}=2\pi-(2+\pi)=\pi-2$. So the inside part is $2(\pi+\tfrac23)=2\pi+\tfrac43$, and the required area is $8\pi-2\pi-\tfrac43=6\pi-\tfrac43$. A numerical integration agrees ($\approx17.516$).