The area of the region common to the circle x^2+y^2=8 and the parabola y^2=2x (that is, points satisfying both x^2+y^2≤8 and y^2≤2x) is

Substituting $y^2=2x$ in the circle gives $x^2+2x-8=0$, so $x=2$ and $y=\pm2$. The region is symmetric about the $x$-axis. For $0\le x\le2$ it is bounded by the parabola, and for $2\le x\le2\sqrt2$ by the circle. Area $=2\left[\int_0^2\sqrt{2x}\,dx+\int_2^{2\sqrt2}\sqrt{8-x^2}\,dx\right]$. The first integral is $\sqrt2\cdot\tfrac23\cdot2^{3/2}=\tfrac83$. The second is the circular segment-type area $\left[\tfrac x2\sqrt{8-x^2}+4\sin^{-1}\tfrac{x}{2\sqrt2}\right]_2^{2\sqrt2}=2\pi-(2+\pi)=\pi-2$. So the area is $2\left(\tfrac83+\pi-2\right)=2\pi+\tfrac43$. Numerically, $2\pi+\tfrac43\approx7.62$, which is slightly less than the area of the full circle $8\pi\approx25.1$ times its fraction inside the parabola, and a quick sanity check against the bounding rectangle of width $2\sqrt2$ and height $4$ (area about $11.3$) confirms the value is reasonable.