The area, in square units, of the region common to the circle x^2+y^2=16 and the parabola y^2=6x is

Solving $x^2+6x=16$ gives the intersection at $x=2$, $y=\pm2\sqrt3$. By symmetry about the $x$-axis, the common region is twice the area in the upper half plane: the parabola $y=\sqrt{6x}$ for $0\le x\le2$ and the circle $y=\sqrt{16-x^2}$ for $2\le x\le4$. First, $2\int_0^2\sqrt{6x}\,dx=2\sqrt6\cdot\dfrac23\,2^{3/2}=\dfrac{16\sqrt3}{3}$. Second, $2\int_2^4\sqrt{16-x^2}\,dx=\left[x\sqrt{16-x^2}+16\sin^{-1}\dfrac x4\right]_2^4=\left(8\pi\right)-\left(2\cdot2\sqrt3+\dfrac{8\pi}{3}\right)=\dfrac{16\pi}{3}-4\sqrt3$. Adding, the area is $\dfrac{16\pi}3+\dfrac{16\sqrt3}3-4\sqrt3=\dfrac{16\pi+4\sqrt3}3$. The numerical value is about $19.06$ square units, which is plausible for a region contained in a circle of area $16\pi\approx50.3$.