Find the area lying above x-axis and included between the circle x² + y² = 8x and inside of the parabola y² = 4x.

$4\pi-\frac{32}{3}$

The correct answer is Option (2) → $4\pi-\frac{32}{3}

$A=\int\limits_{0}^{4}|y_2-y_1|dx$

$=\int\limits_{0}^{4}\left[\sqrt{16-(x-4)^2}-\sqrt{4x}\right]dx$

$\left[\frac{(x-4)}{2}\sqrt{16-(x-4)^2}+\frac{16}{2}\sin^{-1}\left(\frac{x-4}{4}\right)\right]_{0}^{4}-\left[\frac{4x^{3/2}}{3}\right]_{0}^{4}$

$=\left[0+0-0-8\sin^{-1}(\frac{-4}{4})\right]-\frac{4}{3}×4^{3/2}=4\pi-\frac{32}{3}$