Find the area lying above $x$-axis and included between the circle $x^2 + y^2 = 8x$ and inside of the parabola $y^2 = 4x$.

$\frac{4}{3}(8 + 3\pi)$

The correct answer is Option (3) → $\frac{4}{3}(8 + 3\pi)$

The given equation of the circle $x^2 + y^2 = 8x$ can be expressed as $(x - 4)^2 + y^2 = 16$. Thus, the centre of the circle is $(4, 0)$ and radius is $4$. Its intersection with the parabola $y^2 = 4x$ gives

$x^2 + 4x = 8x$

$\text{or } x^2 - 4x = 0$

$\text{or } x(x - 4) = 0$

$\text{or } x = 0, x = 4$

Thus, the points of intersection of these two curves are $O(0, 0)$ and $P(4, 4)$ above the $x$-axis.

From the Figure, the required area of the region $OPQCO$ included between these two curves above $x$-axis is

$= (\text{area of the region } OCPO) + (\text{area of the region } PCQP)$

$= \int\limits_{0}^{4} y \, dx + \int_{4}^{8} y \, dx$

$= 2 \int\limits_{0}^{4} \sqrt{x} \, dx + \int_{4}^{8} \sqrt{4^2 - (x - 4)^2} \, dx$

$= 2 \times \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_0^4 + \int_0^4 \sqrt{4^2 - t^2} \, dt, \text{ where, } x-4=t$

$= \frac{32}{3} + \left[ \frac{t}{2} \sqrt{4^2 - t^2} + \frac{1}{2} \times 4^2 \times \sin^{-1} \frac{t}{4} \right]_0^4$

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