Find the area lying in the first quadrant enclosed by the X axis, the circle $x^2 + y^2 = 8x$ and the parabola $y^2 = 4x$.
The circle and the parabola intersect at the points $A \left(4,4\right)$ and $B \left(4,-4\right)$.
Equation of the circle is $x^2 + y^2 = 8x$
i.e. $y = \sqrt{8x - x^2}$
Let $y = f\left(x_1\right) = \sqrt{8x - x^2}$ $\;\;\; \cdots \; (1)$
Equation of the parabola is $y^2 = 4x$
i.e. $y = 2 \sqrt{x}$
Let $y = f\left(x_2\right) = 2 \sqrt{x}$ $\;\;\; \cdots \; (2)$
Let the area enclosed by the X axis, the circle and the parabola in the first quadrant $= A = \left|I\right|$ where
$I = \displaystyle \int \limits_{0}^{4} f \left(x_2\right) \; dx + \int \limits_{4}^{8} f \left(x_1\right) \; dx$ $\;\;\; \cdots \; (3)$
Since the circle is symmetric about its radius $AD$,
$\displaystyle \int \limits_{4}^{8} f \left(x_1\right) \; dx = \int \limits_{0}^{4} f \left(x_1\right) \; dx$ $\;\;\; \cdots \; (3a)$
$\therefore$ $\;$ In view of equation $(3a)$, equation $(3)$ can be written as
$I = \displaystyle \int \limits_{0}^{4} \left\{f \left(x_1\right) + f \left(x_2\right) \right\} \; dx$ $\;\;\; \cdots \; (3b)$
$\therefore$ $\;$ By equations $(1)$ and $(2)$, equation $(3b)$ becomes
$\begin{aligned}
I & = \int \limits_{0}^{4} \left\{\sqrt{8x - x^2} + 2 \sqrt{x} \right\} \; dx \\\\
& = \int \limits_{0}^{4} \sqrt{8x - x^2} \; dx + 2 \int \limits_{0}^{4} \sqrt{x} \; dx \\\\
& = I_1 + I_2 \;\;\; \cdots \; (4)
\end{aligned}$
$\begin{aligned}
\text{Now, } I_1 & = \int \limits_{0}^{4} \sqrt{8x - x^2} \; dx \\\\
& = \int \limits_{0}^{4} \sqrt{-\left[\left(x^2 - 8x + 16\right) - 16\right]} \; dx \\\\
& = \int \limits_{0}^{4} \sqrt{\left(4\right)^2 - \left(x - 4\right)^2} \; dx \\\\
& \left[\text{Note: } \int \limits_{p}^{q} \sqrt{a^2 - x^2} \; dx = \left[\dfrac{x}{2} \sqrt{a^2 - x^2} + \dfrac{a^2}{2} \sin^{-1} \left(\dfrac{x}{a}\right)\right]_{p}^{q} \right] \\\\
& = \left[\left(\dfrac{x - 4}{2}\right)\sqrt{16 - \left(x-4\right)^2} + \dfrac{16}{2} \sin^{-1} \left(\dfrac{x - 4}{4}\right)\right]_{0}^{4} \\\\
& = 0 + 8 \sin^{-1} \left(0\right) - \left(-2\right) \sqrt{16 - 16} - 8 \sin^{-1} \left(-1\right) \\\\
& = 8 \times \dfrac{\pi}{2} = 4 \pi \;\;\; \cdots \; (5)
\end{aligned}$
$\begin{aligned}
I_2 & = \int \limits_{0}^{4} \sqrt{x} \; dx \\\\
& = \dfrac{2}{3} \left[\left(x\right)^{3/2}\right]_{0}^{4} = \dfrac{2}{3} \times \left(4\right)^{3/2} = \dfrac{16}{3} \;\;\; \cdots \; (6)
\end{aligned}$
$\therefore$ $\;$ In view of equations $(5)$ and $(6)$, equation $(4)$ becomes
$I = 4 \pi + 2 \times \dfrac{16}{3} = 4 \pi + \dfrac{32}{3}$
$\therefore$ $\;$ Required area $= 4 \pi + \dfrac{32}{3}$ sq units