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

The equation of circle is

 ${\mathrm{x}}^2+{\mathrm{y}}^2=8\mathrm{x} ...... \left(\mathrm{i}\right)$

and the equation of parabola is

 ${\mathrm{y}}^2=4\mathrm{x} ........ \left(\mathrm{ii}\right)$

Eq. (i) can be written as

 $$\begin{gathered} \left({\mathrm{x}}^2-8\mathrm{x}\right)+{\mathrm{y}}^2=0 \\ \Rightarrow \left({\mathrm{x}}^2-8\mathrm{x}+16\right)+{\mathrm{y}}^2=16 \\ \Rightarrow {\left(\mathrm{x}-4\right)}^2+{\mathrm{y}}^2{\left(4\right)}^2 ....... \left(\mathrm{iii}\right) \end{gathered}$$

Which is a circle wit centre C(4, .0) and radius=4.

From Eq. (i) and (ii), we get

$$\begin{gathered} {\mathrm{x}}^2+4\mathrm{x}=8\mathrm{x} \\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}=0 \\ \Rightarrow \mathrm{x}(\mathrm{x}-4)=0 \\ \Rightarrow \mathrm{x}=0, 4 \end{gathered}$$

Now, from Eq. (ii), we get 

y = 0, 4

∴ Points of intersection of circle (i) and parabola (ii), above the A-axis, are

0(0, 0) and P(4, 4).
Now, required area = area of region OPQCO ==(Area of region OCPO + (area of region PCQP)

$$\begin{gathered} ={\int }_0^4\mathrm{y}\left(\mathrm{Parabola}\right)\mathrm{dx}+{\int }_4^8\mathrm{y}\left(\mathrm{Circle}\right) \mathrm{dx} \\ =1=2{\int }_0^4 \sqrt{\mathrm{x}} \mathrm{dx}+{\int }_4^8 \sqrt{{\left(4\right)}^2-{(\mathrm{x}-4)}^2} \mathrm{dx} \\ [\mathrm{from} \mathrm{Eqs}. (\mathrm{ii}), (\mathrm{iii}\left)\right] \\ = {2\left[\frac{{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]}_0^4 \\ +{\left[\frac{(\mathrm{x}-4)}{2} \sqrt{{\left(4\right)}^2-{(\mathrm{x}-4)}^2}+\frac{{\left(4\right)}^2}{2}.{\sin }^{-1} \frac{\mathrm{x}-4}{4}\right]}_4^8 \\ \left[\because \int \sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2} \mathrm{ds}=\frac{\mathrm{x}}{2}\sqrt{{\mathrm{a}}^2-{\mathrm{x}}^2}+\frac{{\mathrm{a}}^2}{2}{\sin }^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}\right] \\ =\frac{4}{3} {\left[{\mathrm{x}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^4 \\ +\left\{\left[\left(\frac{8-4}{2}\right)\sqrt{16-16}+8 {\sin }^{-1}\left(1\right)\right]-[0+8{\sin }^{-1}0]\right\} \\ =\frac{4}{3}\left[{\left(4\right)}^{3/2}-0\right]+\left[0+8\times \frac{\mathrm{\pi }}{2}\right]-[0+0] \\ =\frac{4}{3}\times 8+4\mathrm{\pi }\frac{32}{3}+4\mathrm{\pi }=\frac{4}{3}(8+3\mathrm{\pi }) \mathrm{wq} \mathrm{units} \end{gathered}$$