Answer:
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 \[{{x}^{2}}+4x=8x\Rightarrow {{x}^{2}}-4x=0\] \[\Rightarrow \] \[x\,(x-4)=0\Rightarrow x=0\]or \[x=4\] 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 = (Area of the region OCPO) + (Area of the region PCQP) \[=\int_{0}^{4}{ydx}+\int_{4}^{8}{ydx}\] \[=2\int_{0}^{4}{\sqrt{xdx}}+\int_{4}^{8}{\sqrt{{{4}^{2}}-{{(x-4)}^{2}}}dx}\]\[=2\times \frac{2}{3}\left[ {{x}^{\frac{3}{2}}} \right]_{0}^{4}+\int_{0}^{4}{\sqrt{{{4}^{2}}-{{t}^{2}}}dt,}\] where \[x-4=t\] \[=\frac{32}{3}+\left[ \frac{t}{2}\sqrt{{{4}^{2}}-{{t}^{2}}}+\frac{1}{2}\times {{4}^{2}}\times {{\sin }^{-\,1}}\frac{t}{4} \right]_{0}^{4}\] \[=\frac{32}{3}+\left[ \frac{4}{2}\times 0+\frac{1}{2}\times {{4}^{2}}\times {{\sin }^{-\,1}} \right]\] \[=\frac{32}{3}+\left[ 0+8\times \frac{\pi }{2} \right]=\frac{32}{3}+4\pi \] \[=\frac{4}{3}(8+3\pi )\,\,sq.\,\,units\]
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