The point of intersection of $$y^2=4 x$$ and $$x^2+y^2=5$$ are $$(1,2)$$ and $$(1,-2)$$.

$$\because$$ Area of smaller region bounded by $$y^2=4 x$$ and $$x^2+y^2=5$$

$$=2\{\text {area of } O A C O+\text { area of } C A B C\}$$

$$\begin{aligned} & =2\left[\int_\limits0^1 2 \sqrt{x} d x+\int_\limits1^{\sqrt{5}} \sqrt{5-x^2} d x\right. \\ & =2\left[\left|\frac{4}{3} x^{\frac{3}{2}}\right|_0^1+\left(\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right)\right]_1^{\sqrt{5}} \\ & =2\left[\left(\frac{4}{3}-0\right)+\left(0+\frac{5 \pi}{4}\right)-\left(1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right)\right] \\ & =2\left[\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right]=\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right) \end{aligned}$$