The parabola $$y^2 = 4x$$ divides the area of the circle $$x^2 + y^2 = 5$$ in two parts. The area of the smaller part is equal to:

 Find Intersection. Substitute $$y^2=4x$$ into circle: $$x^2 + 4x - 5 = 0 \Rightarrow (x+5)(x-1)=0$$. Since $$x \ge 0$$ for the parabola, $$x=1$$.

Setup Area. The area is symmetric about the x-axis.

$$\text{Area} = 2 \left[ \int_0^1 \sqrt{4x} dx + \int_1^{\sqrt{5}} \sqrt{5-x^2} dx \right]$$

 Integrate.

o Part 1: $$2 \int_0^1 2x^{1/2} dx = 4 [\frac{2}{3}x^{3/2}]_0^1 = \frac{8}{3}$$.

o Part 2: $$2 \int_1^{\sqrt{5}} \sqrt{5-x^2} dx$$. Using $$\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$$.

o Result: $$[x\sqrt{5-x^2} + 5\sin^{-1}\frac{x}{\sqrt{5}}]_1^{\sqrt{5}} = (0 + 5\frac{\pi}{2}) - (2 + 5\sin^{-1}\frac{1}{\sqrt{5}})$$.

 Simplify. Total Area = $$\frac{8}{3} - 2 + 5(\frac{\pi}{2} - \sin^{-1}\frac{1}{\sqrt{5}})$$. Using $$\cos^{-1}\theta = \frac{\pi}{2} - \sin^{-1}\theta$$ and $$\cos^{-1}(1/\sqrt{5}) = \sin^{-1}(2/\sqrt{5})$$:

$$\text{Area} = \frac{2}{3} + 5\sin^{-1}\left(\frac{2}{\sqrt{5}}\right)$$