1. Curves and intersection points

We have: $y^2=4x \quad \text{and} \quad x^2+y^2=5.$

From the parabola, $x=\frac{y^2}{4}.$ Substitute into the circle: $\left(\frac{y^2}{4}\right)^2+y^2=5$ $\frac{y^4}{16}+y^2=5$ $y^4+16y^2-80=0.$

Let $t=y^2$. Then $t^2+16t-80=0.$ So, $t=4 \quad (\text{reject } t=-20).$ Hence, $y=\pm 2, \qquad x=\frac{y^2}{4}=1.$

So the curves intersect at $(1,2),\ (1,-2).$

2. Identify the smaller region

Inside the circle, the parabola cuts off the region lying between the parabola and the right arc of the circle.

For $y \in [-2,2]$:

• left boundary: parabola $x=\dfrac{y^2}{4}$
• right boundary: circle $x=\sqrt{5-y^2}$

Thus the smaller enclosed area is $A=\int_{-2}^{2}\left(\sqrt{5-y^2}-\frac{y^2}{4}\right)dy.$

By symmetry, $A=2\int_0^2\left(\sqrt{5-y^2}-\frac{y^2}{4}\right)dy.$

3. Evaluate the integrals

So, $A=2\int_0^2\sqrt{5-y^2}\,dy-2\int_0^2\frac{y^2}{4}\,dy.$

That is, $A=2I_1-\frac12\int_0^2 y^2\,dy.$

First, $\frac12\int_0^2 y^2\,dy=\frac12\left[\frac{y^3}{3}\right]_0^2=\frac12\cdot\frac{8}{3}=\frac{4}{3}.$

Now compute $I_1=\int_0^2\sqrt{5-y^2}\,dy.$

Use the standard formula: $\int \sqrt{a^2-y^2}\,dy=\frac{y}{2}\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right)+C.$

Here $a=\sqrt5$, so $I_1=\left[\frac{y}{2}\sqrt{5-y^2}+\frac{5}{2}\sin^{-1}\left(\frac{y}{\sqrt5}\right)\right]_0^2.$

At $y=2$: $\frac{2}{2}\sqrt{5-4}=1,$ so $I_1=1+\frac{5}{2}\sin^{-1}\left(\frac{2}{\sqrt5}\right).$

Hence, $2I_1=2+5\sin^{-1}\left(\frac{2}{\sqrt5}\right).$

Therefore, $A=2+5\sin^{-1}\left(\frac{2}{\sqrt5}\right)-\frac{4}{3}$ $A=\frac{2}{3}+5\sin^{-1}\left(\frac{2}{\sqrt5}\right).$

4. Match with options

Thus the smaller part is $\boxed{\frac{2}{3}+5\sin^{-1}\left(\frac{2}{\sqrt5}\right)}.$

This matches Option A.

5. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.