The area (in sq. units) of the smaller portion enclosed between the curves, $$x^2 + y^2 = 4$$ and $$y^2 = 3x$$, is:

We have two curves.

The first curve is a circle whose equation is $$x^{2}+y^{2}=4$$. We recognise this as a circle centred at the origin with radius $$2$$ because a general circle $$x^{2}+y^{2}=r^{2}$$ has radius $$r$$.

The second curve is the right-opening parabola $$y^{2}=3x$$. In the standard form $$y^{2}=4ax$$ a parabola opens towards the right with focus on the positive $$x$$-axis; here $$4a=3$$ so $$a=\dfrac34$$, but that constant is not needed immediately.

To obtain the common region, we first find the points of intersection of the two curves. We substitute $$x=\dfrac{y^{2}}{3}$$ (from the parabola) into the equation of the circle.

Substituting gives

$$\left(\dfrac{y^{2}}{3}\right)^{2}+y^{2}=4.$$

Now we simplify every algebraic step.

$$\dfrac{y^{4}}{9}+y^{2}=4.$$

Multiply by $$9$$ to clear the denominator:

$$y^{4}+9y^{2}-36=0.$$

Put $$t=y^{2}$$ so that the equation becomes a quadratic in $$t$$:

$$t^{2}+9t-36=0.$$

For a quadratic $$at^{2}+bt+c=0,$$ the roots are given by the quadratic formula $$t=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Here $$a=1,\; b=9,\; c=-36.$$

So

$$t=\dfrac{-9\pm\sqrt{9^{2}-4(1)(-36)}}{2(1)} =\dfrac{-9\pm\sqrt{81+144}}{2} =\dfrac{-9\pm\sqrt{225}}{2} =\dfrac{-9\pm15}{2}.$$

This yields two values:

$$t_{1}=\dfrac{-9+15}{2}=\dfrac{6}{2}=3,\qquad t_{2}=\dfrac{-9-15}{2}=\dfrac{-24}{2}=-12.$$

But $$t=y^{2}\ge0,$$ hence the negative root is rejected. Therefore

$$y^{2}=3\;\Longrightarrow\; y=\pm\sqrt3.$$

Substituting back in $$x=\dfrac{y^{2}}{3}$$ gives

$$x=\dfrac{3}{3}=1.$$

Thus the two intersection points are $$\left(1,\sqrt3\right)$$ and $$\left(1,-\sqrt3\right).$$

Next we decide which curve is on the right and which on the left between these $$y$$-limits. For a fixed $$y,$$ the parabola gives $$x=\dfrac{y^{2}}{3}$$ while the circle (taking the right semicircle) gives $$x=\sqrt{4-y^{2}}$$ because from $$x^{2}+y^{2}=4$$ we obtain $$x=\pm\sqrt{4-y^{2}},$$ and the positive sign corresponds to the right half.

At $$y=0,\; x_{\text{parabola}}=0,\; x_{\text{circle}}=2,$$ so the circle is to the right of the parabola. The same order holds for every $$y$$ between $$-\sqrt3$$ and $$\sqrt3$$. Hence, for those $$y$$-values, the horizontal width of the required strip is

$$\sqrt{4-y^{2}}-\dfrac{y^{2}}{3}.$$

Because the region is symmetric about the $$x$$-axis, we calculate the area for $$y\ge0$$ and simply double it.

The desired area is therefore

$$A=2\int_{0}^{\sqrt3}\left(\sqrt{4-y^{2}}-\dfrac{y^{2}}{3}\right)\,dy.$$

We treat the two integrals separately.

First integral $$I_{1}=\int_{0}^{\sqrt3}\sqrt{4-y^{2}}\,dy.$$

We use the trigonometric substitution $$y=2\sin\theta.$$ Then $$dy=2\cos\theta\,d\theta,\qquad \sqrt{4-y^{2}}=\sqrt{4-4\sin^{2}\theta}=2\cos\theta.$$ When $$y=0,\;\theta=0;\quad y=\sqrt3,\;\theta=\arcsin\!\left(\dfrac{\sqrt3}{2}\right)=\dfrac{\pi}{3}.$$

So

$$I_{1}=\int_{0}^{\pi/3}\left(2\cos\theta\right)\left(2\cos\theta\,d\theta\right) =\int_{0}^{\pi/3}4\cos^{2}\theta\,d\theta.$$

Recall the identity $$\cos^{2}\theta=\dfrac{1+\cos2\theta}{2}.$$ Hence

$$I_{1}=4\int_{0}^{\pi/3}\dfrac{1+\cos2\theta}{2}\,d\theta =2\int_{0}^{\pi/3}\left(1+\cos2\theta\right)\,d\theta =2\left[\theta+\dfrac{\sin2\theta}{2}\right]_{0}^{\pi/3}.$$

Evaluating: $$\theta\Big|_{0}^{\pi/3}= \dfrac{\pi}{3}-0=\dfrac{\pi}{3},$$ $$\sin2\theta\Big|_{0}^{\pi/3}= \sin\!\left(\dfrac{2\pi}{3}\right)-0=\dfrac{\sqrt3}{2}.$$

Thus $$I_{1}=2\left(\dfrac{\pi}{3}+\dfrac{1}{2}\cdot\dfrac{\sqrt3}{2}\right) =2\left(\dfrac{\pi}{3}+\dfrac{\sqrt3}{4}\right) =\dfrac{2\pi}{3}+\dfrac{\sqrt3}{2}.$$

Second integral $$I_{2}=\int_{0}^{\sqrt3}\dfrac{y^{2}}{3}\,dy =\dfrac13\int_{0}^{\sqrt3}y^{2}\,dy =\dfrac13\left[\dfrac{y^{3}}{3}\right]_{0}^{\sqrt3} =\dfrac13\cdot\dfrac{(\sqrt3)^{3}}{3} =\dfrac13\cdot\dfrac{3\sqrt3}{3} =\dfrac{\sqrt3}{3}.$$

Now we assemble the area:

$$A=2\bigl(I_{1}-I_{2}\bigr) =2\left(\dfrac{2\pi}{3}+\dfrac{\sqrt3}{2}-\dfrac{\sqrt3}{3}\right).$$

Combine the $$\sqrt3$$ terms. $$\dfrac{\sqrt3}{2}-\dfrac{\sqrt3}{3} =\sqrt3\left(\dfrac12-\dfrac13\right) =\sqrt3\left(\dfrac{3-2}{6}\right) =\dfrac{\sqrt3}{6}.$$

Therefore

$$A=2\left(\dfrac{2\pi}{3}+\dfrac{\sqrt3}{6}\right) =\dfrac{4\pi}{3}+\dfrac{2\sqrt3}{6} =\dfrac{4\pi}{3}+\dfrac{\sqrt3}{3}.$$

Notice that $$\dfrac{\sqrt3}{3}=\dfrac1{\sqrt3}.$$ Hence

$$A=\dfrac1{\sqrt3}+\dfrac{4\pi}{3}.$$

This expression matches Option A exactly.

Hence, the correct answer is Option A.