Given curves

$x^2+y^2=4$ is a circle of radius $2$ centered at origin, and

$y^2=3x$ is a right-opening parabola.

Find points of intersection

From the parabola, $x=\frac{y^2}{3}$

Substitute into the circle: $\left(\frac{y^2}{3}\right)^2+y^2=4$ $\frac{y^4}{9}+y^2=4$ $y^4+9y^2-36=0$

Let $t=y^2$. Then $t^2+9t-36=0$ $t=3 \quad (\text{since } t\ge 0)$

Hence, $y^2=3 \Rightarrow y=\pm \sqrt{3}, \qquad x=\frac{3}{3}=1$

So the curves intersect at $(1,\sqrt{3}) \text{ and } (1,-\sqrt{3}).$

Identify the smaller enclosed region

Between $y=-\sqrt{3}$ and $y=\sqrt{3}$,

• parabola gives left boundary: $x=\frac{y^2}{3}$
• circle gives right boundary: $x=\sqrt{4-y^2}$

So the smaller enclosed area is $A=\int_{-\sqrt{3}}^{\sqrt{3}}\left(\sqrt{4-y^2}-\frac{y^2}{3}\right)dy$

By symmetry, $A=2\int_0^{\sqrt{3}}\left(\sqrt{4-y^2}-\frac{y^2}{3}\right)dy$

Evaluate the integrals

First, $\int \sqrt{4-y^2}\,dy=\frac{y}{2}\sqrt{4-y^2}+2\sin^{-1}\left(\frac{y}{2}\right)$

Therefore,

Compute the area

$A=2\left(\frac{\sqrt{3}}{2}+\frac{2\pi}{3}-\frac{\sqrt{3}}{3}\right)$

$=2\left(\frac{\sqrt{3}}{6}+\frac{2\pi}{3}\right)$

$=\frac{\sqrt{3}}{3}+\frac{4\pi}{3}$

Since $\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}},$ we get $A=\frac{1}{\sqrt{3}}+\frac{4\pi}{3}$

Match with the options

This is Option D.

Compare with stored correct answer

Stored correct answer is D, which matches our derived result.