1. Write the two curves in standard form

• Circle: $(x-2\sqrt{3})^2+y^2=12$ This is a circle with center $(2\sqrt{3},0)$ and radius $r=\sqrt{12}=2\sqrt{3}.$

• Parabola: $y^2=2\sqrt{3}\,x \quad \Rightarrow \quad x=\frac{y^2}{2\sqrt{3}}.$ This is a right-opening parabola with vertex at the origin.

We need the area inside the circle and outside the parabola.

2. Find the points of intersection

Substitute $x=\dfrac{y^2}{2\sqrt{3}}$ into the circle:

$\left(\frac{y^2}{2\sqrt{3}}-2\sqrt{3}\right)^2+y^2=12.$

Now simplify:

$\frac{(y^2-12)^2}{12}+y^2=12.$

Multiply by $12$:

$(y^2-12)^2+12y^2=144.$

Expand:

$y^4-24y^2+144+12y^2=144$ $y^4-12y^2=0$ $y^2(y^2-12)=0.$

So, $y=0,\quad y=\pm 2\sqrt{3}.$

Corresponding $x$-values:

• For $y=0$: $x=0$
• For $y=\pm 2\sqrt{3}$: $x=\frac{12}{2\sqrt{3}}=2\sqrt{3}.$

Hence intersections are: $(0,0),\quad (2\sqrt{3},2\sqrt{3}),\quad (2\sqrt{3},-2\sqrt{3}).$

3. Determine the required region

For a fixed $y$ between $-2\sqrt{3}$ and $2\sqrt{3}$:

• Left boundary of the circle is $x=2\sqrt{3}-\sqrt{12-y^2},$
• Right boundary of the circle is $x=2\sqrt{3}+\sqrt{12-y^2},$
• Parabola gives $x=\frac{y^2}{2\sqrt{3}}.$

Now observe at $y=0$:

• circle spans from $x=0$ to $x=4\sqrt{3}$,
• parabola is at $x=0$.

So the portion inside circle and outside parabola is the part of the circle lying to the left of the parabola excluded, i.e. the region between $x=2\sqrt{3}-\sqrt{12-y^2}$ and $x=\frac{y^2}{2\sqrt{3}}.$

Thus the area is

$A=\int_{-2\sqrt{3}}^{2\sqrt{3}}\left(\frac{y^2}{2\sqrt{3}}-\left(2\sqrt{3}-\sqrt{12-y^2}\right)\right)dy.$

That is,

$A=\int_{-2\sqrt{3}}^{2\sqrt{3}}\left(\sqrt{12-y^2}+\frac{y^2}{2\sqrt{3}}-2\sqrt{3}\right)dy.$

4. Evaluate the integral

Break it into three parts: