The area of the smaller region enclosed by the curves $$y^2 = 8x + 4$$ and $$x^2 + y^2 + 4\sqrt{3}x - 4 = 0$$ is equal to

Given curves are

$$y^2=8x+4$$

and

$$x^2+y^2+4\sqrt3\,x-4=0$$

First rewrite them.

Parabola:

$$y^2=8\left(x+\frac12\right)$$

Hence,

$$x=\frac{y^2-4}{8}$$

Circle:

$$x^2+4\sqrt3\,x+y^2-4=0$$

Completing the square,

$$x^2+4\sqrt3\,x+12+y^2=16$$

$$(x+2\sqrt3)^2+y^2=16$$

So the circle has centre

$$(-2\sqrt3,0)$$

and radius

$$4$$

Now find points of intersection.

Substitute

$$x=\frac{y^2-4}{8}$$

in the circle:

$$\left(\frac{y^2-4}{8}\right)^2+y^2+4\sqrt3\left(\frac{y^2-4}{8}\right)-4=0$$

Solving gives

$$y=\pm2$$

and corresponding

$$x=0$$

Hence, intersection points are

$$(0,2)\quad \text{and}\quad (0,-2)$$

Now the smaller enclosed region lies between:

- parabola on the right

- left arc of the circle on the left

For the circle,

$$x=-2\sqrt3+\sqrt{16-y^2}$$

Hence required area is

$$A=\int_{-2}^{2}\left[\frac{y^2-4}{8}-\left(-2\sqrt3+\sqrt{16-y^2}\right)\right]dy$$

Using symmetry,

$$A=2\int_0^2\left(\frac{y^2-4}{8}+2\sqrt3-\sqrt{16-y^2}\right)dy$$

Now,

$$\int_0^2\frac{y^2-4}{8}\,dy =\frac18\left[\frac{y^3}{3}-4y\right]_0^2 =-\frac23$$

Also,

$$\int_0^22\sqrt3\,dy=4\sqrt3$$

Now evaluate

$$\int_0^2\sqrt{16-y^2}\,dy$$

Using standard formula,

$$\int\sqrt{a^2-y^2}\,dy =\frac y2\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\frac ya$$

with

$$a=4,$$

$$\int_0^2\sqrt{16-y^2}\,dy =\left[\frac y2\sqrt{16-y^2}+8\sin^{-1}\frac y4\right]_0^2$$

$$=\sqrt{12}+8\cdot\frac{\pi}{6}$$

$$=2\sqrt3+\frac{4\pi}{3}$$

Therefore,

$$A=2\left(-\frac23+4\sqrt3-2\sqrt3-\frac{4\pi}{3}\right)$$

$$=2\left(2\sqrt3-\frac23-\frac{4\pi}{3}\right)$$

$$=4\sqrt3-\frac43-\frac{8\pi}{3}$$

Rearranging,

$$A=\frac{12\sqrt3-4-8\pi}{3}$$

Since area is positive,

$$A=\frac{8\pi+4-12\sqrt3}{3}$$

Hence, the required area is

$$\boxed{\frac13\left(4-12\sqrt3+8\pi\right)}$$.