The area enclosed between the curves $$y^2 + 4x = 4$$ and $$y - 2x = 2$$ is

We need to find the area enclosed between $$y^2 + 4x = 4$$ and $$y - 2x = 2$$.

Parabola: $$x = \frac{4 - y^2}{4} = 1 - \frac{y^2}{4}$$ (opens left)

Line: $$x = \frac{y - 2}{2}$$

$$1 - \frac{y^2}{4} = \frac{y-2}{2}$$

$$4 - y^2 = 2(y-2) = 2y - 4$$

$$y^2 + 2y - 8 = 0$$

$$(y+4)(y-2) = 0$$

$$y = -4$$ or $$y = 2$$

For $$-4 \leq y \leq 2$$, the parabola is to the right of the line (check at $$y = 0$$: parabola gives $$x = 1$$, line gives $$x = -1$$).

$$A = \int_{-4}^{2} \left[\left(1 - \frac{y^2}{4}\right) - \frac{y-2}{2}\right] dy$$

$$= \int_{-4}^{2} \left(1 - \frac{y^2}{4} - \frac{y}{2} + 1\right) dy$$

$$= \int_{-4}^{2} \left(2 - \frac{y}{2} - \frac{y^2}{4}\right) dy$$

$$= \left[2y - \frac{y^2}{4} - \frac{y^3}{12}\right]_{-4}^{2}$$

At $$y = 2$$: $$4 - 1 - \frac{8}{12} = 3 - \frac{2}{3} = \frac{7}{3}$$

At $$y = -4$$: $$-8 - 4 + \frac{64}{12} = -12 + \frac{16}{3} = \frac{-36+16}{3} = \frac{-20}{3}$$

$$A = \frac{7}{3} - \frac{-20}{3} = \frac{27}{3} = 9$$

The correct answer is Option C: $$9$$.