The area of the region enclosed by the curves $$y=x^{2}-4x+4\text{ and }y^{2}=16-8x$$ is :

The given curves are $$y = x^2 - 4x + 4$$ and $$y^2 = 16 - 8x$$.

Since the first curve simplifies to $$y = (x - 2)^2$$ and the second becomes $$y^2 = -8(x - 2)\,,\,$$ the first is a parabola opening upward with vertex at $$(2,0)$$, while the second is a parabola opening to the left with the same vertex.

To determine the points of intersection, substituting $$y = (x - 2)^2$$ into $$y^2 = 16 - 8x$$ gives $$ [(x - 2)^2]^2 = 16 - 8x, $$ which simplifies to $$ (x - 2)^4 = 16 - 8x. $$ Then setting $$u = x - 2$$ transforms the equation into $$ u^4 = 16 - 8(u + 2), $$ so $$ u^4 = -8u,\quad u^4 + 8u = 0,\quad u(u^3 + 8) = 0. $$ Hence $$u = 0$$ or $$u^3 = -8$$, giving $$u = -2$$ and therefore $$x = 2$$ or $$x = 0$$.

For each value of $$x$$ we find the corresponding $$y$$ from $$y = (x - 2)^2$$. When $$x = 2$$, $$y = (2 - 2)^2 = 0$$ and one checks that $$y^2 = 16 - 8\cdot 2 = 0$$, so the intersection point is $$(2, 0)$$. When $$x = 0$$, $$y = (0 - 2)^2 = 4$$ while the second curve gives $$y^2 = 16 - 8\cdot 0 = 16$$, so $$y = \pm 4$$, but only $$y = 4$$ satisfies the first curve. Thus the intersection points are $$(0, 4)$$ and $$(2, 0)$$.

The upper branch of the second curve is $$y = \sqrt{16 - 8x}$$ and the lower branch is $$y = -\sqrt{16 - 8x}$$, whereas the first curve $$y = (x - 2)^2$$ is always non-negative. Since the lower branch is non-positive and the first curve non-negative, they meet only at $$(2,0)$$, confirming that no other intersections occur.

The enclosed region lies between $$x = 0$$ and $$x = 2$$, bounded above by $$y = \sqrt{16 - 8x}$$ and below by $$y = (x - 2)^2$$. For example at $$x = 1$$, the upper branch gives $$y = \sqrt{8} \approx 2.828$$ while the first curve gives $$y = 1$$, so the region is well-defined. The area is therefore $$ A = \int_{0}^{2} \bigl(\sqrt{16 - 8x} - (x - 2)^2\bigr)\,dx. $$

Since $$\sqrt{16 - 8x} = \sqrt{8(2 - x)} = 2\sqrt{2}\,\sqrt{2 - x}$$ and $$(x - 2)^2 = (2 - x)^2$$, the integral becomes $$ A = \int_{0}^{2} \Bigl(2\sqrt{2}\,\sqrt{2 - x} - (2 - x)^2\Bigr)\,dx. $$ Substituting $$t = 2 - x$$ gives $$dx = -dt$$ and transforms the limits from $$x = 0 \to 2$$ into $$t = 2 \to 0$$, so that $$ A = \int_{2}^{0} \bigl(2\sqrt{2}\,t^{1/2} - t^2\bigr)(-dt) = \int_{0}^{2} \bigl(2\sqrt{2}\,t^{1/2} - t^2\bigr)\,dt. $$

Integrating term by term, since $$\int t^{1/2}\,dt = \tfrac{2}{3}t^{3/2}$$ and $$\int t^2\,dt = \tfrac{1}{3}t^3$$, we have $$ A = \left[2\sqrt{2}\cdot\tfrac{2}{3}t^{3/2} - \tfrac{1}{3}t^3\right]_{0}^{2} = \left[\tfrac{4\sqrt{2}}{3}t^{3/2} - \tfrac{1}{3}t^3\right]_{0}^{2}. $$ At $$t = 2$$, $$t^{3/2} = 2\sqrt{2}$$ and $$t^3 = 8$$, so $$ \tfrac{4\sqrt{2}}{3}\cdot2\sqrt{2} = \tfrac{16}{3},\quad \tfrac{1}{3}\cdot8 = \tfrac{8}{3}, $$ giving $$A = \tfrac{16}{3} - \tfrac{8}{3} = \tfrac{8}{3}$$, and at $$t = 0$$ both terms vanish.

Thus, the area of the region enclosed by the curves is $$\frac{8}{3}$$. The correct option is A. $$\frac{8}{3}$$.