1. Write the curves in simpler form

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

Rewrite the second curve in terms of $x$: $8x=16-y^2$ $x=2-\frac{y^2}{8}.$

Also, from $y=(x-2)^2,$ we can write $x=2\pm \sqrt{y}.$ Since this parabola has $y\ge 0$, only the upper half is relevant.

2. Find the points of intersection

From $y=(x-2)^2$ and $y^2=16-8x,$ substitute $x=2\pm\sqrt y$ into the second form. But an easier way is to use the second curve as $x=2-\frac{y^2}{8}$ and compare with the left branch of the first curve: $x=2-\sqrt y.$

At intersection, $2-\sqrt y=2-\frac{y^2}{8}$ $\sqrt y=\frac{y^2}{8}.$ Let $t=\sqrt y$ so that $y=t^2$. Then $t=\frac{t^4}{8}$ $t^4-8t=0$ $t(t^3-8)=0$ $t=0 \text{ or } t=2.$ Thus, $y=0 \text{ or } y=4.$

Now find corresponding $x$:

• For $y=0$, from $y=(x-2)^2$, we get $x=2$.
• For $y=4$, from $y=(x-2)^2$, we get $(x-2)^2=4$, so $x=0$ or $4$. But on the parabola $x=2-\frac{y^2}{8}=2-2=0$, so the common point is $(0,4)$.

Hence the curves intersect at $(2,0) \quad \text{and} \quad (0,4).$

3. Identify the enclosed region

For $0\le y\le 4$:

• Left boundary comes from $y^2=16-8x$: $x=2-\frac{y^2}{8}.$
• Right boundary comes from the left branch of $y=(x-2)^2$: $x=2-\sqrt y.$

So area is $A=\int_0^4 \left[\left(2-\sqrt y\right)-\left(2-\frac{y^2}{8}\right)\right]dy.$

Simplify: $A=\int_0^4 \left(\frac{y^2}{8}-\sqrt y\right)dy.$

But check sign at, say, $y=1$: $2-\frac{1}{8}=1.875, \quad 2-1=1.$ So actually the curve $x=2-\frac{y^2}{8}$ lies to the right of $x=2-\sqrt y$.

Therefore, $A=\int_0^4 \left[\left(2-\frac{y^2}{8}\right)-\left(2-\sqrt y\right)\right]dy$ $=\int_0^4 \left(\sqrt y-\frac{y^2}{8}\right)dy.$

4. Evaluate the integral

$A=\int_0^4 y^{1/2}\,dy-\frac{1}{8}\int_0^4 y^2\,dy.$

Now, $\int_0^4 y^{1/2}\,dy=\left[\frac{2}{3}y^{3/2}\right]_0^4=\frac{2}{3}\cdot 8=\frac{16}{3}.$

And $\frac{1}{8}\int_0^4 y^2\,dy=\frac{1}{8}\left[\frac{y^3}{3}\right]_0^4=\frac{1}{8}\cdot\frac{64}{3}=\frac{8}{3}.$

Thus, $A=\frac{16}{3}-\frac{8}{3}=\frac{8}{3}.$

5. Compare with options

The area is $\boxed{\frac{8}{3}}$ which corresponds to Option A.