Consider the curves

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

and

$$ y^2 = 16 - 8x. $$

Notice that the second equation can be rewritten in terms of $$x$$:

$$ 8x = 16 - y^2 \quad \Longrightarrow \quad x = 2 - \frac{y^2}{8}. $$

Step 1. Find the Intersection Points

To find the points where the curves intersect, substitute

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

into

$$ y^2 = 16 - 8x. $$

Let

$$ u = x - 2 \quad \text{so that} \quad y = u^2. $$

Then

$$ y^2 = u^4 $$

and

$$ x = u + 2. $$

Substitute into the second curve:

$$ u^4 = 16 - 8(u+2). $$

Simplify the right side:

$$ 16 - 8(u+2) = 16 - 8u - 16 = -8u. $$

Thus, the equation becomes

$$ u^4 + 8u = 0 \quad \Longrightarrow \quad u(u^4/ u? \,\, **{correcting factor}**). $$

In fact, factor by taking out a common factor $$u$$:

$$ u(u^3 + 8) = 0. $$

Thus, either

$$ u = 0 \quad \text{or} \quad u^3 = -8. $$

For $$u = 0$$:

$$ x = u + 2 = 2, \quad y = u^2 = 0. $$

For $$u^3 = -8$$:

$$ u = -2, \quad \text{so} \quad x = -2 + 2 = 0, \quad y = (-2)^2 = 4. $$

The curves intersect at the points $$ (2, 0) $$ and $$ (0, 4). $$

Step 2. Express the Curves in Terms of $$y$$

It is easier to integrate horizontally by expressing $$x$$ as a function of $$y$$.

From the second curve:

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

From the first curve, solving

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

for $$x$$ gives

$$ x-2 = \pm\sqrt{y}. $$

Since at the intersection $$ (0, 4) $$ the $$x$$–value is less than 2, we take the negative branch:

$$ x = 2 - \sqrt{y}. $$

Thus, for a fixed $$y$$ between 0 and 4, the left boundary is

$$ x_{\text{left}} = 2 - \sqrt{y}, $$

and the right boundary is

$$ x_{\text{right}} = 2 - \frac{y^2}{8}. $$

Step 3. Set Up the Integral for the Area

The horizontal distance between the curves at a given $$y$$ is

$$ \Delta x = x_{\text{right}} - x_{\text{left}} = \left(2 - \frac{y^2}{8}\right) - \left(2 - \sqrt{y}\right) = \sqrt{y} - \frac{y^2}{8}. $$

Integrate with respect to $$y$$ from $$y = 0$$ to $$y = 4$$:

$$ A = \int_{0}^{4} \left(\sqrt{y} - \frac{y^2}{8}\right) \, dy. $$

Step 4. Evaluate the Integral

Write the integral as

$$ A = \int_{0}^{4} y^{1/2}\, dy - \frac{1}{8} \int_{0}^{4} y^2\, dy. $$

Compute each term:

For the first integral:

$$ \int y^{1/2}\, dy = \frac{2}{3} y^{3/2}. $$

For the second integral:

$$ \int y^2\, dy = \frac{y^3}{3}. $$

Thus,

$$ A = \left[\frac{2}{3} y^{3/2}\right]_0^4 - \frac{1}{8}\left[\frac{y^3}{3}\right]_0^4. $$

Substitute $$y = 4$$ (note that at $$y = 0$$ both terms vanish):

Compute $$4^{3/2}$$:

$$ 4^{3/2} = \left( \sqrt{4} \right)^3 = 2^3 = 8. $$

Compute the first term:

$$ \frac{2}{3} \times 8 = \frac{16}{3}. $$

Compute the second term:

$$ \frac{1}{8} \cdot \frac{64}{3} = \frac{64}{24} = \frac{8}{3}. $$

Therefore, the area is

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

Final Answer

The area of the region enclosed by the curves is

$$ \frac{8}{3}. $$