The area of the region enclosed by the curves $y = x^2-4x+4$ and $y^2 = 16-8x$ is:
Step-by-Step Solution
Key Concept: Rewrite as $y = (x-2)^2$ (upward parabola) and $y^2 = 16-8x$ (leftward parabola opening left with vertex $(2,0)$); find intersections, then integrate $\sqrt{16-8x}-(x-2)^2$ between the $x$-limits.
The curves: $y=(x-2)^2$ and $y^2=16-8x \Rightarrow y=\pm\sqrt{16-8x}$.
Intersection: $(x-2)^4 = 16-8x$. Let $u=x-2$: $u^4+8u=0 \Rightarrow u(u^3+8)=0 \Rightarrow u=0$ or $u=-2$, i.e., $x=2$ and $x=0$.
$$\text{Area} = \int_0^2\!\left(\sqrt{16-8x}-(x-2)^2\right)dx = \left[-\frac{(16-8x)^{3/2}}{12}-\frac{(x-2)^3}{3}\right]_0^2 = \frac{8}{3}.$$
Correct Answer: 1