1. Understand the region

The region is $R=\{(x,y): x^2\le y\le 2x\}.$

For this region to exist, we need $x^2\le 2x \implies x(x-2)\le 0 \implies 0\le x\le 2.$

So the curves are:

• lower curve: $y=x^2$
• upper curve: $y=2x$

They intersect at: $x^2=2x \implies x=0,2,$ which gives points $(0,0)$ and $(2,4)$.

2. Find total area of the region

$A=\int_0^2 (2x-x^2)\,dx$

$=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}.$

Hence half the area is $\frac{A}{2}=\frac{2}{3}.$

3. Use horizontal slicing

Since the dividing line is $y=\alpha$, it is convenient to express the region in terms of $y$.

From $y=x^2 \implies x=\sqrt y,$ and $y=2x \implies x=\frac y2.$

For a fixed $y\in[0,4]$, the horizontal strip runs from $x=\frac y2 \quad \text{to} \quad x=\sqrt y,$ because for $0\le y\le 4$, $\frac y2 \le \sqrt y.$

So width of strip at height $y$ is $\sqrt y-\frac y2.$

4. Area below the line $y=\alpha$

If $y=\alpha$ divides the region into two equal parts, then area from $y=0$ to $y=\alpha$ must be $\frac{2}{3}$:

$\int_0^{\alpha}\left(\sqrt y-\frac y2\right)dy=\frac{2}{3}.$

Now evaluate:

$\int \sqrt y\,dy=\frac{2}{3}y^{3/2}, \qquad \int \frac y2\,dy=\frac{y^2}{4}.$

Thus, $\left[\frac{2}{3}y^{3/2}-\frac{y^2}{4}\right]_0^{\alpha}=\frac{2}{3}.$

So, $\frac{2}{3}\alpha^{3/2}-\frac{\alpha^2}{4}=\frac{2}{3}.$

Multiply by $12$:

$8\alpha^{3/2}-3\alpha^2=8.$

Rearrange:

$3\alpha^2-8\alpha^{3/2}+8=0.$

5. Match with options

This is exactly Option C:

$3\alpha^2-8\alpha^{3/2}+8=0.$

6. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.