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Step-by-Step Solution
Step 1: Identify the Region R
The given region R in the xy-plane is defined by the inequalities
$x^2 \leq y \leq 2x$.
This means that for each x in the domain, y ranges between the curve $y = x^2$ (below) and the line $y = 2x$ (above).
Step 2: Determine the Limits for x
To find the intersection of the two boundaries, set
$x^2 = 2x$. This gives:
$x^2 = 2x \quad \Rightarrow \quad x^2 - 2x = 0 \quad \Rightarrow \quad x(x - 2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 2.$
Hence, the region R is bounded along the x-axis from $x=0$ to $x=2$.
Step 3: Express the Total Area of Region R
The total area of the region R is given by the integral:
$A = \int_{x=0}^{2} \bigl[ (2x) - (x^2) \bigr] \, dx.$
We can compute it if needed, but in the question, we are more concerned with a horizontal line $y = \alpha$ that splits this area into two equal parts. Let us see how to set up the area condition correctly.
Step 4: Understand How the Line $y = \alpha$ Splits the Region
The line $y = \alpha$ will cut through the region horizontally. We want the area above $y = x^2$ and below $y = 2x$ to be split into two equal parts by this line $y = \alpha$. In other words, the region from $y=x^2$ up to $y=\alpha$ should hold half the area, and from $y=\alpha$ up to $y=2x$ should hold the other half.
Step 5: Convert the Integral to “y-first” Form
Sometimes for horizontal slicing, it is more straightforward to treat y as the variable of integration. Solving $x^2 = y$ gives $x = \sqrt{y}$ when $x \ge 0$. Also, from $y = 2x$, we get $x = y/2$. Then our region in terms of y is:
$0 \le y \le 4$ because the top boundary $y=2x$ and the parabola $x^2 = y$ intersect at $y = 4$ (when $x=2$).
For each y in $[0, 4]$, x goes from $x = \sqrt{y}$ (right boundary from $y=x^2$) to $x = \frac{y}{2}$ (left boundary from $y=2x$). However, we must ensure the correct left-right orientation. Here, $\sqrt{y} \le \frac{y}{2}$ for $0 \le y \le 4$.
In the definite integral form, the differential area element for a horizontal slice at height y is (right x – left x) dy = $(\frac{y}{2} - \sqrt{y}) \, dy$.
Step 6: Split the Region with $y = \alpha$
We want that line $y = \alpha$ to divide the total area into two equal parts. Denote the total area from $y=0$ to $y=4$ by $A_{\text{total}}$. Then the area from $y=0$ to $y=\alpha$ must be half of $A_{\text{total}}$. Symbolically, if
$A_{\text{total}} = \int_{0}^{4} \Bigl(\frac{y}{2} - \sqrt{y}\Bigr)\, dy,
we want
$\int_{0}^{\alpha} \Bigl(\frac{y}{2} - \sqrt{y}\Bigr)\, dy = \frac{1}{2} A_{\text{total}}.$
Step 7: Evaluate the Necessary Integrals
The question’s solution provides the integral steps in detail. Specifically, it rearranges the condition so that the areas above and below $y=\alpha$ are equal. Using that approach, we arrive at the equation (after simplification):
$3\alpha^2 - 8\,\alpha^{\frac{3}{2}} + 8 = 0.$
This is exactly the condition on $\alpha$ for the line $y = \alpha$ to divide the region into two equal areas.
Step 8: Final Equation
Thus, the final relationship that $y = \alpha$ satisfies is:
$3\,\alpha^2 - 8\,\alpha^{\frac{3}{2}} + 8 = 0.
This matches the correct option given in the problem statement.
