Question

Consider a region \(R = \{(x, y) \in \mathbb{R}^2 : x^2 \le y \le 2x\}\). If a line \(y = \alpha\) divides the area of region R into two equal parts, then which of the following is true?

• A. \(3\alpha^2 - 8\alpha + 8 = 0\)
• B. \(\alpha^3 - 6\alpha^{3/2} - 16 = 0\)
• C. \(3\alpha^2 - 8\alpha^{3/2} + 8 = 0\)
• D. \(\alpha^3 - 6\alpha^2 + 16 = 0\)

Hint:

The parabola \(y=x^2\) meets \(y=2x\) at \((0,0)\) and \((2,4)\); the whole area is \(4/3\). Integrate with respect to y to find the part below (or above) \(y=\alpha\) and set it equal to half.

Answer 1

The Correct Option is C

Intersections: \(x^2=2x\Rightarrow x=0,2\). Total area \(=\int_0^2(2x-x^2)\,dx = 4-\frac83=\frac43\). Half \(=\frac23\).
Horizontal strip at height y: \(x\) runs from \(\frac y2\) (line) to \(\sqrt y\) (parabola).
Area below \(y=\alpha\): \(\int_0^\alpha\left(\sqrt y-\frac y2\right)dy = \frac23\alpha^{3/2}-\frac{\alpha^2}{4}\).
Set equal to \(\frac23\): \(8\alpha^{3/2}-3\alpha^2 = 8\Rightarrow 3\alpha^2-8\alpha^{3/2}+8=0\).
Answer: (3) \(3\alpha^2 - 8\alpha^{3/2} + 8 = 0\)