Question

If the line \( y = \alpha x, \alpha \geq \sqrt{2} \), divides the area of the region \[ R := \{(x, y) \in \mathbb{R}^2 \mid 0 \leq x \leq \sqrt{y}, 0 \leq y \leq 2\} \] into two equal parts, then the value of \( \alpha \) is equal to

• A. \( \frac{3}{\sqrt{2}} \)
• B. \( 2\sqrt{2} \)
• C. \( \sqrt{2} \)
• D. \( \frac{5}{2\sqrt{2}} \)

Hint:

When dividing a region into two equal areas using a line, set the area under the line equal to half of the total area and solve for the line's equation parameters.

Answer 1

The Correct Option is A

Step 1: Understand the region \( R \)
The region \( R \) is defined by \( 0 \leq x \leq \sqrt{y} \) and \( 0 \leq y \leq 2 \). This implies that for a fixed value of \( y \), the value of \( x \) ranges from 0 to \( \sqrt{y} \).
Step 2: Area of region \( R \)
The total area of the region \( R \) can be computed as: \[ \text{Area of } R = \int_0^2 \int_0^{\sqrt{y}} dx \, dy. \] The inner integral with respect to \( x \) gives: \[ \int_0^{\sqrt{y}} dx = \sqrt{y}. \] Now, integrating with respect to \( y \): \[ \text{Area of } R = \int_0^2 \sqrt{y} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_0^2 = \frac{2}{3} (2^{3/2}) = \frac{2}{3} \times 2\sqrt{2} = \frac{4\sqrt{2}}{3}. \]
Step 3: Area division by the line \( y = \alpha x \)
The line \( y = \alpha x \) divides the region \( R \) into two parts. To find \( \alpha \), we require that the area under the line \( y = \alpha x \) (within the region \( R \)) be half of the total area. For the region under the line \( y = \alpha x \), the value of \( x \) ranges from 0 to \( \frac{y}{\alpha} \) for each \( y \). The area under the line is: \[ \text{Area under the line} = \int_0^2 \int_0^{y/\alpha} dx \, dy = \int_0^2 \frac{y}{\alpha} \, dy = \frac{1}{\alpha} \int_0^2 y \, dy = \frac{1}{\alpha} \left[ \frac{y^2}{2} \right]_0^2 = \frac{1}{\alpha} \times \frac{4}{2} = \frac{2}{\alpha}. \]
Step 4: Set the areas equal
To divide the area into two equal parts, we set the area under the line equal to half the total area: \[ \frac{2}{\alpha} = \frac{1}{2} \times \frac{4\sqrt{2}}{3}. \] Solving for \( \alpha \): \[ \frac{2}{\alpha} = \frac{2\sqrt{2}}{3} \quad \Rightarrow \quad \alpha = \frac{3}{\sqrt{2}}. \]
Step 5: Conclusion
Thus, the value of \( \alpha \) is \( \frac{3}{\sqrt{2}} \), and the correct answer is (A).