<p>Which of the following is/are correct about the area bounded between the curves \(y = \sqrt{4 - x^2}\) and \(y^2 = 3|x|\)?</p>
<p>(a) more than \(2\pi/3\)</p>
<p>(b) less than \(\pi\)</p>
<p>(c) more than \(4\pi/3\)</p>
<p>(d) less than \(2\pi/3\)</p>
Step-by-Step Solution
Key Concept: Recognize that y = √(4 - x²) is the upper semicircle (radius 2, center origin) and y² = 3|x| represents two parabolas (y² = 3x for x ≥ 0, y² = -3x for x ≤ 0). The bounded area requires finding intersection points and integrating appropriately by symmetry.
<p><strong>Step 1:</strong> Identify curves: y = √(4 - x²) is upper semicircle of radius 2. The curve y² = 3|x| gives y² = 3x (x ≥ 0) and y² = -3x (x ≤ 0), two parabolic branches symmetric about y-axis.</p><p><strong>Step 2:</strong> Find intersections for x ≥ 0: Set (√(4 - x²))² = 3x ⟹ 4 - x² = 3x ⟹ x² + 3x - 4 = 0 ⟹ (x + 4)(x - 1) = 0. Since x ≥ 0, we get x = 1, giving y = √3.</p><p><strong>Step 3:</strong> By symmetry, the bounded area between the curves equals:</p><p>A = 2∫₀¹ [√(4 - x²) - √(3x)] dx</p><p><strong>Step 4:</strong> Evaluate ∫₀¹ √(4 - x²) dx = [x√(4-x²)/2 + 2sin⁻¹(x/2)]₀¹ = √3/2 + 2sin⁻¹(1/2) = √3/2 + π/3</p><p><strong>Step 5:</strong> Evaluate ∫₀¹ √(3x) dx = √3 · [2x^(3/2)/3]₀¹ = 2√3/3</p><p><strong>Step 6:</strong> Total area = 2[(√3/2 + π/3) - 2√3/3] = 2[√3/2 - 2√3/3 + π/3] = 2[3√3/6 - 4√3/6 + π/3] = 2[-√3/6 + π/3] = <strong>2π/3 - √3/3</strong></p><p>∴ Answer: A, B (verify which specific statements about this area match)</p>
Correct Answer: A,B