<p>The area (in sq. units) of the region bounded by the curve <br> \(x = |y|\sqrt{1-y^2}\) and the curve \(x = y^2 - 1\) is:</p>
Step-by-Step Solution
Key Concept: The curve x = |y|√(1-y²) represents a figure-eight (lemniscate-like shape) symmetric about both axes, while x = y² - 1 is a parabola opening rightward. The bounded region exists where the parabola intersects the lemniscate, and symmetry about the x-axis allows computing half the area and doubling it.
<p><strong>Step 1: Identify the curves</strong></p><p>Curve 1: x = |y|√(1-y²) is defined for |y| ≤ 1 (lemniscate-like, symmetric about x and y axes)</p><p>Curve 2: x = y² - 1 is a parabola with vertex at (-1, 0), opening rightward</p><p><strong>Step 2: Find intersection points</strong></p><p>At intersections: |y|√(1-y²) = y² - 1</p><p>For y ∈ [0,1]: y√(1-y²) = y² - 1</p><p>Testing boundary: at y = 0: 0 = -1 (no); at y = 1: 0 = 0 ✓</p><p>At y = 0: lemniscate gives x = 0, parabola gives x = -1</p><p>Solving numerically or by substitution: intersection points are at y = ±1 and y = 0 (where curves meet)</p><p><strong>Step 3: Set up the integral using symmetry</strong></p><p>By symmetry about x-axis, Area = 2∫₀¹ [|y|√(1-y²) - (y² - 1)] dy</p><p>= 2∫₀¹ [y√(1-y²) - y² + 1] dy</p><p><strong>Step 4: Evaluate the integral</strong></p><p>∫₀¹ y√(1-y²) dy = [-⅓(1-y²)^(3/2)]₀¹ = ⅓</p><p>∫₀¹ y² dy = ⅓</p><p>∫₀¹ 1 dy = 1</p><p>Area = 2(⅓ - ⅓ + 1) = 2(1) = 2</p><p><strong>∴ Answer: B (2 square units)</strong></p>
Correct Answer: B