<p>The area of the region enclosed between the curves \(x = y^2 - 1\) and \(x = |y|\sqrt{1 - y^2}\) is</p>
Step-by-Step Solution
Key Concept: Recognize that x = |y|√(1-y²) represents the right half of a circle (x² + y² = 1), and find intersection points by solving y² - 1 = |y|√(1-y²). Use symmetry about the x-axis to simplify integration.
<p><strong>Step 1: Identify the curves</strong></p><p>• Curve 1: x = y² - 1 (rightward-opening parabola, vertex at (-1, 0))</p><p>• Curve 2: x = |y|√(1-y²) (right semicircle of x² + y² = 1)</p><p><strong>Step 2: Find intersection points</strong></p><p>At intersections: y² - 1 = |y|√(1-y²)</p><p>Let y ≥ 0. Then: y² - 1 = y√(1-y²)</p><p>Square both sides: (y²-1)² = y²(1-y²)</p><p>y⁴ - 2y² + 1 = y² - y⁴</p><p>2y⁴ - 3y² + 1 = 0</p><p>(2y² - 1)(y² - 1) = 0</p><p>y = 1/√2 or y = 1</p><p>By symmetry, intersections are at (0, 1), (0, -1), and (1/2, ±1/√2)</p><p><strong>Step 3: Set up the integral using symmetry</strong></p><p>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 each integral</strong></p><p>∫₀¹ y√(1-y²) dy = [-⅓(1-y²)^(3/2)]₀¹ = 1/3</p><p>∫₀¹ y² dy = 1/3</p><p>∫₀¹ 1 dy = 1</p><p><strong>Step 5: Combine</strong></p><p>Area = 2[1/3 - 1/3 + 1] = 2(1) = <strong>2</strong></p><p>∴ Answer: <strong>2</strong> or <strong>π/2 + 1/3</strong> (verify against given options)</p>
Correct Answer: 2