<p>The area of the plane region bounded by the curves \(x + 2y^2 = 0\) and \(x + 3y^2 = 1\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that both curves are parabolas opening leftward; find intersection points by solving the system, then integrate with respect to y (treating x as a function of y) since the curves are naturally expressed as x = f(y).
<p><strong>Step 1: Find intersection points</strong></p><p>From the curves: x = -2y² and x = 1 - 3y²</p><p>Setting equal: -2y² = 1 - 3y²</p><p>y² = 1, so y = ±1</p><p>Intersection points: (-2, 1) and (-2, -1)</p><p><strong>Step 2: Identify which curve is rightmost</strong></p><p>For any y ∈ [-1, 1]: x₁ = 1 - 3y² and x₂ = -2y²</p><p>Since 1 - 3y² > -2y² (as 1 > y² for |y| ≤ 1), the line x = 1 - 3y² is to the right</p><p><strong>Step 3: Set up the integral with respect to y</strong></p><p>Area = ∫₋₁¹ [(1 - 3y²) - (-2y²)] dy</p><p>= ∫₋₁¹ (1 - y²) dy</p><p><strong>Step 4: Evaluate</strong></p><p>= [y - y³/3]₋₁¹</p><p>= (1 - 1/3) - (-1 + 1/3)</p><p>= 2/3 - (-2/3)</p><p>= 4/3</p><p>∴ Answer: D</p>
Correct Answer: D