<p>Find the area enclosed between the curves <span>\( x + 2y^2 = 0 \)</span> and <span>\( x + 3y^2 = 1 \)</span>.</p>
Step-by-Step Solution
Key Concept: Recognize these are parabolas opening leftward; find intersection points by solving simultaneously, then integrate with respect to y (treating x as a function of y) since the curves are better expressed as x = f(y).
<p><strong>Step 1: Find intersection points</strong></p><p>From x + 2y² = 0: x = -2y²</p><p>From x + 3y² = 1: x = 1 - 3y²</p><p>Setting equal: -2y² = 1 - 3y²</p><p>⟹ y² = 1 ⟹ y = ±1</p><p>Intersection points: (−2, 1) and (−2, −1)</p><p><strong>Step 2: Determine which curve is to the right</strong></p><p>At y = 0: First curve gives x = 0; Second curve gives x = 1</p><p>So x + 3y² = 1 is to the right of x + 2y² = 0</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 - 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>= 2/3 + 2/3</p><p><strong>∴ Answer: 4/3</strong></p>
Correct Answer: 4/3