Area Under the Curve
Area bounded by curves and circles
Grade 12
Question:
<p>The area (in sq. units) of the region bounded by \(y^2 - 2x \geq 0\), \(x^2 + y^2 - 4x \leq 0\), \(x \geq 0\), \(y \geq 0\) is</p>
<p>\(\pi - \dfrac{4}{3}\)</p>
<p>\(\pi - \dfrac{8}{3}\)</p>
<p>\(\pi - \dfrac{2}{3}\)</p>
<p>\(2\pi - \dfrac{8}{3}\)</p>
Step-by-Step Solution
Key Concept: The region is bounded by a parabola y² = 2x and a circle (x-2)² + y² = 4 in the first quadrant. Find intersection points, then integrate to get the area between curves.
<p><strong>Step 1: Identify the boundaries</strong></p><p>• Parabola: y² = 2x (or x = y²/2) where y² - 2x ≥ 0 means x ≤ y²/2</p><p>• Circle: (x-2)² + y² = 4, centered at (2,0) with radius 2</p><p>• Constraints: First quadrant (x ≥ 0, y ≥ 0)</p><p><strong>Step 2: Find intersection points</strong></p><p>Substituting x = y²/2 into circle equation:</p><p>(y²/2 - 2)² + y² = 4</p><p>y⁴/4 - 2y² + 4 + y² = 4</p><p>y⁴/4 - y² = 0</p><p>y²(y² - 4) = 0</p><p>So y = 0 or y = 2</p><p>At y = 2: x = 2; at y = 0: x = 0</p><p>Intersection points: (0,0) and (2,2)</p><p><strong>Step 3: Set up the integral</strong></p><p>For 0 ≤ y ≤ 2, the region is bounded on the left by parabola x = y²/2 and on the right by circle x = 2 - √(4-y²)</p><p>Area = ∫₀² [2 - √(4-y²) - y²/2] dy</p><p><strong>Step 4: Evaluate</strong></p><p>= [2y]₀² - ∫₀² √(4-y²) dy - [y³/6]₀²</p><p>= 4 - [y√(4-y²)/2 + 2sin⁻¹(y/2)]₀² - 8/6</p><p>= 4 - [0 + 2·π/2] - 4/3</p><p>= 4 - π - 4/3 = 8/3 - π</p><p>∴ Answer: B</p>
Correct Answer: B