Area Under the Curve
Area between curves
Grade 12
Question:
<p>The area of the region described by \(A - \{(x, y) : x^2 + y^2 \leq 1\) and \(y^2 \leq 1 - x\}\) is</p>
<p>\(\dfrac{\pi}{2} - \dfrac{2}{3}\)</p>
<p>\(\dfrac{\pi}{2} + \dfrac{2}{3}\)</p>
<p>\(\dfrac{\pi}{2} + \dfrac{4}{3}\)</p>
<p>\(\dfrac{\pi}{2} - \dfrac{4}{3}\)</p>
Step-by-Step Solution
Key Concept: The region A is the complement of the intersection of a disk and a parabolic region. Find the area of intersection first, then subtract from the disk's area π.
<p><strong>Step 1:</strong> Identify the regions. Region A is the disk x² + y² ≤ 1 (area = π). We need A − {intersection of disk and parabolic region y² ≤ 1−x}.</p><p><strong>Step 2:</strong> Find intersection points. At the boundary y² = 1−x and x² + y² = 1, substitute: x² + (1−x) = 1, so x² − x = 0, giving x = 0 or x = 1. Points: (1, 0) and (0, ±1).</p><p><strong>Step 3:</strong> Set up the intersection area. The parabola y² = 1−x opens leftward. For the intersection, integrate from x = 0 to x = 1:<br>Intersection area = ∫₀¹ 2√(1−x) dx = 2[−(2/3)(1−x)^(3/2)]₀¹ = 2 · (2/3) = 4/3</p><p><strong>Step 4:</strong> Calculate the required area.<br>A − intersection = π − 4/3</p><p>∴ Answer: C (π − 4/3)</p>
Correct Answer: C