Area Under the Curve
Area between curves (circle and parabola)
Grade 12
Question:
<p>The area of the smaller portion enclosed between the curves \(x^2 + y^2 = 4\) and \(y^2 = 3x\) is</p>
<p>\(\left(\dfrac{1}{\sqrt{3}} + \dfrac{4\pi}{3}\right)\) sq. units</p>
<p>\(\dfrac{\sqrt{3}}{3} + \dfrac{4\pi}{3}\) sq. units</p>
<p>\(\dfrac{2\sqrt{3}}{3} + \dfrac{2\pi}{3}\) sq. units</p>
<p>\(\dfrac{\sqrt{3}}{2} + \dfrac{\pi}{3}\) sq. units</p>
Step-by-Step Solution
Key Concept: Find intersection points of the circle and parabola, then integrate the difference of the outer and inner curves. The smaller area is bounded by the parabola on the left and the circle on the right.
<p><strong>Step 1: Find intersection points</strong></p><p>From x² + y² = 4 and y² = 3x:</p><p>x² + 3x = 4</p><p>x² + 3x - 4 = 0</p><p>(x + 4)(x - 1) = 0</p><p>Since y² = 3x ≥ 0, we need x ≥ 0, so x = 1</p><p>When x = 1: y² = 3, so y = ±√3</p><p>Intersection points: (1, √3) and (1, -√3)</p><p><strong>Step 2: Set up the area integral</strong></p><p>By symmetry about x-axis, Area = 2∫₀¹ √(3x) dx + 2∫₁² √(4-x²) dx</p><p>The smaller region: left boundary is parabola y² = 3x (from x=0 to x=1), right boundary is circle (from x=1 to x=2)</p><p><strong>Step 3: Evaluate first integral</strong></p><p>2∫₀¹ √(3x) dx = 2√3 ∫₀¹ √x dx = 2√3 · [⅔x^(3/2)]₀¹ = 2√3 · ⅔ = 4√3/3</p><p><strong>Step 4: Evaluate second integral</strong></p><p>2∫₁² √(4-x²) dx: Use x = 2sin(θ), dx = 2cos(θ)dθ</p><p>At x=1: θ = π/6; at x=2: θ = π/2</p><p>2∫_{π/6}^{π/2} 2cos(θ) · 2cos(θ) dθ = 8∫_{π/6}^{π/2} cos²(θ) dθ</p><p>= 8∫_{π/6}^{π/2} (1+cos(2θ))/2 dθ = 4[θ + sin(2θ)/2]_{π/6}^{π/2}</p><p>= 4[(π/2 + 0) - (π/6 + √3/4)] = 4[π/3 - √3/4] = 4π/3 - √3</p><p><strong>Step 5: Total area</strong></p><p>Area = 4√3/3 + 4π/3 - √3 = 4π/3 + √3/3</p><p>∴ Answer: <strong>4π/3 + √3/3</strong> or <strong>(4π + √3)/3</strong></p>
Correct Answer: A