Area Under the Curve
Area using parametric curves
Grade 12
Question:
<p>The area enclosed by the curve \(x = a\cos^3 t\), \(y = b\sin^3 t\), is</p>
<p>\(\dfrac{\pi ab}{8}\)</p>
<p>\(\dfrac{3\pi ab}{8}\)</p>
<p>\(\dfrac{\pi ab}{8} + 4\)</p>
<p>\(ab\left(\dfrac{\pi}{4} + 4\right)\)</p>
Step-by-Step Solution
Key Concept: Use parametric form with the formula A = (1/2)|∮(x·dy - y·dx)| or substitute parametric equations into the standard area integral, recognizing that one complete curve traversal (t: 0→2π) encloses the region.
<p><strong>Step 1:</strong> Write the parametric derivatives.</p><p>x = a·cos³t ⟹ dx/dt = -3a·cos²t·sin t</p><p>y = b·sin³t ⟹ dy/dt = 3b·sin²t·cos t</p><p><strong>Step 2:</strong> Use the parametric area formula A = (1/2)|∮(x·dy - y·dx)|.</p><p>x·dy - y·dx = a·cos³t · 3b·sin²t·cos t dt - b·sin³t · (-3a·cos²t·sin t) dt</p><p>= 3ab·cos⁴t·sin²t dt + 3ab·sin⁴t·cos²t dt</p><p>= 3ab·cos²t·sin²t(cos²t + sin²t) dt = 3ab·cos²t·sin²t dt</p><p><strong>Step 3:</strong> Integrate from 0 to 2π.</p><p>A = (1/2) ∫₀²π 3ab·cos²t·sin²t dt = (3ab/2) ∫₀²π cos²t·sin²t dt</p><p>Using cos²t·sin²t = (1/4)sin²(2t) = (1/8)(1 - cos 4t):</p><p>A = (3ab/2) · (1/8) ∫₀²π (1 - cos 4t) dt = (3ab/16)[t - (sin 4t)/4]₀²π</p><p>= (3ab/16) · 2π = <strong>(3πab)/8</strong></p><p>∴ Answer: B</p>
Correct Answer: B