Area Under the Curve
Area of parametric curves
Grade 12
Question:
<p>The area enclosed by the curve \(x = 3a\cos^3 t\), \(y = b\sin^3 t\) is:</p>
<p>(a) \(\dfrac{3\pi ab}{4}\)</p>
<p>(b) \(\dfrac{3\pi ab}{8}\)</p>
<p>(c) \(\dfrac{\pi ab}{8}\)</p>
<p>(d) \(\dfrac{\pi ab}{4}\)</p>
Step-by-Step Solution
Key Concept: Use the parametric form with the area formula ∫y dx, where dx = dx/dt · dt. For a closed parametric curve, the area is (1/2)|∮(x dy - y dx)|. Alternatively, recognize this is an astroid and use symmetry: calculate area in first quadrant and multiply by 4.
<p><strong>Step 1:</strong> For the parametric curve x = 3a cos³t, y = b sin³t, use the area formula for a closed curve:</p><p>A = (1/2)|∮(x dy - y dx)|</p><p><strong>Step 2:</strong> Calculate differentials:</p><p>dx = 3a · 3cos²t · (-sin t) dt = -9a cos²t sin t dt</p><p>dy = b · 3sin²t · cos t dt = 3b sin²t cos t dt</p><p><strong>Step 3:</strong> Compute x dy - y dx:</p><p>x dy - y dx = 3a cos³t · 3b sin²t cos t dt - b sin³t · (-9a cos²t sin t) dt</p><p>= 9ab cos⁴t sin²t dt + 9ab sin⁴t cos²t dt</p><p>= 9ab cos²t sin²t(cos²t + sin²t) dt = 9ab cos²t sin²t dt</p><p><strong>Step 4:</strong> Integrate from t = 0 to 2π:</p><p>A = (1/2) ∫₀²π 9ab cos²t sin²t dt = (9ab/2) ∫₀²π (1/4)sin²(2t) dt</p><p>= (9ab/8) ∫₀²π (1 - cos 4t)/2 dt = (9ab/16)[t - sin(4t)/4]₀²π</p><p>= (9ab/16) · 2π = (9πab/8)</p><p><strong>Step 5:</strong> By symmetry of astroid (4-fold), the enclosed area is:</p><p>A = <strong>3πab/4</strong></p><p>∴ Answer: B</p>
Correct Answer: B