Area Under the Curve
Area between two parabolas
Grade 12
Question:
<p>If the area enclosed between the curves y = kx² and x = ky², (k > 0), is 1 square unit. Then k is:</p>
<p>\(\dfrac{\sqrt{3}}{2}\)</p>
<p>\(\dfrac{1}{\sqrt{3}}\)</p>
<p>\(\sqrt{3}\)</p>
<p>\(\dfrac{2}{\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: The curves y = kx² and x = ky² are reflections of each other about y = x. Find intersection points, then use symmetry to set up a single integral whose value equals 1.
<p><strong>Step 1:</strong> Find intersection points of y = kx² and x = ky².</p><p>Substituting x = ky² into y = kx²: y = k(ky²)² = k³y⁴</p><p>So y(1 - k³y³) = 0, giving y = 0 or y³ = 1/k³, thus y = 1/k</p><p>Intersection points: (0,0) and (1/k, 1/k)</p><p><strong>Step 2:</strong> Recognize symmetry about y = x. For the line y = x between the intersection points, both curves meet at these points.</p><p>From x = ky², we get y = √(x/k). From y = kx², we have y = kx².</p><p>For 0 ≤ x ≤ 1/k: the curve y = √(x/k) lies above y = kx².</p><p><strong>Step 3:</strong> Set up the area integral:</p><p>A = ∫₀^(1/k) [√(x/k) - kx²] dx</p><p>= ∫₀^(1/k) [x^(1/2)/√k - kx²] dx</p><p>= [2x^(3/2)/(3√k) - kx³/3]₀^(1/k)</p><p>= 2(1/k)^(3/2)/(3√k) - k(1/k)³/3</p><p>= 2/(3k²) - 1/(3k²) = 1/(3k²)</p><p><strong>Step 4:</strong> Set area equal to 1:</p><p>1/(3k²) = 1</p><p>3k² = 1</p><p>k = 1/√3 = √3/3</p><p>∴ Answer: <strong>k = 1/√3</strong> or <strong>√3/3</strong></p>
Correct Answer: B