<p>The area bounded between the parabolas \(x^2 = \dfrac{y}{4}\) and \(x^2 = 9y\), and the straight line \(y = 2\) is</p>
Step-by-Step Solution
Key Concept: Find intersection points of both parabolas with y=2, then integrate the difference of the two parabolic curves. The parabola x²=y/4 is wider (opens slower) than x²=9y, so it forms the outer boundary.
<p><strong>Step 1:</strong> Find intersection points with y=2.</p><p>For x²=y/4: x²=2/4=1/2, so x=±1/√2</p><p>For x²=9y: x²=18, so x=±3√2</p><p><strong>Step 2:</strong> Identify which parabola is outer. At y=2: x²=y/4 gives x²=1/2 (inner), x²=9y gives x²=18 (outer). Wait—recalculate: for x²=9y at y=2: x²=18. For x²=y/4 at y=2: x²=1/2. So x²=9y is outer.</p><p><strong>Step 3:</strong> Find where parabolas intersect: x²/4=9x² is impossible for x≠0. They intersect at origin only.</p><p><strong>Step 4:</strong> By symmetry, Area = 2∫₀² (√(9y) - √(y/4))dy = 2∫₀² (3√y - √y/2)dy</p><p>= 2∫₀² (5√y/2)dy = 2·(5/2)·(2y^(3/2)/3)|₀² = (5/3)·2^(3/2)·2 = (20√2)/3</p><p>∴ Answer: C</p>
Correct Answer: C