The area (in sq units) of the region and is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
The region is bounded by y² = 2x (parabola) and x² + y² = 4x, i.e. (x-2)² + y² = 4 (circle centered at (2,0) with radius 2), with x ≥ 0, y ≥ 0.
Find intersection: substitute y² = 2x into x² + y² = 4x → x² + 2x = 4x → x² - 2x = 0 → x = 0 or x = 2. At x = 2, y = 2.
The required region satisfies y² ≥ 2x (outside/above parabola) AND inside the circle, in the first quadrant. So we take the quarter-circle area (from x=0 to x=4 in first quadrant) and subtract the area under the parabola (y = √(2x)) from x=0 to x=2, plus account for the geometry correctly.
Area = (area of quarter circle) − (area between x-axis and parabola from 0 to 2) − (area of circular segment from x=2 to x=4... wait, let me be precise).
Quarter circle area = π(4)/4 = π. The area inside the circle but below the parabola (y ≤ √(2x)) in first quadrant runs from x=0 to x=2:
∫₀² √(2x) dx = √2 · (2/3)x^(3/2)|₀² = √2 · (2/3)·2√2 = 8/3.
So the required area = π − 8/3
Find intersection: substitute y² = 2x into x² + y² = 4x → x² + 2x = 4x → x² - 2x = 0 → x = 0 or x = 2. At x = 2, y = 2.
The required region satisfies y² ≥ 2x (outside/above parabola) AND inside the circle, in the first quadrant. So we take the quarter-circle area (from x=0 to x=4 in first quadrant) and subtract the area under the parabola (y = √(2x)) from x=0 to x=2, plus account for the geometry correctly.
Area = (area of quarter circle) − (area between x-axis and parabola from 0 to 2) − (area of circular segment from x=2 to x=4... wait, let me be precise).
Quarter circle area = π(4)/4 = π. The area inside the circle but below the parabola (y ≤ √(2x)) in first quadrant runs from x=0 to x=2:
∫₀² √(2x) dx = √2 · (2/3)x^(3/2)|₀² = √2 · (2/3)·2√2 = 8/3.
So the required area = π − 8/3
[JEE Main 2016]