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Find the area (in square units) of the region defined by: y² >= 2x, x² + y² <= 4x, x >= 0, and y >= 0.
- pi/2 - 2*sqrt(2)/3
- pi - 4/3
- pi - 8/3
- pi - 4*sqrt(2)/3
Correct answer: pi - 8/3
Solution
The circle x² + y² = 4x rewrites as (x-2)² + y² = 4 (center (2,0), radius 2). In first quadrant (x>=0, y>=0), the quarter-circle goes from x=0 to x=4. The parabola y² = 2x passes through origin. Intersection: x² + 2x = 4x => x² - 2x = 0 => x=0 or x=2. At x=2, y²=4, y=2. The required region satisfies y²>=2x (outside/on parabola) AND inside circle, in first quadrant. Area = Area of quarter circle - Area between parabola and circle (in first quadrant from x=0 to x=2). Quarter circle area = pi*4/4 = pi. Subtracting the region where y² < 2x inside circle gives area = pi - 8/3.
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