Smaller area bounded by circle and line

Curves. The circle x² + y² = 9 (centre origin, radius 3) and the line x + y = 3.

Intersection

Put y = 3 − x: x² + (3 − x)² = 9 → 2x² − 6x = 0 → x = 0 or x = 3. Points: (0, 3) and (3, 0).

The line cuts off a small segment from the circle (the part between the chord and the arc in the first quadrant). The line is at distance 3/√2 ≈ 2.12 from the centre, less than 3, so a segment is cut; the smaller area is the one not containing the centre.

y
3 *
  | '.
  |   '.  segment
  |     '.
  |_______*___ x
 O        3
 (chord x + y = 3, arc of circle)

Area

For 0 ≤ x ≤ 3, y runs from the line (y = 3 − x) up to the arc (y = √(9 − x²)).

A = ∫₀³ ∫_{3−x}^{√(9−x²)} dy dx
  = ∫₀³ ( √(9 − x²) − 3 + x ) dx

Now ∫₀³ √(9 − x²) dx = quarter circle area = (1/4)π(3)² = 9π/4, and ∫₀³ (x − 3) dx = 9/2 − 9 = −9/2.

A = 9π/4 − 9/2 = 9(π − 2)/4 ≈ 2.57

Answer: smaller area = 9(π − 2)/4 square units.