Double integral xy over ellipse in first quadrant

Short answer

Region: first quadrant inside x²/a² + y²/b² = 1. Integrate xy dy dx: inner integral yields (x b²/2)(1−x²/a²), outer gives a²b²/8.

Evaluate: ∬_R xy dx dy where R is the region over the area of the ellipse x²/a² + y²/b² = 1 in the first quadrant.

Region. The first quadrant part of the ellipse x²/a² + y²/b² = 1: x from 0 to a, and y from 0 to y_max = (b/a)√(a² − x²) = b√(1 − x²/a²).

I = ∫₀^a ∫₀^{b√(1−x²/a²)} xy dy dx

∫₀^{y_max} xy dy = x · y_max²/2 = (x b²/2)(1 − x²/a²)

I = (b²/2) ∫₀^a ( x − x³/a² ) dx
  = (b²/2) [ x²/2 − x⁴/(4a²) ]₀^a
  = (b²/2) ( a²/2 − a²/4 )
  = (b²/2)(a²/4)
  = a²b²/8

Answer: a²b²/8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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