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Double integral xy over region bounded by x²=4ay

Short answer

Integrate over 0≤x≤2a, 0≤y≤x²/(4a). Inner integral yields x⁵/(32a²); outer integral gives a⁴/3.

Evaluate: ∬_R xy dx dy where R is the region bounded by the x-axis, the ordinate x = 2a and the curve x² = 4ay.

Region. Bounded by the x-axis (y = 0), the line x = 2a and the parabola x² = 4ay, i.e. y = x²/(4a). For each x in [0, 2a], y runs from 0 up to the parabola.

I = ∫₀^{2a} ∫₀^{x²/4a} xy dy dx

Inner integral

∫₀^{x²/4a} xy dy = x · (1/2)(x²/4a)² = x⁵/(32a²)

Outer integral

I = (1/32a²) ∫₀^{2a} x⁵ dx
  = (1/32a²) · (2a)⁶/6
  = (1/32a²) · 64a⁶/6
  = a⁴/3

Answer: a⁴/3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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