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Double integral y over region between two parabolas

Short answer

The parabolas intersect at (0,0) and (4,4). For 0≤x≤4, y from x²/4 to 2√x. Integrating y dy dx yields I = 48/5.

Evaluate: ∬_R y dy dx where R is the region bounded by the parabolas y² = 4x and x² = 4y.

Region. Bounded by the parabolas y² = 4x and x² = 4y.

Intersection

From x² = 4y, y = x²/4. Put in y² = 4x: x⁴/16 = 4x, so x(x³ − 64) = 0, giving x = 0 and x = 4. Points: (0,0) and (4,4).

For 0 ≤ x ≤ 4, the upper curve is y = 2√x (from y² = 4x) and the lower curve is y = x²/4.

y
4 |        .*(4,4)
  |     . ' /
  |   .  ' /   y² = 4x (upper)
  |  . '  /    x² = 4y (lower)
  | .'__/______ x
 0           4

Integral

I = ∫₀⁴ ∫_{x²/4}^{2√x} y dy dx
  = ∫₀⁴ (1/2) [ 4x − x⁴/16 ] dx
  = (1/2) [ 2x² − x⁵/80 ]₀⁴
  = (1/2) ( 32 − 1024/80 )
  = (1/2)(32 − 12.8)
  = 9.6 = 48/5

Answer: I = 48/5.

✓ Answer checked by Claude Opus 5.5 (high effort)

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