Triple integral (2x+y) over cylinder region

Region. The solid under the cylinder z = 4 − x² (a parabolic cylinder, parallel to the y-axis), above z = 0, between y = 0 and y = 2, and with x ≥ 0. On z = 0, 4 − x² = 0 gives x = 2. So the limits are

0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 4 − x².

I = ∫₀² ∫₀² ∫₀^{4−x²} (2x + y) dz dy dx

z-integration

∫₀^{4−x²} (2x + y) dz = (2x + y)(4 − x²).

y-integration

∫₀² (2x + y) dy = 4x + 2
→ (4x + 2)(4 − x²) = 16x − 4x³ + 8 − 2x²

x-integration

I = ∫₀² (16x − 4x³ − 2x² + 8) dx
  = [ 8x² − x⁴ − (2/3)x³ + 8x ]₀²
  = 32 − 16 − 16/3 + 16
  = 32 − 16/3
  = 80/3

Answer: 80/3.