Volume integral over a region F=2xzi−xj+y²k

The integral of a vector function over a volume is taken component by component: ∭_V F dv = i ∭ 2xz dv − j ∭ x dv + k ∭ y² dv.

Limits

The region is bounded by x = 0, y = 0, y = 6, z = x² and z = 4. The parabolic cylinder z = x² meets z = 4 at x = 2. So

0 ≤ x ≤ 2, 0 ≤ y ≤ 6, x² ≤ z ≤ 4.

Components

∫_{x²}⁴ 2xz dz = x (16 − x⁴)
∫₀² x(16 − x⁴) dx = [8x² − x⁶/6]₀² = 32 − 64/6 = 64/3
∭ 2xz dv = 6 × 64/3 = 128

∭ x dv = 6 ∫₀² x (4 − x²) dx = 6 [2x² − x⁴/4]₀² = 6 (8 − 4) = 24

∭ y² dv = (∫₀⁶ y² dy) ∫₀² (4 − x²) dx = 72 × (8 − 8/3) = 72 × 16/3 = 384

Answer: ∭_V F dv = 128 i − 24 j + 384 k.