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Surface integral over a surface F=3i+xj−yzk

Short answer

Parameterize the quarter-cylinder with x=3cosθ, y=3sinθ, dS=3dθdz. F·n = 3cosθ + 3cosθ sinθ. Integrate θ from 0 to π/2 and z from 0 to 4 to obtain 54.

Evaluate ∬_S F·n ds where F = 3 i + x j − yz k and S is the surface of the cylinder x² + y² = 9 included in the first octant between the planes z = 0, z = 4.

S is the quarter of the cylinder x² + y² = 9 with x, y ≥ 0 and 0 ≤ z ≤ 4.

Outward normal: n = (x i + y j)/3. F = 3 i + x j − yz k, so

F·n = (3x + xy)/3

Use x = 3cos θ, y = 3sin θ, 0 ≤ θ ≤ π/2, 0 ≤ z ≤ 4, and dS = 3 dθ dz.

∬_S F·n dS = ∫₀⁴ ∫₀^(π/2) (3x + xy) dθ dz
           = ∫₀⁴ ∫₀^(π/2) (9 cos θ + 9 cos θ sin θ) dθ dz
inner: [9 sin θ + (9/2) sin²θ]₀^(π/2) = 9 + 9/2 = 27/2
outer: 4 × 27/2 = 54

Answer: ∬_S F·n dS = 54.

✓ Answer checked by Claude Opus 5.5 (high effort)

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