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Surface integral over a surface F=xi−yj+zk

Short answer

For the curved surface, outward unit normal is (x i + y j)/a, so F·n = (x²−y²)/a. Using x = a cosθ, y = a sinθ, dS = a dθ dz, the integral becomes a²∫₀^b∫₀^{2π} cos2θ dθ dz = 0. If the closed cylinder is intended, the total flux is πa²b.

Evaluate ∬_S F·ds where F = x i − y j + z k and S is the surface of the cylinder x² + y² = a², 0 < z < b.

S is the curved surface of the cylinder x² + y² = a², 0 < z < b. For this surface the outward unit normal is n = (x i + y j)/a (no k component).

F = x i − y j + z k
F·n = (x² − y²)/a

Use x = a cos θ, y = a sin θ, dS = a dθ dz, with 0 ≤ θ ≤ 2π, 0 ≤ z ≤ b.

∬_S F·n dS = ∫₀^b ∫₀^2π [(a² cos²θ − a² sin²θ)/a] · a dθ dz
           = a² ∫₀^b ∫₀^2π cos 2θ dθ dz
           = a² · b · [sin 2θ / 2]₀^2π
           = 0

Answer: ∬_S F·n dS = 0 for the curved surface.

Remark. If the two circular ends are also counted (closed cylinder), div F = 1 − 1 + 1 = 1, so by Gauss' theorem the total flux is the volume = πa²b. This comes entirely from the top (z = b): F·k = b over area πa²; the bottom gives 0 since z = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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