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Surface integral over a surface F=xi+yj+zk
Short answer
Outward normal n = (x i + y j)/a gives F·n = a. Lateral area = 2πa × 4 = 8πa. Flux = a × 8πa = 8πa².
Evaluate ∬_S F·n ds where F = x i + y j + z k and S is the outside of the lateral surface of the circular cylinder x² + y² = a² between the planes z = 0 and z = 4.
S is the curved (lateral) surface of the cylinder x² + y² = a², 0 ≤ z ≤ 4, with outward normal n = (x i + y j)/a.
F = x i + y j + z k
F·n = (x² + y²)/a = a²/a = a
On S the integrand is constant. The area of the lateral surface is circumference × height = 2πa × 4 = 8πa.
∬_S F·n dS = a × 8πa = 8πa²
Check with parameters: x = a cos θ, y = a sin θ, dS = a dθ dz:
∫₀⁴ ∫₀^2π a · a dθ dz = a² · 2π · 4 = 8πa²
Answer: ∬_S F·n dS = 8πa².
✓ Answer checked by Claude Opus 5.5 (high effort)
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