Evaluate using Stokes' theorem F=yi+x(1−2z)j−xyk

Stokes' theorem: ∬_S (∇×F)·n dS = ∮_C F·dr, where C is the boundary of S.

F = y i + x(1 − 2z) j − xy k. S is the upper hemisphere x² + y² + z² = a², z ≥ 0, whose boundary C is the circle x² + y² = a², z = 0.

Line integral on C

On C, z = 0 and dz = 0, so F·dr = y dx + x dy = d(xy). With x = a cos t, y = a sin t, 0 ≤ t ≤ 2π:

∮_C (y dx + x dy) = [xy]  taken once round the closed curve = 0

Explicitly: ∫₀^{2π} (−a² sin²t + a² cos²t) dt = a² ∫₀^{2π} cos 2t dt = 0.

Check by direct surface integral

∇×F = i(−x + 2x) + j(0 + y) + k((1 − 2z) − 1) = x i + y j − 2z k

On the sphere n = (x i + y j + z k)/a, so

(∇×F)·n = (x² + y² − 2z²)/a

With x = a sin φ cos θ, y = a sin φ sin θ, z = a cos φ, we get (∇×F)·n = a(sin²φ − 2cos²φ) and dS = a² sin φ dφ dθ:

∬ = a³ ∫₀^{2π} ∫₀^{π/2} (sin²φ − 2cos²φ) sin φ dφ dθ
  = a³ · 2π · [ ∫ sin³φ dφ − 2∫ cos²φ sin φ dφ ]
  = 2πa³ ( 2/3 − 2/3 ) = 0

Answer: ∬_S (∇×F)·n dS = 0.