Work done by a force field F=(2x−y+z)i+(x+y−z²)j…

Work done = ∮_C F·dr around the closed circle. Use Stokes' theorem to avoid parametrising the three components, then check directly.

By Stokes' theorem (C counter-clockwise, S = disc x² + y² ≤ 9 in the plane z = 0, n = k):

F = (2x − y + z) i + (x + y − z²) j + (3x − 2y + 4z) k
curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (−2 + 2z) i + (1 − 3) j + (1 + 1) k
       = (2z − 2) i − 2 j + 2 k
curl F · k = 2
Work = ∬_S (curl F)·k dA = 2 × (area of disc) = 2 × π(3)² = 18π

Direct check. On the circle z = 0: x = 3cos t, y = 3sin t, 0 ≤ t ≤ 2π, dz = 0.

F·dr = (2x − y) dx + (x + y) dy
     = (6cos t − 3sin t)(−3sin t) dt + (3cos t + 3sin t)(3cos t) dt
     = (−18 sin t cos t + 9 sin²t + 9 cos²t + 9 sin t cos t) dt
     = (9 − 9 sin t cos t) dt
∫₀^2π (9 − 9 sin t cos t) dt = 18π − 0 = 18π

Answer: Work done = 18π ≈ 56.55 units (anticlockwise direction).