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

Short answer

On the circle z=0, parametrize x=2cos t, y=2sin t. Then F·dr = (4−4 sin t cos t) dt, and work = ∫₀^{2π} (4−2 sin 2t) dt = 8π. Alternatively, by Stokes, curl F·k = 2, area = 4π, giving 8π.

Using the line integral, compute the work done by the force F = (2x − y + 2z) i + (x + y − z) j + (3x − 2y − 5z) k when it moves once around the circle x² + y² = 4, z = 0.

Work = ∮_C F·dr around the circle x² + y² = 4, z = 0. Evaluate it as a line integral.

Parametrise: x = 2cos t, y = 2sin t, z = 0, 0 ≤ t ≤ 2π. Then dx = −2sin t dt, dy = 2cos t dt, dz = 0.

F·dr = (2x − y + 2z) dx + (x + y − z) dy + (3x − 2y − 5z) dz
     = (2x − y) dx + (x + y) dy       (z = 0, dz = 0)
     = (4cos t − 2sin t)(−2sin t) dt + (2cos t + 2sin t)(2cos t) dt
     = (−8 sin t cos t + 4 sin²t + 4 cos²t + 4 sin t cos t) dt
     = (4 − 4 sin t cos t) dt
W = ∫₀^2π (4 − 2 sin 2t) dt = 8π − 0 = 8π

Check by Stokes' theorem: (curl F)·k = ∂Q/∂x − ∂P/∂y = 1 − (−1) = 2. Area of circle = π(2)² = 4π, so W = 2 × 4π = 8π.

Answer: Work done = 8π ≈ 25.13 units (anticlockwise).

✓ Answer checked by Claude Opus 5.5 (high effort)

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