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Irrotational field and scalar potential F=2xyz²i+(x²z²+zcosyz)j…

Short answer

Compute curl: ∂R/∂y = ∂Q/∂z, etc., giving curl F = 0, so F is irrotational. Scalar potential: φ = x^2 y z^2 + sin(yz). Work = φ(1,π/4,2) − φ(0,0,1) = π + 1.

Prove that the force field F = 2xyz²i + (x²z² + z cos yz)j + (2x²yz + y cos yz)k is irrotational. Also find the work done by the force F along the path C from (0,0,1) to (1, π/4, 2).

A field is irrotational if curl F = 0. Here P = 2xyz², Q = x²z² + z cos yz, R = 2x²yz + y cos yz.

Irrotational

∂R/∂y = 2x²z + cos yz − yz sin yz
∂Q/∂z = 2x²z + cos yz − yz sin yz      → i component = 0
∂P/∂z = 4xyz
∂R/∂x = 4xyz                            → j component = 0
∂Q/∂x = 2xz²
∂P/∂y = 2xz²                            → k component = 0

So curl F = 0 and F is irrotational (conservative).

Work done

Find φ with F = ∇φ.

∂φ/∂x = 2xyz²            → φ = x²yz² + f(y,z)
∂φ/∂y = x²z² + ∂f/∂y = x²z² + z cos yz   → ∂f/∂y = z cos yz → f = sin yz + g(z)
∂φ/∂z = 2x²yz + y cos yz + g'(z) = R     → g' = 0
φ = x²yz² + sin(yz)

Work is independent of the path:

W = φ(1, π/4, 2) − φ(0, 0, 1)
  = [1·(π/4)·4 + sin(π/2)] − [0 + sin 0]
  = π + 1

Answer: W = π + 1 ≈ 4.14.

✓ Answer checked by Claude Opus 5.5 (high effort)

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