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Irrotational field and scalar potential F=(2x+z²)i+zj+(y+2xz)k

Short answer

Curl F = 0, so it is irrotational. Integrating ∂φ/∂x = 2x+z² gives φ = x²+xz²+f(y,z); from ∂φ/∂y = z we get f = yz+g(z); then ∂φ/∂z = 2xz+y+g' matches R, so g'=0. Potential: φ = x²+xz²+yz+c.

Show that F = (2x + z²) i + z j + (y + 2xz) k is irrotational and find its scalar potential.

F is irrotational if curl F = 0. Here P = 2x + z², Q = z, R = y + 2xz.

curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (1 − 1) i + (2z − 2z) j + (0 − 0) k
       = 0

So F is irrotational, and a scalar potential φ exists with F = ∇φ.

∂φ/∂x = 2x + z²   → φ = x² + xz² + f(y,z)
∂φ/∂y = f_y = z    → f = yz + g(z)
∂φ/∂z = 2xz + y + g'(z) = y + 2xz  → g' = 0

Answer: φ = x² + xz² + yz + c (c is an arbitrary constant). Check: ∇φ = (2x + z²) i + z j + (2xz + y) k = F.

✓ Answer checked by Claude Opus 5.5 (high effort)

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