Conservative fields and potential functions
Problem 12.257 · medium
Show that \( \displaystyle \mathbf F = \left(2 x z\right)\mathbf i + \left(3 y^{2}\right)\mathbf j + \left(x^{2}\right)\mathbf k \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (1, 2, 0) \) to \( \displaystyle (-1, 1, -1) \).
- \[ \left[\begin{matrix}\frac{d}{d x} 3 y^{2}\\\frac{\partial}{\partial y} 2 x z\end{matrix}\right] = \left[\begin{matrix}0\\0\end{matrix}\right] \]∂Q/∂x = ∂P/∂y (and likewise for the other pairs): F is conservative.✓ Proved
- Integrate the first component in x, then fix the 'constant' (a function of the other variables) by matching the other components.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} z + y^{3}\right)\\\frac{\partial}{\partial y} \left(x^{2} z + y^{3}\right)\\\frac{\partial}{\partial z} \left(x^{2} z + y^{3}\right)\end{matrix}\right] = \left[\begin{matrix}2 x z\\3 y^{2}\\x^{2}\end{matrix}\right] \]f = x**2*z + y**3 has gradient F.✓ Proved
- \[ -8 + 0 = -8 \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = x^{2} z + y^{3} + C,\quad \int_C \mathbf F\cdot d\mathbf r = -8 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the line integral computed numerically along two different paths |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution claims F is conservative based only on the equality of mixed partials for P and Q, ignoring the necessary checks for the z-component (e.g., ∂R/∂y vs ∂Q/∂z). While the field is indeed conservative, the reasoning provided is incomplete and mathematically insufficient for 3D vector fields.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution claims F is conservative based only on the equality of mixed partials for P and Q, ignoring the necessary checks for the z-component (e.g., ∂R/∂y vs ∂Q/∂z). While the field is indeed conservative, the reasoning provided is incomplete and mathematically insufficient for 3D vector fields.qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution claims F is conservative based on checking only one pair of partial derivatives (∂Q/∂x = ∂P/∂y), but for a 3D vector field, all three curl components must vanish (e.g., ∂R/∂y = ∂Q/∂z must also be checked). Although the field is indeed conservative, the justification provided is incomplete and mathematically insufficient.gpt-oss:20b: pass 2026-10-09
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/conservative_fields, checked 2026-10-09 with SymPy 1.14.0.