Conservative fields and potential functions
Problem 12.221 · medium
Show that \( \displaystyle \mathbf F = \left(y z\right)\mathbf i + \left(x z\right)\mathbf j + \left(x y\right)\mathbf k \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (0, 0, 0) \) to \( \displaystyle (1, 1, 2) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} x z\\\frac{\partial}{\partial y} y z\end{matrix}\right] = \left[\begin{matrix}z\\z\end{matrix}\right] \]∂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.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} x y z\\\frac{\partial}{\partial y} x y z\\\frac{\partial}{\partial z} x y z\end{matrix}\right] = \left[\begin{matrix}y z\\x z\\x y\end{matrix}\right] \]f = x*y*z has gradient F.✓ Proved
- \[ 0 + 2 = 2 \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = x y z + C,\quad \int_C \mathbf F\cdot d\mathbf r = 2 \)
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 to show F is conservative by checking only one pair of partial derivatives (∂Q/∂x = ∂P/∂y) and stating 'and likewise for the other pairs' without actually verifying them. In 3D, all three cross-partial conditions must hold. While the result is correct, the proof is incomplete and logically flawed.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution claims to show F is conservative by checking only one pair of partial derivatives (∂Q/∂x = ∂P/∂y) and stating 'and likewise for the other pairs' without actually verifying them. In 3D, all three cross-partial conditions must hold. While the result is correct, the proof is incomplete and logically flawed.gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution claims F is conservative by checking only one pair of partial derivatives (∂Q/∂x = ∂P/∂y) and stating 'and likewise for the other pairs' without showing them. In 3D, all three curl components must vanish. Additionally, the notation in step 1 is confusing and incomplete for a 3D vector field.
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 2026-10-07 with SymPy 1.14.0.