Conservative fields and potential functions
Problem 12.190 · medium
Show that \( \displaystyle \mathbf F = \left(y z e^{x}\right)\mathbf i + \left(z e^{x}\right)\mathbf j + \left(y e^{x}\right)\mathbf k \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (-1, 1, 0) \) to \( \displaystyle (1, 2, 1) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} z e^{x}\\\frac{\partial}{\partial y} y z e^{x}\end{matrix}\right] = \left[\begin{matrix}z e^{x}\\z e^{x}\end{matrix}\right] \]∂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.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} y z e^{x}\\\frac{\partial}{\partial y} y z e^{x}\\\frac{\partial}{\partial z} y z e^{x}\end{matrix}\right] = \left[\begin{matrix}y z e^{x}\\z e^{x}\\y e^{x}\end{matrix}\right] \]f = y*z*exp(x) has gradient F.✓ Proved
- \[ 0 + 2 e = 2 e \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = y z e^{x} + C,\quad \int_C \mathbf F\cdot d\mathbf r = 2 e \)
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 on checking only one pair of partial derivatives (∂Q/∂x = ∂P/∂y), ignoring the other two necessary conditions for a 3D vector field. Additionally, the final evaluation step is presented as an algebraic identity '0 + 2*E = 2*E' rather than explicitly showing the substitution of the endpoints into the potential function.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims F is conservative based on checking only one pair of partial derivatives (∂Q/∂x = ∂P/∂y), ignoring the other two necessary conditions for a 3D vector field. Additionally, the final evaluation step is presented as an algebraic identity '0 + 2*E = 2*E' rather than explicitly showing the substitution of the endpoints into the potential function.gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution claims the potential function is f = y z e^{x}, but the gradient of this function is <y z e^{x}, z e^{x}, y e^{x}>, which matches F. However, the evaluation step '0 + 2*E = 2*E' is nonsensical and does not show the calculation f(1,2,1) - f(-1,1,0). Specifically, f(1,2,1) = 2e and f(-1,1,0) = 0, so the integral is 2e. The line 4 is not a valid derivation.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.