Conservative fields and potential functions
Problem 12.165 · medium
Show that \( \displaystyle \mathbf F = \left(y e^{x y}\right)\mathbf i + \left(x e^{x y}\right)\mathbf j \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (0, 2) \) to \( \displaystyle (1, 2) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} x e^{x y}\\\frac{\partial}{\partial y} y e^{x y}\end{matrix}\right] = \left[\begin{matrix}x y e^{x y} + e^{x y}\\x y e^{x y} + e^{x y}\end{matrix}\right] \]∂Q/∂x = ∂P/∂y: 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.Reviewed
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} e^{x y}\\\frac{\partial}{\partial y} e^{x y}\end{matrix}\right] = \left[\begin{matrix}y e^{x y}\\x e^{x y}\end{matrix}\right] \]f = exp(x*y) has gradient F.✓ Proved
- \[ -1 + e^{2} \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = e^{x y} + C,\quad \int_C \mathbf F\cdot d\mathbf r = -1 + e^{2} \)
✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the potential function and applies the Fundamental Theorem of Line Integrals. The setup and calculations are sound.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the potential function and applies the Fundamental Theorem of Line Integrals. The setup and calculations are sound.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the vector field as conservative, finds the potential function, and applies the Fundamental Theorem of Line Integrals correctly.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.