Conservative fields and potential functions
Problem 12.251 · 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 \( \displaystyle (2, 2, 0) \) to \( \displaystyle (1, 2, 2) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} z e^{x}\\\frac{\partial}{\partial y} y z e^{x}\end{matrix}\right] = \left[\begin{matrix}z e^{x}\\z e^{x}\end{matrix}\right] \]∂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
- \[ 0 + 4 e = 4 e \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = y z e^{x} + C,\quad \int_C \mathbf F\cdot d\mathbf r = 4 e \)
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 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 is incorrect: f(1, 2, 2) = 2*2*e^1 = 4e and f(2, 2, 0) = 2*0*e^2 = 0, so the integral is 4e - 0 = 4e. Wait, the equation says '0 + 4*E = 4*E'. This implies f(start)=0 and f(end)=4e. Let's re-read carefully. f(2,2,0) = 2*0*e^2 = 0. f(1,2,2) = 2*2*e^1 = 4e. The result 4e is correct. The equation line 4 is '0 + 4*E = 4*E'. This is a tautology that doesn't show the subtraction f(end) - f(start). It looks like it's just stating the final answer. But is there an error? The prompt asks to find f and evaluate. The solution provides f and the value. The equation in line 4 is weirdly formatted but mathematically consistent with the result. Let's look closer at line 1. It checks partials. P=yze^x, Q=ze^x, R=ye^x. dQ/dx = ze^x. dP/dy = ze^x. Match. dR/dx = ye^x. dP/dz = ye^x. Match. dR/dy = e^x. dQ/dz = e^x. Match. Conservative. Correct. Line 3 confirms gradient. Correct. Line 4 evaluates. f(1,2,2) - f(2,2,0) = 4e - 0 = 4e. The equation '0 + 4*E = 4*E' is a very poor representation of 'f(end) - f(start) = 4e - 0 = 4e'. It might be interpreted as '0 (start) + 4e (end) = 4e', which is wrong logic (it's subtraction). Or it might be '4e - 0 = 4e' but written as '0 + 4e' by mistake? Actually, usually these systems output 'f(end) - f(start)'. If it outputs '0 + 4*E', it suggests the solver added them or messed up the sign. However, the final answer 4e is correct. Is it misleading? Yes, because the arithmetic operation shown is addition (or identity) rather than the difference of potentials. A student might think the integral is f(start) + f(end).
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: fail (error) 2026-10-09 — 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 is incorrect: f(1, 2, 2) = 2*2*e^1 = 4e and f(2, 2, 0) = 2*0*e^2 = 0, so the integral is 4e - 0 = 4e. Wait, the equation says '0 + 4*E = 4*E'. This implies f(start)=0 and f(end)=4e. Let's re-read carefully. f(2,2,0) = 2*0*e^2 = 0. f(1,2,2) = 2*2*e^1 = 4e. The result 4e is correct. The equation line 4 is '0 + 4*E = 4*E'. This is a tautology that doesn't show the subtraction f(end) - f(start). It looks like it's just stating the final answer. But is there an error? The prompt asks to find f and evaluate. The solution provides f and the value. The equation in line 4 is weirdly formatted but mathematically consistent with the result. Let's look closer at line 1. It checks partials. P=yze^x, Q=ze^x, R=ye^x. dQ/dx = ze^x. dP/dy = ze^x. Match. dR/dx = ye^x. dP/dz = ye^x. Match. dR/dy = e^x. dQ/dz = e^x. Match. Conservative. Correct. Line 3 confirms gradient. Correct. Line 4 evaluates. f(1,2,2) - f(2,2,0) = 4e - 0 = 4e. The equation '0 + 4*E = 4*E' is a very poor representation of 'f(end) - f(start) = 4e - 0 = 4e'. It might be interpreted as '0 (start) + 4e (end) = 4e', which is wrong logic (it's subtraction). Or it might be '4e - 0 = 4e' but written as '0 + 4e' by mistake? Actually, usually these systems output 'f(end) - f(start)'. If it outputs '0 + 4*E', it suggests the solver added them or messed up the sign. However, the final answer 4e is correct. Is it misleading? Yes, because the arithmetic operation shown is addition (or identity) rather than the difference of potentials. A student might think the integral is f(start) + f(end).gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution fails to explicitly calculate the potential function values at the start and end points, making the final numerical result appear without derivation. Additionally, the check for conservativeness is incomplete as it only shows one pair of partial derivatives.
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.