Conservative fields and potential functions
Problem 12.282 · medium
Show that \( \displaystyle \mathbf F = \left(3 x^{2} + y^{2}\right)\mathbf i + \left(2 x y\right)\mathbf j \) is conservative, find a potential function \( \displaystyle f \), and evaluate \( \displaystyle \int_C \mathbf F \cdot d\mathbf r \) along any path from \( \displaystyle (0, 0) \) to \( \displaystyle (0, 2) \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} 2 x y\\\frac{\partial}{\partial y} \left(3 x^{2} + y^{2}\right)\end{matrix}\right] = \left[\begin{matrix}2 y\\2 y\end{matrix}\right] \]∂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.
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{3} + x y^{2}\right)\\\frac{\partial}{\partial y} \left(x^{3} + x y^{2}\right)\end{matrix}\right] = \left[\begin{matrix}3 x^{2} + y^{2}\\2 x y\end{matrix}\right] \]f = x**3 + x*y**2 has gradient F.✓ Proved
- \[ 0 + 0 = 0 \]∫_C F·dr = f(end) − f(start), for any path.✓ Proved
Answer \( f = x^{3} + x y^{2} + C,\quad \int_C \mathbf F\cdot d\mathbf r = 0 \)
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 integral is 0, but f(0,2) - f(0,0) = 0 - 0 = 0 is correct only if the potential is evaluated correctly. Wait, f(0,2) = 0^3 + 0*2^2 = 0. f(0,0) = 0. So the integral is indeed 0. However, the sentence in step 4 is misleadingly vague: '0 + 0 = 0' does not show the evaluation f(end) - f(start). It just shows arithmetic. More importantly, step 1 checks ∂Q/∂x = ∂P/∂y but labels it '∂Q/∂x = ∂P/∂y: F is conservative'. Usually P is i-component and Q is j-component. ∂P/∂y = 2y, ∂Q/∂x = 2y. The check is correct. But the final answer calculation is hidden inside an equation '0+0=0' which is not a proper derivation. The main issue is that the solution does not explicitly state f(0,2) and f(0,0). It just asserts the result. Is this an error? The prompt asks to 'evaluate'. The solution provides the answer but the 'work' is just '0+0=0'. This is poor style but is it an error? Let's look closer. Step 4 says '∫_C F·dr = f(end) − f(start), for any path.' and then shows '0 + 0 = 0'. This implies f(end) - f(start) = 0 - 0 = 0. The logic is sound, but the presentation is terrible. However, looking at the provided answer: f = x^3 + xy^2 + C. The integral is 0. The solution is technically correct in its conclusion. But is there a subtle error? No. Is it misleading? A student might think '0+0=0' is the method. It's not. It's just the arithmetic. I will mark it as misleading because it fails to demonstrate the evaluation step clearly, potentially teaching that one can just write '0+0=0' as the justification for the Fundamental Theorem of Line Integrals.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution claims the integral is 0, but f(0,2) - f(0,0) = 0 - 0 = 0 is correct only if the potential is evaluated correctly. Wait, f(0,2) = 0^3 + 0*2^2 = 0. f(0,0) = 0. So the integral is indeed 0. However, the sentence in step 4 is misleadingly vague: '0 + 0 = 0' does not show the evaluation f(end) - f(start). It just shows arithmetic. More importantly, step 1 checks ∂Q/∂x = ∂P/∂y but labels it '∂Q/∂x = ∂P/∂y: F is conservative'. Usually P is i-component and Q is j-component. ∂P/∂y = 2y, ∂Q/∂x = 2y. The check is correct. But the final answer calculation is hidden inside an equation '0+0=0' which is not a proper derivation. The main issue is that the solution does not explicitly state f(0,2) and f(0,0). It just asserts the result. Is this an error? The prompt asks to 'evaluate'. The solution provides the answer but the 'work' is just '0+0=0'. This is poor style but is it an error? Let's look closer. Step 4 says '∫_C F·dr = f(end) − f(start), for any path.' and then shows '0 + 0 = 0'. This implies f(end) - f(start) = 0 - 0 = 0. The logic is sound, but the presentation is terrible. However, looking at the provided answer: f = x^3 + xy^2 + C. The integral is 0. The solution is technically correct in its conclusion. But is there a subtle error? No. Is it misleading? A student might think '0+0=0' is the method. It's not. It's just the arithmetic. I will mark it as misleading because it fails to demonstrate the evaluation step clearly, potentially teaching that one can just write '0+0=0' as the justification for the Fundamental Theorem of Line Integrals.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly verifies the conservative nature of the field, finds the potential function, and applies the Fundamental Theorem of Line Integrals to evaluate the integral.gpt-oss:20b: pass 2026-10-11
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 2026-10-11 with SymPy 1.14.0.