Flux across a plane curve
Problem 12.332 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle x^{2}, y \right\rangle \) across the parabola y = x² from (0, 0) to (1, 1).
- With r(t) = <t, t**2>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ - \frac{d}{d t} t \left. y \right|_{\substack{ x=t\\ y=t^{2} }} + \frac{d}{d t} t^{2} \left. x^{2} \right|_{\substack{ x=t\\ y=t^{2} }} = t^{2} \left(2 t - 1\right) \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{1} t^{2} \left(2 t - 1\right)\, dt = \frac{1}{6} \]Integrate.✓ Proved
Answer \( \frac{1}{6} \)
Lines: 2 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. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 | F · n summed over a 20,000-segment polyline |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The normal vector formula used, <y', -x'>, corresponds to the outward normal for a counter-clockwise traversal of a closed curve (or 'right' of trave
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The normal vector formula used, <y', -x'>, corresponds to the outward normal for a counter-clockwise traversal of a closed curve (or 'right' of travegpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the formula for flux across a plane curve and applies it with the correct parameterization and bounds.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The normal vector formula used, <y', -x'>, points to the right of the direction of travel. For the curve y=x^2 from (0,0) to (1,1), 'right' correspongpt-oss:20b: pass 2026-10-14
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/flux_plane, checked 2026-10-17 with SymPy 1.14.0.