Flux across a plane curve
Problem 12.462 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle x^{2}, y \right\rangle \) across the circle of radius 1, counterclockwise (outward normal).
- With r(t) = <cos(t), sin(t)>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ \frac{d}{d t} \sin{\left(t \right)} \left. x^{2} \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)} }} - \frac{d}{d t} \cos{\left(t \right)} \left. y \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)} }} = \sin^{2}{\left(t \right)} + \cos^{3}{\left(t \right)} \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{2 \pi} \left(\sin^{2}{\left(t \right)} + \cos^{3}{\left(t \right)}\right)\, dt = \pi \]Integrate.✓ Proved
Answer \( \pi \)
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: fail (error) — The normal vector is taken to the right of the direction of travel, which for a counter‑clockwise parametrization points inward, not outward. Thus the flux sign is incorrect.qwen3.6:27b-mlx: fail (error) — The normal vector formula is incorrect for the specified outward normal. For a counterclockwise parametrization, the outward normal is <x', y'>, not <y', -x'> (which points inward). Consequently, the sign of the flux is reversed.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-19 — The normal vector formula is incorrect for the specified outward normal. For a counterclockwise parametrization, the outward normal is <x', y'>, not <y', -x'> (which points inward). Consequently, the sign of the flux is reversed.gpt-oss:20b: fail (error) 2026-10-19 — The normal vector is taken to the right of the direction of travel, which for a counter‑clockwise parametrization points inward, not outward. Thus the flux sign is incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-16 — The normal vector formula used, <y', -x'>, corresponds to the inward normal for a counterclockwise traversal, whereas the problem specifies the outward normal. The correct outward normal vector is <-y', x'>, which would change the sign of the flux integral.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-16 — The normal vector formula provided, <y', -x'>, points inward for a counterclockwise traversal, whereas the problem requires the outward normal. The correct outward normal is <-y', x'>, which would change the sign of the flux integral.gpt-oss:20b: pass 2026-10-16
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 2026-10-19 with SymPy 1.14.0.