Flux across a plane curve
Problem 12.327 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle - y, x \right\rangle \) across the circle of radius 2, counterclockwise (outward normal).
- With r(t) = <2*cos(t), 2*sin(t)>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ \frac{d}{d t} 2 \sin{\left(t \right)} \left. - y \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)} }} - \frac{d}{d t} 2 \cos{\left(t \right)} \left. x \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)} }} = 0 \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{2 \pi} 0\, dt = 0 \]Integrate.✓ Proved
Answer \( 0 \)
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: fail (error) — The normal vector formula used, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies the outward normal. The correct outward normal vector is <-y', x'>, which would yield a non-zero flux of 8π.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The normal vector formula used, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies the outward normal. The correct outward normal vector is <-y', x'>, which would yield a non-zero flux of 8π.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — The formula for the normal vector component is incorrect. For a counterclockwise parametrization, the outward normal vector scaled by arc length is <x', y'>, not <y', -x'>. The latter corresponds to the inward normal (or the right-hand normal for CCW travel, which points inward for a circle). Consequently, the flux integral should be P x' + Q y', not P y' - Q x'. Although the final answer happens to be 0 for this specific field and region (by symmetry or Divergence Theorem), the method described is fundamentally wrong and would yield incorrect results for other fields (e.g., F = <x, y>).gpt-oss:20b: fail (error) 2026-10-15 — The normal vector used is incorrect: for a counter‑clockwise traversal the outward normal points to the left of the direction of travel, not to the right. Consequently the flux integral is not computed with the correct orientation, leading to an incorrect conclusion.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The normal vector formula used, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies the outward normal. The correct outward normal is <-y', x'>, which would yield a non-zero flux (8π).gpt-oss:20b: fail (misleading) 2026-10-14 — The sentence claims that the outward normal for a counter‑clockwise traversal is \(\langle y',-x'\rangle\), which is actually the normal to the right of the tangent. For a circle the outward normal is to the left of the tangent, i.e. \(\langle -y',x'\rangle\). The sign error is not detected because the integrand happens to be zero, but the reasoning is incorrect and could mislead a student.
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.