Flux across a plane curve
Problem 12.325 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle x + y, x - y \right\rangle \) 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.
- \[ \frac{d}{d t} 2 \sin{\left(t \right)} \left. x + y \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)} }} - \frac{d}{d t} 2 \cos{\left(t \right)} \left. x - y \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)} }} = 4 \sqrt{2} \sin{\left(2 t + \frac{\pi}{4} \right)} \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{2 \pi} 4 \sqrt{2} \sin{\left(2 t + \frac{\pi}{4} \right)}\, dt = 0 \]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 is incorrect for the outward normal. For a counterclockwise parametrization, the outward normal is <x', y'>, not <y', -x'> (which is the inward normal). This sign error leads to the correct numerical answer by coincidence (since the flux is 0), but the method described is wrong.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The normal vector formula is incorrect for the outward normal. For a counterclockwise parametrization, the outward normal is <x', y'>, not <y', -x'> (which is the inward normal). This sign error leads to the correct numerical answer by coincidence (since the flux is 0), but the method described is wrong.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — 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 change the sign of the integrand (though the final answer of 0 happens to be correct due to symmetry, the method described is wrong).gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The normal vector formula is incorrect for the outward normal. For a counterclockwise parametrization, the outward normal is <x', y'>, not <y', -x'> (which points inward). Using the wrong normal direction yields the negative of the correct flux.gpt-oss:20b: fail (misleading) 2026-10-14 — The sentence claims the outward normal is to the right of the direction of travel, which is incorrect for a counter‑clockwise traversal of a circle. The outward normal should be to the left, so the sign of the flux integral is wrong. Even though the integral evaluates to zero, the reasoning is misleading.
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.