Flux across a plane curve
Problem 12.330 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle x + y, x - y \right\rangle \) across the circle of radius 3, counterclockwise (outward normal).
- With r(t) = <3*cos(t), 3*sin(t)>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ \frac{d}{d t} 3 \sin{\left(t \right)} \left. x + y \right|_{\substack{ x=3 \cos{\left(t \right)}\\ y=3 \sin{\left(t \right)} }} - \frac{d}{d t} 3 \cos{\left(t \right)} \left. x - y \right|_{\substack{ x=3 \cos{\left(t \right)}\\ y=3 \sin{\left(t \right)} }} = 9 \sqrt{2} \sin{\left(2 t + \frac{\pi}{4} \right)} \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{2 \pi} 9 \sqrt{2} \sin{\left(2 t + \frac{\pi}{4} \right)}\, dt = 0 \]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: fail (error) — The normal vector used is incorrect. For a counter‑clockwise traversal the outward normal is the left normal of the tangent, not ⟨y′,−x′⟩. This leads to an incorrect integrand and the conclusion that the flux is zero is false.qwen3.6:27b-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 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 for inward flux).
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 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 for inward flux).gpt-oss:20b: fail (error) 2026-10-17 — The normal vector used is incorrect. For a counter‑clockwise traversal the outward normal is the left normal of the tangent, not ⟨y′,−x′⟩. This leads to an incorrect integrand and the conclusion that the flux is zero is false.qwen3.6:27b-mlx: fail (error) 2026-10-15 — The normal vector formula used, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies an outward normal. The correct outward normal is <-y', x'>, which would yield a non-zero flux (specifically, 18pi).gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The normal vector formula used, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies an outward normal. The correct outward normal is <-y', x'>, which would yield a non-zero flux of 54pi.gpt-oss:20b: fail (misleading) 2026-10-14 — The solution incorrectly identifies the outward normal for a counter‑clockwise circle. For r(t)=<3cos t,3sin t>, the outward normal is <−y′,x′> dt, not <y′,−x′> dt. Although the final integral still evaluates to zero, the reasoning is incorrect and could mislead a student about normal orientation.
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.