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Flux across a plane curve

Problem 12.464 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle x, y \right\rangle \) across the segment from (0, 0) to (1, 3).

With r(t) = <t, 3*t>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
Reviewed
\[ - \frac{d}{d t} t \left. y \right|_{\substack{ x=t\\ y=3 t }} + \frac{d}{d t} 3 t \left. x \right|_{\substack{ x=t\\ y=3 t }} = 0 \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} 0\, dt = 0 \]
Integrate.✓ Proved

Answer \( 0 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the normal vector for flux across a 2D curve and sets up the integral properly. The calculation is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the normal vector for flux across a 2D curve and sets up the integral properly. The calculation is correct.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the parameterization, the appropriate normal vector for flux (right-hand normal), and the resulting integrand. The calculation is correct.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The normal vector formula used, <y', -x'>, points to the right of the direction of travel. For a standard flux integral across a curve in the plane, the outward or upward normal is typically required, or the problem must specify orientation. More critically, the vector field F = <x, y> is radial. The segment goes from (0,0) to (1,3). The normal <y', -x'> = <3, -1> is perpendicular to the tangent <1, 3>. The dot product F . n = <t, 3t> . <3, -1> = 3t - 3t = 0. So the calculation is actually correct for that specific normal. However, 'flux across a segment' is ambiguous without specifying which side is 'outward' or 'positive'. Usually, for a closed curve, it's outward. For an open curve, it's often the normal obtained by rotating the tangent 90 degrees counter-clockwise < -y', x' > or clockwise < y', -x' >. If the standard convention is counter-clockwise normal <-3, 1>, the flux is still 0. Wait, let's re-read carefully. The solution claims n ds = <y', -x'> dt. This is a valid normal. The integral is 0. Is the answer 0? Yes, because the vector field is parallel to the position vector, and the normal is perpendicular to the position vector (since the path is a straight line through the origin, the position vector is parallel to the tangent vector, so it is perpendicular to the normal vector). Thus F is perpendicular to n everywhere. Flux is 0. The solution is correct. Why did I think it was wrong? Let me double check the 'right of direction of travel' part. Tangent is <1, 3>. Right normal is <3, -1>. F = <t, 3t>. Dot product is 3t - 3t = 0. The math is correct. The setup is valid for *a* normal. Is it the *standard* normal? Usually, flux is defined with respect to the unit normal. The magnitude of <y', -x'> is sqrt(10). ds = sqrt(10) dt. The unit normal is <3/sqrt(10), -1/sqrt(10)> or <-3/sqrt(10), 1/sqrt(10)>. In either case, the dot product is 0. So the flux is 0 regardless of orientation. The solution is sound.
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.