Flux across a plane curve
Problem 12.425 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle - y, x \right\rangle \) across the segment from (0, 0) to (1, 1).
- With r(t) = <t, t>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ - \frac{d}{d t} t \left. x \right|_{\substack{ x=t\\ y=t }} + \frac{d}{d t} t \left. - y \right|_{\substack{ x=t\\ y=t }} = - 2 t \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{1} \left(- 2 t\right)\, dt = -1 \]Integrate.✓ Proved
Answer \( -1 \)
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 used, <y', -x'>, points to the left of the direction of travel, not the right. For r(t)=<t,t>, the tangent is <1,1> and <1,-1> points to the left (dot product with <1,1> is 0, and cross product k-component is negative). Consequently, the sign of the flux is incorrect.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The normal vector used, <y', -x'>, points to the left of the direction of travel, not the right. For r(t)=<t,t>, the tangent is <1,1> and <1,-1> points to the left (dot product with <1,1> is 0, and cross product k-component is negative). Consequently, the sign of the flux is incorrect.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The formula for flux across a curve in the plane is typically defined with the normal vector pointing to the left (outward for closed curves), given by <x', y'> or <-y', x'> depending on convention, but the solution uses <y', -x'> which points to the right. More critically, the standard flux integral is \int (P dy - Q dx) or \int (-Q dx + P dy). The solution computes \int (P y' - Q x') dt. Let's check the vector field F = <-y, x>. P = -y, Q = x. The normal vector n ds for a curve r(t) = <x(t), y(t)> is usually <y', -x'> dt for the right-hand normal or <-y', x'> dt for the left-hand normal. The flux is \int F \cdot n ds. If n = <y', -x'>, then F \cdot n = (-y)(y') + (x)(-x') = -y y' - x x'. The solution calculates P y' - Q x' = (-y)y' - (x)x' = -y y' - x x'. This matches the dot product with the right-hand normal. However, standard convention for "flux across a segment" often implies a specific orientation or simply the net flow. If no orientation is specified, the sign is ambiguous, but usually, flux is defined relative to the left normal (standard outward for counter-clockwise). Using the right normal gives the negative of the standard flux. The problem asks for "the flux", implying a standard convention. The standard convention for flux \iint div F dA or \oint F \cdot n ds uses the outward normal. For an open curve, it's ambiguous, but typically one uses the normal obtained by rotating the tangent vector 90 degrees counter-clockwise (left normal). The left normal is <-y', x'>. F \cdot n_left = (-y)(-y') + (x)(x') = y y' + x x'. With x=t, y=t, this is t + t = 2t. Integral is 1. The solution uses the right normal and gets -1. While mathematically consistent with its own choice, it likely violates the standard convention expected in calculus courses unless "right" is specified. However, the bigger issue is that the problem doesn't specify the side. But wait, is the formula P y' - Q x' correct for the right normal? F = <P, Q>. n_right = <y', -x'>. F \cdot n_right = P y' - Q x'. Yes. Is the standard flux defined with the left normal? Yes, usually. So the sign is likely wrong relative to standard convention. But is it an error? If the problem doesn't specify, it's ambiguous. However, looking at the vector field <-y, x>, this is a rotation field. The flow is counter-clockwise. The segment is from (0,0) to (1,1). The vector field at (0.5, 0.5) is <-0.5, 0.5>. The tangent is <1, 1>. The left normal is <-1, 1>. The dot product is 0.5 + 0.5 = 1. The right normal is <1, -1>. The dot product is -0.5 - 0.5 = -1. The solution gets -1. If the standard answer is 1, this is an error. Given the stated answer is -1, the solution matches the stated answer. But is the stated answer correct? Usually, flux is not defined for open segments without a specified normal direction. However, in many contexts, if not specified, one might assume the normal pointing to the "left" of the direction of travel. If so, the answer should be 1. If the answer key says -1, it might be using the right normal. But without specification, assuming the right normal is non-standard. Let's look closer. Is there a convention? In Green's theorem, the boundary is traversed counter-clockwise, and the normal is outward (left). For an open curve, there is no universal standard. However, most textbooks define the flux integral \int_C F \cdot n ds where n is the unit normal. If the curve is parameterized, n ds = <-y', x'> dt is the standard choice corresponding to the outward normal of a region to the left. Using <y', -x'> corresponds to the inward normal. It is highly likely that the standard expectation is the left normal, yielding 1. The solution yields -1. This is a sign error relative to standard convention.gpt-oss:20b: fail (error) 2026-10-16 — The integrand is computed incorrectly. For r(t)=⟨t,t⟩, r′=⟨1,1⟩, so n ds=⟨1,−1⟩dt. Then F·n=⟨−y,x⟩·⟨1,−1⟩=−y+x, which with y=x=t gives 0, not −2t. Thus the flux is 0, not −1.qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The formula for flux across a curve in the plane is typically defined using the outward or specified normal. For a curve parameterized by r(t) = <x(gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.