Home›Calculus 3›Flux across a plane curve›Problem 12.427

Flux across a plane curve

Problem 12.427 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle x + y, x - y \right\rangle \) across the segment from (0, 0) to (1, 2).

With r(t) = <t, 2*t>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
\[ - \frac{d}{d t} t \left. x - y \right|_{\substack{ x=t\\ y=2 t }} + \frac{d}{d t} 2 t \left. x + y \right|_{\substack{ x=t\\ y=2 t }} = 7 t \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} 7 t\, dt = \frac{7}{2} \]
Integrate.✓ Proved

Answer \( \frac{7}{2} \)

Lines: 2 proved, 1 not checked. 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	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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The formula used, P y' - Q x', calculates the circulation (line integral of F · T ds), not the flux. The flux integral requires F · n ds, which corresponds to P y' - Q x' only if the normal is chosen as <y', -x'>, but the standard flux across a curve from A to B usually implies the normal pointing to the left (<-y', x'>) or requires explicit specification. More critically, the text claims n ds = <y', -x'> is 'to the right', but for r(t) = <t, 2t>, the tangent is <1, 2>. A vector to the right (clockwise normal) would be <2, -1>. The vector <y', -x'> = <2, -1> is indeed to the right. However, the standard definition of flux across a curve usually implies the outward normal for closed curves or a specific orientation. For an open segment, 'flux across' is ambiguous without specifying which side is 'positive'. But the bigger issue is that the standard formula for flux is ∫(P dy - Q dx) or ∫(P y' - Q x') dt if n = <y', -x'>. Wait, let's re-evaluate. F = <P, Q>. n ds = <dy, -dx> = <y', -x'> dt. This normal points to the right of the direction of travel. The problem asks for flux 'across' the segment. Usually, if not specified, one might assume the normal pointing to the left (standard orientation for Green's theorem context) or the problem is ill-posed. However, the solution explicitly states it uses the normal to the right. If the problem implies a standard orientation (left), the sign would be flipped. But often 'flux across' without orientation implies magnitude or a specific convention. Let's look at the result. If we use the left normal <-y', x'> = <-2, 1>, the integrand is P(-y') + Q(x') = (3t)(-2) + (t)(1) = -6t + t = -5t. Integral is -5/2. If we use the right normal, it is 7/2. The solution is internally consistent with its choice of normal. Is the choice of normal justified? The problem does not specify orientation. In many contexts, 'flux across' an open curve is not well-defined without a specified normal direction. However, if we assume the standard convention for line integrals of vector fields, flux is often associated with the normal obtained by rotating the tangent 90 degrees counter-clockwise (left). If so, the answer should be -5/2. If the problem implies the normal pointing into the first quadrant or similar, it's ambiguous. But the most common 'error' in these adversarial reviews is mixing up circulation and flux formulas. Here, the formula P y' - Q x' IS the flux formula for the normal <y', -x'>. So the algebra is correct for that normal. The issue is whether that normal is the 'correct' one. Without specification, assuming the 'right' normal is arbitrary. However, looking at typical textbook problems, if orientation isn't given, they might expect the magnitude or a specific side. But wait, look at the sentence: 'n ds = <y', -x'> dt (to the right of the direction of travel)'. This is a valid choice. Is it the *standard* choice? Usually, the standard normal for flux in Green's theorem context is the outward normal. For an open curve, there is no 'outward'. Often, the normal is taken to be the left normal. If the student assumes the right normal without justification, it's a potential error. But let's look closer. Is there a simpler error? No. The calculation is correct for the stated normal. The ambiguity of the problem statement makes the 'right' normal a guess. However, often 'flux across' implies the normal pointing to the left (standard mathematical orientation). If so, the sign is wrong. Let's assume the standard convention is the left normal. Then the answer is -5/2. The solution gives 7/2. This is a sign error relative to the standard convention.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The formula used, P y' - Q x', calculates the circulation (line integral of F · T ds), not the flux. The flux integral requires F · n ds, which corresponds to P y' - Q x' only if the normal is chosen as <y', -x'>, but the standard flux across a curve from A to B usually implies the normal pointing to the left (<-y', x'>) or requires explicit specification. More critically, the text claims n ds = <y', -x'> is 'to the right', but for r(t) = <t, 2t>, the tangent is <1, 2>. A vector to the right (clockwise normal) would be <2, -1>. The vector <y', -x'> = <2, -1> is indeed to the right. However, the standard definition of flux across a curve usually implies the outward normal for closed curves or a specific orientation. For an open segment, 'flux across' is ambiguous without specifying which side is 'positive'. But the bigger issue is that the standard formula for flux is ∫(P dy - Q dx) or ∫(P y' - Q x') dt if n = <y', -x'>. Wait, let's re-evaluate. F = <P, Q>. n ds = <dy, -dx> = <y', -x'> dt. This normal points to the right of the direction of travel. The problem asks for flux 'across' the segment. Usually, if not specified, one might assume the normal pointing to the left (standard orientation for Green's theorem context) or the problem is ill-posed. However, the solution explicitly states it uses the normal to the right. If the problem implies a standard orientation (left), the sign would be flipped. But often 'flux across' without orientation implies magnitude or a specific convention. Let's look at the result. If we use the left normal <-y', x'> = <-2, 1>, the integrand is P(-y') + Q(x') = (3t)(-2) + (t)(1) = -6t + t = -5t. Integral is -5/2. If we use the right normal, it is 7/2. The solution is internally consistent with its choice of normal. Is the choice of normal justified? The problem does not specify orientation. In many contexts, 'flux across' an open curve is not well-defined without a specified normal direction. However, if we assume the standard convention for line integrals of vector fields, flux is often associated with the normal obtained by rotating the tangent 90 degrees counter-clockwise (left). If so, the answer should be -5/2. If the problem implies the normal pointing into the first quadrant or similar, it's ambiguous. But the most common 'error' in these adversarial reviews is mixing up circulation and flux formulas. Here, the formula P y' - Q x' IS the flux formula for the normal <y', -x'>. So the algebra is correct for that normal. The issue is whether that normal is the 'correct' one. Without specification, assuming the 'right' normal is arbitrary. However, looking at typical textbook problems, if orientation isn't given, they might expect the magnitude or a specific side. But wait, look at the sentence: 'n ds = <y', -x'> dt (to the right of the direction of travel)'. This is a valid choice. Is it the *standard* choice? Usually, the standard normal for flux in Green's theorem context is the outward normal. For an open curve, there is no 'outward'. Often, the normal is taken to be the left normal. If the student assumes the right normal without justification, it's a potential error. But let's look closer. Is there a simpler error? No. The calculation is correct for the stated normal. The ambiguity of the problem statement makes the 'right' normal a guess. However, often 'flux across' implies the normal pointing to the left (standard mathematical orientation). If so, the sign is wrong. Let's assume the standard convention is the left normal. Then the answer is -5/2. The solution gives 7/2. This is a sign error relative to the standard convention.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the formula for flux across a 2D curve segment, applies the parameterization and derivatives correctly, and computes the integral accurately.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: pass 2026-10-15
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.