Home›Calculus 3›Flux across a plane curve›Problem 12.463

Flux across a plane curve

Problem 12.463 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle x^{2}, y \right\rangle \) 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
\[ - \frac{d}{d t} t \left. y \right|_{\substack{ x=t\\ y=3 t }} + \frac{d}{d t} 3 t \left. x^{2} \right|_{\substack{ x=t\\ y=3 t }} = 3 t \left(t - 1\right) \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} 3 t \left(t - 1\right)\, dt = - \frac{1}{2} \]
Integrate.✓ Proved

Answer \( - \frac{1}{2} \)

Lines: 2 proved, 1 reviewed. 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	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

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: fail (style) 2026-10-16 — [domain objection, downgraded to style] The normal vector used, <y', -x'>, points to the right of the direction of travel. For a curve in the first quadrant going from (0,0) to (1,3), 'right' points into the fourth quadrant (away from the origin), which is not the standard convention for 'flux across a segment' unless specified. Standard flux usually implies the normal pointing to the left (or outward from a region if part of a boundary). More critically, the standard formula for flux is \int (P dy - Q dx) or \int \mathbf{F} \cdot \mathbf{n} ds. The vector <y', -x'> is indeed a normal, but one must ensure the orientation matches the problem's implicit convention. However, the bigger issue is that 'flux across a segment' is ambiguous without a specified normal direction. If we assume the standard 'left' normal <-y', x'>, the sign would be flipped. But typically, for an open curve, 'flux' isn't well-defined without a specified side. Let's look at the calculation. \mathbf{F} = <x^2, y>. r(t) = <t, 3t>. r'(t) = <1, 3>. Tangent T = <1, 3>. Left normal N = <-3, 1>. Right normal N = <3, -1>. The solution uses <y', -x'> = <3, -1>, which is the right normal. The integral is \int_0^1 (<t^2, 3t> \cdot <3, -1>) dt = \int_0^1 (3t^2 - 3t) dt = [t^3 - 1.5t^2]_0^1 = 1 - 1.5 = -0.5. The calculation is consistent with the chosen normal. Is the choice of normal justified? The problem does not specify 'right' or 'left'. In many contexts, if no orientation is given for the normal, the problem is ill-posed, or one assumes the normal pointing 'outward' from a region, but this is just a segment. However, often in 2D line integrals, if not specified, one might assume the normal that points to the 'left' of the path (standard orientation for Green's theorem boundaries). If the standard convention is the left normal, the answer should be +1/2. The solution arbitrarily chooses the right normal without justification, leading to a sign error relative to standard conventions where 'flux' often implies the left-hand normal or outward normal from a bounded region (if this segment were part of a boundary). Given the ambiguity, stating the normal is 'to the right' is a specific choice that might be wrong if the standard is 'left'. Let's check if there's a more fundamental error. The formula \int (P dy - Q dx) corresponds to the normal <-dy, dx> normalized? No. \mathbf{F} \cdot \mathbf{n} ds. If \mathbf{n} ds = <dy, -dx>, then \mathbf{F} \cdot <dy, -dx> = P dy - Q dx. This normal <dy, -dx> corresponds to <y', -x'> dt. This is the right normal. The standard flux integral in Green's theorem context is \oint P dy - Q dx, which uses the outward normal. For a counter-clockwise loop, outward is to the right of the tangent? No, outward is to the right of the tangent if you are walking along the boundary? Let's visualize. Circle CCW. Tangent is CCW. Outward is Radial. Right of Tangent is Outward. Yes. So for a CCW boundary, the 'right' normal is the outward normal. If this segment is part of a boundary enclosing a region to its left, the outward normal is to the right. But we don't know the region. Without context, the sign is ambiguous. However, usually, if a problem asks for 'flux across a curve', and doesn't specify, it's often considered ill-posed or assumes a specific convention. But wait, looking at similar problems online, often the 'flux' is defined with the normal pointing to the left (standard mathematical orientation for line integrals of vector fields, analogous to work). Or, it might be asking for the net flow. Let's look at the result. -1/2. If the answer key says -1/2, then the 'right' normal was intended. If the answer key says 1/2, then 'left'. The prompt says 'Stated answer: -1/2'. So the solution matches the stated answer. Is the reasoning sound? It explicitly states 'to the right'. If the problem implies a standard orientation (left), then it's an error. If the problem is from a context where 'right' is standard (e.g. some engineering texts?), it might be okay. But in standard calculus, the 'flux' form \int P dy - Q dx is associated with the normal <-Q, P>? No. \int \mathbf{F} \cdot \mathbf{n} ds. Standard normal for \int P dy - Q dx is <-dy, dx>? No. \int P dy - Q dx = \int <P, -Q> \cdot <dx, dy>? No. \int P dy - Q dx = \int <P, -Q> \cdot <dy, dx>? No. Let's stick to vectors. \mathbf{n} ds = <dy, -dx>. This is the right normal. The standard 'flux' integral in Green's Theorem is \oint_C \mathbf{F} \cdot \mathbf{n} ds = \iint_D div \mathbf{F} dA. Here \mathbf{n} is the outward normal. For a CCW curve, outward is to the right. So if this segment is part of a CCW boundary, the right normal is correct. But it's just a segment. The ambiguity is the main issue. However, the solution *states* its choice. Is it wrong? It's not 'wrong' to choose a normal, but is it the *correct* normal for the problem? Without further context, we can't be sure. But often, 'flux across a curve' without qualification is ambiguous. Let's look for a clearer error. Ah, the normal vector magnitude. \mathbf{n} ds = <y', -x'> dt. The magnitude of <y', -x'> is \sqrt{y'^2 + x'^2} = ds/dt. So \mathbf{n} = <y', -x'> / \sqrt{...}. Then \mathbf{F} \cdot \mathbf{n} ds = \mathbf{F} \cdot <y', -x'> dt. This is correct. The setup is mathematically consistent with the stated normal. The only potential failure is if the 'standard' normal is the left one. In many textbooks (e.g., Stewart), flux is defined with the outward normal for closed curves. For open curves, it's often not defined or specified. If forced to choose, the 'left' normal is the standard orientation for line integrals (like work). If the problem implies the standard orientation, the sign is flipped. Given the stated answer is -1/2, and the solution derives -1/2 using the right normal, the solution is internally consistent. Is it 'misleading'? It teaches that you can pick the right normal. If the standard is left, it's an error. Let's assume the standard convention for 'flux across a curve' in the absence of a region is the normal pointing to the left of the direction of travel (consistent with the orientation of the plane). In that case, the normal should be <-y', x'> = <-3, 1>. The integral would be \int_0^1 (<t^2, 3t> \cdot <-3, 1>) dt = \int_0^1 (-3t^2 + 3t) dt = [-t^3 + 1.5t^2]_0^1 = -1 + 1.5 = 0.5. The stated answer is -1/2. This suggests the problem *did* intend the right normal, or the stated answer is based on the right normal. If the stated answer is correct, the solution is correct. If the stated answer is just a target, and the convention is left, then the solution is wrong. Usually, these problems come from a context where the normal is specified or implied. Without specification, the solution's explicit choice is a valid interpretation, but potentially non-standard. However, there is a more subtle point. Is the normal <y', -x'> actually the normal? Yes. Is it unit? No, but it's multiplied by dt, so it represents \mathbf{n} ds. This is correct. I will mark it as 'pass' because the solution explicitly states the convention it is using ('to the right'), and the calculation follows. The ambiguity of the problem statement is not the solution's fault, and it clarifies its assumption. Wait, if the problem is from a standard calculus course, 'flux' often defaults to the normal pointing to the *left* (or outward from the region to the left). If I mark it pass, I accept 'right' as a valid default. It is not. The standard default for orientation is counter-clockwise/left. Using the right normal without justification is likely an error in convention. Let's look at the severity. If a student learns 'flux uses the right normal', they will get signs wrong on standard problems. Therefore, it is misleading/error. I will fail it for using the non-standard normal direction without justification, leading to a sign difference from the standard convention.
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The normal vector used, <y', -x'>, points to the right of the direction of travel. For a curve in the xy-plane, the standard outward normal for flux
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.