Home›Calculus 3›Flux across a plane curve›Problem 12.423

Flux across a plane curve

Problem 12.423 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle - y, x \right\rangle \) 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.
\[ - \frac{d}{d t} t \left. x \right|_{\substack{ x=t\\ y=2 t }} + \frac{d}{d t} 2 t \left. - y \right|_{\substack{ x=t\\ y=2 t }} = - 5 t \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} \left(- 5 t\right)\, dt = - \frac{5}{2} \]
Integrate.✓ Proved

Answer \( - \frac{5}{2} \)

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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The formula for flux across a curve in 2D is typically defined with the normal vector pointing to the left (outward for closed curves) or specified by context. The standard flux integral is \int (P dy - Q dx) or \int \mathbf{F} \cdot \mathbf{n} ds. The solution uses \langle y', -x' \rangle which is a normal vector, but the sign convention for 'flux across a segment' usually implies the normal pointing to the left of the direction of travel (\langle -y', x' \rangle) or requires explicit specification. More critically, the standard definition of flux \Phi = \int_C \mathbf{F} \cdot \mathbf{n} ds where \mathbf{n} is the unit normal. If we take the normal to the right \langle y', -x' \rangle / |r'|, then \mathbf{F} \cdot \mathbf{n} ds = \mathbf{F} \cdot \langle y', -x' \rangle dt. Here \mathbf{F} = \langle -y, x \rangle. So \mathbf{F} \cdot \langle y', -x' \rangle = -y y' + x (-x') = -y y' - x x'. The solution calculates P y' - Q x' = (-y)(2) - (x)(1) = -2y - x = -4t - t = -5t. Wait, the solution says P y' - Q x'. Let's check the vector dot product. \mathbf{F} = \langle P, Q \rangle = \langle -y, x \rangle. Normal \mathbf{N} = \langle y', -x' \rangle. Dot product: P y' + Q (-x') = P y' - Q x'. This matches the formula used. However, is the normal \langle y', -x' \rangle the correct one for 'flux'? Usually, flux is defined with the normal pointing to the left, \langle -y', x' \rangle, giving \int (-P y' + Q x') dt. Or it might be defined as \int \mathbf{F} \cdot \mathbf{n} ds where \mathbf{n} is the outward normal. For an open segment, the direction of the normal must be specified. Without specification, 'flux across' is ambiguous. However, looking at the result -5/2. If we used the left normal \langle -y', x' \rangle, the integrand would be P(-y') + Q(x') = -(-y)(2) + (x)(1) = 2y + x = 5t, integral 5/2. The solution gets -5/2. Is there a standard convention? Often, flux is \int P dy - Q dx. This corresponds to the normal \langle dy, -dx \rangle = \langle y', -x' \rangle dt. This is the normal to the *right*. Many textbooks define flux as \int \mathbf{F} \cdot \mathbf{n} ds with \mathbf{n} being the *outward* normal for closed curves, which is left for counter-clockwise. For an open curve, it's ambiguous. But let's look closer. The problem asks for flux. The standard line integral for flux is \int_C \mathbf{F} \cdot \mathbf{n} ds. If no direction is specified, it's often assumed to be the normal pointing to the left of the path (consistent with Green's theorem orientation). If so, the answer should be 5/2. The solution uses the right normal. This is a sign error relative to the standard convention unless 'right' is specified. Given the ambiguity, is it an error? In many contexts (e.g., Stewart Calculus), flux across a curve C is \int_C \mathbf{F} \cdot \mathbf{n} ds where \mathbf{n} is the unit normal vector. If C is oriented, \mathbf{n} is usually the normal obtained by rotating the tangent vector 90 degrees counterclockwise (left). Rotating \langle 1, 2 \rangle 90 deg CCW gives \langle -2, 1 \rangle. The solution uses \langle 2, -1 \rangle (right). Thus, the sign is likely wrong based on standard conventions.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-18 — [domain objection, downgraded to style] The formula for flux across a curve in 2D is typically defined with the normal vector pointing to the left (outward for closed curves) or specified by context. The standard flux integral is \int (P dy - Q dx) or \int \mathbf{F} \cdot \mathbf{n} ds. The solution uses \langle y', -x' \rangle which is a normal vector, but the sign convention for 'flux across a segment' usually implies the normal pointing to the left of the direction of travel (\langle -y', x' \rangle) or requires explicit specification. More critically, the standard definition of flux \Phi = \int_C \mathbf{F} \cdot \mathbf{n} ds where \mathbf{n} is the unit normal. If we take the normal to the right \langle y', -x' \rangle / |r'|, then \mathbf{F} \cdot \mathbf{n} ds = \mathbf{F} \cdot \langle y', -x' \rangle dt. Here \mathbf{F} = \langle -y, x \rangle. So \mathbf{F} \cdot \langle y', -x' \rangle = -y y' + x (-x') = -y y' - x x'. The solution calculates P y' - Q x' = (-y)(2) - (x)(1) = -2y - x = -4t - t = -5t. Wait, the solution says P y' - Q x'. Let's check the vector dot product. \mathbf{F} = \langle P, Q \rangle = \langle -y, x \rangle. Normal \mathbf{N} = \langle y', -x' \rangle. Dot product: P y' + Q (-x') = P y' - Q x'. This matches the formula used. However, is the normal \langle y', -x' \rangle the correct one for 'flux'? Usually, flux is defined with the normal pointing to the left, \langle -y', x' \rangle, giving \int (-P y' + Q x') dt. Or it might be defined as \int \mathbf{F} \cdot \mathbf{n} ds where \mathbf{n} is the outward normal. For an open segment, the direction of the normal must be specified. Without specification, 'flux across' is ambiguous. However, looking at the result -5/2. If we used the left normal \langle -y', x' \rangle, the integrand would be P(-y') + Q(x') = -(-y)(2) + (x)(1) = 2y + x = 5t, integral 5/2. The solution gets -5/2. Is there a standard convention? Often, flux is \int P dy - Q dx. This corresponds to the normal \langle dy, -dx \rangle = \langle y', -x' \rangle dt. This is the normal to the *right*. Many textbooks define flux as \int \mathbf{F} \cdot \mathbf{n} ds with \mathbf{n} being the *outward* normal for closed curves, which is left for counter-clockwise. For an open curve, it's ambiguous. But let's look closer. The problem asks for flux. The standard line integral for flux is \int_C \mathbf{F} \cdot \mathbf{n} ds. If no direction is specified, it's often assumed to be the normal pointing to the left of the path (consistent with Green's theorem orientation). If so, the answer should be 5/2. The solution uses the right normal. This is a sign error relative to the standard convention unless 'right' is specified. Given the ambiguity, is it an error? In many contexts (e.g., Stewart Calculus), flux across a curve C is \int_C \mathbf{F} \cdot \mathbf{n} ds where \mathbf{n} is the unit normal vector. If C is oriented, \mathbf{n} is usually the normal obtained by rotating the tangent vector 90 degrees counterclockwise (left). Rotating \langle 1, 2 \rangle 90 deg CCW gives \langle -2, 1 \rangle. The solution uses \langle 2, -1 \rangle (right). Thus, the sign is likely wrong based on standard conventions.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The formula for flux across a curve in 2D is typically defined as the integral of F · n ds, where n is the outward or specified normal. The standard
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The normal vector 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 specified normal must be consistent with the problem context. However, the primary error is in the formula application: the flux of F = <P, Q> across a curve is ∫ F · n ds. If n ds = <dy, -dx>, then F · n ds = P dy - Q dx. The solution uses P y' - Q x', which corresponds to P dy - Q dx. Let's check the sign. r(t) = <t, 2t>, so x'=1, y'=2. n = <2, -1>. F = <-2t, t>. F · n = -4t - t = -5t. The integral is -5/2. Wait, is the normal direction specified? 'Flux across the segment' usually implies a specific orientation or requires specifying the normal. If no normal is specified, the sign is ambiguous. However, the sentence claims n ds = <y', -x'> is 'to the right'. Let's check: tangent T = <1, 2>. Right normal is <2, -1> (rotate T 90 deg clockwise: <y, -x>). Yes, <2, -1> is to the right. So the calculation is consistent with the 'right' normal. Is 'right' the standard convention? Usually, for a closed curve, it's outward. For an open curve, it's ambiguous. But the problem asks for 'the flux', implying a specific answer. Often, if not specified, one might assume the normal pointing in a certain way, or the problem implies a boundary of a region. Without further context, assuming 'right' is arbitrary. However, looking at the stated answer -5/2, it matches the calculation. Is there a more standard interpretation? If we used the left normal <-2, 1>, the flux would be 5/2. Many textbooks define flux as ∫ F · n ds where n is the unit normal. If the problem doesn't specify the normal, the answer is ambiguous. But often in such online problems, 'flux across' might imply the normal obtained by rotating the tangent 90 degrees counter-clockwise (left) or clockwise (right). Let's look at the formula again. Flux = ∫ M dy - N dx. Here P=-y, Q=x. Flux = ∫ -y dy - x dx? No, F=<P,Q>. Flux = ∫ P dy - Q dx. P=-y, Q=x. ∫ -y dy - x dx. y=2t, dy=2dt. x=t, dx=dt. ∫_0^1 (-2t)(2) - (t)(1) dt = ∫ (-4t - t) dt = -5/2. This matches. The sentence says n ds = <y', -x'>. This is the normal corresponding to the form P dy - Q dx. This is the normal pointing to the RIGHT of the tangent vector <x', y'>. Is this the standard 'flux'? Usually, for a curve oriented from A to B, the 'positive' normal is often taken to be the one pointing to the LEFT (outward for a region on the left). If the standard convention is the LEFT normal, the answer should be 5/2. The solution arbitrarily chooses the RIGHT normal without justification, leading to a sign error relative to the standard convention (where flux is often defined with the normal pointing to the left of the direction of traversal, i.e., n = <-y', x'>). If the standard answer is -5/2, then the 'right' normal was intended. But typically, 'flux across' without qualification is ambiguous. However, checking common conventions: Green's theorem relates ∮ P dx + Q dy to double integral of (Q_x - P_y). Flux form is ∮ P dy - Q dx. This corresponds to the normal n = <dy, -dx>/ds = <y', -x'>/|r'|. This is the RIGHT normal. So the calculation is internally consistent with the 'right' normal. Is the 'right' normal the standard? In many contexts, the normal is chosen to be outward. For a segment, there is no 'outward'. If this is part of a boundary, say, of a region below the segment, the outward normal might be down/right. If above, up/left. Without context, the sign is a guess. However, the prompt asks to review the solution. The solution states 'n ds = <y', -x'> (to the right...)'. It explicitly defines its normal. The calculation follows. The result matches the stated answer. Is it 'wrong'? If the problem implies a standard orientation (e.g., counter-clockwise around a region containing the segment), the normal might be different. But for a single segment, 'flux' is not well-defined without specifying the normal direction. The solution *chooses* a direction and calculates it. Is this an error? It's a modeling choice. If the stated answer is -5/2, then the 'right' normal is the intended one. The solution is consistent with the answer. I will mark it pass because it is internally consistent and matches the stated answer, assuming the 'right' normal is the intended convention for this specific problem source.
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.