Flux across a plane curve
Problem 12.337 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle x + y, x - y \right\rangle \) across the parabola y = x² from (0, 0) to (1, 1).
- With r(t) = <t, t**2>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ - \frac{d}{d t} t \left. x - y \right|_{\substack{ x=t\\ y=t^{2} }} + \frac{d}{d t} t^{2} \left. x + y \right|_{\substack{ x=t\\ y=t^{2} }} = t \left(2 t \left(t + 1\right) + t - 1\right) \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{1} t \left(2 t \left(t + 1\right) + t - 1\right)\, dt = 1 \]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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The normal vector formula is incorrect for the standard flux definition. The outward/right normal to the tangent vector <x', y'> is <y', -x'>, so the
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The normal vector formula is incorrect for the standard flux definition. The outward/right normal to the tangent vector <x', y'> is <y', -x'>, so thegpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the formula for flux across a parametric curve and applies it with the correct bounds and direction. The algebraic simplification and integration are verified as correct.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The normal vector formula used, <y', -x'>, points to the right of the direction of travel. For a curve moving from (0,0) to (1,1) in the first quadrant, 'right' points into the fourth quadrant (downwards/outwards relative to the parabola's concavity), whereas standard flux problems typically imply the normal pointing towards the 'inside' or consistent with the standard outward normal for a closed region, or simply require specifying the orientation. More critically, the problem does not specify the orientation of the normal (left/right or inward/outward). However, the primary error is likely in the interpretation of 'flux across a curve' without a specified normal direction. Usually, for an open curve, one must specify the normal. If we assume the standard convention where the normal is obtained by rotating the tangent vector 90 degrees counter-clockwise (pointing 'left'), the normal would be <-y', x'>. The solution uses <y', -x'> (pointing 'right'). Without a specified direction in the problem statement, the sign is ambiguous, but typically 'flux across' implies a specific orientation or the problem is ill-posed. However, looking at the calculation: P=y'=2t, Q=x'=1. Flux = int (P dy - Q dx) = int ( (x+y)2t - (x-y)1 ) dt. The solution calculates int ( (t+t^2)2t - (t-t^2) ) dt = int ( 2t^2 + 2t^3 - t + t^2 ) dt = int ( 2t^3 + 3t^2 - t ) dt. The solution's integrand is t*(2*t*(t+1) + t - 1) = t*(2t^2+2t+t-1) = 2t^3+3t^2-t. This matches. The integral is [t^4/2 + t^3 - t^2/2] from 0 to 1 = 1/2 + 1 - 1/2 = 1. The calculation is correct for the normal <y', -x'>. The issue is whether this normal is the 'correct' one. In many contexts, the normal is taken to be the one pointing to the left of the tangent vector (standard orientation for Green's theorem boundary). If the problem implies the standard 'outward' normal for the region bounded by the parabola and y=x, the direction matters. But for an open curve, 'flux' is ambiguous without a specified normal. However, often 'flux across' implies the normal obtained by rotating the tangent vector 90 degrees counter-clockwise, i.e., <-y', x'>. If so, the sign should be negative. Let's check the wording. 'Flux ... across the parabola'. Usually, if no direction is specified, it's ambiguous. But let's look at the sentence: 'normal n ds = <y', -x'> dt (to the right of the direction of travel)'. This is a valid choice of normal. Is it the *standard* choice? Often, the standard normal for a curve y=f(x) is < -f'(x), 1 > or similar. Let's look at the vector field. F = <x+y, x-y>. At (0.5, 0.25), F = <0.75, 0.25>. Tangent T = <1, 1>. Normal N_right = <1, -1>. Dot product F.N = 0.75 - 0.25 = 0.5 > 0. Normal N_left = <-1, 1>. Dot product F.N = -0.75 + 0.25 = -0.5. The solution calculates positive flux. If the standard convention is 'left' normal, the answer should be -1. Many textbooks define the normal vector for flux integrals as the one pointing to the left of the direction of motion (consistent with Green's Theorem orientation). Using the right-pointing normal is non-standard unless specified. Therefore, the solution is misleading because it assumes a non-standard orientation without justification, likely leading to the wrong sign compared to standard conventions.gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.