Home›Calculus 3›Flux across a plane curve›Problem 12.397

Flux across a plane curve

Problem 12.397 · medium

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

Answer \( 1 \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. 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	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 2026-10-18
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the normal vector for the flux integral and applies the formula accurately. The parametrization and integration are correct.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The normal vector used, <y', -x'>, points to the left of the direction of travel, not the right. For a curve moving in the first quadrant, the right normal should point into the fourth quadrant (positive x, negative y), but <1, -1> is indeed the right normal for <1,1>. Wait, let's re-evaluate. Direction is <1,1>. Right normal is obtained by rotating <1,1> by -90 degrees, which gives <1, -1>. The formula <y', -x'> with x'=1, y'=1 gives <1, -1>. So the vector is correct. However, the standard flux integral is usually defined with the outward or specific normal. The problem does not specify orientation. Usually, 'flux across a segment' implies a specific orientation or requires clarification. But more importantly, the standard formula for flux is integral of F . n ds. If n ds = <dy, -dx>, then flux is integral of (P dy - Q dx). Here P=x+y, Q=x-y. dy=dt, dx=dt. Integral ( (t+t)dt - (t-t)dt ) = Integral 2t dt = 1. The calculation is correct for the normal <y', -x'>. Is this the 'right' normal? Yes. Is it the standard 'outward' normal? There is no closed curve. The problem is ambiguous about which side is 'positive' flux. However, usually, if not specified, one might assume the normal pointing to the right of the path. The solution states 'to the right'. The calculation follows. Why is it wrong? Let's check the sign convention. Often flux is defined as integral F . n ds where n is the unit normal. Here n ds = <dy, -dx>. This corresponds to the normal pointing to the right. The result 1 is consistent with this choice. Is there a subtle error? The prompt asks to find what is wrong. Maybe the issue is that 'flux across a segment' is ill-defined without specifying the normal direction, and the solution arbitrarily chooses 'right' without justification or stating that the answer depends on this choice? Or perhaps the standard convention for 'flux' in 2D line integrals often assumes the normal pointing 'outward' from a region, which doesn't exist here. Let's look closer at the sentence: 'the normal n ds = <y', -x'> dt (to the right of the direction of travel)'. This is a correct statement of fact. The calculation follows. Is the answer 1 correct? Yes, for that normal. Is it misleading? If the standard convention is different, it might be. But usually, problems specify 'outward' or 'left/right'. Without specification, the problem is ambiguous. However, in many calculus contexts, if orientation isn't specified, the 'right' normal is a common default for open curves, or the problem implies a specific orientation. Let's look for a harder error. What if the vector field is conservative? Curl is -2 - 1 = -3. Not conservative. What if the question implies the normal pointing 'up' or 'out' from the origin? No. Let's re-read the normal definition. n ds = <dy, -dx>. This is the normal obtained by rotating the tangent vector <dx, dy> by -90 degrees. This points to the right. The solution says 'to the right'. This is consistent. Is there an error in the integral setup? Flux = int F . n ds. F = <x+y, x-y>. n ds = <dy, -dx>. Dot product: (x+y)dy - (x-y)dx. On the line y=x, dy=dx. So (2x)dx - (0)dx = 2x dx. Integral from 0 to 1 of 2t dt = 1. The math is correct. Why would it fail? Perhaps the 'standard' normal for a curve from A to B is not defined, and the solution presents a specific choice as *the* answer without noting the ambiguity? Or is there a sign error in the 'right' normal? Tangent T = <1, 1>. Right normal N = <1, -1>. Dot product F . N = (2t)(1) + (0)(-1) = 2t. Integral 2t dt = 1. Everything seems consistent. Let's look at the 'unchecked' equations. They are just algebraic verifications. The sentence is the target. 'With r(t) = <t, t>, the normal n ds = <y', -x'> dt (to the right of the direction of travel), so flux = int (P y' - Q x') dt.' This is a correct derivation. Is it possible the problem expects the normal pointing to the *left*? Or is the 'flux' defined as integral F . n ds where n is the *unit* normal? The formula n ds = <y', -x'> dt uses the non-unit normal scaled by ds/dt? No, n ds is the vector differential. If n is unit, n ds = n |r'| dt. Here <y', -x'> has length sqrt(2). |r'| is sqrt(2). So n = <1/sqrt(2), -1/sqrt(2>). n ds = <1/sqrt(2), -1/sqrt(2)> * sqrt(2) dt = <1, -1> dt. This matches <y', -x'> dt. So the scaling is correct. I suspect the 'error' is that the problem statement 'Find the flux... across the segment' is ambiguous regarding the direction of the normal, and the solution arbitrarily picks one without stating that the sign depends on this choice, potentially misleading a student into thinking the direction is fixed by the segment endpoints alone. However, 'right' is a standard convention. Let's look for a more definitive error. What if the normal should be < -y', x' >? That would be the left normal. If the problem implies a standard orientation (e.g., counter-clockwise around a region), it's not applicable. Let's assume the 'error' is the lack of specification leading to a potentially wrong sign if the grader expects the other normal. But 'right' is explicitly stated. Is 'right' the standard? In Green's theorem, the boundary is traversed counter-clockwise, and the normal points outward. For a segment from (0,0) to (1,1), if it were part of a CCW loop enclosing the area below it, the outward normal would point down/right. If it enclosed the area above, it would point up/left. Without context, 'right' is a reasonable choice. Is there a mistake in the vector field components? F = <x+y, x-y>. P=x+y, Q=x-y. Formula P dy - Q dx. Correct. I will mark it as 'misleading' because the direction of the normal is not specified in the problem, and while 'right' is a valid choice, presenting it as *the* answer without qualification is poor practice. Actually, looking at similar problems, often the normal is taken to be the one pointing 'outward' from the origin or similar. But here, the most likely 'error' intended by an adversarial reviewer is that the normal vector <y', -x'> points to the *right*, but standard conventions for 'flux across a curve' often default to the normal pointing to the *left* (consistent with Green's theorem orientation where the region is on the left) or require explicit specification. If the standard convention is 'left', the answer should be -1. Given the ambiguity, stating the answer is definitively 1 is misleading. Let's go with misleading/error due to arbitrary sign choice presented as fact. Actually, let's look at the phrasing 'n ds = <y', -x'> dt'. This vector <y', -x'> is perpendicular to <x', y'>. Dot product x'y' - y'x' = 0. It is a normal. Does it point right? <1,1> rotated -90 deg is <1,-1>. <y', -x'> is <1, -1>. Yes. So the statement is factually correct. The only issue is the problem's ambiguity. I will flag it as 'misleading' because the sign of the flux depends on the choice of normal, which is not specified in the problem statement, and the solution presents one arbitrary choice as the unique answer.
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 2026-10-18 with SymPy 1.14.0.