Home›Calculus 3›Flux across a plane curve›Problem 12.401

Flux across a plane curve

Problem 12.401 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle - y, x \right\rangle \) 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.
\[ - \frac{d}{d t} t \left. x \right|_{\substack{ x=t\\ y=t^{2} }} + \frac{d}{d t} t^{2} \left. - y \right|_{\substack{ x=t\\ y=t^{2} }} = - 2 t^{3} - t \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} \left(- 2 t^{3} - t\right)\, dt = -1 \]
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: pass
qwen3.6:27b-mlx: fail (error) — The normal vector formula used, <y', -x'>, corresponds to the left normal (inward for a convex curve like y=x^2), not the right normal as claimed. The standard right normal for a curve traversed from left to right is <y', -x'> only if the orientation is reversed or the definition of 'right' is non-standard; typically, the outward/right normal for y=f(x) is <f'(x), -1> scaled, which is <2t, -1>. The solution uses <-2t, 1> effectively (via P y' - Q x' with signs swapped or misidentified), leading to the wrong sign for the flux. Specifically, for F=<-y, x>, the flux integral should be ∫(F · n) ds. If n is the right normal <y', -x'> = <2t, -1>, then F·n = -y(2t) + x(-1) = -2t^3 - t. Wait, let's re-evaluate. r(t) = <t, t^2>. r'(t) = <1, 2t>. A normal vector is <-2t, 1> (left) or <2t, -1> (right). The solution claims n ds = <y', -x'> dt = <2t, -1> dt. This IS the right normal. Then Flux = ∫ F · <2t, -1> dt = ∫ (-t^2)(2t) + (t)(-1) dt = ∫ (-2t^3 - t) dt = [-t^4/2 - t^2/2]_0^1 = -1/2 - 1/2 = -1. The calculation is actually correct for the right normal. However, the text says 'normal n ds = <y', -x'> dt (to the right...)'. Let's check the direction. Travel is (0,0) to (1,1). Tangent is <1, 2t>. Right normal is obtained by rotating tangent 90 deg clockwise: <2t, -1>. The formula <y', -x'> gives <2t, -1>. So the vector is correct. The integral is correct. The result is -1. Why did I think it was wrong? Let's re-read carefully. 'flux = ∫ (P y′ − Q x′) dt'. P=-y, Q=x. P y' - Q x' = (-y)(2t) - (x)(1) = -2t^3 - t. This matches the equation. The result -1 is correct for the right normal. Is there a convention issue? Usually 'flux across a curve' implies a specific orientation. If the problem doesn't specify 'outward' or 'right', it's ambiguous, but 'right of direction of travel' is a valid specification. The solution is internally consistent and mathematically correct for the stated orientation. Wait, is <y', -x'> actually the right normal? Tangent T = <x', y'>. Right normal N_right = <y', -x'>. Left normal N_left = <-y', x'>. Yes. So the setup is correct. The answer -1 is correct. I will mark it as pass.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-18 — The normal vector formula used, <y', -x'>, corresponds to the left normal (inward for a convex curve like y=x^2), not the right normal as claimed. The standard right normal for a curve traversed from left to right is <y', -x'> only if the orientation is reversed or the definition of 'right' is non-standard; typically, the outward/right normal for y=f(x) is <f'(x), -1> scaled, which is <2t, -1>. The solution uses <-2t, 1> effectively (via P y' - Q x' with signs swapped or misidentified), leading to the wrong sign for the flux. Specifically, for F=<-y, x>, the flux integral should be ∫(F · n) ds. If n is the right normal <y', -x'> = <2t, -1>, then F·n = -y(2t) + x(-1) = -2t^3 - t. Wait, let's re-evaluate. r(t) = <t, t^2>. r'(t) = <1, 2t>. A normal vector is <-2t, 1> (left) or <2t, -1> (right). The solution claims n ds = <y', -x'> dt = <2t, -1> dt. This IS the right normal. Then Flux = ∫ F · <2t, -1> dt = ∫ (-t^2)(2t) + (t)(-1) dt = ∫ (-2t^3 - t) dt = [-t^4/2 - t^2/2]_0^1 = -1/2 - 1/2 = -1. The calculation is actually correct for the right normal. However, the text says 'normal n ds = <y', -x'> dt (to the right...)'. Let's check the direction. Travel is (0,0) to (1,1). Tangent is <1, 2t>. Right normal is obtained by rotating tangent 90 deg clockwise: <2t, -1>. The formula <y', -x'> gives <2t, -1>. So the vector is correct. The integral is correct. The result is -1. Why did I think it was wrong? Let's re-read carefully. 'flux = ∫ (P y′ − Q x′) dt'. P=-y, Q=x. P y' - Q x' = (-y)(2t) - (x)(1) = -2t^3 - t. This matches the equation. The result -1 is correct for the right normal. Is there a convention issue? Usually 'flux across a curve' implies a specific orientation. If the problem doesn't specify 'outward' or 'right', it's ambiguous, but 'right of direction of travel' is a valid specification. The solution is internally consistent and mathematically correct for the stated orientation. Wait, is <y', -x'> actually the right normal? Tangent T = <x', y'>. Right normal N_right = <y', -x'>. Left normal N_left = <-y', x'>. Yes. So the setup is correct. The answer -1 is correct. I will mark it as pass.
gpt-oss:20b: pass 2026-10-18
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The normal vector used, <y', -x'>, points to the left of the direction of travel (inward for this curve), not to the right. The standard outward/right normal is <-y', x'>. Using the wrong normal sign leads to the incorrect answer -1 instead of the correct flux 1.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The normal vector used, <y', -x'>, points to the right of the direction of travel. For the curve y=x^2 from (0,0) to (1,1), 'right' points into the region below the parabola (away from the concave side). While 'flux across a curve' is ambiguous without a specified orientation, standard convention for open curves often implies the normal pointing towards the 'interior' or consistent with a closed loop, or simply requires specifying the side. More critically, the problem asks for flux 'across' the parabola. Usually, if no side is specified, one might assume the normal pointing towards the origin or the 'inside' of the bend, or the problem is ill-posed. However, the bigger issue is that the standard formula for flux is \int \mathbf{F} \cdot \mathbf{n} ds. If \mathbf{n} ds = \langle dy, -dx \rangle, this corresponds to the normal pointing to the right. Let's check the sign. \mathbf{F} = \langle -y, x \rangle. \mathbf{n} ds = \langle 2t, -1 \rangle dt. \mathbf{F} \cdot \mathbf{n} ds = (-t^2)(2t) + (t)(-1) = -2t^3 - t. Integral is -1. If the normal was to the left \langle -dy, dx \rangle, the answer would be 1. Without a specified orientation (e.g., 'outward' from a region, or 'upward'), the sign is ambiguous. However, in many calculus contexts, 'flux across a curve' from A to B implies the normal is obtained by rotating the tangent vector 90 degrees clockwise (right) or counter-clockwise (left). The solution explicitly states 'to the right'. Is this the standard convention? Often, for a curve in the plane, if not specified, the 'positive' normal might be assumed to be the one pointing to the left (counter-clockwise rotation of tangent) to match the standard orientation of boundaries. If the standard convention is left, the answer should be 1. If the problem implies a specific region (e.g., bounded by the parabola and axes), the 'outward' normal would vary. Given the ambiguity, stating the answer is definitively -1 based on an arbitrary choice of 'right' without justification or problem statement support is misleading/error-prone. But wait, looking at similar problems, often 'flux across' without qualification might imply the normal pointing in the positive y-direction component or similar? No. Let's look at the vector field. It's a rotation. The flow is counter-clockwise. The curve goes from (0,0) to (1,1). The vector field pushes 'out' from the origin. The normal to the right points 'in' towards the origin side? At (0.5, 0.25), tangent is <1, 1>. Right normal is <1, -1>. This points down/right. The vector field is <-0.25, 0.5>. Dot product is -0.25 - 0.5 = -0.75. Negative flux. If we took the left normal <-1, 1>, dot product is 0.25 + 0.5 = 0.75. Positive flux. Which is 'correct'? Usually, for an open curve, the orientation of the normal must be specified. If not, the problem is ill-posed. However, many textbooks define the flux integral \int P dy - Q dx as the flux across the curve with the normal pointing to the *left* of the direction of motion (standard orientation). The solution uses the normal to the *right*. This reverses the sign. If the standard convention is left, the answer should be 1. The solution claims -1. This is likely an error in convention application.
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.