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Flux across a plane curve

Problem 12.336 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle - y, x \right\rangle \) across the circle of radius 1, counterclockwise (outward normal).

With r(t) = <cos(t), sin(t)>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
\[ \frac{d}{d t} \sin{\left(t \right)} \left. - y \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)} }} - \frac{d}{d t} \cos{\left(t \right)} \left. x \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)} }} = 0 \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{2 \pi} 0\, dt = 0 \]
Integrate.✓ Proved

Answer \( 0 \)

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 provided, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem requires the outward normal. This sign error leads to the incorrect conclusion that the flux is 0 instead of 2π.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-10-17 — The normal vector formula provided, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem requires the outward normal. This sign error leads to the incorrect conclusion that the flux is 0 instead of 2π.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (error) 2026-10-15 — The normal vector formula provided, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies the outward normal. The correct outward normal is <-y', x'>, which would yield a non-zero flux of 2pi.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — The formula for the normal vector component is incorrect. For a counterclockwise parametrization, the outward normal vector scaled by ds is <y', -x'>, so the flux integral should be P*y' - Q*x'. However, the standard formula for flux is often written as P*dy - Q*dx or <P, Q> . <y', -x'> dt. Let's check the sign. If r(t) = <cos t, sin t>, then r'(t) = <-sin t, cos t>. The outward normal is <cos t, sin t> = <x, y>. Note that <y', -x'> = <cos t, sin t>, which is indeed the outward normal. So the vector <y', -x'> is correct for the outward normal. The integrand is P*y' - Q*x'. P = -y, Q = x. P*y' = -y*cos(t) = -sin(t)*cos(t). Q*x' = x*(-sin(t)) = cos(t)*(-sin(t)) = -sin(t)*cos(t). So P*y' - Q*x' = -sin(t)cos(t) - (-sin(t)cos(t)) = 0. The calculation yields 0. Wait, let's re-read the sentence carefully. 'normal n ds = <y', -x'> dt'. This is correct for outward normal on CCW curve. 'flux = int (P y' - Q x') dt'. This is correct. The result is 0. Is the answer 0? Div F = 0. By Green's theorem, flux = int div F dA = 0. So the answer is correct. Why did I think it was wrong? Let me re-read the prompt's instruction. 'Find the flux... across the circle... counterclockwise (outward normal)'. The solution calculates 0. The answer is 0. The steps seem correct. Is there a subtle error? 'n ds = <y', -x'> dt'. For r(t) = <x(t), y(t)>, the tangent is <x', y'>. The outward normal is <y', -x'>? Let's check. Tangent T = <-sin, cos>. Outward N = <cos, sin>. <y', -x'> = <cos, -(-sin)> = <cos, sin>. Yes. So the vector is correct. The integrand P y' - Q x' is correct. The evaluation is correct. The verdict should be pass? Let me double check the 'unchecked' equations. Line 2: Derivative(sin(t), t) is cos(t). Subs(-y, ...) is -sin(t). Product is -sin(t)cos(t). Derivative(cos(t), t) is -sin(t). Subs(x, ...) is cos(t). Product is -sin(t)cos(t). Subtracting: -sin(t)cos(t) - (-sin(t)cos(t)) = 0. Correct. Line 3: Integral of 0 is 0. Correct. Sentence 1: 'normal n ds = <y', -x'> dt (to the right of the direction of travel)'. If you travel CCW, 'right' is outward. Correct. So why would this fail? Maybe the notation 'n ds' is slightly abusive but standard. Or maybe the prompt implies I should find an error. Let me look closer. 'P y' - Q x''. Usually flux is int F . n ds. F = <P, Q>. n ds = <dy, -dx> = <y' dt, -x' dt>. So F . n ds = P y' dt - Q x' dt. This matches. Is it possible the user considers the formula P dy - Q dx to be for circulation? No, circulation is P dx + Q dy. Flux is P dy - Q dx. This is standard. I see no error. I will mark it pass.
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.