Flux across a plane curve
Problem 12.403 · medium
Find the flux of \( \displaystyle \mathbf F = \left\langle x, y \right\rangle \) across the circle of radius 2, counterclockwise (outward normal).
- With r(t) = <2*cos(t), 2*sin(t)>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
- \[ \frac{d}{d t} 2 \sin{\left(t \right)} \left. x \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)} }} - \frac{d}{d t} 2 \cos{\left(t \right)} \left. y \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)} }} = 4 \]P y′ − Q x′.✓ Proved
- \[ \int\limits_{0}^{2 \pi} 4\, dt = 8 \pi \]Integrate.✓ Proved
Answer \( 8 \pi \)
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: fail (error) — The normal vector formula provided, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies an outward normal. The correct outward normal is <-y', x'>, which would yield a negative integrand and a flux of -8\pi.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-18 — The normal vector formula provided, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies an outward normal. The correct outward normal is <-y', x'>, which would yield a negative integrand and a flux of -8\pi.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: fail (error) 2026-10-15 — The normal vector formula given, <y', -x'>, points inward for a counterclockwise parametrization, whereas the problem specifies an outward normal. The correct outward normal is <-y', x'>, which would yield a negative integrand and a flux of -8\pi.gpt-oss:20b: fail (misleading) 2026-10-15 — The sentence claiming the outward normal is given by ⟨y′,−x′⟩ for a counter‑clockwise traversal is incorrect. For a circle centered at the origin, the outward normal points radially outward, not to the right of the tangent. The subsequent computation still yields the correct value, but the explanation would mislead a student about normal orientation.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The formula for flux across a plane curve is \int (P dy - Q dx) = \int (P y' - Q x') dt. The solution uses this correct formula. However, the text claims the normal vector is \langle y', -x' \rangle, which points to the *right* of the direction of travel (inward for a counterclockwise circle). The outward normal is \langle -y', x' \rangle, corresponding to the integrand Q x' - P y' or P(-y') + Q(x')? No, standard flux is \int \mathbf{F} \cdot \mathbf{n} ds. For CCW, outward normal is \langle y, -x \rangle / R? No, for r(t) = \langle x(t), y(t) \rangle, tangent is \langle x', y' \rangle. Outward normal is \langle y', -x' \rangle? Let's check. r = \langle 2\cos t, 2\sin t \rangle. r' = \langle -2\sin t, 2\cos t \rangle. Outward normal should be radial \langle \cos t, \sin t \rangle. \langle y', -x' \rangle = \langle 2\cos t, 2\sin t \rangle. This IS the outward normal. Wait. \langle y', -x' \rangle = \langle 2\cos t, 2\sin t \rangle. This points outward. The text says "to the right of the direction of travel". If you walk CCW, right is outward. So the vector \langle y', -x' \rangle is indeed the outward normal. The formula used is P y' - Q x'. Let's check the dot product. \mathbf{F} \cdot \mathbf{n} ds = \langle x, y \rangle \cdot \langle y', -x' \rangle dt = (x y' - y x') dt. The solution writes P y' - Q x'. Since P=x, Q=y, this is x y' - y x'. This matches. The calculation yields 4. Integral is 8\pi. The solution is actually correct. Why did I think it was wrong? Let me re-read carefully. "normal n ds = \langle y', -x' \rangle dt". This vector is \langle 2\cos t, 2\sin t \rangle. Its magnitude is 2. ds = |r'| dt = 2 dt. The unit normal is \langle \cos t, \sin t \rangle. \mathbf{n} ds = \langle \cos t, \sin t \rangle 2 dt = \langle 2\cos t, 2\sin t \rangle dt. This matches \langle y', -x' \rangle dt. So the vector part is correct. The formula P y' - Q x' is correct for outward flux with CCW parametrization. The result is correct. Is there a subtle error? "to the right of the direction of travel". For CCW, right is outward. Correct. The solution seems sound. Wait, let me double check the sign convention. Standard Green's theorem for flux: \oint P dy - Q dx. Here P=x, Q=y. \oint x dy - y dx. x=2\cos t, y=2\sin t. dx = -2\sin t dt, dy = 2\cos t dt. x dy - y dx = (2\cos t)(2\cos t) - (2\sin t)(-2\sin t) = 4\cos^2 t + 4\sin^2 t = 4. Integral 0 to 2\pi is 8\pi. The solution is correct. I will mark it as 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 2026-10-18 with SymPy 1.14.0.