Stokes' theorem
Problem 12.371 · hard
Use Stokes' theorem to evaluate \( \displaystyle \oint_C \mathbf F \cdot d\mathbf r \) for \( \displaystyle \mathbf F = \left\langle z, x, y \right\rangle \), where \( \displaystyle C \) is the circle \( \displaystyle x^2 + y^2 = 4 \) in the plane \( \displaystyle z = 0 \), counterclockwise seen from above.
- Take S to be the flat disk the circle bounds, with upward normal k. Stokes: ∮_C F · dr = ∬_S (curl F) · k dA.Reviewed
- \[ \left[\begin{matrix}- \frac{d}{d z} x + \frac{d}{d y} y\\- \frac{d}{d x} y + \frac{d}{d z} z\\\frac{d}{d x} x - \frac{d}{d y} z\end{matrix}\right] = \left[\begin{matrix}1\\1\\1\end{matrix}\right] \]curl F.✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r \left. 1 \right|_{\substack{ x=r \cos{\left(\theta \right)}\\ y=r \sin{\left(\theta \right)} }}\, dr\, d\theta = 4 \pi \]The k-component of the curl over the disk, in polar coordinates.✓ Proved
- \[ \int\limits_{0}^{2 \pi} 4 \cos^{2}{\left(t \right)}\, dt = 4 \pi \]The circulation computed directly around C agrees.✓ Proved
Answer \( 4 \pi \)
Lines: 3 proved, 1 reviewed. 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 | 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 |
| 4 | ✓ 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 · Δr summed around a 20,000-sided polygon on the circle |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies Stokes' theorem, computes the curl, and evaluates the surface integral. The final line confirms the result via direct line integration, which is consistent and correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies Stokes' theorem, computes the curl, and evaluates the surface integral. The final line confirms the result via direct line integration, which is consistent and correct.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies Stokes' theorem, computes the curl, and evaluates the surface integral. The final line confirms the result via direct line integration, which is consistent and correct.gpt-oss:20b: fail (error) 2026-10-15 — The direct line‑integral computation in step 4 is incorrect: for F=<z,x,y> on the circle z=0 the integrand vanishes, so the circulation is 0, not 4π. The earlier Stokes’ theorem calculation is correct, but the concluding sentence misleads the reader.qwen3.6:27b-mlx: fail (misleading) 2026-10-14 — The solution claims that the direct line integral equals 4*pi, but the direct calculation of \oint_C \mathbf{F} \cdot d\mathbf{r} for \mathbf{F}=\langle z, x, y \rangle on z=0 yields 0 (since \mathbf{F}=\langle 0, x, y \rangle and \oint x dx + y dy = 0). The Stokes' theorem calculation is correct (yielding 4\pi), so the 'direct' check is factually wrong and misleading.gpt-oss:20b: fail (error) 2026-10-14 — The computed curl of F is incorrect; ∇×F = 0, not <1,1,1>, so the surface integral and resulting circulation should be zero, not 4π.
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/stokes_theorem, checked 2026-10-17 with SymPy 1.14.0.