Stokes' theorem
Problem 12.383 · hard
Use Stokes' theorem to evaluate \( \displaystyle \oint_C \mathbf F \cdot d\mathbf r \) for \( \displaystyle \mathbf F = \left\langle x z, x - y, x y \right\rangle \), 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.
- \[ \left[\begin{matrix}\frac{\partial}{\partial y} x y - \frac{\partial}{\partial z} \left(x - y\right)\\- \frac{\partial}{\partial x} x y + \frac{\partial}{\partial z} x z\\- \frac{\partial}{\partial y} x z + \frac{\partial}{\partial x} \left(x - y\right)\end{matrix}\right] = \left[\begin{matrix}x\\x - y\\1\end{matrix}\right] \]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
- \[ \int\limits_{0}^{2 \pi} \left(2 \sqrt{2} \cos{\left(2 t + \frac{\pi}{4} \right)} + 2\right)\, dt = 4 \pi \]The circulation computed directly around C agrees.✓ Proved
Answer \( 4 \pi \)
Lines: 3 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 |
| 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: fail (error) — The computed curl is incorrect: the second component should be –z, not x–y. This invalidates the subsequent integration and the final result.qwen3.6:27b-mlx: pass — The solution correctly applies Stokes' theorem. The setup identifies the correct surface and normal vector, the curl calculation is correct, and the resulting surface integral is evaluated properly. The direct line integral check in step 4 is consistent with the result.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies Stokes' theorem. The setup identifies the correct surface and normal vector, the curl calculation is correct, and the resulting surface integral is evaluated properly. The direct line integral check in step 4 is consistent with the result.gpt-oss:20b: fail (error) 2026-10-17 — The computed curl is incorrect: the second component should be –z, not x–y. This invalidates the subsequent integration and the final result.qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies Stokes' theorem. The setup identifies the correct surface and normal vector, the curl calculation is correct, and the resulting surface integral is evaluated properly. The direct line integral check confirms the result.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (misleading) 2026-10-14 — The solution includes an unnecessary and potentially confusing direct line integral calculation (Step 4) to verify the result, which is not part of using Stokes' theorem as requested. More importantly, Step 2 presents the curl calculation in a confusing matrix format that obscures the standard vector component derivation, and Step 3 skips the explicit dot product step, jumping straight to the integral of the k-component without clearly stating that (curl F) · k = 1. While the final answer is correct, the presentation is sloppy and the inclusion of the direct method as part of the 'solution' is misleading regarding the problem's constraint to use Stokes' theorem.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/stokes_theorem, checked 2026-10-17 with SymPy 1.14.0.