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Stokes' theorem

Problem 12.373 · 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 \( \displaystyle x^2 + y^2 = 1 \) 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
\[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{1} r \left. 1 \right|_{\substack{ x=r \cos{\left(\theta \right)}\\ y=r \sin{\left(\theta \right)} }}\, dr\, d\theta = \pi \]
The k-component of the curl over the disk, in polar coordinates.✓ Proved
\[ \int\limits_{0}^{2 \pi} \cos^{2}{\left(t \right)}\, dt = \pi \]
The circulation computed directly around C agrees.✓ Proved

Answer \( \pi \)

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: pass
qwen3.6:27b-mlx: pass — The solution correctly applies Stokes' theorem, computes the curl, and evaluates the surface integral. 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, computes the curl, and evaluates the surface integral. The direct line integral check in step 4 is consistent with the result.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly applies Stokes' theorem by identifying the surface, normal vector, and curl. The final line confirms the result via direct computation, which is valid and helpful, though not strictly required by the prompt.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (misleading) 2026-10-14 — The solution claims that the direct line integral equals pi, but the actual circulation of F=<z,x,y> around the unit circle in the z=0 plane is 2pi (since F=<0,x,y> on C, and integral of x dx + y dy is 0, wait. Let's re-evaluate. On C, z=0, so F=<0, x, y>. dr=<dx, dy, 0>. F.dr = x dx + y dy. Integral of x dx + y dy around closed loop is 0? No. Let's check Stokes again. Curl F = <1, 1, 1>. Normal k=<0,0,1>. Curl F . k = 1. Integral of 1 dA over unit disk is pi. So Stokes gives pi. Direct calculation: F=<0, x, y>. dr=<-sin t, cos t, 0> dt. F.dr = 0*(-sin t) + x*(cos t) + y*0 = x cos t = cos^2 t. Integral 0 to 2pi of cos^2 t dt = pi. Okay, the direct calculation is correct. Wait, did I make a mistake in my head? x=cos t, y=sin t. F=<0, cos t, sin t>. dr=<-sin t, cos t, 0>dt. Dot product: 0 + cos^2 t + 0 = cos^2 t. Integral is pi. So the answer is correct. Why did I think it was 2pi? Maybe confusing with x dy - y dx? That would be 2pi. But here it is x dx + y dy? No, F_y is x, F_z is y. dr_y is dy. So F_y dy = x dy. x=cos t, dy=cos t dt. So cos^2 t. Correct. The solution is actually correct. Let me re-read the prompt constraints. 'Find what is wrong'. Is there a subtle error? Step 1: 'upward normal k'. Correct for counterclockwise. Step 2: Curl calculation. F=<z, x, y>. Curl = <d/dy(y)-d/dz(x), d/dz(z)-d/dx(y), d/dx(x)-d/dy(z)> = <1-0, 1-0, 1-0> = <1,1,1>. Correct. Step 3: Integral of k-component (1) over disk. Area is pi. Correct. Step 4: Direct check. Correct. Is there a style issue? 'Matrix(...)' is not a sentence. The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. The review targets SENTENCES. Sentence 1 is fine. Are there other sentences? No. The equations are marked 'unchecked' but the prompt says 'do not re-check the algebra of a line marked proved'. Wait, the prompt says 'marked proved'. These are marked 'unchecked'. Does that mean I should check them? 'a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. These lines are NOT marked proved. They are marked 'unchecked'. This implies I SHOULD check them? Or does it mean the CAS hasn't checked them? The prompt says 'Each line... is either an EQUATION, which a computer algebra system has already checked... or a SENTENCE'. This implies ALL equations provided are checked by CAS. The label 'unchecked' in the text might be a distractor or a typo in the problem description provided to me. However, looking closely at the provided text: '2. [equation, unchecked] ...'. If the prompt says 'do not re-check the algebra of a line marked proved', and these are marked 'unchecked', maybe I am supposed to check them? But the instructions say 'The sentences are what you are here for.' and 'Check that... each conclusion follows from the lines before it'. If the equations are wrong, the conclusion fails. But I established they are correct. Is there a misleading statement? 'Take S to be the flat disk...'. This is a valid choice. 'Stokes: ...'. Valid. I see no error. The verdict should be 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/stokes_theorem, checked 2026-10-17 with SymPy 1.14.0.