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Area with Green's theorem

Problem 12.189 · easy

Use Green's theorem to evaluate \( \displaystyle \oint_C \left(y^{2}\right) dx + \left(2 x\right) dy \), where \( \displaystyle C \) is the circle \( \displaystyle x^2 + y^2 = 4 \), oriented counterclockwise.

\[ \frac{d}{d x} 2 x - \frac{d}{d y} y^{2} = 2 - 2 y \]
∂Q/∂x − ∂P/∂y.✓ Proved
\[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r \left(- 2 r \sin{\left(\theta \right)} + 2\right)\, dr\, d\theta = 8 \pi \]
∬_D (∂Q/∂x − ∂P/∂y) dA in polar coordinates.✓ Proved

Answer \( 8 \pi \)

Every line of this solution was proved by the computer algebra system SymPy. 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.

The full receipt

Line	Status	Checked by	Detail
1	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
2	✓ 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	the line integral around the boundary, computed directly

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The partial derivative of P = y^2 with respect to y is 2, not 2y. Consequently, the integrand should be 2 - 2 = 0, making the integral 0, not 8\pi.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-04 — The partial derivative of P = y^2 with respect to y is 2, not 2y. Consequently, the integrand should be 2 - 2 = 0, making the integral 0, not 8\pi.
gpt-oss:20b: pass 2026-10-04
qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The partial derivative of P with respect to y is 2y, not 2. The integrand should be 2 - 2y, but the integral calculation in line 2 incorrectly uses a
gpt-oss:20b: pass 2026-10-04

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/greens_area, checked 2026-10-04 with SymPy 1.14.0.