Area with Green's theorem
Problem 12.187 · easy
Use Green's theorem to evaluate \( \displaystyle \oint_C \left(x y\right) dx + \left(x^{2}\right) dy \), where \( \displaystyle C \) is the boundary of the rectangle \( \displaystyle [0, 1] \times [0, 1] \), oriented counterclockwise.
- \[ \frac{d}{d x} x^{2} - \frac{\partial}{\partial y} x y = x \]∂Q/∂x − ∂P/∂y.✓ Proved
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1} x\, dy\, dx = \frac{1}{2} \]∬_D (∂Q/∂x − ∂P/∂y) dA.✓ Proved
Answer \( \frac{1}{2} \)
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: passqwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer returned a non-objectgpt-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.