Green's theorem
Problem 12.90 · easy
Use Green's theorem to evaluate \( \displaystyle \oint_C (x^{2} - y)\,dx + (x + y^{2})\,dy \), where C is the boundary of the rectangle \( \displaystyle [0, 1] \times [0, 3] \), counterclockwise.
- Green's theorem: ∮ P dx + Q dy = ∬ (∂Q/∂x − ∂P/∂y) dA.Reviewed
- \[ \frac{\partial}{\partial x} \left(x + y^{2}\right) - \frac{\partial}{\partial y} \left(x^{2} - y\right) = 2 \]∂Q/∂x − ∂P/∂y.✓ Proved
- \[ \int\limits_{0}^{3}\int\limits_{0}^{1} 2\, dx\, dy = 6 \]Integrate over the rectangle.✓ Proved
Answer \( 6 \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the line integral around the four edges, done directly, gives the same value |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-26gpt-oss:20b: pass 2026-09-26qwen3.6:27b-mlx: pass 2026-09-26gpt-oss:20b: pass 2026-09-26
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_theorem, checked 2026-09-26 with SymPy 1.14.0.