∫Calc Practice

Area with Green's theorem

Problem 12.188 · easy

Use Green's theorem to evaluate \( \displaystyle \oint_C \left(x y\right) dx + \left(3 x\right) dy \), where \( \displaystyle C \) is the boundary of the rectangle \( \displaystyle [0, 2] \times [0, 1] \), oriented counterclockwise.
  1. \[ \frac{d}{d x} 3 x - \frac{\partial}{\partial y} x y = 3 - x \]
    ∂Q/∂x − ∂P/∂y.✓ Proved
  2. \[ \int\limits_{0}^{2}\int\limits_{0}^{1} \left(3 - x\right)\, dy\, dx = 4 \]
    ∬_D (∂Q/∂x − ∂P/∂y) dA.✓ Proved
Answer \( 4 \)

✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the line integral around the boundary, computed directly

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly applies Green's Theorem, computing the partial derivatives and evaluating the double integral over the specified rectangular region.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Green's Theorem, computing the partial derivatives and evaluating the double integral over the specified rectangular region.
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Green's Theorem, computing the partial derivatives and evaluating the double integral over the specified rectangular region to obtain the correct result.
  • 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.