Area with Green's theorem
Problem 12.309 · 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 circle \( \displaystyle x^2 + y^2 = 4 \), oriented counterclockwise.
- \[ \frac{d}{d x} x^{2} - \frac{\partial}{\partial y} x y = x \]∂Q/∂x − ∂P/∂y.✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r^{2} \cos{\left(\theta \right)}\, dr\, d\theta = 0 \]∬_D (∂Q/∂x − ∂P/∂y) dA in polar coordinates.✓ Proved
Answer \( 0 \)
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: fail (error) — The integrand for the double integral is incorrect. The partial derivative difference is x, which converts to r*cos(theta) in polar coordinates, not r^2*cos(theta). The extra factor of r suggests a confusion between the function value and the area element, or an incorrect conversion of x.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The integrand for the double integral is incorrect. The partial derivative difference is x, which converts to r*cos(theta) in polar coordinates, not r^2*cos(theta). The extra factor of r suggests a confusion between the function value and the area element, or an incorrect conversion of x.gpt-oss:20b: fail (misleading) 2026-10-12 — The integrand in polar coordinates is written as r^2 cosθ, which incorrectly mixes the Jacobian factor with the function x=r cosθ. The correct integrand should be r cosθ, and the Jacobian r is then applied separately. Although the final value is still 0, the explanation misleads a student about how to convert the integrand to polar form.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The integrand for the double integral is incorrect. Green's Theorem requires integrating (∂Q/∂x - ∂P/∂y) over the region. Here ∂Q/∂x = 2x and ∂P/∂y = x, so the integrand should be 2x - x = x. In polar coordinates, this is r cos(θ), not r^2 cos(θ). The extra factor of r likely comes from confusing the integrand with the area element dA = r dr dθ, but the solution line implies the integrand itself is r^2 cos(θ).
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 2026-10-12 with SymPy 1.14.0.