Area with Green's theorem
Problem 12.215 · easy
Use Green's theorem to evaluate \( \displaystyle \oint_C \left(- y^{3}\right) dx + \left(x y\right) dy \), where \( \displaystyle C \) is the triangle with vertices \( \displaystyle (0, 0) \), \( \displaystyle (1, 0) \), \( \displaystyle (0, 1) \), oriented counterclockwise.
- \[ - \frac{d}{d y} \left(- y^{3}\right) + \frac{\partial}{\partial x} x y = 3 y^{2} + y \]∂Q/∂x − ∂P/∂y.✓ Proved
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1 - x} \left(3 y^{2} + y\right)\, dy\, dx = \frac{5}{12} \]∬_D (∂Q/∂x − ∂P/∂y) dA.✓ Proved
Answer \( \frac{5}{12} \)
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: pass — The solution correctly applies Green's Theorem, computing the integrand as ∂Q/∂x - ∂P/∂y = y + 3y² and setting up the double integral over the triangular region with correct bounds. The final result is correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly applies Green's Theorem, computing the integrand as ∂Q/∂x - ∂P/∂y = y + 3y² and setting up the double integral over the triangular region with correct bounds. The final result is correct.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution computes the integral of (3y^2 + y) but the integrand should be (∂Q/∂x - ∂P/∂y) = y - (-3y^2) = y + 3y^2. Wait, the equation says 3*y**2 + y, which is correct. Let me re-read carefully. P = -y^3, Q = xy. ∂Q/∂x = y. ∂P/∂y = -3y^2. ∂Q/∂x - ∂P/∂y = y - (-3y^2) = y + 3y^2. The equation says 3*y**2 + y. This is correct. The integral bounds are y from 0 to 1-x, x from 0 to 1. This describes the triangle correctly. The result 5/12 is correct. Why did I think it was wrong? Let me check the calculation. Integral_0^1 Integral_0^(1-x) (3y^2 + y) dy dx. Inner: [y^3 + y^2/2]_0^(1-x) = (1-x)^3 + (1-x)^2/2. Outer: Integral_0^1 ((1-x)^3 + (1-x)^2/2) dx. Let u = 1-x, du = -dx. Integral_1^0 (u^3 + u^2/2) (-du) = Integral_0^1 (u^3 + u^2/2) du = [u^4/4 + u^3/6]_0^1 = 1/4 + 1/6 = 5/12. The solution is actually correct. The prompt asks me to find what is wrong. If nothing is wrong, I should say pass. But usually these adversarial prompts have a trick. Is there a sign error in Green's theorem? Green's theorem: ∮ P dx + Q dy = ∬ (∂Q/∂x - ∂P/∂y) dA. Orientation is counterclockwise, which is standard positive orientation. So the sign is correct. The vertices (0,0), (1,0), (0,1) form a triangle in the first quadrant. The bounds are correct. The algebra is correct. The answer is correct. I will mark it as pass.
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 2026-10-06 with SymPy 1.14.0.