Area with Green's theorem
Problem 12.271 · easy
Use Green's theorem to evaluate \( \displaystyle \oint_C \left(y^{2}\right) dx + \left(x y^{2}\right) dy \), where \( \displaystyle C \) is the boundary of the rectangle \( \displaystyle [0, 3] \times [0, 1] \), oriented counterclockwise.
- \[ - \frac{d}{d y} y^{2} + \frac{\partial}{\partial x} x y^{2} = y^{2} - 2 y \]∂Q/∂x − ∂P/∂y.✓ Proved
- \[ \int\limits_{0}^{3}\int\limits_{0}^{1} \left(y^{2} - 2 y\right)\, dy\, dx = -2 \]∬_D (∂Q/∂x − ∂P/∂y) dA.✓ Proved
Answer \( -2 \)
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: fail (error) — The solution computes the double integral of (y^2 - 2y) over the region, but fails to multiply by the length of the x-interval (3). The integral with respect to y is -1/3, so the total area integral is 3 * (-1/3) = -1, not -2.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution computes the double integral of (y^2 - 2y) over the region, but fails to multiply by the length of the x-interval (3). The integral with respect to y is -1/3, so the total area integral is 3 * (-1/3) = -1, not -2.qwen3.6:27b-mlx: fail (error) 2026-10-09 — The solution computes the integral of (y^2 - 2y) over the region, but the integrand for Green's Theorem is (∂Q/∂x - ∂P/∂y). Here P=y^2 and Q=xy^2, so ∂Q/∂x = y^2 and ∂P/∂y = 2y. The integrand is indeed y^2 - 2y. However, the double integral of (y^2 - 2y) dy from 0 to 1 is [y^3/3 - y^2]_0^1 = 1/3 - 1 = -2/3. Then integrating -2/3 dx from 0 to 3 gives -2. Wait, let me re-evaluate. Integral of (y^2 - 2y) dy from 0 to 1 is -2/3. Integral of -2/3 dx from 0 to 3 is -2. The calculation is actually correct. Why did I think it was wrong? Let me re-read the prompt. "Derivative(y**2, y) + Derivative(x*y**2, x)". The comment says "∂Q/∂x − ∂P/∂y". The equation shows `Derivative(y**2, y) + Derivative(x*y**2, x)`. This evaluates to `2y + y^2`. But the text says the result is `y**2 - 2*y`. The equation line is internally inconsistent: the LHS expression `Derivative(y**2, y) + Derivative(x*y**2, x)` equals `2y + y^2`, but the RHS is `y**2 - 2*y`. The comment claims it is `∂Q/∂x − ∂P/∂y`, which is `y^2 - 2y`. The equation line has a sign error in the operator (plus instead of minus) or a typo in the derivative term. Specifically, it writes `Derivative(y**2, y)` which is `∂P/∂y`, and adds it. It should be subtracting it. The result `y**2 - 2*y` is correct for the theorem, but the equation line `Derivative(y**2, y) + Derivative(x*y**2, x) = y**2 - 2*y` is mathematically false because `2y + y^2 != y^2 - 2y`. This is an error in the stated equation.gpt-oss:20b: pass 2026-10-09
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 2026-10-09 with SymPy 1.14.0.