Area with Green's theorem
Problem 12.301 · easy
Use Green's theorem to evaluate \( \displaystyle \oint_C \left(x y\right) dx + \left(x y\right) dy \), where \( \displaystyle C \) is the circle \( \displaystyle x^2 + y^2 = 4 \), oriented counterclockwise.
- \[ \frac{\partial}{\partial x} x y - \frac{\partial}{\partial y} x y = - x + y \]∂Q/∂x − ∂P/∂y.✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r \left(r \sin{\left(\theta \right)} - r \cos{\left(\theta \right)}\right)\, dr\, d\theta = 0 \]∬_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 solution computes ∂Q/∂x - ∂P/∂y as -x + y, but Green's Theorem requires ∂Q/∂x - ∂P/∂y. Here P=xy and Q=xy, so ∂Q/∂x = y and ∂P/∂y = x. The integrand should be y - x. The solution has the sign reversed (computing ∂P/∂y - ∂Q/∂x or similar error), although the final answer happens to be 0 due to symmetry, the intermediate step is mathematically incorrect.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution computes ∂Q/∂x - ∂P/∂y as -x + y, but Green's Theorem requires ∂Q/∂x - ∂P/∂y. Here P=xy and Q=xy, so ∂Q/∂x = y and ∂P/∂y = x. The integrand should be y - x. The solution has the sign reversed (computing ∂P/∂y - ∂Q/∂x or similar error), although the final answer happens to be 0 due to symmetry, the intermediate step is mathematically incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution computes ∂Q/∂x - ∂P/∂y as (-x + y), but Green's Theorem requires ∂Q/∂x - ∂P/∂y. Here P=xy and Q=xy, so ∂Q/∂x = y and ∂P/∂y = x. The integrand should be (y - x). The solution has the sign reversed (it computed ∂P/∂y - ∂Q/∂x or similar error in the formula application, though coincidentally the integral of both (y-x) and (-x+y) over the symmetric disk is 0, the setup is formally incorrect regarding the standard formula ∂Q/∂x - ∂P/∂y). Wait, let's re-read carefully. Green's theorem: ∮ P dx + Q dy = ∬ (∂Q/∂x - ∂P/∂y) dA. P=xy, Q=xy. ∂Q/∂x = y. ∂P/∂y = x. Integrand is y - x. The solution states the derivative expression is -x + y. This is equal to y - x. So the algebra is correct. The integral of (y-x) over a disk centered at origin is 0 by symmetry. The solution is actually correct. Let me double check the 'unchecked' tag. The prompt says 'equation, unchecked' but then says 'computer algebra system has already checked... do not re-check the algebra'. However, the line is labeled 'unchecked' in the text provided by the user? No, the user prompt says 'Each line... is either an EQUATION... (its result is shown; do not re-check...)'. The text provided is '1. [equation, unchecked] ...'. This is contradictory. Usually, these prompts imply the algebraic result is trusted. Let's look at the sentence part. There are no sentences. Just two equations. The prompt asks to review 'sentences'. If there are no sentences, is it pass? The prompt says 'The sentences are what you are here for.' But it also says 'Check that... each theorem or test is applied...'. Line 1 is an equation. Line 2 is an equation. There are no explanatory sentences. However, the transition from Line 1 to Line 2 implies the application of Green's Theorem. The setup in Line 2 uses polar coordinates for the disk x^2+y^2<=4. The integrand in Line 2 is r*(r*sin(theta) - r*cos(theta)). This corresponds to (y-x)*r dr dtheta. Since y = r sin theta and x = r cos theta, y-x = r(sin theta - cos theta). So the integrand is r * r(sin theta - cos theta) = r^2(sin theta - cos theta). The integral of sin theta and cos theta from 0 to 2pi is 0. So the result 0 is correct. The algebraic sign in Line 1 is -x+y which is y-x. This matches. So the math is correct. Is there a missing sentence? The prompt format usually includes sentences like 'By Green's Theorem...'. Here it is absent. But the instructions say 'Reply with JSON...'. If the solution is just equations and they are correct, it should pass. However, often these 'adversarial' reviews look for missing justifications. But the instructions say 'The sentences are what you are here for.' If there are no sentences, there is nothing to fail on? Or is the lack of explanation a 'style' issue? Or is the fact that it jumps from Cartesian derivative to Polar integral without stating Green's Theorem a 'misleading' omission? Given the strict 'adversarial' nature, skipping the statement of the theorem might be considered 'style' or 'misleading' if a student doesn't know why the double integral appears. But usually, 'pass' is for sound solutions. The solution is sound. I will mark it pass.gpt-oss:20b: pass 2026-10-12
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.