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Area with Green's theorem

Problem 12.241 · easy

Use Green's theorem to evaluate \( \displaystyle \oint_C \left(- x y^{2}\right) dx + \left(x^{2}\right) dy \), where \( \displaystyle C \) is the circle \( \displaystyle x^2 + y^2 = 4 \), oriented counterclockwise.
  1. \[ \frac{d}{d x} x^{2} - \frac{\partial}{\partial y} \left(- x y^{2}\right) = 2 x y + 2 x \]
    ∂Q/∂x − ∂P/∂y.✓ Proved
  2. \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r \left(2 r^{2} \sin{\left(\theta \right)} \cos{\left(\theta \right)} + 2 r \cos{\left(\theta \right)}\right)\, dr\, d\theta = 0 \]
    ∬_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
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, computes the partial derivatives accurately, and sets up the polar integral with correct bounds and Jacobian. The evaluation to 0 is correct due to symmetry.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-07
  • qwen3.6:27b-mlx: pass 2026-10-07 — The solution correctly applies Green's Theorem, computes the partial derivatives accurately, and sets up the polar integral with correct bounds and Jacobian. The evaluation to 0 is correct due to symmetry.
  • qwen3.6:27b-mlx: fail (error) 2026-10-07 — The setup incorrectly identifies the integrand for Green's Theorem. The problem asks for \oint (-xy^2) dx + (x^2) dy, so P = -xy^2 and Q = x^2. The correct integrand is \partial Q/\partial x - \partial P/\partial y = 2x - (-2xy) = 2x + 2xy. The solution computes 2xy + 2x, which is algebraically equivalent, but the description says 'Derivative(x**2, x) - Derivative(-x*y**2, y) = 2*x*y + 2*x'. Wait, \partial/\partial x (x^2) = 2x. \partial/\partial y (-xy^2) = -2xy. So 2x - (-2xy) = 2x + 2xy. The equation line says the result is 2xy + 2x. This is correct. However, the integral setup in line 2 is wrong. The integrand is 2x + 2xy. In polar, x = r cos \theta, y = r sin \theta. So 2(r cos \theta) + 2(r cos \theta)(r sin \theta) = 2r cos \theta + 2r^2 sin \theta cos \theta. The solution integrates r*(2*r**2*sin(theta)*cos(theta) + 2*r*cos(theta)). This matches 2r^2 sin \theta cos \theta + 2r cos \theta. Wait, the term 2x becomes 2r cos \theta. Multiplied by r (from dA), it is 2r^2 cos \theta. The solution has 2*r*cos(theta) inside the parenthesis, multiplied by r outside, giving 2r^2 cos \theta. The term 2xy becomes 2r^2 sin \theta cos \theta. Multiplied by r, it is 2r^3 sin \theta cos \theta. The solution has 2*r**2*sin(theta)*cos(theta) inside, multiplied by r, giving 2r^3 sin \theta cos \theta. So the integrand in the integral is correct. The integral of 2r^2 cos \theta from 0 to 2\pi is 0. The integral of 2r^3 sin \theta cos \theta from 0 to 2\pi is 0. So the result 0 is correct. Why did I think it was wrong? Let me re-read carefully. Line 1: Derivative(x**2, x) is 2x. Derivative(-x*y**2, y) is -2xy. 2x - (-2xy) = 2x + 2xy. The line says = 2*x*y + 2*x. This is correct. Line 2: Integral(r*(2*r**2*sin(theta)*cos(theta) + 2*r*cos(theta)), ...). The term 2xy in polar is 2(r cos \theta)(r sin \theta) = 2r^2 sin \theta cos \theta. The term 2x in polar is 2r cos \theta. The sum is 2r^2 sin \theta cos \theta + 2r cos \theta. Multiplying by r for dA gives 2r^3 sin \theta cos \theta + 2r^2 cos \theta. The code has r*(2*r**2*sin(theta)*cos(theta) + 2*r*cos(theta)). This expands to 2r^3 sin \theta cos \theta + 2r^2 cos \theta. This is correct. The integral is 0. The answer is 0. Is there a subtle error? The problem is straightforward. The solution seems correct. Let me check the orientation. Counterclockwise is standard for Green's theorem. The region is x^2+y^2 <= 4. Bounds r: 0 to 2, theta: 0 to 2pi. Correct. The calculation is correct. The verdict should be pass? Wait, look at line 1 again. `Derivative(x**2, x) - Derivative(-x*y**2, y)`. This is Q_x - P_y. P = -xy^2, Q = x^2. Correct. The result is 2x + 2xy. Correct. The integral is 0. Correct. Why would this fail? Maybe the prompt implies I should find an error if there is one. Is there a style issue? The solution is very brief. But it is correct. Let me double check the integral of 2r^2 cos(theta) from 0 to 2pi. Integral of cos is sin. sin(2pi) - sin(0) = 0. Integral of sin(theta)cos(theta) is 1/2 sin^2(theta). sin^2(2pi) - sin^2(0) = 0. So the integral is indeed 0. The solution is correct. I will mark it as pass.
  • gpt-oss:20b: pass 2026-10-07

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 2026-10-07 with SymPy 1.14.0.

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