Area with Green's theorem
Problem 12.274 · easy
Use Green's theorem to find the area enclosed by the ellipse \( \displaystyle \frac{x^2}{1} + \frac{y^2}{16} = 1 \).
- Area = ½∮ (x dy − y dx). Parametrize x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.Reviewed
- \[ 4 \sin^{2}{\left(t \right)} + 4 \cos^{2}{\left(t \right)} = 4 \]x·y′ − y·x′ simplifies to ab.✓ Proved
- \[ \frac{\int\limits_{0}^{2 \pi} \left(4 \sin^{2}{\left(t \right)} + 4 \cos^{2}{\left(t \right)}\right)\, dt}{2} = 4 \pi \]½∫₀^{2π} ab dt.✓ Proved
Answer \( 4 \pi \)
Lines: 2 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ 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 | shoelace area of a 20000-gon inscribed in the ellipse (agrees to 1 part in 10⁵) |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies Green's theorem for area using the standard parametrization of the ellipse. The algebraic simplification and integration are correct, leading to the right answer.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly applies Green's theorem for area using the standard parametrization of the ellipse. The algebraic simplification and integration are correct, leading to the right answer.qwen3.6:27b-mlx: inconclusive 2026-10-09 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The parametrization uses a=1 and b=4, so x dy - y dx should simplify to ab = 4. However, line 2 shows 4*sin(t)^2 + 4*cos(t)^2, which implies the stugpt-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.