Area with Green's theorem
Problem 12.278 · easy
Use Green's theorem to find the area enclosed by the ellipse \( \displaystyle \frac{x^2}{9} + \frac{y^2}{9} = 1 \).
- Area = ½∮ (x dy − y dx). Parametrize x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.Reviewed
- \[ 9 \sin^{2}{\left(t \right)} + 9 \cos^{2}{\left(t \right)} = 9 \]x·y′ − y·x′ simplifies to ab.✓ Proved
- \[ \frac{\int\limits_{0}^{2 \pi} \left(9 \sin^{2}{\left(t \right)} + 9 \cos^{2}{\left(t \right)}\right)\, dt}{2} = 9 \pi \]½∫₀^{2π} ab dt.✓ Proved
Answer \( 9 \pi \)
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 identifies the curve as a circle of radius 3, applies Green's theorem for area, and computes the integral correctly.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-09qwen3.6:27b-mlx: pass 2026-10-09 — The solution correctly identifies the curve as a circle of radius 3, applies Green's theorem for area, and computes the integral correctly.qwen3.6:27b-mlx: inconclusive 2026-10-09 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The parametrization uses a=3 and b=3, so x dy - y dx should simplify to 9. However, the intermediate equation shows 9*sin(t)^2 + 9*cos(t)^2, which isgpt-oss:20b: fail (misleading) 2026-10-09 — The solution introduces parameters a and b but never specifies that for the given ellipse a=b=3. This omission could mislead a student into thinking the result holds for any a,b, whereas the area formula used requires the specific ellipse parameters.
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.