Area with Green's theorem
Problem 12.181 · easy
Use Green's theorem to find the area enclosed by the ellipse \( \displaystyle \frac{x^2}{16} + \frac{y^2}{16} = 1 \).
- Area = ½∮ (x dy − y dx). Parametrize x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.Reviewed
- \[ 16 \sin^{2}{\left(t \right)} + 16 \cos^{2}{\left(t \right)} = 16 \]x·y′ − y·x′ simplifies to ab.✓ Proved
- \[ \frac{\int\limits_{0}^{2 \pi} \left(16 \sin^{2}{\left(t \right)} + 16 \cos^{2}{\left(t \right)}\right)\, dt}{2} = 16 \pi \]½∫₀^{2π} ab dt.✓ Proved
Answer \( 16 \pi \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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. The algebraic simplification and integration are correct, yielding the expected area of the circle (which is a special case of an ellipse with a=b=4).
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly applies Green's theorem for area using the standard parametrization. The algebraic simplification and integration are correct, yielding the expected area of the circle (which is a special case of an ellipse with a=b=4).gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the curve as a circle of radius 4 (since a=b=4) and applies the standard area formula derived from Green's theorem. The parametrization and integration steps are correct.gpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.