Area with Green's theorem
Problem 12.308 · easy
Use Green's theorem to find the area enclosed by the ellipse \( \displaystyle \frac{x^2}{16} + \frac{y^2}{25} = 1 \).
- Area = ½∮ (x dy − y dx). Parametrize x = a cos t, y = b sin t, 0 ≤ t ≤ 2π.
- \[ 20 \sin^{2}{\left(t \right)} + 20 \cos^{2}{\left(t \right)} = 20 \]x·y′ − y·x′ simplifies to ab.✓ Proved
- \[ \frac{\int\limits_{0}^{2 \pi} \left(20 \sin^{2}{\left(t \right)} + 20 \cos^{2}{\left(t \right)}\right)\, dt}{2} = 20 \pi \]½∫₀^{2π} ab dt.✓ Proved
Answer \( 20 \pi \)
Lines: 2 proved, 1 not checked. 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 | Not checked | — | 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: fail (error) — The solution never substitutes the actual ellipse parameters a=4 and b=5. It claims x·y′−y·x′ simplifies to ab and then writes 20 sin²+20 cos², but the value 20 comes from ab=20 without justification. Consequently the setup is incomplete and the conclusion is not fully supported.qwen3.6:27b-mlx: pass — The solution correctly applies Green's theorem for area, uses the standard parametrization for the ellipse, and correctly computes the integral. The logic is sound and the result is correct.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-12 — The solution never substitutes the actual ellipse parameters a=4 and b=5. It claims x·y′−y·x′ simplifies to ab and then writes 20 sin²+20 cos², but the value 20 comes from ab=20 without justification. Consequently the setup is incomplete and the conclusion is not fully supported.qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly applies Green's theorem for area, uses the standard parametrization for the ellipse, and correctly computes the integral. The logic is sound and the result is correct.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly applies Green's theorem for area using the standard parametrization of an ellipse. The algebraic simplification and integration are correct, leading to the right answer.
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.