Surface area by double integrals
Problem 11.487 · hard
Find the area of the part of the surface \( \displaystyle z = - 2 x - y + 1 \) that lies above the rectangle \( \displaystyle [0, 1] \times [0, 3] \).
- Surface area = ∬ √(1 + f_x² + f_y²) dA.Reviewed
- \[ \left(\frac{\partial}{\partial x} \left(- 2 x - y + 1\right)\right)^{2} + \left(\frac{\partial}{\partial y} \left(- 2 x - y + 1\right)\right)^{2} + 1 = 6 \]1 + f_x² + f_y².✓ Proved
- \[ \int\limits_{0}^{3}\int\limits_{0}^{1} \sqrt{6}\, dx\, dy = 3 \sqrt{6} \]Integrate.✓ Proved
Answer \( 3 \sqrt{6} \approx 7.34847 \)
✓ 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. How we verify
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 | the surface as a mesh of 28,800 triangles |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the surface area formula, computes the partial derivatives and the integrand constant, and evaluates the double integral over the specified rectangular region.gpt-oss:20b: pass 2026-10-15
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/surface_area_double, checked 2026-10-18 with SymPy 1.14.0.