Surface integrals
Problem 12.439 · hard
Evaluate \( \displaystyle \iint_S x^{2} \, dS \), where \( \displaystyle S \) is the cylinder \( \displaystyle x^2 + y^2 = 4,\ 0 \le z \le 1 \).
- Parametrize S by r(u, v) = <2*cos(u), 2*sin(u), v>; dS = ‖r_u × r_v‖ du dv.Reviewed
- \[ 4 \sin^{2}{\left(u \right)} + 4 \cos^{2}{\left(u \right)} = 4 \]‖r_u × r_v‖² = (2)².✓ Proved
- \[ \int\limits_{0}^{1}\int\limits_{0}^{2 \pi} 8 \cos^{2}{\left(u \right)}\, du\, dv = 8 \pi \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( 8 \pi \)
✓ 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. 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 | f times area summed over a triangulated mesh of the surface |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The parametrization correctly models the cylinder of radius 2 and height 1. The surface element calculation is correct, and the integration limits and result are accurate.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The parametrization correctly models the cylinder of radius 2 and height 1. The surface element calculation is correct, and the integration limits and result are accurate.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The parametrization correctly models the cylinder of radius 2 and height 1. The surface element calculation is correct, and the integration limits and result are accurate.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization, bounds, and surface element calculation are correct. The integration steps are valid and yield the correct result.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_integral, checked 2026-10-18 with SymPy 1.14.0.