Surface integrals
Problem 12.452 · hard
Evaluate \( \displaystyle \iint_S x^{2} \, dS \), where \( \displaystyle S \) is the cylinder \( \displaystyle x^2 + y^2 = 1,\ 0 \le z \le 3 \).
- Parametrize S by r(u, v) = <cos(u), sin(u), v>; dS = ‖r_u × r_v‖ du dv.Reviewed
- \[ \sin^{2}{\left(u \right)} + \cos^{2}{\left(u \right)} = 1 \]‖r_u × r_v‖² = (1)².✓ Proved
- \[ \int\limits_{0}^{3}\int\limits_{0}^{2 \pi} \cos^{2}{\left(u \right)}\, du\, dv = 3 \pi \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( 3 \pi \)
✓ 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 is correct for the unit cylinder, the bounds cover the specified region, and the surface element magnitude is correctly calculated as 1. The integration steps are accurate.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-19 — The parametrization is correct for the unit cylinder, the bounds cover the specified region, and the surface element magnitude is correctly calculated as 1. The integration steps are accurate.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-16 — The parametrization is correct for the unit cylinder, and the bounds for u and v correctly cover the specified surface. The magnitude of the normal vector is correctly calculated as 1, and the resulting integral evaluates to the stated answer.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization is correct for the given cylinder, and the bounds for u (0 to 2π) and v (0 to 3) correctly cover the surface S. The magnitude of the normal vector is correctly calculated as 1, and the resulting integral evaluation is correct.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 2026-10-19 with SymPy 1.14.0.