∫Calc Practice
Home›Calculus 3›Surface integrals›Problem 12.450

Surface integrals

Problem 12.450 · hard

Evaluate \( \displaystyle \iint_S z \, dS \), where \( \displaystyle S \) is the cylinder \( \displaystyle x^2 + y^2 = 1,\ 0 \le z \le 4 \).
  1. Parametrize S by r(u, v) = <cos(u), sin(u), v>; dS = ‖r_u × r_v‖ du dv.
    Reviewed
  2. \[ \sin^{2}{\left(u \right)} + \cos^{2}{\left(u \right)} = 1 \]
    ‖r_u × r_v‖² = (1)².✓ Proved
  3. \[ \int\limits_{0}^{4}\int\limits_{0}^{2 \pi} v\, du\, dv = 16 \pi \]
    ∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ 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. How we verify

The full receipt
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0f times area summed over a triangulated mesh of the surface

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The parametrization is correct for the specified cylinder, and the bounds for u and v correctly cover the surface S. The calculation of the surface element and the final integral are accurate.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-19 — The parametrization is correct for the specified cylinder, and the bounds for u and v correctly cover the surface S. The calculation of the surface element and the final integral are accurate.
  • gpt-oss:20b: pass 2026-10-19
  • qwen3.6:27b-mlx: pass 2026-10-16 — The parametrization is correct for the given cylinder, the magnitude of the normal vector is correctly calculated as 1, and the integration limits and result are correct.
  • gpt-oss:20b: pass 2026-10-16
  • qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization is correct for the cylinder, the surface element magnitude is correctly calculated as 1, and the integration limits and result are 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.