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Surface integrals

Problem 12.521 · hard

Evaluate \( \displaystyle \iint_S z \, dS \), where \( \displaystyle S \) is the cylinder \( \displaystyle x^2 + y^2 = 1,\ 0 \le z \le 2 \).

Parametrize S by r(u, v) = <cos(u), sin(u), v>; dS = ‖r_u × r_v‖ du dv.
\[ \sin^{2}{\left(u \right)} + \cos^{2}{\left(u \right)} = 1 \]
‖r_u × r_v‖² = (1)².✓ Proved
\[ \int\limits_{0}^{2}\int\limits_{0}^{2 \pi} v\, du\, dv = 4 \pi \]
∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved

Answer \( 4 \pi \)

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. How we verify

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	f times area summed over a triangulated mesh of the surface

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (error) — The parameter bounds in step 3 are swapped: the integral is written as Integral(v, (u, 0, 2*pi), (v, 0, 2)), implying u ranges from 0 to 2 and v from 0 to 2pi. Since u corresponds to the angle (0 to 2pi) and v to z (0 to 2), the bounds are incorrect, although the final numerical answer happens to be correct due to symmetry.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-11-09 — The parameter bounds in step 3 are swapped: the integral is written as Integral(v, (u, 0, 2*pi), (v, 0, 2)), implying u ranges from 0 to 2 and v from 0 to 2pi. Since u corresponds to the angle (0 to 2pi) and v to z (0 to 2), the bounds are incorrect, although the final numerical answer happens to be correct due to symmetry.
gpt-oss:20b: pass 2026-11-09
qwen3.6:27b-mlx: pass 2026-11-08 — The parametrization correctly models the cylinder surface, the magnitude of the normal vector is correctly calculated as 1, and the integration bounds and result are correct.
gpt-oss:20b: pass 2026-11-08
qwen3.6:27b-mlx: pass 2026-11-08 — The parametrization is correct for the specified cylinder, and the bounds for u (0 to 2π) and v (0 to 2) correctly cover the surface S. The magnitude of the normal vector is correctly calculated as 1, leading to the correct integral evaluation.
gpt-oss:20b: pass 2026-11-08

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 2026-11-09 with SymPy 1.14.0.