Surface integrals
Problem 12.486 · hard
Evaluate \( \displaystyle \iint_S x^{2} + z \, dS \), where \( \displaystyle S \) is the cylinder \( \displaystyle x^2 + y^2 = 4,\ 0 \le z \le 4 \).
- Parametrize S by r(u, v) = <2*cos(u), 2*sin(u), v>; dS = ‖r_u × r_v‖ du dv.
- \[ 4 \sin^{2}{\left(u \right)} + 4 \cos^{2}{\left(u \right)} = 4 \]‖r_u × r_v‖² = (2)².✓ Proved
- \[ \int\limits_{0}^{4}\int\limits_{0}^{2 \pi} \left(2 v + 8 \cos^{2}{\left(u \right)}\right)\, du\, dv = 64 \pi \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( 64 \pi \)
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: fail (error) — The integrand should be (4cos²u+v)·2=8cos²u+2v. Integrating over u∈[0,2π] and v∈[0,4] gives 40π, not 64π. The stated answer is incorrect and the solution’s conclusion is wrong.qwen3.6:27b-mlx: pass — The parametrization, surface element calculation, and integration limits are all correct. The final result matches the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-21 — The parametrization, surface element calculation, and integration limits are all correct. The final result matches the stated answer.gpt-oss:20b: fail (error) 2026-10-21 — The integrand should be (4cos²u+v)·2=8cos²u+2v. Integrating over u∈[0,2π] and v∈[0,4] gives 40π, not 64π. The stated answer is incorrect and the solution’s conclusion is wrong.qwen3.6:27b-mlx: pass 2026-10-20 — The parametrization, surface element calculation, and integration bounds are all correct. The solution correctly models the problem and arrives at the right answer.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: pass 2026-10-20 — The parametrization correctly models the cylinder and the bounds are correct. The integrand substitution and magnitude of the normal vector are handled correctly, leading to the right answer.gpt-oss:20b: pass 2026-10-20
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 2026-10-21 with SymPy 1.14.0.