Surface integrals
Problem 12.481 · hard
Evaluate \( \displaystyle \iint_S z^{2} \, dS \), where \( \displaystyle S \) is the upper hemisphere \( \displaystyle x^2 + y^2 + z^2 = 1,\ z \ge 0 \).
- Parametrize S by r(u, v) = <sin(u)*cos(v), sin(u)*sin(v), cos(u)>; dS = ‖r_u × r_v‖ du dv.Reviewed
- \[ \left(\sin{\left(u \right)} \sin^{2}{\left(v \right)} \cos{\left(u \right)} + \sin{\left(u \right)} \cos{\left(u \right)} \cos^{2}{\left(v \right)}\right)^{2} + \sin^{4}{\left(u \right)} \sin^{2}{\left(v \right)} + \sin^{4}{\left(u \right)} \cos^{2}{\left(v \right)} = \sin^{2}{\left(u \right)} \]‖r_u × r_v‖² = (sin(u))².✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{\frac{\pi}{2}} \sin{\left(u \right)} \cos^{2}{\left(u \right)}\, du\, dv = \frac{2 \pi}{3} \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( \frac{2 \pi}{3} \)
✓ 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 covers the upper hemisphere with appropriate bounds. The surface element calculation is correct, and the resulting integral evaluates to the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-21 — The parametrization correctly covers the upper hemisphere with appropriate bounds. The surface element calculation is correct, and the resulting integral evaluates to the stated answer.gpt-oss:20b: pass 2026-10-21qwen3.6:27b-mlx: pass 2026-10-20 — The parametrization, surface element calculation, and integration limits are correct for the upper hemisphere. The final result matches the stated answer.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: pass 2026-10-20 — The parametrization, surface element calculation, and integration bounds are all correct for the upper hemisphere.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.