Surface integrals
Problem 12.360 · hard
Evaluate \( \displaystyle \iint_S z^{2} \, dS \), where \( \displaystyle S \) is the upper hemisphere \( \displaystyle x^2 + y^2 + z^2 = 4,\ z \ge 0 \).
- Parametrize S by r(u, v) = <2*sin(u)*cos(v), 2*sin(u)*sin(v), 2*cos(u)>; dS = ‖r_u × r_v‖ du dv.Reviewed
- \[ \left(4 \sin{\left(u \right)} \sin^{2}{\left(v \right)} \cos{\left(u \right)} + 4 \sin{\left(u \right)} \cos{\left(u \right)} \cos^{2}{\left(v \right)}\right)^{2} + 16 \sin^{4}{\left(u \right)} \sin^{2}{\left(v \right)} + 16 \sin^{4}{\left(u \right)} \cos^{2}{\left(v \right)} = 16 \sin^{2}{\left(u \right)} \]‖r_u × r_v‖² = (4*sin(u))².✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{\frac{\pi}{2}} 16 \sin{\left(u \right)} \cos^{2}{\left(u \right)}\, du\, dv = \frac{32 \pi}{3} \]∬ f(r(u, v)) ‖r_u × r_v‖ du dv.✓ Proved
Answer \( \frac{32 \pi}{3} \)
✓ 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
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The parametrization correctly models the upper hemisphere with appropriate bounds. The surface element calculation and the resulting integral evaluation are correct.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The parametrization, surface element calculation, and integration limits are all correct for the upper hemisphere. The final result matches the stated answer.gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.