Triple integrals in spherical coordinates
Problem 11.225 · easy
Use spherical coordinates to find the volume of the upper half of the ball \( \displaystyle x^2 + y^2 + z^2 \le 4 \).
- ρ runs from 0 to 2, φ from 0 to pi/2, θ around the full circle; dV = ρ² sin φ dρ dφ dθ.Reviewed
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{2} \rho^{2} \sin{\left(\phi \right)}\, d\rho\, d\phi\, d\theta = \frac{16 \pi}{3} \]The iterated integral factors into three one-variable integrals.✓ Proved
Answer \( \frac{16 \pi}{3} \approx 16.75516 \)
✓ Nihil obstat Lines: 1 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | nested quadrature in rectangular coordinates over one symmetric piece |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05 — The setup correctly identifies the bounds for the upper hemisphere in spherical coordinates and uses the correct volume element. The resulting integral and its evaluation are correct.gpt-oss:20b: pass 2026-10-05
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/spherical_integral, checked 2026-10-05 with SymPy 1.14.0.