Triple integrals in cylindrical coordinates
Problem 11.198 · easy
Use cylindrical coordinates to find the volume of the solid inside \( \displaystyle x^2 + y^2 = 4 \), above \( \displaystyle z = 0 \) and below the cone \( \displaystyle z = \sqrt{x^2 + y^2} \).
- In cylindrical coordinates x² + y² = r² and dV = r dz dr dθ; the region is 0 ≤ θ ≤ 2π, 0 ≤ r ≤ R and z between the surfaces.Reviewed
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2}\int\limits_{0}^{r} r\, dz\, dr\, d\theta = \frac{16 \pi}{3} \]The iterated integral.✓ Proved
Answer \( \frac{16 \pi}{3} \approx 16.75516 \)
Lines: 1 proved, 1 reviewed. 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.
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 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The setup correctly identifies the bounds for cylindrical coordinates and the integrand. The calculation is correct.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The setup correctly identifies the bounds for cylindrical coordinates and the integrand. The calculation is correct.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution incorrectly sets the upper limit of integration for r to 2. The solid is bounded above by the cone z = r and laterally by the cylinder r = 2. The intersection occurs at z = 2, so the region is split: for 0 <= z <= 2, r goes from 0 to z (if integrating dz first) or the bounds for r must account for the fact that the cone is the upper bound for z, not a bound for r in the outer integral in this specific order. Specifically, if integrating dz first, z goes from 0 to r, but r only goes up to 2. However, the cone z=r is below the plane z=2 only for r<=2. Wait, the problem says 'below the cone'. The cone opens upward. The region is inside the cylinder r=2, above z=0, and below z=r. This implies z goes from 0 to r. But for r > 2, we are outside the cylinder. So r goes from 0 to 2. For each r, z goes from 0 to r. The integral is int_0^{2pi} int_0^2 int_0^r r dz dr dtheta. This yields 16pi/3. The stated answer is correct. The sentence says 'z between the surfaces'. The surfaces are z=0 and z=r. The sentence is vague but not strictly wrong. However, the equation line is marked 'unchecked' in the prompt description but the instructions say 'EQUATION... checked... do not re-check'. The prompt says '2. [equation, unchecked]'. This is a contradiction in the prompt format vs instructions. Assuming the equation is correct, the sentence is acceptable. But wait, is the region actually bounded by the cone? 'Below the cone z=sqrt(x^2+y^2)' means z <= r. Inside x^2+y^2=4 means r<=2. Above z=0. So 0<=z<=r and 0<=r<=2. The integral calculates this volume. The result 16pi/3 is correct. Why would it fail? Let's re-read carefully. 'below the cone'. Usually, 'below the cone' z=r means the region containing the z-axis? No, z=r is a cone opening up. 'Below' it means z < r. This is the region outside the cone? No, z < r is the region 'outside' the cone in terms of angle, but 'below' in terms of z. For a fixed r, z is between 0 and r. This is the region inside the cone? No, inside the cone usually means the volume containing the z-axis, which is z > r. 'Below the cone' means z < r. This is the region between the xy-plane and the cone. This is an infinite region unless bounded. It is bounded by r=2. So it is a cylinder with a conical hole? No, z < r means for a given r, z is small. At r=0, z=0. At r=2, z=2. The volume is the space under the cone surface. This is correct. The calculation is correct. The sentence is slightly imprecise ('z between the surfaces' implies two surfaces bounding z, which are z=0 and z=r). It is acceptable. Is there a subtle error? The prompt asks to find what is wrong. If nothing is wrong, pass. But usually these have a trick. Let's check the volume of the cone itself. Volume of cone height 2, radius 2 is 1/3 * pi * 2^2 * 2 = 8pi/3. The volume calculated is 16pi/3. This is twice the volume of the cone. The region 0<=z<=r inside r=2 is the volume *under* the cone. The volume of the cylinder is pi*2^2*2 = 8pi. The volume above the cone (inside cylinder) is 8pi - 8pi/3 = 16pi/3? No. Volume of cone (z>r) is 8pi/3. Volume of cylinder is 8pi. Volume below cone (z<r) is 8pi - 8pi/3 = 16pi/3. Yes. The math is correct. The sentence is fine. Verdict: pass.gpt-oss:20b: pass 2026-10-04
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/cylindrical_integral, checked 2026-10-04 with SymPy 1.14.0.