Triple integrals in cylindrical coordinates
Problem 11.197 · easy
Use cylindrical coordinates to find the volume of the solid inside \( \displaystyle x^2 + y^2 = 1 \), 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.
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{1}\int\limits_{0}^{r} r\, dz\, dr\, d\theta = \frac{2 \pi}{3} \]The iterated integral.✓ Proved
Answer \( \frac{2 \pi}{3} \approx 2.09440 \)
Lines: 1 proved, 1 not checked. 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 | Not checked | — | 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: fail (style) — [domain objection, downgraded to style] The setup incorrectly uses an undefined upper bound R for r; it should be 0 ≤ r ≤ 1. The rest of the description is otherwise correct.qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution incorrectly sets the upper bound for r as R (implied 1) instead of the intersection radius 1/sqrt(2). The cone z=r intersects the cylinder r=1 at z=1, but the problem asks for the volume *inside* the cylinder and *below* the cone. Wait, let's re-read carefully. 'Inside x^2+y^2=1' means r<=1. 'Below z=sqrt(x^2+y^2)' means z<=r. 'Above z=0'. The intersection of the cone and the cylinder is at r=1, z=1. The region is bounded by r=0 to 1? No. The cone z=r goes up. The cylinder is vertical. The volume is inside the cylinder (r<=1) and below the cone (z<=r). This describes an infinite region if z is not bounded above by something else, or rather, the cone opens upward. 'Below the cone' usually means the region containing the z-axis if the cone is z=sqrt(x^2+y^2)? No, z=sqrt(x^2+y^2) is the upper nappe. 'Below' it means z < r. But z >= 0. So 0 <= z <= r. And r <= 1. This is a valid finite volume. Let's check the integral in line 2. Integral(r, (z, 0, r), (r, 0, 1), (theta, 0, 2*pi)). This integrates z from 0 to r, r from 0 to 1, theta 0 to 2pi. This calculates the volume under the cone z=r inside the cylinder r=1. Is this the correct region? Yes. The solid is bounded by z=0, z=r, and r=1. The intersection of z=r and r=1 is z=1. So the height at the edge is 1. The volume is indeed integral_0^{2pi} integral_0^1 integral_0^r r dz dr dtheta = 2pi * integral_0^1 r^2 dr = 2pi/3. The stated answer is 2pi/3. The setup in sentence 1 says '0 <= r <= R'. It does not explicitly define R=1, but line 2 uses 1. Sentence 1 is vague but not strictly wrong if R is defined as 1 implicitly or in the next step. However, usually 'R' implies a generic radius. The main issue is whether the problem implies the standard 'ice cream cone' shape (inside cone, below plane) or this shape (inside cylinder, below cone). 'Inside x^2+y^2=1' is the cylinder. 'Below z=sqrt(x^2+y^2)' is below the cone. The region 0<=z<=r, 0<=r<=1 is correct. The calculation is correct. Why did I think it was wrong? Often these problems ask for the volume *inside* the cone and *below* a plane z=h, or *inside* the cylinder and *above* the cone. Let's re-read: 'inside x^2+y^2=1', 'above z=0', 'below z=sqrt(x^2+y^2)'. This is the region under the cone, inside the cylinder. The volume is 2pi/3. The solution is actually correct. Wait, is there a subtle error? Sentence 1 says '0 <= r <= R'. It doesn't specify R=1. Line 2 uses 1. This is a minor style/clarity issue, not an error. Is the answer 2pi/3 correct? Yes. Is the method correct? Yes. Is the sentence misleading? It says 'the region is ... 0 <= r <= R'. It fails to specify that R=1 for this specific problem, leaving it ambiguous. However, line 2 clarifies it. I will mark it as 'style' or 'pass'. Let's look closer. Is it possible the problem meant the volume *inside* the cone and *below* the plane z=1? No, it says 'inside x^2+y^2=1'. That is a cylinder. Okay, I will pass it.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (style) 2026-10-04 — [domain objection, downgraded to style] The solution incorrectly sets the upper bound for r as R (implied 1) instead of the intersection radius 1/sqrt(2). The cone z=r intersects the cylinder r=1 at z=1, but the problem asks for the volume *inside* the cylinder and *below* the cone. Wait, let's re-read carefully. 'Inside x^2+y^2=1' means r<=1. 'Below z=sqrt(x^2+y^2)' means z<=r. 'Above z=0'. The intersection of the cone and the cylinder is at r=1, z=1. The region is bounded by r=0 to 1? No. The cone z=r goes up. The cylinder is vertical. The volume is inside the cylinder (r<=1) and below the cone (z<=r). This describes an infinite region if z is not bounded above by something else, or rather, the cone opens upward. 'Below the cone' usually means the region containing the z-axis if the cone is z=sqrt(x^2+y^2)? No, z=sqrt(x^2+y^2) is the upper nappe. 'Below' it means z < r. But z >= 0. So 0 <= z <= r. And r <= 1. This is a valid finite volume. Let's check the integral in line 2. Integral(r, (z, 0, r), (r, 0, 1), (theta, 0, 2*pi)). This integrates z from 0 to r, r from 0 to 1, theta 0 to 2pi. This calculates the volume under the cone z=r inside the cylinder r=1. Is this the correct region? Yes. The solid is bounded by z=0, z=r, and r=1. The intersection of z=r and r=1 is z=1. So the height at the edge is 1. The volume is indeed integral_0^{2pi} integral_0^1 integral_0^r r dz dr dtheta = 2pi * integral_0^1 r^2 dr = 2pi/3. The stated answer is 2pi/3. The setup in sentence 1 says '0 <= r <= R'. It does not explicitly define R=1, but line 2 uses 1. Sentence 1 is vague but not strictly wrong if R is defined as 1 implicitly or in the next step. However, usually 'R' implies a generic radius. The main issue is whether the problem implies the standard 'ice cream cone' shape (inside cone, below plane) or this shape (inside cylinder, below cone). 'Inside x^2+y^2=1' is the cylinder. 'Below z=sqrt(x^2+y^2)' is below the cone. The region 0<=z<=r, 0<=r<=1 is correct. The calculation is correct. Why did I think it was wrong? Often these problems ask for the volume *inside* the cone and *below* a plane z=h, or *inside* the cylinder and *above* the cone. Let's re-read: 'inside x^2+y^2=1', 'above z=0', 'below z=sqrt(x^2+y^2)'. This is the region under the cone, inside the cylinder. The volume is 2pi/3. The solution is actually correct. Wait, is there a subtle error? Sentence 1 says '0 <= r <= R'. It doesn't specify R=1. Line 2 uses 1. This is a minor style/clarity issue, not an error. Is the answer 2pi/3 correct? Yes. Is the method correct? Yes. Is the sentence misleading? It says 'the region is ... 0 <= r <= R'. It fails to specify that R=1 for this specific problem, leaving it ambiguous. However, line 2 clarifies it. I will mark it as 'style' or 'pass'. Let's look closer. Is it possible the problem meant the volume *inside* the cone and *below* the plane z=1? No, it says 'inside x^2+y^2=1'. That is a cylinder. Okay, I will pass it.gpt-oss:20b: fail (style) 2026-10-04 — [domain objection, downgraded to style] The setup incorrectly uses an undefined upper bound R for r; it should be 0 ≤ r ≤ 1. The rest of the description is otherwise correct.qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the bounds for cylindrical coordinates and sets up the integral properly. The calculation is correct.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.