Double integrals in polar coordinates
Problem 11.42 · easy
Evaluate \( \displaystyle \iint_D \sqrt{x^{2} + y^{2}} \, dA \) over the disk \( \displaystyle x^2 + y^2 \le 4 \) using polar coordinates.
- In polar coordinates the disk is 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and dA = r dr dθ.
- \[ \sqrt{r^{2} \sin^{2}{\left(\theta \right)} + r^{2} \cos^{2}{\left(\theta \right)}} = \left|{r}\right| \]The integrand in r and θ.✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r \left|{r}\right|\, dr\, d\theta = \frac{16 \pi}{3} \]Integrate, remembering the extra r.✓ Proved
Answer \( \frac{16 \pi}{3} \)
Lines: 2 proved, 1 not checked. 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. How we verify
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 |
| 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 | the integral done in x and y, numerically, without polar coordinates, agrees |
Reviewers
gpt-oss:20b: fail (error) — The setup incorrectly uses a generic radius R instead of the specific radius 2 for the disk x^2+y^2≤4. The bounds should be 0≤r≤2, not 0≤r≤R.qwen3.6:27b-mlx: pass — The solution correctly identifies the region and integrand. Although it uses Abs(r) for the square root, this is mathematically correct and simplifies to r in the first quadrant, leading to the correct result.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the region and integrand. Although it uses Abs(r) for the square root, this is mathematically correct and simplifies to r in the first quadrant, leading to the correct result.gpt-oss:20b: fail (error) 2026-09-28 — The setup incorrectly uses a generic radius R instead of the specific radius 2 for the disk x^2+y^2≤4. The bounds should be 0≤r≤2, not 0≤r≤R.qwen3.6:27b-mlx: inconclusive 2026-09-27 — reviewer returned a non-objectgpt-oss:20b: fail (misleading) 2026-09-27 — The first sentence states the disk as 0 ≤ r ≤ R but never specifies that R = 2 for the given disk x² + y² ≤ 4. This omission could mislead a student into thinking the radius is arbitrary.qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the polar bounds and the Jacobian. The use of Abs(r) is technically precise, and since r is non-negative in the integration range, it simplifies correctly to r^2, yielding the correct result.gpt-oss:20b: pass 2026-09-26
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/double_integral_polar, checked 2026-09-28 with SymPy 1.14.0.