Double integrals in polar coordinates
Problem 11.37 · easy
Evaluate \( \displaystyle \iint_D x^{2} + y^{2} \, dA \) 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.
- \[ r^{2} \sin^{2}{\left(\theta \right)} + r^{2} \cos^{2}{\left(\theta \right)} = r^{2} \]The integrand in r and θ.✓ Proved
- \[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{2} r^{3}\, dr\, d\theta = 8 \pi \]Integrate, remembering the extra r.✓ Proved
Answer \( 8 \pi \)
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 first sentence incorrectly describes the radial bound as "0 ≤ r ≤ R" instead of the specific value R=2 for the disk x²+y²≤4. This omission misstates the region and could mislead a student about the correct limits.qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to explicitly substitute the radius R=2 into the integration bounds, leaving the upper limit as the undefined variable R in the setup sentence. While the final calculation uses 2, the logical gap between the generic setup and the specific evaluation is an error in rigor.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-09-28 — [domain objection, downgraded to style] The solution fails to explicitly substitute the radius R=2 into the integration bounds, leaving the upper limit as the undefined variable R in the setup sentence. While the final calculation uses 2, the logical gap between the generic setup and the specific evaluation is an error in rigor.gpt-oss:20b: fail (error) 2026-09-28 — The first sentence incorrectly describes the radial bound as "0 ≤ r ≤ R" instead of the specific value R=2 for the disk x²+y²≤4. This omission misstates the region and could mislead a student about the correct limits.qwen3.6:27b-mlx: fail (error) 2026-09-27 — Line 1 defines the radius bound as R but fails to specify that R=2 for the given disk x^2 + y^2 <= 4. Line 3 then implicitly uses 2 without establishing this connection from the problem statement.gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: fail (style) 2026-09-26 — [domain objection, downgraded to style] Line 1 defines a generic radius R but fails to substitute R=2 from the problem statement (x^2 + y^2 <= 4) before setting up the integral in Line 3. This leaves the bounds undefined in the text, relying on the equation line to implicitly fix the value.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.