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Double integrals in polar coordinates

Problem 11.34 · easy

Evaluate \( \displaystyle \iint_D \sqrt{x^{2} + y^{2}} \, dA \) over the disk \( \displaystyle x^2 + y^2 \le 9 \) using polar coordinates.

In polar coordinates the disk is 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and dA = r dr dθ.
Reviewed
\[ \sqrt{r^{2} \sin^{2}{\left(\theta \right)} + r^{2} \cos^{2}{\left(\theta \right)}} = \left|{r}\right| \]
The integrand in r and θ.✓ Proved
\[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{3} r \left|{r}\right|\, dr\, d\theta = 18 \pi \]
Integrate, remembering the extra r.✓ Proved

Answer \( 18 \pi \)

Lines: 2 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. How we verify

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
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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the polar bounds for the disk of radius 3, simplifies the integrand to |r| (which is r in the first quadrant/general context of radial distance), and includes the Jacobian factor r. The final integration is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-09-28 — The solution correctly identifies the polar bounds for the disk of radius 3, simplifies the integrand to |r| (which is r in the first quadrant/general context of radial distance), and includes the Jacobian factor r. The final integration is correct.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to simplify the integrand Abs(r) to r, which is valid since r ≥ 0 in polar coordinates. While the final numerical answer is correct, the step skips the necessary justification for removing the absolute value, and presenting 'Abs(r)' as the final integrand without simplification is mathematically incomplete and potentially confusing for a student learning the method.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: pass 2026-09-26 — The solution correctly identifies the polar bounds and the Jacobian. Although it retains the absolute value |r| in the integrand, this is mathematically correct and simplifies to r within the domain r ≥ 0, leading to 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.