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Double integrals in polar coordinates

Problem 11.41 · easy

Evaluate \( \displaystyle \iint_D x^{2} + y^{2} + 1 \, dA \) over the disk \( \displaystyle x^2 + y^2 \le 1 \) using polar coordinates.

In polar coordinates the disk is 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and dA = r dr dθ.
\[ r^{2} \sin^{2}{\left(\theta \right)} + r^{2} \cos^{2}{\left(\theta \right)} + 1 = r^{2} + 1 \]
The integrand in r and θ.✓ Proved
\[ \int\limits_{0}^{2 \pi}\int\limits_{0}^{1} r \left(r^{2} + 1\right)\, dr\, d\theta = \frac{3 \pi}{2} \]
Integrate, remembering the extra r.✓ Proved

Answer \( \frac{3 \pi}{2} \)

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: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution uses an undefined variable R in the bounds description (line 1) instead of the specific value 1 given in the problem statement. While the subsequent integration uses the correct bound, the setup sentence is mathematically imprecise and fails to model the specific problem as stated.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-09-28 — [domain objection, downgraded to style] The solution uses an undefined variable R in the bounds description (line 1) instead of the specific value 1 given in the problem statement. While the subsequent integration uses the correct bound, the setup sentence is mathematically imprecise and fails to model the specific problem as stated.
gpt-oss:20b: pass 2026-09-28
qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution setup in step 1 uses a generic variable R for the radius bound but fails to specify that R=1 for the given disk x^2 + y^2 <= 1. While the subsequent integration uses the correct bound 1, the initial statement is incomplete and technically incorrect as a model of the specific problem.
gpt-oss:20b: pass 2026-09-27
qwen3.6:27b-mlx: fail (style) 2026-09-26 — [domain objection, downgraded to style] The solution sets the radial bound R without defining it, but the problem specifies the disk x^2 + y^2 <= 1, implying R=1. While the final calculation uses r from 0 to 1, the initial sentence is incomplete and technically incorrect as stated because R is an undefined variable in that context. A student might think R is a general parameter rather than specifically 1 for this problem.
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.