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Evaluate using Stokes' theorem

Short answer

By Stokes, ∮ F·dr = ∬ curl F·n dS. Here curl F = (y²−x)k, and S is the square |x|+|y|≤1. The x term vanishes by symmetry. Integral of y² gives 4∫₀^1 y²(1−y)dy = 1/3.

Evaluate ∮_C (xy dx + xy² dy) by applying Stokes' theorem where C is the square in the xy-plane with vertices (1,0), (−1,0), (0,1), (0,−1).

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS.

Write the integrand as F·dr with F = xy i + xy² j (no k component, z = 0 on C).

curl F = (∂Q/∂x − ∂P/∂y) k = (y² − x) k

C is the square with vertices (1,0), (0,1), (−1,0), (0,−1): the region |x| + |y| ≤ 1 in the xy-plane, traversed anticlockwise. Take S as this flat region, n = k, dS = dx dy.

∬ (y² − x) dx dy
−x term: odd in x over a symmetric region → 0
y² term: 2 ∫₀¹ ∫₋(1−y)^(1−y) y² dx dy
       = 2 ∫₀¹ 2y²(1 − y) dy
       = 4 (1/3 − 1/4)
       = 1/3

Answer: ∮_C (xy dx + xy² dy) = 1/3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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