Evaluate using Stokes' theorem

Statement

If S is an open surface bounded by a closed curve C, and F has continuous first partial derivatives, then

∮_C F·dr = ∬_S (curl F)·n dS,

where C is traversed anticlockwise when seen from the tip of n (right-hand rule).

Evaluation

F = xy i + xy² j, so

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

The square with vertices (1,0), (0,1), (−1,0), (0,−1) is a diamond |x| + |y| ≤ 1 in the xy-plane, so n = k. For a given y (0 ≤ y ≤ 1) x runs from −(1 − y) to (1 − y); the lower half is symmetrical.

∬ (y² − x) dx dy
The x term: odd in x over a symmetric interval → 0.
∬ y² dx dy = 2 ∫₀¹ y² · 2(1 − y) dy
           = 4 [y³/3 − y⁴/4]₀¹
           = 4 (1/3 − 1/4)
           = 1/3

Answer: ∮_C (xy dx + xy² dy) = 1/3.