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Evaluate using Stokes' theorem

Short answer

Curl F = (y^2 − x) k. Surface integral over the square (−1≤x,y≤1) gives ∬(y^2 − x) dxdy = 4/3.

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

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS. Write the integral as ∮_C (xy dx + xy² dy + 0 dz), so F = xy i + xy² j.

curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = 0 i + 0 j + (y² − x) k

C is the square with vertices (1,1), (−1,1), (−1,−1), (1,−1), traversed anticlockwise. Take S as the flat square region in the xy-plane, so n = k and dS = dx dy.

∮_C = ∬_S (y² − x) dx dy
    = ∫₋₁¹ ∫₋₁¹ (y² − x) dx dy
    = ∫₋₁¹ [y² x − x²/2]₋₁¹ dy
    = ∫₋₁¹ 2y² dy
    = 2 · (2/3)
    = 4/3

Answer: ∮_C (xy dx + xy² dy) = 4/3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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