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Apply Green's theorem F=(x²−xy³)i+(y²−2xy)j

Short answer

Apply Green's theorem: ∂Q/∂x − ∂P/∂y = −2y + 3xy². The double integral over the square 0≤x,y≤2 equals 8, so ∮_C F·dr = 8.

Apply Green's theorem in plane to evaluate ∮_C F·dr, where F = (x² − xy³) i + (y² − 2xy) j and C is a square with vertices (0,0), (2,0), (2,2), (0,2).

Green's theorem: ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy.

Here F·dr = (x² − xy³) dx + (y² − 2xy) dy.

P = x² − xy³   → ∂P/∂y = −3xy²
Q = y² − 2xy   → ∂Q/∂x = −2y
∂Q/∂x − ∂P/∂y = −2y + 3xy²

Region R: the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.

∬_R (−2y + 3xy²) dx dy = ∫₀² ∫₀² (−2y + 3xy²) dx dy
inner (x): [−2xy + 3x²y²/2]₀² = −4y + 6y²
outer (y): ∫₀² (−4y + 6y²) dy = [−2y² + 2y³]₀² = −8 + 16 = 8

Answer: ∮_C F·dr = 8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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