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Evaluate a line integral F=(x²+y²)i−2xyj

Short answer

Using Green's theorem, ∂Q/∂x − ∂P/∂y = −4y. The double integral over the rectangle −a≤x≤a, 0≤y≤b gives −4ab², so ∮_C F·dr = −4ab².

Evaluate ∮_C F·dr, where C is the rectangle bounded by the lines x = ±a, y = 0, y = b and F = (x² + y²) i − 2xy j.

Let P = x² + y², Q = −2xy. C is the closed rectangle −a ≤ x ≤ a, 0 ≤ y ≤ b, taken anticlockwise. Use Green's theorem:

∮_C F·dr = ∬_R (∂Q/∂x − ∂P/∂y) dx dy
∂Q/∂x = −2y,   ∂P/∂y = 2y
∂Q/∂x − ∂P/∂y = −4y

∬_R (−4y) dx dy = ∫₋ₐ^a ∫₀^b (−4y) dy dx
                = ∫₋ₐ^a (−2b²) dx
                = −2b² · 2a
                = −4ab²

Direct check (anticlockwise: bottom, right, top, left):

Bottom y = 0, x: −a→a:   ∫ x² dx = 2a³/3
Right  x = a, y: 0→b:    ∫ −2ay dy = −ab²
Top    y = b, x: a→−a:   ∫ (x² + b²) dx = −(2a³/3 + 2ab²)
Left   x = −a, y: b→0:   ∫ 2ay dy (since −2xy = 2ay) = −ab²
Sum = 2a³/3 − ab² − 2a³/3 − 2ab² − ab² = −4ab²

Answer: ∮_C F·dr = −4ab² (anticlockwise; reverse the sign for clockwise).

✓ Answer checked by Claude Opus 5.5 (high effort)

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