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Verify Stokes' theorem F=(x²+y²)i−2xyj

Short answer

Line integration along the rectangle edges yields −4ab². Curl F = −4y k, whose double integral over the rectangle gives −4ab², confirming the theorem.

Verify Stokes' theorem for F = (x² + y²) i − 2xy j taken round the rectangle bounded by the lines x = ±a, y = 0, y = b.

Stokes' theorem (plane form): ∮_C F·dr = ∬_S (∇×F)·k dx dy, where C bounds S in the xy-plane.

F = (x² + y²) i − 2xy j. The rectangle has vertices A(−a, 0), B(a, 0), C(a, b), D(−a, b).

Left side: line integral (counter-clockwise)

F·dr = (x² + y²) dx − 2xy dy.

AB : y = 0, x from −a to a, dy = 0
     ∫ x² dx = [x³/3]₋ₐᵃ = 2a³/3

BC : x = a, y from 0 to b, dx = 0
     ∫ −2a y dy = −a b²

CD : y = b, x from a to −a, dy = 0
     ∫ (x² + b²) dx = [x³/3 + b²x]ₐ^{−a} = −(2a³/3 + 2ab²)

DA : x = −a, y from b to 0, dx = 0
     ∫ −2(−a) y dy = ∫_b^0 2a y dy = −a b²

Total = 2a³/3 − ab² − 2a³/3 − 2ab² − ab² = −4ab²

Right side: surface integral

∇×F = k ( ∂Q/∂x − ∂P/∂y ) = k ( −2y − 2y ) = −4y k

∬ (∇×F)·k dx dy = ∫₀^b ∫₋ₐᵃ (−4y) dx dy
                = −4 · 2a · [y²/2]₀^b = −4 · 2a · b²/2 = −4ab²

Both sides equal −4ab², so Stokes' theorem is verified. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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