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

Short answer

curl F = 4y k. The surface integral over the rectangle 0≤x≤a, 0≤y≤b gives ∬ 4y dxdy = 2ab². The line integral around the boundary also yields 2ab², verifying Stokes' theorem.

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

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS. Here F = (x² − y²) i + 2xy j, and the rectangle with vertices O(0,0), A(a,0), B(a,b), D(0,b) lies in the xy-plane, so n = k.

Surface integral

curl F · k = ∂Q/∂x − ∂P/∂y = 2y − (−2y) = 4y
∬_S 4y dx dy = ∫₀^a ∫₀^b 4y dy dx = ∫₀^a 2b² dx = 2ab²

Line integral (anticlockwise: OA → AB → BD → DO)

F·dr = (x² − y²) dx + 2xy dy.

OA: y = 0, dy = 0:       ∫₀^a x² dx = a³/3
AB: x = a, dx = 0:       ∫₀^b 2a y dy = ab²
BD: y = b, dy = 0, x: a→0: ∫ₐ⁰ (x² − b²) dx = −a³/3 + ab²
DO: x = 0, dx = 0, and 2xy = 0: 0
Sum = a³/3 + ab² − a³/3 + ab² = 2ab²

Both sides equal 2ab², so Stokes' theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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