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

Short answer

The surface integral of curl F over the rectangle is −2ab². The line integral along the boundary also gives −2ab², confirming Stokes’ theorem.

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

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

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, −2xy = 0: 0
Total = a³/3 − ab² − a³/3 − ab² = −2ab²

Both are −2ab², so Stokes' theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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