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Verify Stokes' theorem F=2yi+3xj−z²k

Short answer

Line integral over boundary circle (x=3cos t, y=3sin t) yields 9π. Surface integral of curl F·n dS reduces to ∬ dxdy over disk of radius 3, also 9π, verifying Stokes' theorem.

Verify Stokes' theorem for F = 2y i + 3x j − z² k and the surface S is the upper half of the sphere x² + y² + z² = 9, C being the boundary.

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS. The boundary C of the upper hemisphere x² + y² + z² = 9, z ≥ 0 is the circle x² + y² = 9 in the plane z = 0, taken anticlockwise.

Line integral

F = 2y i + 3x j − z² k. On C: x = 3cos t, y = 3sin t, z = 0, dz = 0.

F·dr = 2y dx + 3x dy
     = 2(3 sin t)(−3 sin t) dt + 3(3 cos t)(3 cos t) dt
     = (−18 sin²t + 27 cos²t) dt
∮_C F·dr = −18π + 27π = 9π

Surface integral

curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = 0 i + 0 j + (3 − 2) k = k

For the sphere, n = (x i + y j + z k)/3 and n·k = z/3. Project S on the xy-plane: dS = dx dy/|n·k|, so (curl F)·n dS = (z/3)(3/z) dx dy = dx dy.

∬_S (curl F)·n dS = ∬ dx dy over x² + y² ≤ 9 = π(3)² = 9π

Both sides equal 9π, so Stokes' theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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