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Verify Stokes' theorem F=2yi+3xj−z²k

Short answer

Line integral around the circle yields 9π. Curl F = k, surface integral over the disc of radius 3 gives 9π. Both sides agree, verifying the theorem.

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

Stokes' theorem: ∮_C F·dr = ∬_S (∇×F)·n dS.

F = 2y i + 3x j − z² k. S: upper half of x² + y² + z² = 9 (radius 3). Boundary C: the circle x² + y² = 9, z = 0.

Left side: line integral

On C: x = 3 cos t, y = 3 sin t, z = 0, 0 ≤ t ≤ 2π; dx = −3 sin t dt, dy = 3 cos t dt, dz = 0.

F·dr = 2y dx + 3x dy − z² dz
     = 2(3 sin t)(−3 sin t) dt + 3(3 cos t)(3 cos t) dt
     = (−18 sin²t + 27 cos²t) dt

∮ F·dr = −18π + 27π = 9π

Right side: surface integral

∇×F = i(0 − 0) − j(0 − 0) + k(∂(3x)/∂x − ∂(2y)/∂y) = (3 − 2) k = k

So (∇×F)·n dS = n_z dS = dx dy, the projection on the xy-plane, which is the disc x² + y² ≤ 9:

∬_S (∇×F)·n dS = area of disc = 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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