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Verify Stokes' theorem F=(2x−y)i−yz²j−y²zk

Short answer

The line integral around the boundary circle equals 4π, and the surface integral of curl F = k over the hemisphere yields 4π. Both sides match, verifying Stokes’ theorem.

Verify Stokes' theorem for F(x, y, z) = (2x − y) i − yz² j − y²z k where S is the upper half of the sphere x² + y² + z² = 4 and C is its boundary.

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS. Here S is the upper half of x² + y² + z² = 4 and its boundary C is the circle x² + y² = 4, z = 0 (anticlockwise).

F = (2x − y) i − yz² j − y²z k, so P = 2x − y, Q = −yz², R = −y²z.

Line integral

On C: x = 2cos t, y = 2sin t, z = 0, dz = 0, and Q = 0.

∮_C F·dr = ∮ (2x − y) dx
         = ∫₀^2π (4cos t − 2sin t)(−2 sin t) dt
         = ∫₀^2π (−8 sin t cos t + 4 sin²t) dt
         = 0 + 4π = 4π

Surface integral

curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (−2yz + 2yz) i + 0 j + (0 + 1) k = k

For the hemisphere, n·k = z/2 and dS = dx dy/(z/2), so (curl F)·n dS = dx dy:

∬_S (curl F)·n dS = ∬ dx dy over x² + y² ≤ 4 = π(2)² = 4π

Both sides equal 4π, so Stokes' theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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