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

Stokes' theorem: ∮_C F·dr = ∬_S (∇×F)·n dS, where C is the boundary of the open surface S, traversed counter-clockwise when seen from the tip of n.

F = (2x − y) i − yz² j − y²z k. S: upper hemisphere x² + y² + z² = 1, z ≥ 0. Boundary C: the circle x² + y² = 1, z = 0.

Left side: line integral around C

On C: x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π. Then F·dr = (2x − y)dx − yz² dy − y²z dz = (2x − y)dx (the other terms vanish since z = 0, dz = 0).

∮ F·dr = ∫₀^{2π} (2cos t − sin t)(−sin t) dt
       = ∫₀^{2π} (−2 sin t cos t + sin²t) dt
       = 0 + π = π

Right side: surface integral

∇×F = | i       j       k    |
      | ∂/∂x   ∂/∂y   ∂/∂z  |
      | 2x−y   −yz²   −y²z  |
    = i(−2yz + 2yz) − j(0 − 0) + k(0 + 1) = k

So (∇×F)·n dS = n_z dS = dx dy (projection onto the xy-plane). The projection of the hemisphere is the unit disc, of area π:

∬_S (∇×F)·n dS = ∬_{x²+y²≤1} dx dy = π

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