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

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

F = (2x − y) i − yz² j − y²z k, S: upper hemisphere x² + y² + z² = a², z ≥ 0. Boundary C: the circle x² + y² = a², z = 0.

Left side: line integral

On C, z = 0 and dz = 0, so F·dr = (2x − y) dx. Put x = a cos t, y = a sin t, 0 ≤ t ≤ 2π, dx = −a sin t dt:

∮ F·dr = ∫₀^{2π} (2a cos t − a sin t)(−a sin t) dt
       = a² ∫₀^{2π} (−2 sin t cos t + sin²t) dt
       = a² (0 + π) = πa²

Right side: surface integral

∇×F = i(∂R/∂y − ∂Q/∂z) + j(∂P/∂z − ∂R/∂x) + k(∂Q/∂x − ∂P/∂y)
    = 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 disc x² + y² ≤ a²:

∬_S (∇×F)·n dS = ∬_{x²+y²≤a²} dx dy = πa²

Both sides equal πa², so Stokes' theorem is verified. ∎