Stokes' Theorem tells us that this integral is equal to
∫C F•dr
Where C is the boundary circle of the hemisphere, given by setting x=0.
Along C, F(0,y,z) = (cos(z),0,0). This is perpendicular to the yz-plane. But dr lives in the yz-plane!
More precisely, if we parametrize C as r(t) = (0, cos t, sin t), then
F(r(t))•r'(t) = (cos (sin t)), 0, 0)•(0, -sin t, cos t) = 0, so
∫CF•dr = ∫F(r(t))•r'(t)dt = 0
Therefore ∫CF•dr =0