how to find the flux of F across the surface S?

Since z is not negative, solving for z from the equation of the hemisphere yields z = √(4 - x² - y²).

So, the flux ∫∫s F · dS equals
∫∫ <xz, y, z> · <-z_x, -z_y, 1> dA
= ∫∫ <x√(4 - x² - y²), y, √(4 - x² - y²)> · <x/√(4 - x² - y²), y/√(4 - x² - y²), 1> dA

= ∫∫ [x² + (y²/√(4 - x² - y²)) + √(4 - x² - y²)] dA.

Now, convert to polar coordinates (since we are integrating over the region bounded by x² + y² = 4):

∫(r = 0 to 2) ∫(θ = 0 to 2π) [(r cos θ)² + (r sin θ)²/√(4 - r²) + √(4 - r²)] * (r dθ dr)

= ∫(r = 0 to 2) ∫(θ = 0 to 2π) [r³ cos²θ + r³ sin²θ/√(4 - r²) + r√(4 - r²)] dθ dr

= ∫(r = 0 to 2) ∫(θ = 0 to 2π) (1/2) [r³ (1 + cos(2θ)) + r³ (1 - cos(2θ))/√(4 - r²) + 2r√(4 - r²)] dθ dr

= ∫(r = 0 to 2) (1/2) [r³ (2π + 0) + r³ (2π - 0)/√(4 - r²) + 4πr√(4 - r²)] dr
= π ∫(r = 0 to 2) [r³ + r³/√(4 - r²) + 2r√(4 - r²)] dr
= π {∫(r = 0 to 2) r³ dr + ∫(r = 0 to 2) [r²/√(4 - r²) + 2√(4 - r²)] r dr}
= π {(1/4) r⁴ {for r = 0 to 2} + ∫(r = 0 to 2) [r²/√(4 - r²) + 2√(4 - r²)] r dr}
= π {4 + ∫(u = 4 to 0) [(4 - u)/√u + 2√u] * (-1/2) du}, letting u = 4 - r²
= π {4 + (1/2) ∫(u = 0 to 4) [(4u^(-1/2) + u^(1/2)] du}
= 4π + (π/2) [(8u^(1/2) + (2/3)u^(3/2)) {for u = 0 to 4}]
= 44π/3.

I hope this helps!

it helps a lot. Thank you very much~