Surface integral over a surface F=y²z²i+z²x²j+x²y²k

F = y²z² i + z²x² j + x²y² k and S is the upper hemisphere x² + y² + z² = 1, z ≥ 0 (curved surface only).

Use the divergence theorem on the closed surface (hemisphere + base disc)

∇·F = ∂(y²z²)/∂x + ∂(z²x²)/∂y + ∂(x²y²)/∂z = 0

So the total flux through the closed surface (curved part + base) is zero:

∬_{curved} F·n dS + ∬_{base} F·n dS = 0

Flux through the base

The base is the disc x² + y² ≤ 1 at z = 0, with n = −k:

F·n = −x²y²
∬_{base} F·n dS = −∬_{x²+y²≤1} x²y² dx dy

In polar form x = r cos θ, y = r sin θ:

∬ x²y² dx dy = ∫₀^{2π} cos²θ sin²θ dθ · ∫₀¹ r⁵ dr
             = (π/4)(1/6) = π/24

so the base flux is −π/24.

Curved surface

∬_{curved} F·n dS = 0 − (−π/24) = π/24

Answer: ∬_S F·n dS = π/24.

(Direct check: on the sphere n = (x, y, z), so F·n = xyz(xy + yz + zx). The terms xy²z² and x²yz² are odd in x or y and integrate to zero over the hemisphere, leaving ∬ x²y²z dS = π/24 ✓.)