Surface integral over a surface F=yzi+zxj+xyk

Here F = yz i + zx j + xy k and S is the part of the unit sphere x² + y² + z² = 1 with x, y, z ≥ 0 (curved surface only).

Unit normal

For the sphere φ = x² + y² + z² = 1, n = ∇φ/|∇φ| = (x i + y j + z k) (since |∇φ| = 2 and the radius is 1).

F·n

F·n = (yz)(x) + (zx)(y) + (xy)(z) = 3xyz

Parametrize the surface

x = sin φ cos θ, y = sin φ sin θ, z = cos φ, with 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ π/2, and dS = sin φ dφ dθ.

xyz = sin²φ cos φ · sin θ cos θ
∬ F·n dS = 3 ∫₀^{π/2} ∫₀^{π/2} sin²φ cos φ sin θ cos θ · sin φ dφ dθ
         = 3 · [∫₀^{π/2} sin³φ cos φ dφ] · [∫₀^{π/2} sin θ cos θ dθ]
         = 3 · (1/4) · (1/2)
         = 3/8

Answer: ∬ F·n dS = 3/8.

Check by the divergence theorem

div F = ∂(yz)/∂x + ∂(zx)/∂y + ∂(xy)/∂z = 0, so the flux through the closed octant surface is 0. The flux out through the three coordinate quarter-discs is, for each, −∬ yz dy dz (or its analogue) = −1/8. Hence the curved part carries +3/8 ✓.