Surface integral over a surface F=xyi−x²j+(x+z)k

S is the part of the plane 2x + 2y + z = 6 in the first octant: a triangle with vertices (3,0,0), (0,3,0), (0,0,6).

∇φ = 2i + 2j + k,  |∇φ| = 3,  n = (2i + 2j + k)/3
F = xy i − x² j + (x + z) k
F·n = (2xy − 2x² + x + z)/3
dS = dx dy/|n·k| = 3 dx dy

On S, z = 6 − 2x − 2y, so F·n dS = (2xy − 2x² + x + 6 − 2x − 2y) dx dy = (2xy − 2x² − x − 2y + 6) dx dy.

The projection R is the triangle x ≥ 0, y ≥ 0, x + y ≤ 3. With s = 3 − x:

∫₀^s (2xy − 2x² − x − 2y + 6) dy = x s² + (6 − x − 2x²) s − s²
                                 = 3 s (1 + x − x²)

∫₀³ 3(3 − x)(1 + x − x²) dx = 3 ∫₀³ (3 + 2x − 4x² + x³) dx
                            = 3 [3x + x² − 4x³/3 + x⁴/4]₀³
                            = 3 (9 + 9 − 36 + 81/4) = 27/4

Answer: ∬_S F·n dS = 27/4 = 6.75.