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

"First quadrant" here means the first octant: S is the part of the plane 2x + 2y + z = 6 with x, y, z ≥ 0 (triangle with vertices (3,0,0), (0,3,0), (0,0,6)).

φ = 2x + 2y + z,   ∇φ = 2i + 2j + k,  |∇φ| = 3
n = (2i + 2j + k)/3   (upward / away from origin)
F = xy i − x² j + (x + z) k
F·n = (2xy − 2x² + x + z)/3

Project onto the xy-plane: dS = dx dy/|n·k| = 3 dx dy. On S, z = 6 − 2x − 2y.

F·n dS = (2xy − 2x² + x + 6 − 2x − 2y) dx dy
       = (2xy − 2x² − x − 2y + 6) dx dy

Region R: triangle x ≥ 0, y ≥ 0, x + y ≤ 3. Integrate y from 0 to 3 − x, then x from 0 to 3.

Let s = 3 − x. Integrating (2xy − 2x² − x − 2y + 6) over y from 0 to s:

Inner = x s² + (6 − x − 2x²) s − s²
      = s [x s + 6 − x − 2x² − s]
      = s [3x − x² + 6 − x − 2x² − 3 + x]
      = 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) = 3 (9/4) = 27/4

Answer: ∬_S F·n dS = 27/4 = 6.75.