Surface integral over a surface F=x²i+y²j+z²k

F = x² i + y² j + z² k, and S is the part of the plane x + y + z = 1 lying in the first octant (between the coordinate planes).

Normal and surface element

For φ = x + y + z, ∇φ = i + j + k, so n = (i + j + k)/√3. Projecting S onto the xy-plane (region R: x ≥ 0, y ≥ 0, x + y ≤ 1):

dS = dx dy / |n·k| = √3 dx dy

F·n

F·n = (x² + y² + z²)/√3 , with z = 1 − x − y

Integral

∬_S F·n dS = ∬_R (x² + y² + (1−x−y)²) dx dy

Over the triangle R, each of the three terms gives the same value 1/12:

• ∬ x² dy dx = ∫₀¹ x²(1−x) dx = 1/3 − 1/4 = 1/12
• ∬ y² dx dy = 1/12 by symmetry
• ∬ (1−x−y)² dx dy = 1/12 by symmetry (it is the same integral in the third variable)

∬_S F·n dS = 1/12 + 1/12 + 1/12 = 1/4

Answer: ∬_S F·n dS = 1/4.

Check by the divergence theorem

∇·F = 2(x + y + z). Over the tetrahedron x + y + z ≤ 1: ∭ (x + y + z) dV = 3 · (1/24) = 1/8, so ∭ ∇·F dV = 1/4. On the three coordinate faces F·n = 0 (for example on x = 0, F·(−i) = −x² = 0), so the flux through the slanted face is 1/4 ✓.