Evaluate using Gauss' divergence theorem F=(2xy+z)i+y²j−(x+3y)k

F = (2xy + z) i + y² j − (x + 3y) k. The region V is bounded by the plane 2x + 2y + z = 6 and the coordinate planes x = 0, y = 0, z = 0 (a tetrahedron).

Divergence

∇·F = ∂(2xy + z)/∂x + ∂(y²)/∂y + ∂(−x − 3y)/∂z = 2y + 2y + 0 = 4y

Limits

The plane meets the axes at x = 3, y = 3, z = 6. So

z : 0 → 6 − 2x − 2y ,   y : 0 → 3 − x ,   x : 0 → 3

Volume integral

∭ 4y dV = ∫₀³ ∫₀^{3−x} 4y (6 − 2x − 2y) dy dx

Let m = 3 − x, so 6 − 2x − 2y = 2(m − y):

∫₀^m 8y (m − y) dy = 8 [ m y²/2 − y³/3 ]₀^m = 8 (m³/2 − m³/3) = 4m³/3

∫₀³ (4/3)(3 − x)³ dx = (4/3) · [ −(3 − x)⁴/4 ]₀³ = (4/3)(81/4) = 27

Answer: ∬_S F·n dS = 27.