Verify Gauss' divergence theorem F=x²i+3j+yzk

Gauss' theorem: ∬_S F·n dS = ∭_V div F dV, with F = x² i + 3 j + yz k over the unit cube 0 ≤ x, y, z ≤ 1.

Volume integral

div F = ∂(x²)/∂x + ∂(3)/∂y + ∂(yz)/∂z = 2x + 0 + y

∭ (2x + y) dV = ∫₀¹∫₀¹∫₀¹ (2x + y) dx dy dz
              = ∫₀¹ ∫₀¹ (1 + y) dy dz  (since ∫₀¹ 2x dx = 1)
              = 1 + 1/2 = 3/2

Surface integral (six faces, outward normals)

Face x = 1 (n = i):   F·n = x² = 1       → ∬ = 1
Face x = 0 (n = −i):  F·n = −x² = 0      → 0
Face y = 1 (n = j):   F·n = 3            → 3
Face y = 0 (n = −j):  F·n = −3           → −3
Face z = 1 (n = k):   F·n = yz = y       → ∫₀¹∫₀¹ y dx dy = 1/2
Face z = 0 (n = −k):  F·n = −yz = 0      → 0
Total = 1 + 0 + 3 − 3 + 1/2 + 0 = 3/2

Both integrals are 3/2, so Gauss' divergence theorem is verified.