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Evaluate using Gauss' divergence theorem F=x²yi+xy²j+2xyzk

Short answer

Divergence ∇·F = 6xy. Integrating over the tetrahedron bounded by the coordinate planes and x+2y+z=2 with limits z = 0→2−x−2y, y = 0→(2−x)/2, x = 0→2 yields a flux of 2/5.

Using Gauss' divergence theorem, evaluate ∬_S F·n ds, where F = x²y i + xy² j + 2xyz k and S is the surface bounded by the planes x = 0, y = 0, z = 0, x + 2y + z = 2.

Gauss' divergence theorem: ∬_S F·n dS = ∭_V (∇·F) dV for a closed surface S enclosing V.

F = x²y i + xy² j + 2xyz k

Divergence

∇·F = ∂(x²y)/∂x + ∂(xy²)/∂y + ∂(2xyz)/∂z = 2xy + 2xy + 2xy = 6xy

Region V

The tetrahedron bounded by x = 0, y = 0, z = 0 and x + 2y + z = 2. Its intercepts are x = 2, y = 1, z = 2. Limits:

z : 0 → 2 − x − 2y
y : 0 → (2 − x)/2
x : 0 → 2

Evaluate

∭ 6xy dV = ∫₀² ∫₀^{(2−x)/2} 6xy (2 − x − 2y) dy dx

Let k = 2 − x. Inner integral:

∫₀^{k/2} 6xy(k − 2y) dy = 6x [ k y²/2 − 2y³/3 ]₀^{k/2}
                        = 6x ( k³/8 − k³/12 ) = x k³/4

Outer integral (put u = 2 − x, x = 2 − u):

∫₀² x(2−x)³/4 dx = (1/4) ∫₀² (2 − u) u³ du
                 = (1/4) [ u⁴/2 − u⁵/5 ]₀² = (1/4)(8 − 32/5) = (1/4)(8/5) = 2/5

Answer: ∬_S F·n dS = 2/5.

✓ Answer checked by Claude Opus 5.5 (high effort)

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