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

Short answer

div F = 4y. By the divergence theorem, the flux is the triple integral of 4y over the tetrahedron bounded by 2x+3y+z=6 and coordinate planes, which evaluates to 12.

Use Gauss' divergence theorem to evaluate ∬_S F·n ds where F = (2xy + z) i + y² j − (x + 3y) k and S is the surface bounded by the plane 2x + 3y + z = 6, x = 0, y = 0, z = 0.

Gauss' divergence theorem: ∬_S F·n dS = ∭_V div F dV. S is the closed surface of the tetrahedron bounded by the plane 2x + 3y + z = 6 and the coordinate planes (vertices (3,0,0), (0,2,0), (0,0,6)).

F = (2xy + z) i + y² j − (x + 3y) k
div F = 2y + 2y + 0 = 4y

Limits: x from 0 to 3; y from 0 to (6 − 2x)/3; z from 0 to 6 − 2x − 3y.

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

Let k = 6 − 2x. The inner integral is

∫₀^(k/3) (4ky − 12y²) dy = [2ky² − 4y³]₀^(k/3)
                         = 2k·k²/9 − 4k³/27
                         = 2k³/9 − 4k³/27 = 2k³/27

Now integrate with respect to x:

∫₀³ (2k³/27) dx = (2/27) ∫₀³ (6 − 2x)³ dx
                = (2/27) · [−(6 − 2x)⁴/8]₀³
                = (2/27) · (1296/8)
                = (2/27)(162) = 12

Answer: ∬_S F·n dS = 12.

✓ Answer checked by Claude Opus 5.5 (high effort)

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