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Surface integral over a surface F=6zi−4j+yk

Short answer

Project the plane onto the xy-plane: dS = (7/6) dxdy, F·n = (12−4x)/7. Then ∬ F·n dS = (1/6)∬ (12−4x) dxdy over the triangle 2x+3y≤12, x,y≥0, giving 8.

Evaluate ∬_S F·n ds where F = 6z i − 4 j + y k and S is the region of the plane 2x + 3y + 6z = 12 bounded in the first octant.

S is the part of the plane 2x + 3y + 6z = 12 in the first octant (a triangle with vertices (6,0,0), (0,4,0), (0,0,2)).

φ = 2x + 3y + 6z
∇φ = 2i + 3j + 6k,  |∇φ| = 7
n = (2i + 3j + 6k)/7   (outward, away from the origin)
F·n = (6z·2 − 4·3 + y·6)/7 = (12z − 12 + 6y)/7

Project S onto the xy-plane (region R: triangle 2x + 3y ≤ 12 with x, y ≥ 0): dS = dx dy/|n·k| = (7/6) dx dy. On S, z = (12 − 2x − 3y)/6, so 12z = 24 − 4x − 6y and

F·n = (24 − 4x − 6y − 12 + 6y)/7 = (12 − 4x)/7
∬_S F·n dS = ∬_R (12 − 4x)/7 · (7/6) dx dy
           = (1/6) ∬_R (12 − 4x) dx dy

R: x from 0 to 6, y from 0 to (12 − 2x)/3.

∫₀⁶ (12 − 4x)(12 − 2x)/3 dx = (1/3) ∫₀⁶ (144 − 72x + 8x²) dx
                            = (1/3)(864 − 1296 + 576)
                            = (1/3)(144) = 48
∬_S F·n dS = 48/6 = 8

Answer: ∬_S F·n dS = 8.

✓ Answer checked by Claude Opus 5.5 (high effort)

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