Flux through a surface F=(x+y²)i−2xj+2yzk

Short answer

Using projection onto the xy-plane, F·n dS = (6y−2xy) dx dy. Integrating over 0≤x≤3, 0≤y≤6−2x yields flux = 81.

Evaluate the flux of F = (x + y²) i − 2x j + 2yz k over the surface of the plane 2x + y + 2z = 6 lying in the first octant.

Flux = ∬_S F·n dS over the part of the plane 2x + y + 2z = 6 in the first octant (triangle with vertices (3,0,0), (0,6,0), (0,0,3)).

∇φ = 2i + j + 2k,  |∇φ| = 3,  n = (2i + j + 2k)/3   (away from origin)
F = (x + y²) i − 2x j + 2yz k
F·n = [2(x + y²) − 2x + 4yz]/3 = (2y² + 4yz)/3

Project on the xy-plane: |n·k| = 2/3, so dS = (3/2) dx dy. Then

F·n dS = (y² + 2yz) dx dy

On the plane z = (6 − 2x − y)/2, so 2yz = y(6 − 2x − y) and y² + 2yz = 6y − 2xy.

Limits: x from 0 to 3, y from 0 to 6 − 2x.

∫₀^(6−2x) (6y − 2xy) dy = (3 − x)(6 − 2x)² = 4(3 − x)³
∫₀³ 4(3 − x)³ dx = 4 · 81/4 = 81

Answer: Flux = 81.

✓ Answer checked by Claude Opus 5.5 (high effort)

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