Skip to main content

Surface integral over a surface F=(x−2z)i+(x+3y+z)j…

Short answer

On plane x+y+z=1, n = (i+j+k)/√3, F·n dS = (8x+5y−1) dx dy over the triangle x,y≥0, x+y≤1. Integration yields 5/3.

Evaluate ∬_S F·n ds where S is the upper side of the triangle with vertices (1,0,0), (0,1,0), (0,0,1) and F = (x − 2z) i + (x + 3y + z) j + (5x + y) k.

S is the triangle on the plane x + y + z = 1 with vertices (1,0,0), (0,1,0), (0,0,1). The upper side has n = (i + j + k)/√3.

F = (x − 2z) i + (x + 3y + z) j + (5x + y) k
F·n = [(x − 2z) + (x + 3y + z) + (5x + y)]/√3
    = (7x + 4y − z)/√3

Project onto the xy-plane: the region R is the triangle x ≥ 0, y ≥ 0, x + y ≤ 1, and dS = dx dy/|n·k| = √3 dx dy. On S, z = 1 − x − y.

F·n dS = (7x + 4y − (1 − x − y)) dx dy = (8x + 5y − 1) dx dy

∬_R (8x + 5y − 1) dx dy = ∫₀¹ ∫₀^(1−x) (8x + 5y − 1) dy dx

Using area(R) = 1/2, ∬ x dA = 1/6, ∬ y dA = 1/6:

= 8(1/6) + 5(1/6) − 1/2 = 13/6 − 3/6 = 10/6 = 5/3

Answer: ∬_S F·n dS = 5/3.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 2 questions →