Surface Integral..........

Evaluate ∫∫ g(x,y,z) dS

g(x,y,z) = x; Σ is the part of the plane 2x+3y+z=6 in the first octant.

∫∫s g(x,y,z) dS
= ∫∫ x √(1 + (z_x)^2 + (z_y)^2) dA
= ∫∫ x √(1 + (-2)^2 + (-3)^2) dA, since z = 6 - 2x - 3y
= √14 ∫(x = 0 to 3) ∫(y = 0 to 2 - 2x/3) x dy dx
= √14 ∫(x = 0 to 3) x(2 - 2x/3) dx
= (1/3)√14 ∫(x = 0 to 3) (6x - 2x^2) dx
= (1/3)√14 (3x^2 - (2/3)x^3) {for x = 0 to 3}
= 3√14.

I hope this helps!