Let S be the part of the plane 1x2yz2 which lies in the firs

Let S be the part of the plane 1x+2y+z=2 which lies in the first octant, oriented upward. Find the flux of the vector field F=4i+1j+2k across the surface S.

Solution

Using rectangular coordinates with z = 2 - 1x - 2y,
n = (-dz/dx, -dz/dy, 1) = (1, 2, 1).

The region R in the xy-plane is given by x + 2y = 2.

Thus, flux
= integral(R) F * n dA
= integral(R) (4,1,2) * (1,2,1) dA

= integral(R) (4*1+1*2+2*1) dA
= integral(R) 8 dA
= 19 * Area(R, which is a triangle)
= 19 * ((1/1)*(1/2)/2) = 19/4.