1 pt Let S be the part of the plane 4x 4y z 2 which lies

(1 pt) Let S be the part of the plane 4x + 4y + z = 2 which lies in the first octant, oriented upward. Find the flux of the vector field F = 3i + 4j + 2k across the surface S.

Solution

First the definition of flux:
Flux = Integral F. dS
F is constant, S direction is <4,4,1> /sqrt(33) so F.S = (12+16+2)/sqrt(33)=30/sqrt(33).
Next integrate over area described:
Integral[sqrt(33)dxdy] with x from 0 to (2-4y)/4 and y from 0 to 1/2
Area is sqrt33/8

now the value is 30/sqrt33 *sqrt33/8 =30/8 =15/4=3.75