Use the Divergence Theorem to compute the net outward flux of the field F=<-2x, y,-2z> across the surface S, where S is the boundary of the tetrahedron in the first octant formed by the planex+y+z=2. The net outward flux across the boundary of the tetrahedron is:.

The net outward flux across the boundary of the tetrahedron is: -4

Step-by-step explanation:

Given vector field F = ( -2x, y, - 2 z )

= -2 + 1 -2

= -3

According to divergence theorem;

Flux =

x+y+z = 2;  Octant

x from 0 to 2

y from 0 to 2 -x

z from 0 to 2-x-y

= -4

Thus; The net outward flux across the boundary of the tetrahedron is: -4

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2