Question

F = 2xi+3yj and σ is the cube with opposite corners at (0,0,0) and (3,3,3), oriented outwards. Find the flux of the flow field F across σ.

Answer 1

the flux of the flow field F across σ = 135

Explanation:

Given that :

F = 2xi + 3yj

and σ is the cube with opposite corners at (0,0,0) and (3,3,3) oriented outwards.

Using divergence theorem,

`\iint \ F.ds = \iiint \ div. f \ dV`

`div \ f = (\partial )/(\partial x)2x + (\partial)/(\partial y )(3y)`

f = 2 +3 = 5

where ;

F = 2xi + 3yj

Thus , the triple integral can now be ;

`= \iiint 5.dV`

`=5 \iiint \ dV`

`= 5 \ \int^(3)_(0)\int^(3)_(0)\int^(3)_(0) \ dV`

= 5(3)(3)(3)

= 135