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

Answer:

the flux of the flow field F across σ = 135

Step-by-step 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,

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

$div \ f = \dfrac{\partial }{\partial x}2x + \dfrac{\partial}{\partial y }(3y)$

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