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,
[tex]\iint \ F.ds = \iiint \ div. f \ dV[/tex]
[tex]div \ f = \dfrac{\partial }{\partial x}2x + \dfrac{\partial}{\partial y }(3y)[/tex]
f = 2 +3 = 5
where ;
F = 2xi + 3yj
Thus , the triple integral can now be ;
[tex]= \iiint 5.dV[/tex]
[tex]=5 \iiint \ dV[/tex]
[tex]= 5 \ \int^{3}_{0}\int^{3}_{0}\int^{3}_{0} \ dV[/tex]
= 5(3)(3)(3)
= 135
