Answer:
Flux = 16Ļ
Step-by-step explanation:
The outward flux of F across the solid cylinder and z = 0 is
ā«ā«F*ds Ā = ā«ā«ā« DivF*dv
F = 2xyĀ²i Ā + Ā 2xĀ²yj + 2xyk
DivF Ā = Ā D/dx (2xyĀ²) Ā + Ā D/dy (2xĀ²y )
DivF Ā = Ā 2yĀ² Ā + Ā 2xĀ²
In cylindrical coordinates Ā dV = rdrdĪødz and as z = 0 the region is a surface ds = rdrdĪø
Parametryzing the surface equation
x = rcosĪø Ā Ā Ā Ā y = Ā r sinĪø Ā Ā Ā and z = z
Div F Ā = 2rĀ²sinĀ²Īø Ā + 2rĀ²cosĀ²Īø
ā«ā«ā« DivF*dv Ā = Ā ā«ā« [2rĀ²sinĀ²Īø Ā + 2rĀ²cosĀ²Īø]* rdrdĪø
ā«ā« 2rĀ² [sinĀ²Īø Ā + cosĀ²Īø]* rdrdĪødz Ā ā Ā ā«ā« 2rĀ³ drdĪø
Integration limits
0 < r < 2 Ā Ā Ā Ā Ā 0 < Īø < 2Ļ
2ā«āĀ² rĀ³ ā«dĪø
(2/4)(2)ā“ 2Ļ
Flux = 16Ļ
