Answer:
Flux = 16π
Step-by-step explanation:
The outward flux of F across the solid cylinder and z = 0 is
∫∫F*ds = ∫∫∫ DivF*dv
F = 2xy²i + 2x²yj + 2xyk
DivF = D/dx (2xy²) + D/dy (2x²y )
DivF = 2y² + 2x²
In cylindrical coordinates dV = rdrdθdz and as z = 0 the region is a surface ds = rdrdθ
Parametryzing the surface equation
x = rcosθ y = r sinθ and z = z
Div F = 2r²sin²θ + 2r²cos²θ
∫∫∫ DivF*dv = ∫∫ [2r²sin²θ + 2r²cos²θ]* rdrdθ
∫∫ 2r² [sin²θ + cos²θ]* rdrdθdz ⇒ ∫∫ 2r³ drdθ
Integration limits
0 < r < 2 0 < θ < 2π
2∫₀² r³ ∫dθ
(2/4)(2)⁴ 2π
Flux = 16π
