Using the Divergence Theorem, find the outward flux of F across the boundary of the region D. F = 2xy2i + 2x2yj + 2xyk ; D: the region cut from the solid cylinder x2 + y2 ≤ 4 by the planes z = 0 and CHEGG

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

1.) 2:3

2.) 64.03

3.) 96.05

4.) 22.81