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

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>