Using the divergence theorem we can compute that the outward flux of the vector field is 16π .
The outward flux of F over the solid cylinder and z = 0 is
∫∫F·ds = ∫∫∫ DivF dv
F = 2xy² i + 2x²y j + 2xy k
Div F = D/dx (2xy²) + D/dy (2x²y )
Div F = 2y² + 2x²
In cylindrical coordinates dV = rdrdθdz and as z = 0 the region is a surface ds = r·dr·dθ
Using the parametric form of the surface equation
x = rcosθ y = r sinθ and z = z
Div F = 2r² sin²θ + 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
The divergence theorem, commonly known as Gauss' theorem or Ostrogradsky's theorem, is a theorem that connects the flow of a vector field across a closed surface to the field's divergence in the volume enclosed.
In more detail, the divergence theorem states that the surface integral of a vector field across a closed surface, sometimes referred to as the "flux" through the surface, is equal to the volume integral of the divergence over the region inside the surface.
Therefore the flux is 16π .
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