The total flux ( $\Phi$ ) of a vector field through a closed surface is critically determined by the parity of the integer $k$ in the vector field's definition. If $k$ is even, the vector field's components are always positive, resulting in a symmetrical field where inward and outward flows cancel each other out, leading to zero net flux. If $k$ is odd, the vector field is perfectly radial, with vectors pointing directly away from the origin, resulting in a positive, non-zero flux quantified by $\Phi=\frac{12 \pi R^{k+2}}{k+2}$. This illustrates how the nature of the vector field, influenced by $k$, dictates the net flow across the surface.
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$\complement\cdots$Counselor
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The most critical factor in determining the flux integral $\Phi$ is whether the integer $k$ is even or odd. This is because the parity of $k$ dictates the fundamental behavior of the vector field $v$.
When $k$ is an even positive integer, the components of the vector field (e.g., $\left(x^1\right)^k$ ) are always positive, regardless of the coordinate's sign. This creates a highly symmetrical vector field that points into certain octants. As a result, the flow of the vector field entering the sphere on one side is perfectly balanced by the flow exiting on the other, leading to a net flux of zero. This outcome is a powerful illustration of physical symmetry causing a cancellation of effects.
When $k$ is an odd positive integer, the components of the vector field (e.g., $\left(x^1\right)^k$ ) retain the sign of their respective coordinates. This means the vector field is perfectly radial, with every vector pointing directly away from the origin. Because the vector field is perfectly aligned with the outward-pointing surface normal of the sphere, the total flow is exclusively outward, resulting in a positive, non-zero flux. The exact value is given by the formula $\Phi=\frac{12 \pi R^{k+2}}{k+2}$.
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