The divergence of the vector field V = x² i + 2y³ j + z⁴ k at x = 1, y = 2, z = 3 is ________

In the range: 133.9 - 134.1

$\mathrm{V}={\mathrm{x}}^2\hat{\mathrm{i}}+2{\mathrm{y}}^3\hat{\mathrm{j}}+{\mathrm{z}}^4\hat{\mathrm{k}}$

Divergence $\left(\nabla V\right)=\frac{\partial }{\partial \mathrm{x}}\left({\mathrm{x}}^2\right)+\frac{\partial }{\partial \mathrm{y}}\left(2{\mathrm{y}}^3\right)+\frac{\partial }{\partial \mathrm{x}}\left({\mathrm{z}}^4\right)$

= 2x + 6y² + 4z³

At x= 1, y=2 and z=3

[Divergence (V)]= 2 × 1 + 6 × 2² + 4 × 3³ = 134