Question

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

Answer 1

\({\rm{V}} = {{\rm{x}}^2}{\rm{\hat i}} + 2{{\rm{y}}^3}{\rm{\hat j}} + {{\rm{z}}^4}{\rm{\hat k}}\)

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

= 2x + 6y² + 4z³

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

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