The divergence of the vector field \[\overrightarrow {\rm{U}} = {{\rm{e}}^{\rm{x}}}\left( {\cos \,{\rm{y\hat i}} + \sin {\rm{y\hat j}}} \right)\] is A. 0 B. ex cos y + ex sin y C. 2ex cos y D. 2ex sin y

The divergence of a vector field is a measure of how much the vector field spreads out or converges at a point. It is defined as the sum of the partial derivatives of the $x$, $y$, and $z$ components of the vector field with respect to their respective coordinates.

In this case, the vector field is $\overrightarrow {\rm{U}} = {{\rm{e}}^{\rm{x}}}\left( {\cos \,{\rm{y\hat i}} + \sin {\rm{y\hat j}}} \right)$. The partial derivatives of the $x$ and $y$ components are $0$, so the divergence is $0$.

The other options are incorrect because they are not equal to $0$.