Given a smooth vector field, $\textbf{A}$, defined on a smooth parameterised surface, $\textbf{r}(s,t)$ say, how do we obtain an expression for its divergence, $\nabla\cdot\textbf{A}(\textbf{r}(s,t))$, in terms of its components, defined through $\textbf{A}=A_{s}(s,t)\partial_{s}\textbf{r}(s,t)+A_{t}(s,t)\partial_{t}\textbf{r}(s,t)$? Thanks in advance for any help.
To address Ted's comment, I mean the usual three dimensional divergence applied to a vector field $\textbf{B}(\textbf{r})\delta(\textbf{r}-\textbf{r}(s,t))$ satisfying $\textbf{B}(\textbf{r}(s,t))=\textbf{A}(\textbf{r}(s,t))$.
Best Answer
I've worked out the answer. For interest it is
$$ \nabla\cdot\textbf{A}=\partial_{s}A_{s}+\partial_{t}A_{t}+\frac{\partial_{s}|\partial_{s}\textbf{r}\times\partial_{t}\textbf{r}|}{|\partial_{s}\textbf{r}\times\partial_{t}\textbf{r}|}A_{s}+\frac{\partial_{t}|\partial_{s}\textbf{r}\times\partial_{t}\textbf{r}|}{|\partial_{s}\textbf{r}\times\partial_{t}\textbf{r}|}A_{t} $$