Consider a vector field $$\overrightarrow {\text{A}} \left( {\overrightarrow {\text{r}} } \right).$$  The closed loop line integral $$\oint {\overrightarrow {\text{A}} \cdot \overrightarrow {{\text{d}}l} } $$  can be expressed as

A. $$\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc} {\left( {\nabla \times \overrightarrow {\text{A}} } \right) \cdot \overrightarrow {{\text{ds}}} } $$    over the closed surface bounded by the loop

B. $$\mathop{{\int\!\!\!\!\!\int\!\!\!\!\!\int}\mkern-31.2mu \bigodot} {\left( {\nabla \cdot \overrightarrow {\text{A}} } \right){\text{dv}}} $$    over the closed volume bounded by the top

C. $$\iiint {\left( {\nabla \cdot \overrightarrow {\text{A}} } \right){\text{dv}}}$$    over the open volume bounded by the loop

D. $$\iint {\left( {\nabla \times \overrightarrow {\text{A}} } \right) \cdot \overrightarrow {{\text{ds}}} }$$    over the open surface bounded by the loop

Answer: Option D