The value of integral \[\mathop{{\int\!\!\!\!\!\int}\mkern-21mu \bigcirc}\limits_{\text{S}} {\overrightarrow {\text{r}} \cdot \overrightarrow {\text{n}} {\text{ds}}} \] over the closed surface S bounding a volume, where \[\overrightarrow {\rm{r}} = {\rm{x\hat i}} + {\rm{y\hat j}} + {\rm{z\hat k}}\] is the position vector and \[{{\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}} \over n} }}}\] is the normal to the surface S, is A. V B. 2V C. 3V D. 4V

The correct answer is $\boxed{\text{A) }V}$.

The given integral is the surface integral of the dot product of the position vector and the surface normal. The surface normal is a vector that is perpendicular to the surface, and its direction is determined by the right-hand rule. The dot product of two vectors is zero if they are perpendicular, so the surface integral will be zero if the surface normal is always perpendicular to the position vector. This is the case if the surface is a sphere, since the surface normal is always pointing directly away from the origin. Therefore, the value of the integral is the volume of the sphere, which is $V$.

The other options are incorrect because they do not take into account the fact that the surface normal is always perpendicular to the position vector. If the surface is not a sphere, then the surface normal will not always be perpendicular to the position vector, and the value of the integral will be different from $V$.