The value of the integral

over the closed surface S bounding a volume V, where   is the position vector and is the normal to the surface S, is

The correct option is 3V.

Step-by-Step Explanation:

We are given the surface integral to evaluate over the closed surface $S$ bounding a volume $V$:
$${∯}_S\overrightarrow{r}\cdot \overrightarrow{n}dS$$
where the position vector is defined as:
$$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
and $\overrightarrow{n}$ is the outward normal vector to the surface $S$.

According to Gauss's Divergence Theorem, the surface integral of a vector field over a closed surface is equal to the volume integral of the divergence of that vector field over the volume enclosed by the surface:
$${∯}_S\overrightarrow{r}\cdot \overrightarrow{n}dS={\iiint }_V(\nabla \cdot \overrightarrow{r})dV$$

First, we calculate the divergence of the position vector $\overrightarrow{r}$:
$$\nabla \cdot \overrightarrow{r}=\frac{\partial }{\partial x}\left(x\right)+\frac{\partial }{\partial y}\left(y\right)+\frac{\partial }{\partial z}\left(z\right)$$
Evaluating each partial derivative:
$$\nabla \cdot \overrightarrow{r}=1+1+1=3$$

Substituting this constant value back into the volume integral yields:
$${∯}_S\overrightarrow{r}\cdot \overrightarrow{n}dS={\iiint }_{V}3dV$$
We factor out the constant 3 from the integral:
$${∯}_S\overrightarrow{r}\cdot \overrightarrow{n}dS=3{\iiint }_{V}dV$$

Since the triple integral of $dV$ over the domain $V$ simply gives the volume $V$:
$${∯}_S\overrightarrow{r}\cdot \overrightarrow{n}dS=3V$$