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 answer is 3V.

Based on the provided images, we identify the following components of the problem:

1. Image 1 shows the position vector:
$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

2. Image 2 shows the unit outward normal vector notation:
$\overrightarrow{n}$

3. Image 0 shows the closed surface integral to be evaluated:
${\iint }_S\overrightarrow{r}·\overrightarrow{n}dS$

To find the value of this surface integral over a closed surface S bounding a volume V, we can apply Gauss's Divergence Theorem.

Gauss's Divergence Theorem states that the flux of a vector field through a closed surface is equal to the volume integral of the divergence of that vector field over the enclosed volume V:

${\iint }_S\overrightarrow{F}·\overrightarrow{n}dS={\iiint }_V\left(\nabla ·\overrightarrow{F}\right)dV$

In this problem, the vector field is the position vector:
$\overrightarrow{F}=\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

Let's calculate the divergence of the position vector:
$\nabla ·\overrightarrow{r}=\left(\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k}\right)·\left(x\hat{i}+y\hat{j}+z\hat{k}\right)$

Applying the dot product operation:
$\nabla ·\overrightarrow{r}=\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}$

Evaluating the partial derivatives:
$\nabla ·\overrightarrow{r}=1+1+1=3$

Now, substitute this divergence back into the Divergence Theorem formula:
${\iint }_S\overrightarrow{r}·\overrightarrow{n}dS={\iiint }_{V}3dV$

Since 3 is a constant, we can factor it outside the integral:
${\iint }_S\overrightarrow{r}·\overrightarrow{n}dS=3{\iiint }_{V}dV$

The volume integral of the differential volume element over the entire region is simply the total volume V bounded by S:
${\iiint }_{V}dV=V$

Substituting this in, we obtain:
${\iint }_S\overrightarrow{r}·\overrightarrow{n}dS=3V$