A vector field is defined as

$\overrightarrow{f}\left(x,y,z\right)=\frac{x}{{\left[x^2+y^2+z^2\right]}^{\frac{3}{2}}}\hat{i}+\frac{y}{{\left[x^2+y^2+z^2\right]}^{\frac{3}{2}}}\hat{j}+\frac{z}{{\left[x^2+y^2+z^2\right]}^{\frac{3}{2}}}\hat{k}$

where î, ĵ, k̂ are unit vectors along the axes of a right-handed rectangular/Cartesian coordinate system. The surface integral $\int \int \overrightarrow{f}.d\overrightarrow{S}$(Where $d\overrightarrow{S}$ is an elemental surface area vector) evaluated over the inner and outer surfaces of a spherical shell formed by two concentric spheres with origin as the centre, and internal and external radii of 1 and 2, respectively, is

The correct option is 0.

1. Understanding the Vector Field and the Region
The given vector field is:
$$\overrightarrow{f}(x,y,z)=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{{[x^2+y^2+z^2]}^{3/2}}$$
Let $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ be the position vector, and $r=\left|\overrightarrow{r}\right|=\sqrt{x^2+y^2+z^2}$ be its magnitude. The vector field can be simplified in spherical coordinates as:
$$\overrightarrow{f}=\frac{\overrightarrow{r}}{r^3}=\frac{\hat{r}}{r^2}$$
where $\hat{r}$ is the unit radial vector.

The region of integration is a spherical shell bounded by:
1. An inner sphere $S_1$ of radius $r_1=1$.
2. An outer sphere $S_2$ of radius $r_2=2$.
The surface integral is evaluated over the boundary $S=S_1\cup S_2$ of this shell. By standard convention, the boundary of a solid region is oriented with unit normals pointing outwards from the region.

2. Method 1: Using the Gauss Divergence Theorem
According to the Gauss Divergence Theorem:
$${\iint }_S\overrightarrow{f}\cdot d\overrightarrow{S}={\iiint }_V(\nabla \cdot \overrightarrow{f})dV$$
where $V$ is the volume of the spherical shell (the region $1\le r\le 2$).

Let us compute the divergence of the vector field:
$$\nabla \cdot \overrightarrow{f}=\frac{\partial }{\partial x}\left(\frac{x}{r^3}\right)+\frac{\partial }{\partial y}\left(\frac{y}{r^3}\right)+\frac{\partial }{\partial z}\left(\frac{z}{r^3}\right)$$
Evaluating the first partial derivative:
$$\frac{\partial }{\partial x}\left(x{(x^2+y^2+z^2)}^{-3/2}\right)=r^{-3}-3x^2{r}^{-5}$$
Summing all three components:
$$\nabla \cdot \overrightarrow{f}=\frac{3}{r^3}-\frac{3(x^2+y^2+z^2)}{r^5}=\frac{3}{r^3}-\frac{3r^2}{r^5}=0$$
Thus, the divergence of $\overrightarrow{f}$ is zero everywhere in space except at the origin $r=0$.

Since the spherical shell $V$ lies between $r=1$ and $r=2$, it does not contain the origin. Therefore, $\nabla \cdot \overrightarrow{f}=0$ at all points inside $V$.
Consequently:
$${\iint }_S\overrightarrow{f}\cdot d\overrightarrow{S}={\iiint }_{V}0dV=0$$

3. Method 2: Direct Integration
For confirmation, we can calculate the flux through both boundaries individually:
- For the outer boundary sphere $S_2$ (radius $r=2$), the outward-pointing normal is $\hat{n}=\hat{r}$.
- For the inner boundary sphere $S_1$ (radius $r=1$), the normal pointing out of the shell region is oriented radially inward towards the origin, i.e., $\hat{n}=-\hat{r}$.

1. Flux through the outer surface $S_2$:
$${\iint }_{S_2}\overrightarrow{f}\cdot d\overrightarrow{S}={\iint }_{S_2}\left(\frac{\hat{r}}{r^2}\right)\cdot (\hat{r}dS)={\iint }_{S_2}\frac{1}{r^2}dS$$
Since $r=2$ on $S_2$:
$${\iint }_{S_2}\overrightarrow{f}\cdot d\overrightarrow{S}=\frac{1}{2^2}{\iint }_{S_2}dS=\frac{1}{4}(4\pi \cdot 2^2)=4\pi$$

2. Flux through the inner surface $S_1$:
$${\iint }_{S_1}\overrightarrow{f}\cdot d\overrightarrow{S}={\iint }_{S_1}\left(\frac{\hat{r}}{r^2}\right)\cdot (-\hat{r}dS)=-{\iint }_{S_1}\frac{1}{r^2}dS$$
Since $r=1$ on $S_1$:
$${\iint }_{S_1}\overrightarrow{f}\cdot d\overrightarrow{S}=-\frac{1}{1^2}{\iint }_{S_1}dS=-1(4\pi \cdot 1^2)=-4\pi$$

3. Summing the total flux:
$${\iint }_S\overrightarrow{f}\cdot d\overrightarrow{S}=4\pi +(-4\pi )=0$$
Both methods yield a result of 0.