Answer :

Solution :

$\mathrm{V}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{r}\mathrm{f}\mathrm{i}\mathrm{e}\mathrm{l}\mathrm{d}=\overrightarrow{\mathrm{f}}\left(x,y,z\right)=\frac{x\hat{i}+y\hat{j}+zk}{{\left[x^2+y^2+z^2\right]}^{\frac{3}{2}}}=\frac{x\hat{i}+y\hat{j}+zk}{{(r^2)}^{\frac{3}{2}}}=\frac{x\hat{i}+y\hat{j}+zk}{(r^3)}$

$\therefore div\overrightarrow{f}=\frac{\partial }{\partial x}\left[\frac{x}{r^3}\right]+\frac{\partial }{\partial y}\left[\frac{y}{r^3}\right]+\frac{\partial }{\partial z}\left[\frac{z}{r^3}\right]$

$\therefore div\overrightarrow{f}=x\left(-3\right)r^{-4}\left(\frac{\partial r}{\partial x}\right)+\frac{1}{r^3}+y\left(-3\right)r^{-4}\left(\frac{\partial r}{\partial y}\right)+z\left(-3\right)r^{-4}\left(\frac{\partial r}{\partial z}\right)+\frac{1}{r^3}$

$\therefore div\overrightarrow{f}=\left(-3\right)x{r}^{-4}\left(\frac{x}{r}\right)+\left(-3\right)y{r}^{-4}\left(\frac{y}{r}\right)+\left(-3\right)z{r}^{-4}\left(\frac{z}{r}\right)+\frac{3}{r^3}$

$\therefore div\overrightarrow{f}=\frac{-3}{r^5}\left[x^2+y^2+z^2\right]+\frac{3}{r^3}=\frac{-3}{r^5}\left[r^2\right]+\frac{3}{r^3}$

$\therefore div\overrightarrow{f}=\frac{-3}{r^3}+\frac{3}{r^3}=0$

Now, according to the theorem

$\iint \overrightarrow{f.}\overrightarrow{ds}=\iiint div\overrightarrow{f}dv=\iiint div\overrightarrow{0}dv=0$

$\iint \overrightarrow{f.}\overrightarrow{ds}=0$