Example 1. Using the invariant definition, calculate the divergence of the vector $a=x \mathbf{i}$ at the point $O(0,0,0)$, choosing as the surface $\sigma$ surrounding the point $O$, a sphere $\sigma_{\varepsilon}$ of radius $\varepsilon$ centered at this point.
Solution. By the definition of divergence at the given point, we have

$$
\operatorname{div} a(0)=\lim _{\left(\sigma_{k}\right) \rightarrow 0} \frac{\int\left(a, n^{0}\right) d \sigma}{v_{\varepsilon}}
$$

where $v_{\varepsilon}$ is the volume of the ball bounded by the sphere $\sigma_{\varepsilon}$, or

$$
\operatorname{div} a(0)=\lim _{k \rightarrow 0} \frac{\prod_{0}\left(a, n^{0}\right) d \theta}{v_{\varepsilon}}
$$

But since the volume of the ball is $v_{\varepsilon}=\frac{4}{3} \pi \varepsilon^{3}$, then

$$
\operatorname{div} a(0)=\lim _{\varepsilon \rightarrow 0} \frac{\sigma_{\varepsilon}\left(a, n^{0}\right) d \sigma}{(4 / 3) \pi \varepsilon^{3}}
$$

We will compute the flux $\oiint_{\sigma_{\varepsilon}}\left(a, n^{\dagger}\right) d \sigma$ of the given vector through the sphere $\sigma_{\varepsilon}$. The normal vector $n^{\dagger}$ to the sphere $\sigma_{\varepsilon}$ is directed along the radius of the sphere, so we can set

$$
\mathbf{n}^{\mathbf{n}}=\mathbf{r}^{\mathbf{0}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\mathbf{r}}{\varepsilon}
$$

where $\mathbf{r}^{\mathbf{0}}$ is the unit vector of the radius vector $\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$, or

$$
\mathbf{n}^{0}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\varepsilon}
$$

The desired flux will be equal to

$$
\oiint_{\sigma_{\varepsilon}}\left(a, n^{0}\right) d \sigma=\oiint_{\sigma_{\varepsilon}} \frac{x^{2}}{\varepsilon} d \sigma
$$

Switching to coordinates on the sphere $\sigma_{\varepsilon}$

$$
x=\varepsilon \cos \varphi \sin \theta, \quad y=\varepsilon \sin \varphi \sin \theta, \quad z=\varepsilon \cos \theta .
$$

we get

$$
\begin{aligned}
\oiint_{\sigma_{\varepsilon}}\left(a, n^{0}\right) d \sigma=\iint_{\sigma_{\varepsilon}} \frac{\varepsilon^{2} \cos ^{2} \varphi \sin ^{2} \theta \varepsilon^{2} \sin \theta d \varphi d \theta}{\varepsilon}= \\
=\varepsilon^{3} \int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3} \pi \varepsilon^{3} .
\end{aligned}
$$

Therefore,

$$
\operatorname{div} a(0)=\lim _{\varepsilon \rightarrow 0} \frac{(4 / 3) \pi \varepsilon^{3}}{(4 / 3) \pi \varepsilon^{3}}=1
$$