2.18. Find the divergence of the vector field:

a) $\vec{A}=x y^{2} \vec{i}+x^{2} y \vec{j}+z^{3} \vec{k}$ at the point $M(1,-1,3)$;

b) the gradient of the function $\varphi=x y^{2} z^{2}$;

c) $\dot{A}=y z(4 x \vec{i}-y \vec{j}-z \vec{k})$.
Solution.

a) By definition (see 2.20)

$$
\operatorname{div} \vec{A}=\nabla \vec{A}=\frac{\partial}{\partial x}\left(x y^{2}\right)+\frac{\partial}{\partial y}\left(x^{2} y\right)+\frac{\partial}{\partial z} z^{3}=y^{2}+x^{2}+3 z^{2}
$$

From this, at $x=1, y=-1, z=3$ we get

$$
\left.\operatorname{div} \vec{A}\right|_{M(1,-1,3)}=29
$$

b) First, we calculate the gradient of the function $\varphi$:

$\nabla \varphi=\operatorname{grad} \varphi=\frac{\partial \varphi}{\partial x} \dot{i}+\frac{\partial \varphi}{\partial y} \dot{j}+\frac{\partial \varphi}{\partial z} \dot{k}=y^{2} z^{2} \dot{i}+2 x y z^{2} \dot{j}+2 x z y^{2} \dot{k}$.

Now we find the divergence of this vector field:

$$
\begin{aligned}
\nabla(\nabla \varphi)= & \nabla^{2} \varphi=\Delta \varphi=\operatorname{grad} \varphi=\frac{\partial \varphi}{\partial x^{2}}+\frac{\partial \varphi}{\partial y^{2}}+\frac{\partial \varphi}{\partial z^{2}}= \\
& =2 x z^{2}+2 x y^{2}=2 x\left(y^{2}+z^{2}\right)
\end{aligned}
$$

c) $\operatorname{div} \dot{A}=\nabla \dot{A}=\frac{\partial(4 x y z)}{\partial x}+\frac{\partial\left(-y^{2} z\right)}{\partial y}+\frac{\partial\left(-z^{2} y\right)}{\partial z}=4 y z-2 y z-2 z y=0$.

The divergence of the field $\dot{A}$ turned out to be zero at every point in space. The vector field $\dot{A}$ is solenoidal.