## Example 3. Find the potential of the vector field $\mathbf{a}=y z \mathbf{i}+x z \mathbf{j}+x y \mathbf{k}$
Solution. It is easy to see that $\operatorname{rot} \mathbf{a} \equiv 0$, i.e., the given vector field is potential. This field is defined throughout the entire three-dimensional space, which is star-shaped with the center at the origin of coordinates $O(0,0,0)$, so to find its potential, we will use formula (10). Since in this case

$$
\mathbf{a}\left(M^{\prime}\right)=\mathbf{a}(t x, t y, t z)=t^{2} y \mathbf{i}+t^{2} x z \mathbf{j}+t^{2} x y \mathbf{k}
$$

the scalar product of the vector $\mathbf{a}\left(M^{\prime}\right)$ and $\mathbf{r}\left(M^{\prime}\right)$ is

$$
\left(\mathbf{a}\left(M^{\prime}\right), \mathbf{r}(M)\right)=t^{2}(x y z+x y z+x y z)=3 t^{2} x y z
$$

The sought potential is

$$
\varphi(M)=\int_{0}^{1}\left(\mathbf{a}\left(M^{\prime}\right), \mathbf{r}(M)\right) d t=x y z \int_{0}^{1} 3 t^{2} d t+C=x y z+C
$$

Thus,

$$
\varphi(M)=x y z+C
$$

## Problems for Independent Solution

Using formula (10), find the potentials of the following vector fields:

234. $\mathbf{a}=\alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k}$, where $\alpha, \beta, \gamma$ are constants.
235. $\mathbf{a}=(y+z) \mathbf{i}+(x+z) \mathbf{j}+(y+x) \mathbf{k}$.
236. $\mathbf{a}=y \mathbf{i}+x \mathbf{j}+e^{*} \mathbf{k}$.
237. $\mathbf{a}=e^{x} \sin y \mathbf{i}+e^{x} \cos y \mathbf{k}$.

## Chapter 5

## 5.20. The Hamilton Operator "Nabla"

Many operations in vector analysis can be written in a shortened and convenient form for calculations using the symbolic Hamilton operator nabla:

$$
\nabla=\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}
$$

In this operator, differential and vector properties are combined. We will understand the formal multiplication of $\frac{\partial}{\partial x}$ by a function $f(x, y, z)$ as partial differentiation $\frac{\partial f}{\partial x}$.

Within the framework of vector algebra, we will perform formal operations with the operator "nabla" as if it were a vector. Using this formalism, we obtain the following.

1. If $u=u(x, y, z)$ is a differentiable scalar function, then by the rule of multiplying a vector by a scalar, we have

$$
\nabla u=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) u=\mathbf{i} \frac{\partial u}{\partial x}+\mathbf{j} \frac{\partial u}{\partial y}+\mathbf{k} \frac{\partial u}{\partial z}=\operatorname{grad} u
$$

2. If $\mathbf{a}=P(x, y, z) \mathbf{i}+Q(x, y, z) \mathbf{j}+R(x, y, z) \mathbf{k}$, where $P, Q, R$ are differentiable functions, then by the known formula for the scalar product, we have

$$
(\nabla, \mathbf{a})=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}, P \mathbf{i}+Q \mathbf{j}+R \mathbf{k}\right)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=\operatorname{div} \mathbf{a}, \text{ (3) }
$$

in particular, $[\nabla, \mathbf{c}]=0$ (where $\mathbf{c}$ is a constant vector).

3. If $\mathbf{a}=P(x, y, z) \mathbf{i}+Q(x, y, z) \mathbf{j}+R(x, y, z) \mathbf{k}$, then

$$
[\nabla, \mathbf{a}]=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R
\end{array}\right|=\operatorname{rot} \mathbf{a}
$$

Continuing the formalism of actions with $\nabla$ as a vector, from the distributive property of scalar and vector products, we obtain

$$
(\nabla, \mathbf{a}+\mathbf{b})=(\nabla, \mathbf{a})+(\nabla, \mathbf{b})
$$

i.e., $\operatorname{div}(\mathbf{a}+\mathbf{b})=\operatorname{div} \mathbf{a}+\operatorname{div} \mathbf{b}$,

$$
[\nabla, \mathbf{a}+\mathbf{b}]=[\nabla, \mathbf{a}]+[\nabla, \mathbf{b}]
$$

i.e., $\operatorname{rot}(\mathbf{a}+\mathbf{b})=\operatorname{rot} \mathbf{a}+\operatorname{rot} \mathbf{b}$.

Formulas (5) and (6) can also be interpreted as manifestations of the differential properties of the operator nabla ($\nabla$ is a linear differential operator).

Using the formalism of actions with the operator $\nabla$ as a vector, it should be remembered that $\nabla$ is not a vector - it has neither magnitude nor direction, so, for example, the vector $[\nabla, \mathbf{a}]$ will not, in general, be perpendicular to the vector $\mathbf{a}$ (however, for a planar field $\mathbf{a}=P(x, y) \mathbf{i}+Q(x, y) \mathbf{j}$, the vector

$$
[\nabla, \mathbf{a}]=\operatorname{rot} \mathbf{a}=\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \mathbf{k}
$$

will be perpendicular to the plane $x O y$, and hence to the vector $\mathbf{a}$). Similarly, the concept of collinearity with respect to the symbolic vector $\nabla$ does not make sense. For example, the expression $[\nabla \varphi, \nabla \psi]$, where $\varphi$ and $\psi$ are scalar functions, formally resembles the vector product of two collinear vectors, which is always zero. But this is not generally the case. Indeed, the vector $\nabla \varphi=\operatorname{grad} \varphi$ is directed along the normal to the surface $\varphi=\text{const}$, and the vector $\nabla \psi=\operatorname{grad} \psi$ defines the normal to the surface $\psi=\text{const}$, and these normals are not generally collinear (Fig. 35). On the other hand, in any differentiable scalar field $\varphi$, we have $[\nabla \varphi, \nabla \varphi]=0$. These examples show that the operator $\nabla$ should be handled with care.

In addition to its vector nature, the Hamilton operator "nabla" has a differential nature. To emphasize the differential character of $\nabla$, we will agree that the operator $\nabla$ acts on all elements written after it. In this sense, for example, $(\nabla, \mathbf{a}) \neq (\mathbf{a}, \nabla)$. Indeed,

$$
(\nabla, \mathbf{a})=\operatorname{div} \mathbf{a}
$$

but

$$
(\mathbf{a}, \nabla)=P \frac{\partial}{\partial x}+Q \frac{\partial}{\partial y}+R \frac{\partial}{\partial z}
$$

is a scalar differential operator.

When applying the operator $\nabla$ to the product of some quantities, we must keep in mind the rule of differentiation of a product

$$
\frac{\partial}{\partial x}(u v)=v \frac{\partial u}{\partial x}+u \frac{\partial v}{\partial x}
$$

From this, it follows that the operator "nabla" must be applied sequentially to each factor, keeping the other factors unchanged, and then taking the sum of the resulting expressions. In doing so, we will follow the following rules:

1. If the operator $\nabla$ acts on some product, its differential nature is taken into account first, and then its vector properties.
2. To indicate that "nabla" does not act on some quantity included in a complex formula, this quantity is marked with a subscript $c$ (const), which can be removed in the final result.
3. All quantities on which the operator "nabla" does not act will be placed in front of "nabla" in the final result, i.e., to the left of it.