Find vector fields A such that V = curl A for each given V.

$\mathbf{V}\mathbf{=}\mathbf{i}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{xz}\mathbf{)}\mathbf{+}\mathbf{j}\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{xy}\mathbf{)}\mathbf{+}\mathbf{k}\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{yz}\mathbf{+}\mathbf{xy}\mathbf{)}$

The given equation is $\mathrm{V}=\mathrm{i}({\mathrm{x}}^2-2\mathrm{xz})+\mathrm{j}({\mathrm{y}}^2-2\mathrm{xy})+\mathrm{k}({\mathrm{z}}^2-2\mathrm{yz}+\mathrm{xy})$

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

In this problem find the vector whose curl gives the vector.

$\mathrm{V}=\mathrm{i}({\mathrm{x}}^2-2\mathrm{xz})+\mathrm{j}({\mathrm{y}}^2-2\mathrm{xy})+\mathrm{k}({\mathrm{z}}^2-2\mathrm{yz}+\mathrm{xy})$

The components of the vector are given in terms of the derivatives of .

$$\begin{gathered} \frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=x^2-2xz \\ \frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=x^2-2xy \\ \frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=z^2-2yz+xy \end{gathered}$$

Observe the first two equation, notice that we can choose${\mathrm{A}}_{\mathrm{z}}={\mathrm{x}}^2\mathrm{y}$ , which gives the first term in the first equation and the second term in the second equation. Then solve for the other two components by integrating the first two equations.

$$\begin{gathered} {\mathrm{A}}_{\mathrm{y}}={\mathrm{xz}}^2+\mathrm{f}(\mathrm{x},\mathrm{y}) \\ {\mathrm{A}}_{\mathrm{x}}={\mathrm{y}}^2\mathrm{z}+\mathrm{g}(\mathrm{x},\mathrm{y}) \end{gathered}$$

By substituting those solutions into the last equation.

role="math" localid="1659244074527" $$\begin{gathered} z^2+\frac{\partial f}{\partial x}-2yz-\frac{\partial g}{\partial y}=z^2-2yz+xy \\ \frac{\partial f}{\partial x}-\frac{\partial g}{\partial y}=xy \end{gathered}$$

Choose$f=0$ and$g=-\frac{1}{2}x{y}^2.$

The total solution is then as mentioned below.

$\mathrm{A}=\left({\mathrm{y}}^2\mathrm{z}-\frac{1}{2}{\mathrm{xy}}^2\right)\mathrm{i}+{\mathrm{xz}}^2\mathrm{j}+{\mathrm{x}}^2\mathrm{yk}+\nabla \mathrm{u}$

Here is an arbitrary function.

Hence, the vector field is derived.

$\mathrm{A}=\left({\mathrm{y}}^2\mathrm{z}-\frac{1}{2}{\mathrm{xy}}^2\right)\mathrm{i}+{\mathrm{xz}}^2\mathrm{j}+{\mathrm{x}}^2\mathrm{yk}+\nabla \mathrm{u}$