Find vector fields $\mathbf{A}$such that role="math" localid="1657346627450" $\mathbf{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{curl}\mathbf{}\mathbf{A}\mathbf{}$for each givenrole="math" localid="1657346639484" $\mathbf{V}\mathbf{.}$

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

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 its magnitude.

In this problem find vector $\mathrm{A}$, whose curl gives the vector

.$\mathrm{V}=\left({\mathrm{x}}^2-\mathrm{yz}+\mathrm{y}\right)\mathrm{i}+(\mathrm{x}-2\mathrm{yz})\mathrm{j}+\left({\mathrm{z}}^2-2\mathrm{zx}+\mathrm{x}+\mathrm{y}\right)\mathrm{k}$

The components of vector are given in terms of the derivatives of $\mathrm{A}$.

$\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{z}}}{\mathrm{\partial }\mathrm{y}}-\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{z}}={\mathrm{x}}^2-\mathrm{yz}+\mathrm{y}$

$\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{y}}-\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{z}}}{\mathrm{\partial }\mathrm{z}}=\mathrm{x}-2\mathrm{yz}$

$\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{y}}-\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{z}}={\mathrm{z}}^2-2\mathrm{zx}+\mathrm{x}+\mathrm{y}$

There are an infinite number of options, try to set ${\mathrm{A}}_{\mathrm{z}}=0$.This gives the equation given below.

$-\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{z}}={\mathrm{x}}^2-\mathrm{yz}+\mathrm{y}$

$\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{z}}=\mathrm{x}-2\mathrm{yz}$

$\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{y}}}{\mathrm{\partial }\mathrm{x}}-\frac{\mathrm{\partial }{\mathrm{A}}_{\mathrm{x}}}{\mathrm{\partial }\mathrm{y}}={\mathrm{z}}^2-2\mathrm{zx}+\mathrm{x}+\mathrm{y}$

Integrate the first equation.

${\mathrm{A}}_{\mathrm{y}}=-{\mathrm{x}}^2\mathrm{z}+\frac{1}{2}{\mathrm{yz}}^2-\mathrm{yz}+\mathrm{f}(\mathrm{x},\mathrm{y})$

Now, integrate the second equation to obtain the equation shown below.

${\mathrm{A}}_{\mathrm{x}}=\mathrm{xz}-{\mathrm{yz}}^2+\mathrm{g}(\mathrm{x},\mathrm{y})$

Plug those two solutions into the last equation.

$-2\mathrm{xz}+\frac{\mathrm{\partial }\mathrm{f}}{\mathrm{\partial }\mathrm{x}}+{\mathrm{z}}^2-\frac{\mathrm{\partial }\mathrm{g}}{\mathrm{\partial }\mathrm{y}}={\mathrm{z}}^2-2\mathrm{zx}+\mathrm{x}+\mathrm{y}$

$\frac{\mathrm{\partial }\mathrm{f}}{\mathrm{\partial }\mathrm{x}}-\frac{\mathrm{\partial }\mathrm{g}}{\mathrm{\partial }\mathrm{y}}=\mathrm{x}+\mathrm{y}$

Take the function f and g as mentioned below.

$\mathrm{f}=\frac{{\mathrm{x}}^2}{2}$

$\mathrm{g}=-\frac{{\mathrm{y}}^2}{2}$

The total solution is derived as shown below.

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

Here,$\mathrm{u}$ is an arbitrary function.

Hence, the vector field derived is

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