91影视

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

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

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

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

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

$\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{x}}}{\mathrm{鈭�}\mathrm{y}}鈭�\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{z}}}{\mathrm{鈭�}\mathrm{z}}=\mathrm{x}鈭�2\mathrm{yz}$

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

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

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

$\frac{\mathrm{鈭�}{\mathrm{A}}_{\mathrm{x}}}{\mathrm{鈭�}\mathrm{z}}=\mathrm{x}鈭�2\mathrm{yz}$

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

Integrate the first equation.

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

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

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

Plug those two solutions into the last equation.

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

$\frac{\mathrm{鈭�}\mathrm{f}}{\mathrm{鈭�}\mathrm{x}}鈭�\frac{\mathrm{鈭�}\mathrm{g}}{\mathrm{鈭�}\mathrm{y}}=\mathrm{x}+\mathrm{y}$

Take the function f and g as mentioned below.

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

$\mathrm{g}=鈭�\frac{{\mathrm{y}}^2}{2}$

The total solution is derived as shown below.

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

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

Hence, the vector field derived is

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