Question

Show that the vector →{v}=(yz-1)i-z(1+x+z)j+y(1+x+2z)k is conservative and find it scalar potential function.

Answer 1

Given:

${\bf v}=(yz-1){\bf i}+z(1+x+z){\bf j}+y(1+x+2z){\bf k}$

For the conservative vector field

$$\rm curl {\bf v}=\bf 0$$

We have

$$\rm curl {\bf v}=\begin{vmatrix} {\bf i} & {\bf j} & {\bf k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ (yz-1)&z(1+x+z)& y(1+x+2z) \end{vmatrix}$$

$$=(1+x+2z-(1+x+2z)){\bf i}$$

$$+(y-y){\bf j}+(z-z){\bf k}=\bf 0$$

Therefore, the given vector field is conservative.

We can put

$${\bf v}=\nabla\psi$$

where

$$\psi=yz(1+x+z)-x$$