Given:
v = ( y z − 1 ) i + z ( 1 + x + z ) j + y ( 1 + x + 2 z ) k {\bf v}=(yz-1){\bf i}+z(1+x+z){\bf j}+y(1+x+2z){\bf k} v = ( yz − 1 ) i + z ( 1 + x + z ) j + y ( 1 + x + 2 z ) k
For the conservative vector field
c u r l v = 0 \rm curl {\bf v}=\bf 0 curl v = 0 We have
c u r l v = ∣ i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z ( y z − 1 ) z ( 1 + x + z ) y ( 1 + x + 2 z ) ∣ \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} curl v = ∣ ∣ i ∂ x ∂ ( yz − 1 ) j ∂ y ∂ z ( 1 + x + z ) k ∂ z ∂ y ( 1 + x + 2 z ) ∣ ∣ = ( 1 + x + 2 z − ( 1 + x + 2 z ) ) i =(1+x+2z-(1+x+2z)){\bf i} = ( 1 + x + 2 z − ( 1 + x + 2 z )) i + ( y − y ) j + ( z − z ) k = 0 +(y-y){\bf j}+(z-z){\bf k}=\bf 0 + ( y − y ) j + ( z − z ) k = 0 Therefore, the given vector field is conservative.
We can put
v = ∇ ψ {\bf v}=\nabla\psi v = ∇ ψ where
ψ = y z ( 1 + x + z ) − x \psi=yz(1+x+z)-x ψ = yz ( 1 + x + z ) − x
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