Given:
v=(yz−1)i+z(1+x+z)j+y(1+x+2z)k
For the conservative vector field
curlv=0We have
curlv=∣∣i∂x∂(yz−1)j∂y∂z(1+x+z)k∂z∂y(1+x+2z)∣∣=(1+x+2z−(1+x+2z))i+(y−y)j+(z−z)k=0Therefore, the given vector field is conservative.
We can put
v=∇ψwhere
ψ=yz(1+x+z)−x
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