Question

Find Curla, where A = 2xi+2yj is the motion
irrotational​

Answer 1

Answer:

We have

∇×v=(∂∂y(4x−y+2z)−∂∂z(2ax+by−z))i⃗ −(∂∂x(4x−y+2z)−∂∂z(x+2y+4z))j⃗ +(∂∂x(2ax+by−z)−∂∂y(x+2y+4z))k⃗ =(−1−(−1))i⃗ −(4−4)j⃗ +(2a−2)k⃗ =(2a−2)k⃗

For this to be irrotational, we require

∇×v⟹a=0=(2a−2)k⃗ =1

For solenoidal, we require

∇⋅v⟹b=0=(∂∂x,∂∂y,∂∂z)⋅(x+2y+4z,2(1)+by−z,4x−y+2z)=1+b+2=3+b=−3