Consider the vector field. f(x, y, z) = xy2z2i + x2yz2j + x2y2zk (a) find the curl of the vector field.
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Answer:
curl of the given vector field is zero.
Step-by-step explanation:
The given vector field is
The formula for the curl of a vector field f(x,y,z)=ai+bj+zk is given by
On substituting these values, we get
Therefore, the curl of the vector field is zero.
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