Question

Consider the vector field. f(x, y, z) = xy2z2i + x2yz2j + x2y2zk (a) find the curl of the vector field.

Answer 1

Final Answer : 0

For Calculation see pic

Answer 2

Answer:

curl of the given vector field is zero.

Step-by-step explanation:

The given vector field is $f(x,y,z)=xy^2z^2i+x^2yz^2j+x^2y^2zk$

The formula for the curl of a vector field f(x,y,z)=ai+bj+zk is given by

$\text{Curl F}=\begin{vmatrix}i&j&k \\\frac{\partial }{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z}\\a&b&c\end{vmatrix}$

On substituting these values, we get

$\text{Curl F}=\begin{vmatrix}i&j&k \\\frac{\partial }{\partial x}&\frac{\partial }{\partial y}&\frac{\partial }{\partial z}\\xy^2z^2&x^2yz^2&x^2y^2z\end{vmatrix}\\\\\text{Curl F}=i(2x^2yz-2x^2yz)-j(2xy^2z-2xy^2x)+k(2xyz^2-2xyz^2)\\\\\text{Curl F}=0$

Therefore, the curl of the vector field is zero.