Question

If xz⁴i-2ײyzj+2yz³k ,find ∆×A at point (1,-1,1)

Answer 1

In cartesian coordinates

$$\vartriangle A=\vartriangle A_x \vec{i}+\vartriangle A_y\vec{j}+\vartriangle A_z\vec{k}$$

$$\dfrac{\partial A_x}{\partial x}=z^4, \ \dfrac{\partial^2 A_x}{\partial x^2}=0$$

$$\dfrac{\partial A_y}{\partial y}=-2x^2z, \ \dfrac{\partial^2 A_y}{\partial y^2}=0$$

$$\dfrac{\partial A_z}{\partial z}=6yz^2, \ \dfrac{\partial^2 A_z}{\partial z^2}=12yz$$

$$\vartriangle A=0\vec{i}+0\vec{j}+12yz\vec{k}$$

Point $(1,-1,1)$

$$\vartriangle A|_{(1,-1,1)}=-12\vec{k}$$