Question

Determine the \\(curl F\\) at the point (2, 0, 3) given that \\(F=xz i+(2x^2-y)j-yz^2 k\\).

Answer 1

$$\text{curl}\vec{F}=\nabla\times\vec{F}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ xz & 2x^2-y & -yz^2 \end{vmatrix}=$$

$$=\vec{i}(-z^2-0)-\vec{j}(0-x)+\vec{k}(4x-0)=$$

$$=-z^2\vec{i}+x\vec{j}+4x\vec{k}$$

$M(2,0,3)$

$$\text{curl}\vec{F}_M=-9\vec{i}+2\vec{j}+8\vec{k}$$