How to calculate ${\rm curl}\, X$ $\def\curl{\mathop{\rm curl}}$This question is just little one part of my homework How to calculate $\curl$
$\curl X=\text?$
where $X=(y^2+yz, xz+z^2, y^2-xy)$
 A: The curl of a vector field $F$ can be calculated by the following 
$$curl \:F  =\begin{vmatrix} \mathbf{i} \quad\quad \mathbf{j}\quad\quad  \mathbf{k} \\  \\
{\frac{\partial}{\partial x}} \quad {\frac{\partial}{\partial y}} \quad {\frac{\partial}{\partial z}} \\
 \\  F_x \quad F_y \quad F_z \end{vmatrix}=\left(\frac{\partial F_z}{\partial y}  - \frac{\partial F_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \mathbf{k}$$
In our case $F=(y^2+yz)\hat{x}+(xz+z^2)\hat{y}+ (y^2-xy)\hat{z}$
$$F_x=y^2+yz,F_y=xz+z^2,F_z=y^2-xy$$
Substitute $F_x,F_y,F_z$ into above equation
$$=\left(\frac{\partial (y^2-xy)}{\partial y}  - \frac{\partial (xz+z^2)}{\partial z}\right) \mathbf{i} + \left(\frac{\partial (y^2+yz)}{\partial z} - \frac{\partial (y^2-xy)}{\partial x}\right) \mathbf{j} + \left(\frac{\partial (xz+z^2)}{\partial x} - \frac{\partial (y^2+yz)}{\partial y}\right) \mathbf{k}=-2 (x-y+z)\mathbf{i}+2 y\mathbf{j}-2 y\mathbf{k}$$
