The curl of [tex]\vec{\bf{F}}[/tex] and the divergence of [tex]\vec{\bf{F}}[/tex] are
[tex]curl \text{ }\vec{\bf{F}}=x(z^2-y^2)\bf{\hat{i}}+y(x^2-z^2)\bf{\hat{j}}+z(y^2-x^2)\bf{\hat{k}}[/tex]
[tex]div\text{ }\vec{\bf{F}}=6xyz[/tex]
Our vector field is given by
[tex]\vec{\bf{F}}(x,y,z)=x^2yz \text{ }\bf{\hat{i}}+xy^2z\text{ }\bf{\hat{j}}+xyz^2\text{ }\bf{\hat{k}}[/tex]
The curl of [tex]\vec{\bf{F}}[/tex] is given by the "determinant"
[tex]curl\text{ }\vec{\bf{F}}=\begin{vmatrix}\bf{ {\hat{i}}} & \bf{ {\hat{j}}} & \bf{ {\hat{k}}} \\\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\\\ x^2yz & xy^2z & xyz^2\end{vmatrix}[/tex]
or
[tex]curl\text{ }\vec{\bf{F}}\\=[\frac{\partial}{\partial y} (xyz^2)-\frac{\partial}{\partial z} (xy^2z)]\bf{ {\hat{i}}} - [\frac{\partial}{\partial x} (xyz^2)-\frac{\partial}{\partial z} (x^2yz)]\bf{ {\hat{j}}} + [\frac{\partial}{\partial x} (xy^2z)-\frac{\partial}{\partial y} (x^2yz)]\bf{ {\hat{k}}}[/tex]
evaluating each partial derivative, we get
[tex]curl \text{ }\vec{\bf{F}}=(xz^2-xy^2)\bf{\hat{i}}-(yz^2-x^2y)\bf{\hat{j}}+(y^2z-x^2z)\bf{\hat{k}}\\=(xz^2-xy^2)\bf{\hat{i}}+(x^2y-yz^2)\bf{\hat{j}}+(y^2z-x^2z)\bf{\hat{k}}\\=x(z^2-y^2)\bf{\hat{i}}+y(x^2-z^2)\bf{\hat{j}}+z(y^2-x^2)\bf{\hat{k}}[/tex]
To find the divergence of [tex]\vec{\bf{F}}[/tex]
[tex]div\text{ }\vec{\bf{F}}=\frac{\partial}{\partial x} (x^2yz )+\frac{\partial}{\partial y}(xy^2z)+\frac{\partial}{\partial z}(xyz^2)\\\\=2xyz+2xyz+2xyz\\=6xyz[/tex]
Another solved example can be found here: https://brainly.com/question/12734135
