Answer
We show that
$curl\left( {\left( {{F_1}\left( x \right),{F_2}\left( y \right),{F_3}\left( z \right)} \right)} \right) = {\bf{0}}$
For ${\bf{F}}\left( {x,y,z} \right) = \left( {{x^2} + {y^2},\ln y + {z^2},{z^3}\sin \left( {{z^2}} \right){{\rm{e}}^{{z^3}}}} \right)$,
we get
$curl\left( {{\bf{F}}\left( {x,y,z} \right)} \right) = - 2z{\bf{i}} - 2y{\bf{k}}$
Work Step by Step
1. Suppose that ${\bf{F}}\left( {x,y,z} \right) = \left( {{F_1}\left( x \right),{F_2}\left( y \right),{F_3}\left( z \right)} \right)$.
Evaluate the curl of ${\bf{F}}$:
$curl\left( {{\bf{F}}\left( {x,y,z} \right)} \right) = curl\left( {\left( {{F_1}\left( x \right),{F_2}\left( y \right),{F_3}\left( z \right)} \right)} \right)$
$curl\left( {\bf{F}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{{F_1}\left( x \right)}&{{F_2}\left( y \right)}&{{F_3}\left( z \right)}
\end{array}} \right| = {\bf{0}}$
So, $curl\left( {\left( {{F_1}\left( x \right),{F_2}\left( y \right),{F_3}\left( z \right)} \right)} \right) = {\bf{0}}$.
2. We are given ${\bf{F}}\left( {x,y,z} \right) = \left( {{x^2} + {y^2},\ln y + {z^2},{z^3}\sin \left( {{z^2}} \right){{\rm{e}}^{{z^3}}}} \right)$
Write
${\bf{F}}\left( {x,y,z} \right) = \left( {{x^2},\ln y,{z^3}\sin \left( {{z^2}} \right){{\rm{e}}^{{z^3}}}} \right) + \left( {{y^2},{z^2},0} \right)$
Evaluate the curl of ${\bf{F}}$:
$curl\left( {{\bf{F}}\left( {x,y,z} \right)} \right) = curl\left( {\left( {{x^2},\ln y,{z^3}\sin \left( {{z^2}} \right){{\rm{e}}^{{z^3}}}} \right)} \right) + curl\left( {\left( {{y^2},{z^2},0} \right)} \right)$
If we consider ${F_1}\left( x \right) = {x^2}$, ${\ \ }$ ${F_2}\left( y \right) = \ln y$, ${\ \ }$ ${F_3}\left( z \right) = {z^3}\sin \left( {{z^2}} \right){{\rm{e}}^{{z^3}}}$, then by the result in part (1):
$curl\left( {\left( {{x^2},\ln y,{z^3}\sin \left( {{z^2}} \right){{\rm{e}}^{{z^3}}}} \right)} \right) = {\bf{0}}$
Thus,
$curl\left( {{\bf{F}}\left( {x,y,z} \right)} \right) = curl\left( {\left( {{y^2},{z^2},0} \right)} \right)$
$curl\left( {{\bf{F}}\left( {x,y,z} \right)} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{{y^2}}&{{z^2}}&0
\end{array}} \right| = - 2z{\bf{i}} - 2y{\bf{k}}$
So, $curl\left( {{\bf{F}}\left( {x,y,z} \right)} \right) = - 2z{\bf{i}} - 2y{\bf{k}}$.
