Set up$,$ but do not evaluate, an integral for the area of the surface obtained by rotating the curve

$y=x{e}^{-x},1x2$

$($a$)$ about the $x-$axis.

${\displaystyle \underset{1}{\overset{2}{}}}2\sqrt[\phantom{2}]{1+e^{-2y}(1-y{)}^2}dy$

${\displaystyle \underset{1}{\overset{2}{}}}2\sqrt[\phantom{2}]{1+e^{-2x}(1-x{)}^2}dx$

${\displaystyle \underset{1}{\overset{2}{}}}2x\sqrt[\phantom{2}]{1+e^{-2x}(1-x{)}^2}dx$

${\displaystyle \underset{1}{\overset{2}{}}}2y\sqrt[\phantom{2}]{1+e^{-2y}(1-y{)}^2}dy$

${\displaystyle \underset{1}{\overset{2}{}}}2x{e}^{-x}\sqrt[\phantom{2}]{1+e^{-2x}(1-x{)}^2}dx$

$($b$)$ about the $y-$axis.

${\displaystyle \underset{1}{\overset{2}{}}}2y\sqrt[\phantom{2}]{1+e^{-2y}(1-y{)}^2}dy$

${\displaystyle \underset{1}{\overset{2}{}}}2x\sqrt[\phantom{2}]{1+e^{-2x}(1-x{)}^2}dx$

${\displaystyle \underset{1}{\overset{2}{}}}2\sqrt[\phantom{2}]{1+e^{-2y}(1-y{)}^2}dy$

${\displaystyle \underset{1}{\overset{2}{}}}2x{e}^{-x}\sqrt[\phantom{2}]{1+e^{-2x}(1-x{)}^2}dx$

${\displaystyle \underset{1}{\overset{2}{}}}2\sqrt[\phantom{2}]{1+e^{-2x}(1-x{)}^2}dx$