Answer
\[
r \cos \theta=x, r \sin \theta=y, \quad e^{-r^{2}}=z
\]
Work Step by Step
\[
\begin{array}{c}
r \sin \theta=y, \quad r \cos \theta=x\\
\Rightarrow z=e^{-\left(x^{2}+y^{2}\right)}=e^{-\left(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta\right)}=e^{-r^{2}} \quad\left(\because1= \cos ^{2} \theta+\sin ^{2} \theta \right)
\end{array}
\]
So, the surface can be represented parametrically in terms of $r$ and $\theta$ as:
\[
r \cos \theta=x, r \sin \theta=y, \quad e^{-r^{2}}=z
\]
