Answer
\[
r \cos \theta=x, r \sin \theta=y, \quad z=\frac{1}{r^{2}+1}
\]
Work Step by Step
\[
\begin{array}{l}
\qquad r \sin \theta=y \theta, \quad r \cos =x \\
\Rightarrow z=\frac{1}{1+(r \cos \theta)^{2}+(r \sin \theta)^{2}}=\frac{1}{1+r^{2}\left[\cos ^{2} \theta+\sin ^{2} \theta\right]}=\frac{1}{1+r^{2}} \quad\left(\cos ^{2} \theta+\sin ^{2} \theta=1\right)
\end{array}
\]
So, the surface can be represented parametrically in terms of $r$ and and theta:
\[
r \cos \theta=x, r \sin \theta=y, \quad z=\frac{1}{r^{2}+1}
\]
