Answer
We show that $G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,1 - {r^2}} \right)$ parametrizes the paraboloid $z = 1 - {x^2} - {y^2}$.
The grid curves of this parametrization (see the figure attached):
1. $\theta $-grid curve: $G\left( {{r_0},\theta } \right) = \left( {{r_0}\cos \theta ,{r_0}\sin \theta ,1 - {r_0}^2} \right)$
This is a circle of radius ${r_0}$ at the height $z = 1 - {r_0}^2$.
2. $r$-grid curve: $G\left( {r,{\theta _0}} \right) = \left( {r\cos {\theta _0},r\sin {\theta _0},1 - {r^2}} \right)$
This is a parabola with maximum point at $\left( {0,0,1} \right)$.

Work Step by Step
We have $x = r\cos \theta $, $y = r\sin \theta $, and $z = 1 - {r^2}$. Since
$z = 1 - {x^2} - {y^2} = 1 - {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta = 1 - {r^2}$
So, $G\left( {r,\theta } \right) = \left( {r\cos \theta ,r\sin \theta ,1 - {r^2}} \right)$ parametrizes the paraboloid $z = 1 - {x^2} - {y^2}$.
The grid curves of this parametrization through $P = \left( {{r_0},{\theta _0}} \right)$ are (see the figure attached)
1. $\theta $-grid curve: $G\left( {{r_0},\theta } \right) = \left( {{r_0}\cos \theta ,{r_0}\sin \theta ,1 - {r_0}^2} \right)$
This is a circle of radius ${r_0}$ at the height $z = 1 - {r_0}^2$.
2. $r$-grid curve: $G\left( {r,{\theta _0}} \right) = \left( {r\cos {\theta _0},r\sin {\theta _0},1 - {r^2}} \right)$
This is a parabola with maximum point at $\left( {0,0,1} \right)$.