Answer
$0 \leq \theta \leq \dfrac{\pi}{6}$ and $1\leq r\leq 2 \sqrt 3 \sec \theta$
and
$\dfrac{\pi}{6} \leq \theta \leq \dfrac{\pi}{2}$ and $1\leq r\leq 2 \csc \theta$
Work Step by Step
Conversion of polar coordinates and Cartesian coordinates are as follows:
a)$r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$
c) $x=r \cos \theta$
d) $y=r \sin \theta$
As $y=r \cos \theta =1\implies r=\csc \theta$
Therefore, the region described in polar coordinates is:
$0 \leq \theta \leq \dfrac{\pi}{6}$ and $1\leq r\leq 2 \sqrt 3 \sec \theta$
and
$\dfrac{\pi}{6} \leq \theta \leq \dfrac{\pi}{2}$ and $1\leq r\leq 2 \csc \theta$
