Answer
$\dfrac{-\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$ and $0 \leq r\leq 2 \cos \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) $y=r \sin \theta$
Here, $r^2=2x$ and $r^2=2 r\cos \theta \implies 2r \cos \theta-r^2=0$
This gives $r=0, 2 \cos \theta$
Therefore, the region described in polar coordinates is:
$\dfrac{-\pi}{2} \leq \theta \leq \dfrac{\pi}{2}$ and $0 \leq r\leq 2 \cos \theta $
