Answer
$0 \leq \theta \leq \dfrac{\pi}{3}$ and $0\leq r\leq \sec \theta$
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, $(x,y)=(1,\sqrt 3)$
The diagonal points to these points are: $(x,y)=P(\sqrt{1+3},\tan^{-1} \dfrac{1}{\sqrt 3}=P(2, \dfrac{\pi}{3})$
As $x=r \cos \theta =1\implies r=\sec \theta$
Therefore, the region described in polar coordinates is:
$0 \leq \theta \leq \dfrac{\pi}{3}$ and $0\leq r\leq \sec \theta$
